Chemistry Notes

Naming Compounds
Balancing Chemical Equations
Stoichiometry
Limiting Reactant
Atomic Structure
Lewis Structures
Valence Shell Electron Pair Repulsion (VSEPR) Model
Hybrid Atomic Orbitals
Covalent Bonds
Bond Energies
Gases
Solubility
Concentration of Solutions
Boiling Point Elevation
Freezing Point Depression
Osmotic Pressure
Chemical Equilibrium
Reaction Rate
Acids and Bases
Titrations
Electrochemistry
Thermochemistry
Gibbs Free Energy
Clausius-Clapeyron Equation
Nuclear Chemistry
Halogens
Organic Chemistry
Nuclear Magnetic Resonance (NMR)

PROBLEM

Name the following ionic compounds:
(a) Ag₃PO₄
(b) Ca₃(PO₄)₂
(c) BaF₂
(d) PbF₂
(e) AgCl
(f) PbBr₂
(g) Hg₂Cl₂
(h) FeCl₃
(i) Ag₂SO₄
(j) BaSO₄
(k) PbSO₄
(l) Co(OH)₂
(m) Fe(OH)₃
(n) Ag₂S
(o) PbCO₃
(p) FeCO₃
(q) PbCrO₄

SOLUTION

Each of the seventeen compounds listed belongs to one of four possible categories:

	Metal Has One Oxidation State	Metal Has Two Oxidation States
Binary Ionic Compounds (two elements, metal + nonmetal)	(c), (e), (n) Combine the name of the metal with the root name of the nonmetal and the suffix "ide".	(d), (f), (g), (h) Combine the name of the metal with the Roman numeral corresponding to its oxidation state1, the root name of the nonmetal, and the suffix "ide".
Ternary Ionic Compounds (three elements, metal + inorganic anion)	(a), (b), (i), (j) Combine the name of the metal with the name of the anion.	(k), (l), (m), (o), (p), (q) Combine the name of the metal with the Roman numeral corresponding to its oxidation state1 and the name of the anion.

¹In an older nomenclature system, the Latin root of the metal name is combined with the suffix ("ous", "ic") when the metal has its (less, more) positively charged oxidation state.

PROBLEM

Name the following compounds or ions:
(a) H₃PO₄, H₂PO₄^-, HPO₄^2-, PO₄^3-
(b) H₂SO₄, HSO₄^-, SO₄^2-
(c) H₂CO₃, HCO₃^-, CO₃^2-

SOLUTION

(a) H₃PO₄ is phosphoric acid, a triprotic acid. H₂PO₄^-, HPO₄^2-, and PO₄^3- are the anions resulting from first, second, and third stage ionization of phosphoric acid.

Formula	Name
H3PO4	phosphoric acid
H2PO4-	dihydrogen phosphate
HPO42-	hydrogen phosphate
PO43-	phosphate

(b) H₂SO₄ is sulfuric acid, a diprotic acid. HSO₄^- and SO₄^2- are the anions resulting from first and second stage ionization of sulfuric acid.

Formula	Name
H2SO4	sulfuric acid
HSO4-	hydrogen sulfate
SO42-	sulfate

(c) H₂CO₃ is carbonic acid, a diprotic acid. HCO₃^- and CO₃^2- are the anions resulting from first and second stage ionization of carbonic acid.

Formula	Name
H2CO3	carbonic acid
HCO3-	hydrogen carbonate
CO32-	carbonate

PROBLEM

Name the following acids:
(a) HBr
(b) HCl
(c) HClO
(d) HClO₂
(e) HClO₃
(f) HClO₄
(g) HF
(h) HI
(i) HNO₂
(j) HNO₃
(k) H₂CO₃
(l) H₂S
(m) H₂SO₃
(n) H₂SO₄
(o) H₃PO₄

SOLUTION

The fifteen acids listed can be broken up into three categories:

PROBLEM

Identify the ions that exist in (a) sodium acetate, (b) aluminum oxide, and (c) potassium phosphate.

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problems 5-7]

SOLUTION

(a) Sodium acetate is NaCH₃COO and consists of the sodium (Na⁺) and acetate (CH₃COO^-) ions.

(b) Aluminum oxide is Al₂O₃ and consists of aluminum (Al³⁺) and oxygen (O^2-) ions.

(c) Potassium phosphate is K₃PO₄ and consists of potassium (K⁺) and phosphate (PO₄^3-) ions.

PROBLEM

Write the balanced equation and state the reaction type for each of the following reactions:
(a) zinc + sulfuric acid => zinc sulfate + hydrogen
(b) iron(III) chloride + sodium hydroxide => iron(III) hydroxide + sodium chloride
(c) aluminum hydroxide + sulfuric acid => aluminum sulfate + water
(d) potassium + water =>
(e) magnesium + oxygen =>
(f) silver nitrate + copper => copper(II) nitrate + silver(s)
(g) magnesium bromide + chlorine => magnesium chloride + bromine

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), problems 6.1-6.7]

SOLUTION

(a) Zn(s) + H₂SO₄(aq) => ZnSO₄(aq) + H₂(g), single replacement, zinc replaces hydrogen

(b) FeCl₃ + 3NaOH => Fe(OH)₃ + 3NaCl, double replacement, hydroxyl and chlorine are interchanged

(c) Al(OH)₃ + H₂SO₄ => Al₂(SO₄)₃ + H₂O

We can rewrite this as

Al(OH)₃ + H₂SO₄ => Al₂(SO₄)₃ + H(OH)

The balanced equation is

2Al(OH)₃ + 3H₂SO₄ => Al₂(SO₄)₃ + 6H(OH)

or

2Al(OH)₃ + 3H₂SO₄ => Al₂(SO₄)₃ + 6H₂O, double replacement, hydroxyl and sulfate are interchanged

(d) 2K + 2H₂O => 2KOH + H₂(g), single replacement, potassium replaces hydrogen

(e) 2Mg + O₂ => 2MgO, synthesis

(f) 2AgNO₃(aq) + Cu(s) => Cu(NO₃)₂(aq) + 2Ag(s), single replacement, copper replaces silver

(g) MgBr₂ + Cl₂ => MgCl₂ + Br₂, single replacement, chlorine replaces bromine

PROBLEM

Balance the following equations:
(a) N₂O₅ => N₂O₄ + O₂
(b) KNO₃ => KNO₂ + O₂
(c) NH₄NO₃ => N₂O + H₂O
(d) NH₄NO₂ => N₂ + H₂O
(e) NaHCO₃ => Na₂CO₃ + H₂O + CO₂
(f) P₄O₁₀ + H₂O => H₃PO₄
(g) HCl + CaCO₃ => CaCl₂ + H₂O + CO₂
(h) Al + H₂SO₄ => Al₂(SO₄)₃ + H₂
(i) CO₂ + KOH => K₂CO₃ + H₂O
(j) CH₄ + O₂ => CO₂ + H₂O
(k) Be₂C + H₂O => Be(OH)₂ + CH₄
(l) Cu + HNO₃ => Cu(NO₃)₂ + NO + H₂O
(m) S + HNO₃ => H₂SO₄ + NO₂ + H₂O
(n) NH₃ + CuO => Cu + N₂ + H₂O

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 3.58]

SOLUTION

(a) 2N₂O₅ => 2N₂O₄ + O₂

(b) 2KNO₃ => 2KNO₂ + O₂

(c) NH₄NO₃ => N₂O + 2H₂O

(d) NH₄NO₂ => N₂ + 2H₂O

(e) 2NaHCO₃ => Na₂CO₃ + H₂O + CO₂

(f) P₄O₁₀ + 6H₂O => 4H₃PO₄

(g) 2HCl + CaCO₃ => CaCl₂ + H₂O + CO₂

(h) 2Al + 3H₂SO₄ => Al₂(SO₄)₃ + 3H₂

(i) CO₂ + 2KOH => K₂CO₃ + H₂O

(j) CH₄ + 2O₂ => CO₂ + 2H₂O

(k) Be₂C + 4H₂O => 2Be(OH)₂ + CH₄

(l) Cu + HNO₃ => Cu(NO₃)₂ + NO + H₂O

This equation is more complex than the previous ones because two elements, nitrogen and oxygen, appear in more than two reactants and/or products. We will assume that

A · Cu + B · HNO₃ => C · Cu(NO₃)₂ + D · NO + E · H₂O

and solve for A, B, C, D, and E by requiring that the numbers of different atoms are equal on both sides of the equation.

Cu: A = C
H: B = 2E
N: B = 2C + D
O: 3B = 6C + D + E

We have four equations in five unknowns, which is sufficient, since all we have to do is get the relative sizes of the coefficients. Set A = C = 1 and substitute into the last three equations.

B = 2E
B = 2 + D
3B = 6 + D + E

Replace B with 2E in the last two equations.

2E = 2 + D
6E = 6 + D + E => D = 5E - 6 => 2E = 2 + 5E - 6 => E = 4/3 => D = 20/3 - 6 = 2/3 => B = 2 + 2/3 = 8/3

We have A = 1, B = 8/3, C = 1, D = 2/3, and E = 4/3. Since we need integral coefficients, we multiply all of these by three and get A = 3, B = 8, C = 3, D = 2, and E = 4. Thus,

3Cu + 8HNO₃ => 3Cu(NO₃)₂ + 2NO + 4H₂O

(m) Assume that

A · S + B · HNO₃ => C · H₂SO₄ + D · NO₂ + E · H₂O

and solve for A, B, C, D, and E by balancing the numbers of different atoms.

S: A = C
H: B = 2C + 2E
N: B = D
O: 3B = 4C + 2D + E

Set A = C = 1 and substitute into the last three equations.

B = 2 + 2E
B = D
3B = 4 + 2D + E

Replace B with D in the first and third equations.

D = 2 + 2E
3D = 4 + 2D + E => E = D - 4 => D = 2 + 2(D - 4) = 2 + 2D - 8 = 2D - 6 => D = 6 => E = D - 4 = 6 - 4 = 2 => B = 2 + 2E = 2 + (2)(2) = 6

We have A = 1, B = 6, C = 1, D = 6, and E = 2. The balanced equation is

S + 6HNO₃ => H₂SO₄ + 6NO₂ + 2H₂O

(n) 2NH₃ + 3CuO => 3Cu + N₂ + 3H₂O

PROBLEM

In an experiment 1.056 g of a metal carbonate, containing an unknown metal M, is heated to give the metal oxide and 0.376 g CO₂:

MCO₃(s) + heat => MO(s) + CO₂(g)

What is the identity of the metal M?

[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.69]

SOLUTION

The molar mass of CO₂ is 12.011 g/mole + (2)(15.9994 g/mole) = 44.0098 g/mole, so 0.376 g CO₂ = 8.544 x 10^-3 mole CO₂. For every mole of CO₂ produced in the above reaction, one mole of MO is produced. Thus, 8.544 x 10^-3 mole MO is produced with a mass of 1.056 g - 0.376 g = 0.68 g. The molar mass of MO is thus (0.68 g) / (8.544 x 10^-3 mole) = 79.59 g/mole, and the molar mass of M is 79.59 g/mole - 15.9994 g/mole = 63.59 g/mole, so M is Cu (copper).

PROBLEM

A compound containing the elements C, H, N, and O is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO₂(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H.
(a) Determine the mass, in grams, of C in the 1.2359 g sample of the compound.
(b) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound.
(c) Determine the mass, in grams, of O in the original 1.2359 g sample of the compound.
(d) Determine the empirical formula of the compound.

[from AP Chemistry 2006 Free-Response Questions, College Board, Question 3(a)]

SOLUTION

(a) The molar mass of carbon dioxide (CO₂) is 44.011 g/mole. It follows that 2.241 g CO₂ = 5.092 x 10^-2 mole CO₂. Thus, there is 5.092 x 10^-2 mole C = 0.6116 g C in the 1.2359 g sample of the compound.

(b) The mass of N in the original sample is (0.2884)(1.2359 g) = 0.3564 g.

(c) The mass of O in the original sample is 1.2359 g - 0.0648 g - 0.6116 g - 0.3564 g = 0.2031 g.

(d) The sample contains

0.0648 g H = 6.429 x 10^-2 mole H
5.092 x 10^-2 mole C
0.3564 g N = 2.545 x 10^-2 mole N
0.2031 g O = 1.269 x 10^-2 mole O

For each mole of O, the sample contains

5.065 mole H
4.012 mole C
2.005 mole N

Therefore, the empirical formula of the compound is C₄H₅N₂O.

PROBLEM

The important industrial chemical carbon tetrachloride, CCl₄, is made by the reaction of carbon disulfide with chlorine according to the following unbalanced equation:

CS₂(g) + Cl₂(g) => S₂Cl₂(l) + CCl₄(l)

What is the theoretical yield of CCl₄ if 125 g of CS₂ is mixed with 435 g of Cl₂? What mass of excess reactant remains when the reaction has gone to completion?

[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 5.73]

SOLUTION

The balanced equation is

CS₂(g) + 3Cl₂(g) => S₂Cl₂(l) + CCl₄(l)

The molar mass of CS₂ is 76.143 g/mole. The molar mass of Cl₂ is 70.9054 g/mole. The molar mass of CCl₄ is 153.8218 g/mole. Initially, there is 125 g CS₂ = 1.642 mole CS₂ and 435 g Cl₂ = 6.135 mole Cl₂. To use up all of the CS₂ would require (3)(1.642 mole) = 4.925 mole Cl₂. Since there is 6.135 mole Cl₂, the limiting reactant is CS₂ and there will be excess Cl₂ remaining. According to the above balanced equation, each mole of CS₂ yields one mole of CCl₄. Thus, 1.642 mole CCl₄ = 252.5 g CCl₄ is produced. The mass of Cl₂ remaining is (6.135 mole - 4.925 mole)(70.9054 g/mole) = 85.79 g.

PROBLEM

Suppose that a stable element with atomic number 119, symbol Q, has been discovered.
(a) Write the ground-state electron configuration for Q.
(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration.
(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure.
(d) What would be the most likely charge of the Q ion in stable ionic compounds?
(e) Write a balanced equation that would represent the reaction of Q with water.
(f) Assume that Q reacts to form a carbonate compound. (i) Write the formula for the compound formed between Q and the carbonate ion, CO₃^2-. (ii) Predict whether or not the compound would be soluble in water. Explain your reasoning.

[from AP Chemistry 2006 Free-Response Questions, College Board, Question 8]

SOLUTION

(a) Using the following diagram, we fill in the electronic structure along diagonals from upper right to lower left beginning with the 1s, 2s, 2p, 3p, 3d, 4d, 4f, and 5f shells.

1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
8s

For element Q (Z = 119), we get

[1s²][2s²2p⁶][3s²3p⁶][4s²3d¹⁰4p⁶][5s²4d¹⁰5p⁶][6s²4f¹⁴5d¹⁰6p⁶][7s²5f¹⁴6d¹⁰7p⁶][8s¹]

Colors identify subshells which belong to the same diagonal. Square brackets indicate periods in the periodic table.

(b) Q has a single valence electron in the 8s subshell and is therefore an alkali metal.

(c) Q would have the largest atomic radius in its group, the alkali metals.

(d) The charge of Q in stable ionic compounds would be +1.

(e) Q would react with water by replacing the hydrogen according to:

2Q + 2H₂O => 2QOH + H₂

This is a hydrogen displacement redox reaction.

(f) (i) The formula of the compound formed between Q and the carbonate ion would be Q₂CO₃. (ii) Carbonates containing alkali metal ions are soluble in aqueous solution, so Q₂CO₃ would be soluble in water.

PROBLEM

Write the Lewis structure(s) or Lewis dot symbol for each of the following molecules and indicate which have resonance structures:
(a) sulfur dioxide (SO₂)
(b) ethane (C₂H₆)
(c) chlorine (Cl₂)
(d) hydrogen bromide (HBr)
(e) sodium chloride (NaCl)

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 64]

SOLUTION

(a) Sulfur dioxide has two Lewis structures, which are resonance structures:

..  .. ..       .. ..  ..
O===S---O: <-> :O---S===O
..     ..       ..     ..

(b) ethane:

   H  H
   |  |
H--C--C--H
   |  |
   H  H

(c) chlorine:

 ..   ..
:Cl---Cl:
 ..   ..

(d) hydrogen bromide:

    ..
H---Br:
    ..

(e) Sodium chloride is an ionic compound:

     ..
Na⁺ :Cl:^-
     ..

PROBLEM

Write Lewis structures for the following four isoelectronic species: (a) CO, (b) NO⁺, (c) CN^-, (d) N₂. Show formal charges.

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 9.104]

SOLUTION

(a) CO (10 valence electrons):

-1   +1
:C///O:     /// = triple bond

(b) NO⁺ (10 valence electrons):

 0   +1
:N///O:     /// = triple bond

(c) CN^- (10 valence electrons):

-1   0
:C///N:     /// = triple bond

(d) N₂ (10 valence electrons)

 0   0
:N///N:     /// = triple bond

PROBLEM

Determine if each of the following species is planar or non-planar:
(a) CO₃^2-
(b) NO₃^-
(c) ClF₃
(d) BF₃
(e) PCl₃

[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 4]

SOLUTION

(a) CO₃^2- (carbonate) has 24 valence electrons. Its Lewis structure is:
..     ..      ..      ..     ..       ..
O===C---O: <=> :O---C===O <=> :O---C---O:
..  |  ..      ..   |  ..     ..   ||  ..
    |               |              ||
   :O:             :O:            :O:
   ..              ..

There are three resonance structures. Their shape is approximately trigonal planar.

(b) NO₃^- (nitrate) has 24 valence electrons. Its Lewis structure is:
..     ..      ..      ..     ..       ..
O===N---O: <=> :O---N===O <=> :O---N---O:
..  |  ..      ..   |  ..     ..   ||  ..
    |               |              ||
   :O:             :O:            :O:
   ..              ..

There are three resonance structures, which are all planar. Their shape is approximately trigonal planar.

(c) ClF₃ (chlorine trifluoride) has 28 valence electrons. Its Lewis structure is:
 .. ....  ..
:F---Cl---F:
 ..  |   ..
     |
    :F:
     ..

There are two lone electron pairs and three bonding pairs surrounding the central chlorine atom. The molecule is T shaped and therefore planar.

(d) BF₃ (boron trifluoride) has 24 valence electrons. Its Lewis structure is:
..     ..      ..      ..     ..       ..
F===B---F: <=> :F---B===F <=> :F---B---F:
..  |  ..      ..   |  ..     ..   ||  ..
    |               |              ||
   :F:             :F:            :F:
   ..              ..

There are three resonance structures, which are all planar. Their shape is approximately trigonal planar.

(e) PCl₃ (phosphorus trichloride) has 26 valence electrons. Its Lewis structure is:
 ..   ..  ..
:Cl---P---Cl:
 ..   |   ..
      |
     :Cl:
      ..

There is one lone electron pair and three bonding pairs surrounding the central phosphorus atom. The molecule is trigonal pyramidal and therefore non-planar.

PROBLEM

Write the Lewis structure for each of the following and use it and the valence-shell electron-pair repulsion (VSEPR) model to predict the geometry of the molecule:
(a) carbon dioxide (CO₂)
(b) carbon tetrachloride (CCl₄)

SOLUTION

(a) The carbon has four valence electrons, while each of the two oxygens has six. Thus, we have a total of sixteen valence electrons to account for. The Lewis structure for carbon dioxide is

..       ..
 O===C===O
..       ..

There is a double bond between the carbon and each of the oxygens. Since the central atom, carbon, has two double bonds and no lone pairs, the VSEPR model predicts that the carbon dioxide molecule is linear, which is consistent with experiment.

(b) The carbon has four valence electrons, while each of the four chlorines has seven. Thus, we have a total of 32 valence electrons to account for. The Lewis structure for carbon tetrachloride is

      ..
     :Cl:
      |
 ..   |   ..
:Cl---C---Cl:
 ..   |   ..
      |
     :Cl:
      ..

There is a single bond between the carbon and each of the chlorines. Since the central atom, carbon, has four single bonds and no lone pairs, the VSEPR model predicts that the carbon tetrachloride molecule is tetrahedral.

PROBLEM

For each of the following hybrid atomic orbitals, identify (i) which pure atomic orbitals are used, (ii) the shape of the hybrid atomic orbitals, (iii) the angle between bonded atoms, and (iv) an example of a molecule which exhibits this type of hybrid atomic orbital, without any lone pairs, and its Lewis structure:
(a) sp
(b) sp²
(c) sp³
(d) sp³d
(e) sp³d²

SOLUTION

(a) sp: one s and one p orbital, linear, 180°, BeCl₂

 ..      ..
:Cl--Be--Cl:
 ..      ..

(b) sp²: one s and two p orbitals, trigonal planar, 120°, BF₃
    ..
   :F:
 .. | ..
:F--B--F:
 ..   ..

(c) sp³: one s and three p orbitals, tetrahedral, 109.5°, CH₄

   H
   |
H--C--H
   |
   H

(d) sp³d: one s, three p, and one d orbital, trigonal bipyramidal, 90° and 120°, PCl₅

      ..
     :Cl:
      |   
 ..   |   ..
:Cl---P---Cl:
..   / \  ..
    /   \
 :Cl:   :Cl:
  ..     ..

(e) sp³d²: one s, three p, and two d orbitals, octahedral, 90°, SF₆

  ..   ..
 :F:   :F:
   \   /  
 .. \ / ..
:F---S---F:
 .. / \ ..
   /   \
 :F:   :F:
  ..   ..

PROBLEM

(a) Draw the complete Lewis structure (electron-dot diagram) for CF₄, PF₅ and SF₄.

(b) On the basis of the Lewis structures drawn, answer the following questions about the particular molecule indicated:
(i) What is the F-C-F bond angle in CF₄?
(ii) What is the hybridization of the valence orbitals of P in PF₅?
(iii) What is the geometric shape formed by the atoms in SF₄?

(c) Two Lewis structures can be drawn for the OPF₃ molecule, as shown below.
                  ..
   :O:           :O:
..  || ..     ..  |  ..
:F--P--F:     :F--P--F:
..  |  ..     ..  |  ..
   :F:           :F:
    ..            ..
structure 1   structure 2

(i) How many sigma bonds and how many pi bonds are in structure 1?
(ii) Which of the two structures best represents a molecule of OPF₃? Justify your answer in terms of formal charge.

[from 2005 AP Chemistry Free-Response Question 6]

SOLUTION

(a) carbon tetrafluoride (CF₄):
     ..
    :F:
     |
 ..  |  ..
:F---C---F:
 ..  |  ..
     |
    :F:
     ..

phosphorus pentafluoride (PF₅):
 ..     ..
:F:     :F:
   \   /
..  \ /  ..
:F---P---F:
..   |   ..
     |
    :F:
     ..

sulfur tetrafluoride (SF₄)
 ..     ..
:F:     :F:
   \   /
..  \ /  ..
:F---S---F:
..   ..  ..

(b)(i) Carbon tetrafluoride has four electron pairs surrounding the central carbon and no lone pairs. The geometry is tetrahedral. The F-C-F bond angle is 109.5°.

(ii) Phosphorus pentafluoride has five electron pairs surrounding the central phosphorus and no lone pairs. The geometry is trigonal bipyramidal. The hybridization that corresponds to this geometry is sp³d.

(iii) Sulfur tetrafluoride has five electron pairs surrounding the central sulfur. One of these is a lone pair. The corresponding geometry is distorted tetrahedral or seesaw.

(c)(i) Structure 1 has four sigma bonds and one pi bond. The double bond consists of one sigma bond and one pi bond.

(ii) In structure 1, the formal charges are all zero. In structure 2, the formal charges are +1 on the phosphorus, -1 on the oxygen, and zero on the fluorines. When a molecule is neutral, like this one, the Lewis structure in which there are no formal charges is preferred, so structure 1 is the better representation.

PROBLEM

(a) From the following data calculate the bond energy of the F₂^- ion.

F₂(g) => 2F(g), DH₁⁰ = 156.9 kJ
F ^-(g) => F(g) + e^-, DH₂⁰ = 333 kJ
F₂^-(g) => F₂(g) + e^-, DH₃⁰ = 290 kJ

(b) Explain the difference between the bond energies of F₂ and F₂^-.

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 9.108]

SOLUTION

(a) The bond energy of F₂^-(g) is the energy required to break it up into F(g) and F ^-(g). F₂^-(g) can be converted into F₂(g) as follows:

      DH⁰             DH₂⁰               -DH₁⁰
F₂^-(g) => F(g) + F^-(g) => F(g) + F(g) + e^- => F₂(g) + e^-

The total enthalpy change for the above process is

DH₃⁰ = DH⁰ + DH₂⁰ - DH₁⁰
=> DH⁰ = DH₃⁰ - DH₂⁰ + DH₁⁰ = 290 kJ - 333 kJ + 156.9 kJ = 113.9 kJ

(b) The bond energy of F₂(g) is the energy required to break it up into 2F(g), which is 156.9 kJ. The additional electron in F ^- increases the repulsion between the electron clouds in F and F ^-, resulting in a weaker bond in F₂^- than in F₂(g).

PROBLEM

Experiments show that it takes 1656 kJ/mole to break all the bonds in methane (CH₄) and 4006 kJ/mole to break all the bonds in propane (C₃H₈). Based on these data, calculate the average bond energy of the C-C bond.

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 9.118]

SOLUTION

The methane (CH₄) molecule has four C-H bonds:

   H
   |
H--C--H
   |
   H

Each mole of CH₄ contains 4 moles of C-H bonds. Thus, the average bond energy of the C-H bond is (1656 kJ/mole)/4 = 414 kJ/mole. The propane (C₃H₈) molecule has eight C-H bonds and two C-C bonds:

   H  H  H
   |  |  |
H--C--C--C--H
   |  |  |
   H  H  H

Each mole of C₃H₈ contains 8 moles of C-H bonds and 2 moles of C-C bonds. Thus, the average bond energy of the C-C bond is [4006 kJ/mole - (8)(414 kJ/mole)] / 2 = 347 kJ/mole

PROBLEM

An unknown gas is one of three possible gases: nitrogen, hydrogen, or oxygen. For each of the three possibilities, describe the result expected when the gas is tested using a glowing splint (a wooden stick with one end that has been ignited and extinguished, but still contains hot, glowing, partially burned wood).

[from 2005 AP Chemistry Free-Response Question 5]

SOLUTION

nitrogen: The splint will completely stop burning and cool.

hydrogen: The gas will ignite.

oxygen: The splint will start burning again with a flame.

PROBLEM

Summarize the following ideal gas laws:
(a) Boyle's law
(b) Charles's law
(c) Gay-Lussac's law
(d) Dalton's law

SOLUTION

(a) Boyle's law, discovered by English scientist Robert Boyle in 1662, states that, for a fixed amount of gas at constant temperature, the product of the pressure P and the volume V is a constant, or PV = constant.

(b) Charles's law, discovered by French chemist Jacques Charles, states that, for a fixed amount of gas at constant pressure, the volume V is proportional to the absolute temperature T or V/T = constant.

(c) Gay-Lussac's law, discovered by French scientist Joseph Gay-Lussac in 1808, states that the ratio between the combining volumes of gases and the volume of the product, if gaseous, can be expressed in small whole numbers.

(d) Dalton's law of partial pressures, discovered by English scientist and school teacher John Dalton in 1801, states that the total gas pressure of a mixture of gases is equal to the sum of the partial pressures of the gases that make up the mixture.

PROBLEM

A compound which has the empirical formula CH₂Br has a vapor density of 6.00 g/L at 375 K and 0.983 atm. Using these data, determine the following:
(a) The molar mass of the compound.
(b) The molecular formula of the compound.

[from AP Chemistry 2006 Free-Response Questions, College Board, Question 3(b)]

SOLUTION

(a) The molar density of an ideal gas is

n/V = p/RT

where p is the pressure, R = 0.0821 L atm / mole K is the gas constant, and T is the temperature. The density of an ideal gas is

r = Mn/V = Mp/RT

where M is the molar mass. Thus, the molar mass of the compound is

M = rRT/p = (6.00 g/L)(0.0821 L atm / mole K)(375 K) / (0.983 atm) = 187.9196 g/mole

(b) The molar mass of CH₂Br is 93.9268 g/mole. Therefore, the molecular formula of the compound is C₂H₄Br₂.

PROBLEM

Each of three beakers contains a 0.1 M solution of one of the following solutes: potassium chloride, silver nitrate, or sodium sulfide. The three beakers are labeled randomly as solution 1, solution 2, and solution 3. Mixing solution 1 with solution 2 results in a black precipitate. Mixing solution 2 with solution 3 produces no reaction.
(a) Write the chemical formula of the black precipitate.
(b) Describe the expected results of mixing solution 1 with solution 3.
(c) Identify each of the solutions 1, 2, and 3.

[from 2005 AP Chemistry Free-Response Question 5]

SOLUTION

KCl, AgNO₃, and Na₂S are all soluble in aqueous solution. The following table shows the solubility of various combinations of the available ions.

	Cl-	NO3-	S2-
K+	KCl soluble	KNO3 soluble	K2S soluble
Ag+	AgCl insoluble	AgNO3 soluble	Ag2S insoluble
Na+	NaCl soluble	NaNO3 soluble	Na2S soluble

(a) There are only two possible precipitates, AgCl and Ag₂S. AgCl is a white precipitate. Ag₂S is a black precipitate. It forms on silver when it is exposed to the hydrogen sulfide of the atmosphere. So the black precipitate is Ag₂S.

(c) In order for Ag₂S to form when solutions 1 and 2 are mixed, one of the solutions must be AgNO₃ and the other must be Na₂S. In order for no reaction to occur when solutions 2 and 3 are mixed, one must be KCl and the other must be Na₂S. Solution 2 is common to both mixtures, so it is Na₂S. Solution 1 is AgNO₃ and solution 3 is KCl.

(b) The result of mixing solutions 1 and 3, AgNO₃ and KCl, is to produce a white precipitate, AgCl.

Molarity is the number of moles of solute per liter of solution.

The gram-equivalent weight or equivalent of a compound is the weight in grams of the compound which yields or reacts with one mole of hydrogen ions or hydroxyl ions.

Normality is the number of equivalents of solute per liter of solution.

PROBLEM

(a) What is the gram-formula mass of AlCl₃?
(b) What is the gram-equivalent mass of AlCl₃?

SOLUTION

(a) The gram-formula mass of AlCl₃ is the mass, in grams, of one mole of AlCl₃, which is 26.98154 g + (3)(35.453 g) = 133.3405 g

(b) The gram-equivalent mass of AlCl₃ is the mass, in grams, of AlCl₃ which yields or reacts with one mole of hydrogen or hydroxyl ions. Since each of the three Cl^- ions in one molecule of AlCl₃ can react with one a hydrogen ion, the gram-equivalent mass of AlCl₃ is 1/3 of its gram-formula mass or (1/3)(133.3405 g) = 44.44685 g.

PROBLEM

Determine the equivalent weights (i.e., the weight of one "equivalent") of each of the following:
(a) H₂SO₄
(b) H₃PO₄
(c) Ca(OH)₂
(d) AlCl₃
(e) K₂SO₄
(f) NaCl
(g) Na₃PO₄
(h) CuSO₄
(i) KOH
(j) Fe₂O₃

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), problem 14.13]

SOLUTION

(a) One mole of H₂SO₄ contains two moles of H⁺ and is thus equal to two equivalents of H₂SO₄. Therefore, one equivalent of H₂SO₄ is 0.5 mole of H₂SO₄ or (0.5 mole)[(2)(1.008 g/mole) + 32.07 g/mole + (4)(16.00 g/mole)] = 49.04 g.

(b) One mole of H₃PO₄ contains three moles of H⁺ and is thus equal to three equivalents of H₃PO₄. Therefore, one equivalent of H₃PO₄ is 1/3 mole of H₃PO₄ or (1/3 mole)[(3)(1.008 g/mole) + 30.97 g/mole + (4)(16.00 g/mole)] = 32.66 g.

(c) One mole of Ca(OH)₂ contains two moles of OH^- and is thus equal to two equivalents of Ca(OH)₂. Therefore, one equivalent of Ca(OH)₂ is 0.5 mole of Ca(OH)₂ or (0.5 mole)[40.08 g/mole + (2)(16.00 g/mole) + (2)(1.008 g/mole)] = 37.048 g.

(d) One mole of AlCl₃ contains three moles of Cl^-, which is capable of reacting with three moles of H⁺, and is thus equal to three equivalents of AlCl₃. Therefore, one equivalent of AlCl₃ is 1/3 mole of AlCl₃ or (1/3 mole)[26.98 g/mole + (3)(35.45 g/mole)] = 44.44 g.

(e) One mole of K₂SO₄ contains one mole of SO₄^2-, which is capable of reacting with two moles of H⁺, and is thus equal to two equivalents of K₂SO₄. Therefore, one equivalent of K₂SO₄ is 0.5 mole of K₂SO₄ or (0.5 mole)[(2)(39.10 g/mole) + 32.07 g/mole + (4)(16.00 g/mole)] = 87.14 g.

(f) One mole of NaCl contains one mole of Cl^-, which is capable of reacting with one mole of H⁺, and is thus equal to one equivalent of NaCl. Therefore, one equivalent of NaCl is (1 mole)(22.99 g/mole + 35.45 g/mole) = 58.44 g.

(g) One mole of Na₃PO₄ contains one mole of PO₄^3-, which is capable of reacting with three moles of H⁺, and is thus equal to three equivalents of Na₃PO₄. Therefore, one equivalent of Na₃PO₄ is 1/3 mole of Na₃PO₄ or (1/3 mole)[(3)(22.99 g/mole) + 30.97 g/mole + (4)(16.00 g/mole)] = 54.65 g.

(h) One mole of CuSO₄ contains one mole of SO₄^2-, which is capable of reacting with two moles of H⁺, and is thus equal to two equivalents of CuSO₄. Therefore, one equivalent of CuSO₄ is 0.5 mole of CuSO₄ or (0.5 mole)[63.55 g/mole + 32.07 g/mole + (4)(16.00 g/mole)] = 79.81 g.

(i) One mole of KOH contains one mole of OH^- and is thus equal to one equivalent KOH. Therefore, one equivalent of KOH is (1 mole)(39.10 g/mole + 16.00 g/mole + 1.008 g/mole) = 56.11 g.

(j) One mole of Fe₂O₃ contains two moles of Fe³⁺, which is capable of reacting with six moles of OH^-, and is thus equal to six equivalents of Fe₂O₃. Therefore, one equivalent of Fe₂O₃ is 1/6 mole of Fe₂O₃ or (1/6 mole)[(2)(55.85 g/mole) + (3)(16.00 g/mole)] = 26.62 g.

PROBLEM

What is the normality of (a) a 1 M HCl solution, (b) a 0.005 M NaOH solution, (c) a 0.05 M H₃PO₄ solution, (d) a 0.4 M Mg(OH)₂ solution?

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.6]

SOLUTION

(a) Each mole of HCl contains one mole of H⁺ ions. Therefore, each mole of HCl is equal to one equivalent of HCl. The normality of a 1 M HCl solution is 1 N.

(b) Each mole of NaOH contains one mole of OH^- ions. Therefore, each mole of NaOH is equal to one equivalent of NaOH. The normality of a 0.005 M NaOH solution is 0.005 N.

(c) Each mole of H₃PO₄ contains three moles of H⁺ ions. Therefore, each mole of H₃PO₄ is equal to three equivalents of H₃PO₄. The normality of a 0.05 M H₃PO₄ solution is 0.15 N.

(d) Each mole of Mg(OH)₂ contains two moles of OH^- ions. Therefore, each mole of Mg(OH)₂ is equal to two equivalents of Mg(OH)₂. The normality of a 0.4 M Mg(OH)₂ solution is 0.8 N.

PROBLEM

If 1 mole of each of the following substances was dissolved in 1000 g of water, what would be the boiling point?
(a) sodium chloride (NaCl)
(b) potassium chloride (KCl)
(c) calcium chloride (CaCl₂)
(d) cellulose (C₆H₁₀O₅)
(e) sucrose (C₁₂H₂₂O₁₁)

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 47]

SOLUTION

(a) The boiling point elevation of a water solution is given by

DT_b = K_bm

where K_b = 0.52°C/m and m is the molality of the solution (the number of moles of solute particles per 1000 g of solvent). Sodium chloride is soluble in aqueous solution and yields two solute particles (Na⁺ and Cl^-) per molecule. Thus, the molality of solute particles is (2 moles) / (1 kg) = 2 moles/kg = 2 m and

DT_b = (0.52°C/m)(2 m) = 1.04°C

The boiling point is

T_b = 100°C + 1.04°C = 101.04°C

(b) Potassium chloride is soluble in aqueous solution and yields two solute particles (K⁺ and Cl^-) per molecule. Thus, the molality of solute particles is 2 m and the boiling point is T_b = 101.04°C, as in part (a).

(c) Calcium chloride is soluble in aqueous solution and yields three solute particles (Ca²⁺ and 2Cl^-) per molecule. Thus, the molality of solute particles is 3 m and

DT_b = (0.52°C/m)(3 m) = 1.56°C => T_b = 100°C + 1.56°C = 101.56°C

(d) Cellulose does not dissociate into more than one particle. Thus, the molality of solute particles is 1 m and

DT_b = (0.52°C/m)(1 m) = 0.52°C => T_b = 100°C + 0.52°C = 100.52°C

(e) Sucrose does not dissociate into more than one particle. Thus, the molality of solute particles is 1 m and

DT_b = (0.52°C/m)(1 m) = 0.52°C => T_b = 100°C + 0.52°C = 100.52°C

PROBLEM

Fishes in the Antarctic Ocean swim in water at about - 2°C. (a) To prevent their blood from freezing, what must be the concentration (in molality) of the blood? Is this a reasonable physiological concentration? (b) In recent years scientists have discovered a special type of protein in these fishes' blood which, although present in quite low concentrations (< 0.001 m), has the ability to prevent the blood from freezing. Suggest a mechanism for its action.

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 12.122]

SOLUTION

(a) The freezing point depression of water is

DT_f = K_fm

where K_f is the molal freezing-point depression constant, which is 1.86 °C/m for water. Thus,

m > DT_f/K_f = (2°C) / (1.86°C/m) = 1.075 m

A list of reference ranges for common human blood tests appears at http://en.wikipedia.org/wiki/Reference_ranges_for_common_blood_tests. If we add up the mean concentrations of all electrolytes, metabolites, ions, and trace metals that are listed, we get 0.154 mole/L = 0.154 M. The corresponding molality is approximately 0.154 m. Thus, the molality needed to depress the freezing point by the required amount is substantially larger than the total molality of the above substances in normal human blood, and appears to be an unreasonable physiological concentration.

(b) The protein could attract water molecules and prevent them from bonding with each other and forming ice crystals.

PROBLEM

In the apparatus shown below, what will happen if the membrane is (a) permeable to both water and the Na⁺ and Cl^- ions, (b) permeable to water and Na⁺ ions but not to Cl^- ions, (c) permeable to water but not to Na⁺ and Cl^- ions?

    | |         | |
    |_|         |_|
    | |         | |
+---+ +---+-+---+ +---+
|         |/|         |
|         |/|         |
| 0.01 M  |/|  0.1 M  |
|  NaCl   |/|  NaCl   |
|         |/|         |
|         |/|         |
+---------+-+---------+
           ^
           |
        membrane

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 12.104]

SOLUTION

(a) If the membrane is permeable to both water and the Na⁺ and Cl^- ions, the Na⁺ and Cl^- ions will diffuse through the membrane until both sides contain a 0.055 M solution of NaCl. The water level will be the same on both sides.

(b) If the membrane is permeable to both water and Na⁺ ions, but not to Cl^- ions, the Na⁺ will not diffuse to the left because, if it did, there would be a net positive charge on the left and a net negative charge on the right. The osmotic pressure will be higher on the right side, causing the water level to increase on the right side and decrease on the left side.

(c) If the membrane is permeable to water but not to Na⁺ or Cl^- ions, the osmotic pressure will be higher on the right, and water will diffuse from the left to the right, causing the water level to decrease on the left and increase on the right.

PROBLEM

Answer the following questions that relate to solubility of salts of lead and barium.
(a) A saturated solution is prepared by adding excess PbI₂(s) to distilled water to form 1.0 L of solution at 25°C. The concentration of Pb²⁺(aq) in the saturated solution is found to be 1.3 x 10^-3 M. The chemical equation for the dissociation of PbI₂(s) in water is shown below.

PbI₂ <=> Pb²⁺(aq) + 2I ^-(aq)

(i) Write the equilibrium constant expression for the equation.
(ii) Calculate the molar concentration of I ^-(aq) in the solution.
(iii) Calculate the value of the equilibrium constant K_sp.

(b) A saturated solution is prepared by adding PbI₂(s) to distilled water to form 2.0 L of solution at 25°C. What are the molar concentrations of Pb²⁺ and I ^- in the solution? Justify your answer.

(c) Solid NaI is added to a saturated solution of PbI₂ at 25°C. Assuming that the volume of the solution does not change, does the molar concentration of Pb²⁺(aq) in the solution increase, decrease, or remain the same? Justify your answer.

(d) The value of K_sp for the salt BaCrO₄ is 1.2 x 10^-10. When a 500. mL sample of 8.2 x 10^-6 M Ba(NO₃)₂ is added to 500. mL of 8.2 x 10^-6 M Na₂CrO₄, no precipitate is observed.
(i) Assuming that volumes are additive, calculate the molar concentrations of Ba²⁺(aq) and CrO₄^2-(aq) in the 1.00 L of solution.
(ii) Use the molar concentrations of Ba²⁺(aq) ions and CrO₄^2-(aq) ions as determined above to show why a precipitate does not form. You must include a calculation as part of your answer.

[from AP Chemistry 2006 Free-Response Questions, College Board, Question 1]

SOLUTION

(a) (i) K_sp = [Pb²⁺][I ^-]²

(ii) For each Pb²⁺ in solution, there are two I ^-. Thus,

[I ^-] = 2[Pb²⁺] = (2)(1.3 x 10^-3 M) = 2.6 x 10^-3 M

(iii) K_sp = (1.3 x 10^-3)(2.6 x 10^-3)² = 8.788 x 10^-9

(b) The PbI₂ will dissolve into Pb²⁺ and I ^- ions in the same proportion as in part (a) until
[Pb²⁺][I ^-]² is equal to K_sp = 8.788 x 10^-9. Thus, the final values of [Pb²⁺] and [I ^-] will be the same as in part (a).

(c) When solid NaI is added to the saturated solution of PbI₂, it breaks up into Na⁺ and I ^- ions. The concentration of I ^- ions increases. In order for the product [Pb²⁺][I ^-]² to remain the same, the concentration of Pb²⁺ must decrease, and some PbI₂ will precipitate.

(d) Nitrates are soluble in aqueous solution. Chromates containing alkali metal ions are also soluble in aqueous solution. Therefore, Ba(NO₃)₂ and Na₂CrO₄ are both soluble in aqueous solution.

(i) The molar concentrations of Ba²⁺(aq) and CrO₄^2-(aq) in the final solution are both 4.1 x 10^-6 M.

(ii) In the final solution, [Ba²⁺][CrO₄^2-] = (4.1 x 10^-6)² = 1.681 x 10^-11 < K_sp, so no precipitate forms.

PROBLEM

Pure nitrosyl chloride (NOCl) gas is heated to 240°C in a 1.00-L container. The NOCl decomposes according to

2NOCl(g) <=> 2NO(g) + Cl₂(g)

At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm.
(a) Calculate the partial pressures of NO and Cl₂ in the system at equilibrium.
(b) Calculate the equilibrium constant K_P.

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 14.62]

SOLUTION

Let P₀ be the initial pressure of the NOCl and let x be the amount (in atm) of NOCl that decomposes. At equilibrium, the partial pressures of NOCl, NO, and Cl₂ are

P(NOCl) = P₀ - x = 0.64 atm (1)
P(NO) = x
P(Cl₂) = x/2

The total pressure is

P(NOCl) + P(NO) + P(Cl₂) = P₀ - x + x + x/2 = 1.00 atm (2)

Combining (1) and (2), we get

0.64 atm + 3x/2 = 1.00 atm => x = (2/3)(1.00 atm - 0.64 atm) = 0.24 atm

(a) The partial pressures of NO and Cl₂ are P(NO) = 0.24 atm and P(Cl₂) = 0.12 atm.

(b) The equilibrium constant is

K_P = [P(NO)]²P(Cl₂) / [P(NOCl)]² = (0.24 atm)²(0.12 atm) / (0.64 atm)² = 0.016875

PROBLEM

A mixture of 0.47 mole of H₂ and 3.59 moles of HCl is heated to 2800°C. Calculate the equilibrium partial pressures of H₂, Cl₂, and HCl if the total pressure is 2.00 atm. For the reaction

H₂(g) + Cl₂(g) <=> 2HCl(g)

K_P is 193 at 2800°C.

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 14.72]

SOLUTION

Let x be the number of moles of H₂ that are produced in the reaction. At equilibrium, the number of moles of H₂, Cl₂, and HCl are

n(H₂) = 0.47 + x
n(Cl₂) = x
n(HCl) = 3.59 - 2x

The partial pressures are

P(H₂) = n(H₂)RT/V = (0.47 + x)RT/V
P(Cl₂) = n(Cl₂)RT/V = xRT/V
P(HCl) = n(HCl)RT/V = (3.59 - 2x)RT/V

The total pressure is

P(H₂) + P(Cl₂) + P(HCl) = (0.47 + x + x + 3.59 - 2x)RT/V = 4.06RT/V = 2.00 atm => RT/V = (2.00/4.06) atm/mole = (1.00/2.03) atm/mole

The equilibrium constant is

K_P = [P(HCl)]² / [P(H₂)P(Cl₂)] = (3.59 - 2x)²(RT/V)² / [(0.47 + x)(RT/V)(xRT/V)]
= (3.59 - 2x)² / [(0.47 + x)x] = 193

Cross multiplying, we get

(3.59 - 2x)² = 193x(0.47 + x) => (3.59)² - 7.18x + 4x² = (193)(0.47)x + 193x²
=> 189x² + [7.18 + (193)(0.47)]x - (3.59)² = 0 => ax² + bx + c = 0

where

a = 189
b = 7.18 + (193)(0.47)
c = - (3.59)²

Solve for x using the quadratic formula.

x = [- b ± sqrt(b² - 4ac)] / 2a => [- b + sqrt(b² - 4ac)] / 2a = 0.1088 mole

The equilibrium partial pressures are

P(H₂) = (0.47 + x)RT/V = (0.47 + 0.1088)(1.00/2.03) atm = 0.2851 atm
P(Cl₂) = xRT/V = (0.1088)(1.00/2.03) atm = 0.0536 atm
P(HCl) = (3.59 - 2x)RT/V = [3.59 - (2)(0.1088)](1.00/2.03) atm = 1.661 atm

PROBLEM

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at 500 °C.

CH₃OCH₃(g) => CH₄(g) + CO(g) + H₂(g)

(a) Starting with 8.00 g, what mass remains (in grams) after 125 min and after 145 min?
(b) Calculate the time in minutes required to decrease 7.60 ng (nanograms) to 2.25 ng.
(c) What fraction of the original dimethyl ether remains after 150 min?

[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 15.72]

SOLUTION

(a) For a first order process,

q(t)/q₀ = (1/2)^t/t_1/2 (1) => q(t) = q₀(1/2)^t/t_1/2

where q₀ is the initial quantity of a reactant, q(t) is the quantity of the reactant that remains at time t, and t_1/2 is the half-life.

q(125 min) = (8.00 g)(1/2)^{(125 min / 25.0 min)} = (8.00 g)/32 = 0.25 g
q(145 min) = (8.00 g)(1/2)^{(145 min / 25.0 min)} = (8.00 g)(1.795 x 10^-2) = 0.1436 g

(b) From equation (1), we get

ln[q(t)/q₀] = (t/t_1/2) ln(1/2)
=> t = t_1/2 ln[q(t)/q₀] / ln(1/2) = (25.0 min) ln(2.25 ng / 7.60 ng) / ln(1/2) = 43.90 min

(c) Using (1), we get

q(150 min)/q₀ = (1/2)^{(150 min / 25.0 min)} = (1/2)⁶ = 1/64

PROBLEM

Iodide ion, I ^-, is oxidized to hypoiodite ion, IO^-, by hypochlorite, ClO^-, in basic solution according to

I ^-(aq) + ClO^-(aq) => IO^-(aq) + Cl^-(aq)

Three initial-rate experiments were conducted; the results are shown in the following table:

Experiment	[I -] (mole/L)	[ClO-] (mole/L)	Initial Rate of Formation of IO- (mole/L s)
1	0.017	0.015	0.156
2	0.052	0.015	0.476
3	0.016	0.061	0.596

(a) Determine the order of the reaction with respect to each reactant.
(b) For the reaction, write the rate law that is consistent with the result of part (a), calculate the value of the specific rate constant, k, and specify units.

[from 2005 AP Chemistry Free-Response Question 3]

SOLUTION

(a) From experiments 1 and 2, for which the initial concentrations of ClO^- are the same, we see that tripling the initial concentration of I ^- triples the rate, so the reaction is first order with respect to I ^-.

From experiments 1 and 3, for which the initial concentrations of I ^- are approximately the same, we see that quadrupling the initial concentration of ClO^- quadruples the rate, so the reaction is first order with respect to ClO^-.

(b) The rate law is of the form

rate = k[I ^-][ClO^-]

The rate constant is

k = rate / [I ^-][ClO^-]

From the three experiments in the table, we get k = 611.8, 610.3, and 610.7 L/mole s. The average of the three values is k = 610.9 L/mole s.

PROBLEM

The catalyzed decomposition of hydrogen peroxide, H₂O₂(aq), is represented by the following equation:

2H₂O₂(aq) => 2H₂O(l) + O₂(g)

The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a first-order reaction. Some of the experimental data are shown in the table below.

[H2O2] (mole/L)	Time (minutes)
1.00	0.0
0.78	5.0
0.61	10.0

During the analysis of the data, the graph below was produced.

^
|\
| \
|  \
|   \
|    \
+-----\-------->
 Time (minutes)

(a) Label the vertical axis of the graph.
(b) What are the units of the rate constant, k, for the decomposition of H₂O₂(aq)?
(c) On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition of 1.00 M H₂O₂(aq).

[from 2005 AP Chemistry Free-Response Question 3]

SOLUTION

(a) For a first-order reaction, the concentration of H₂O₂ at time t is

[H₂O₂](t) = [H₂O₂](0) e^-kt

where [H₂O₂](0) is the concentration at time t = 0, and k is the rate constant. If we take the natural logarithms of both sides, we get

ln [H₂O₂](t) = ln [H₂O₂](t) - kt

A graph of ln [H₂O₂](t) vs. t is a straight line with a negative slope equal to - k and an intercept of ln [H₂O₂](0) with the vertical axis. The vertical axis should be labeled "ln [H₂O₂](t)".

(b) The rate law for the reaction is

rate = - (1/2) d[H₂O₂] / dt = k[H₂O₂]

Since the units of [H₂O₂] are M/s or moles/L s, the units of k are s^-1.

(c) For the uncatalyzed decomposition of 1.00 M H₂O₂, the graph of ln [H₂O₂](t) vs. time would be a straight line with negative slope starting at the same point at t = 0 but decreasing at a slower rate.

The solution of a salt derived from a strong acid and a strong base is neutral. The solution of a salt derived from a strong acid and a weak base is acidic. The solution of a salt derived from a weak acid and a strong base is basic. The solution of a salt derived from a weak acid and a weak base is (acidic, basic, approximately neutral) if K_a (>, <, ≈) K_b.

PROBLEM

Which of the following is true of pure water at a temperature of 40 °C?
(a) It is neutral (i.e., neither acidic nor basic).
(b) It has a pH of 7.00.
(c) Both (a) and (b).
(d) Neither (a) nor (b).

SOLUTION

The ionization constant of water is K_w = 10^-14 at 25 °C. A solution of pure water at this temperature will have [H⁺] = [OH ^-] = 10^-7 => pH = - log [H⁺] = 7.00. At other temperatures, the value of K_w is different and the values of [H⁺] and [OH ^-] will be different from their values at 25 °C, but still equal to each other, since each ionization of H₂O produces one H⁺ and one OH ^-. A neutral solution is a solution in which [H⁺] = [OH ^-], so at 40 °C the water will still be neutral, but its pH will not be exactly 7.00. So only (a) is true.

PROBLEM

Determine whether or not each of the following is a Lewis acid-base reaction:
(a) Al(OH)₃(s) + OH ^-(aq) => Al(OH)₄^-(aq)
(b) Cl₂(g) + H₂O(l) => HOCl(aq) + H⁺(aq) + Cl ^-(aq)
(c) SnCl₄(s) + 2Cl ^-(aq) => SnCl₆^2-(aq)
(d) NH₄⁺(g) + NH₂^-(g) => 2NH₃(g)
(e) H⁺(aq) + NH₃(aq) => NH₄⁺(aq)

[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 13]

SOLUTION

A Lewis acid-base reaction is one in which one substance, the Lewis base, donates an electron pair to another substance, the Lewis acid.

(a) Al(OH)₃(s) + OH ^-(aq) => Al(OH)₄^-(aq)

The Lewis structures for the reactants and product are

      H                   H
      |                   |
     :O:                 :O:
   .. |    ..          .. |  ..
H--O--Al + :O--H => H--O--Al--O--H
   .. |    ..          .. |  ..
     :O:                 :O:
      |                   |
      H                   H

The hydroxide ion (OH ^-) donates an electron pair (shown in pink) and is the Lewis base. The aluminum hydroxide (Al(OH)₃) accepts an electron pair and is the Lewis acid.

(b) Cl₂(g) + H₂O(l) => HOCl(aq) + H⁺(aq) + Cl ^-(aq)

The Lewis structures for the reactants and products are

 ..  ..       ..       ..    ..         ..       ..  ..          ..
:Cl--Cl: + H--O--H => :Cl + :Cl: + H + :O--H => :Cl--O--H + H + :Cl:
 ..  ..       ..       ..    ..         ..       ..  ..          ..

This reaction is more complex than a simple transfer of an electron pair from one of the reactants to the other, so it is not a Lewis acid-base reaction.

(c) SnCl₄(s) + 2Cl ^-(aq) => SnCl₆^2-(aq)

The Lewis structures for the reactants and product are

                                ..      ..
     ..                        :Cl      Cl:
    :Cl:                        ..\    /..
 ..  |   ..     ..     ..      ..  \  /  ..
:Cl--Sn--Cl: + :Cl: + :Cl: => :Cl---Sn---Cl:
 ..  |   ..     ..     ..      ..  /  \  ..
    :Cl:                        ../    \..
     ..                        :Cl      Cl:
                                ..      ..

The two chlorine ions (Cl ^-) each donate an electron pair (shown in pink and blue) and are the Lewis base. The tin(IV) chloride or stannic chloride (SnCl₄) accepts two electron pairs and is the Lewis acid.

(d) NH₄⁺(g) + NH₂^-(g) => 2NH₃(g)

The Lewis structures for the reactants and product are

   H       H        H       H
   |       |        |       |
H--N--H + :N: => H--N: + H--N:
   |       |        |       |
   H       H        H       H

The nitrogen dihyride ion (NH₂^-) donates an electron pair (shown in blue) and is the Lewis base. The hydrogen ion (H⁺) (shown in green) from the ammonium ion (NH₄⁺) accepts an electron pair and is the Lewis acid.

(e) H⁺(aq) + NH₃(aq) => NH₄⁺(aq)

The Lewis structures for the reactants and product are

     H          H
     |          |
H + :N--H => H--N--H
     |          |
     H          H

The ammonia (NH₃) donates an electron pair (shown in pink) and is the Lewis base. The hydrogen ion (H⁺) accepts the electron pair and is the Lewis acid.

PROBLEM

Decide if each of the following substances should be classified as a Lewis acid or base:
(a) Mn²⁺
(b) CH₃NH₂
(c) H₂NOH in the reaction H₂NOH(aq) + HCl => [H₃NOH]Cl(aq)
(d) SO₂ in the reaction SO₂(g) + BF₃(g) <=> O₂S-BF₃(s)
(e) Zn(OH)₂ in the reaction Zn(OH)₂(s) + 2OH ^-(aq) <=> Zn(OH)₄^2-(aq)

[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 17.74]

SOLUTION

(a) Mn²⁺ has two missing valence electrons. It can accept an electron pair, so it is a Lewis acid.

(b) CH₃NH₂:

   H  H
   |  |
H--C--N:
   |  |
   H  H

CH₃NH₂ has a lone electron pair which it can donate, so it is a Lewis base.

(c) H₂NOH + HCl => [H₃NOH]Cl:

      H                    H
  ..  |       ..       ..  |     ..
H--O--N: + H :Cl: => H--O--N--H :Cl:
  ..  |       ..       ..  |     ..
      H                    H

The H₂NOH donates an electron pair, so it is a Lewis base.

(d) SO₂(g) + BF₃(g) <=> O₂S-BF₃(s):
 ..   ..        ..  ..
:O:  :F:       :O: :F:
 |    |  ..     |   |  ..
 S: + B--F: <=> S---B--F:
||    |  ..    ||   |  ..
:O:  :F:       :O: :F:
      ..            ..

The SO₂ donates an electron pair, so it is a Lewis base.

(e) Zn(OH)₂(s) + 2OH ^-(aq) <=> Zn(OH)₄^2-(aq):

        H                    H
        |                    |
       :O:                  :O:
  ..    |     ..         ..  |   ..
H--O: + Zn + :O--H <=> H--O--Zn--O--H
  ..    |     ..         ..  |   ..
       :O:                  :O:
        |                    |
        H                    H

Zn(OH)₂ accepts two electron pairs, so it is a Lewis acid.

PROBLEM

Decide if each of the following substances should be classified as a Lewis acid or base:
(a) BCl₃
(b) H₂N-NH₂, hydrazine
(c) CN ^- in the reaction Au⁺(aq) + 2CN ^-(aq) <=> [Au(CN)₂]^-(aq)
(d) Fe³⁺

[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 17.75]

SOLUTION

(a) BCl₃:

 ..
:Cl:
  |  ..
  B--Cl:
  |  ..
:Cl:
 ..

BCl₃ can accept an electron pair, so it is a Lewis acid.

(b) H₂N-NH₂:

 H  H
 |  |
:N--N:
 |  |
 H  H

H₂N-NH₂ can donate two electron pairs, so it is a Lewis base.

(c) Au⁺(aq) + 2CN ^-(aq) <=> [Au(CN)₂]^-(aq):

:N///C: + Au + :C///N: <=> :N///C--Au--C///N:     /// = triple bond

The two CN ^- each donate an electron pair, so the CN ^- is a Lewis base.

(d) Fe³⁺ has three missing valence electrons. It can accept an electron pair, so it is a Lewis acid.

PROBLEM

Equilibrium constants can be measured for the dissociation of Lewis acid-base complexes such as the dimethyl ether complex of BF₃, (CH₃)₂O-BF₃. The value of K (here K_P) for the reaction

(CH₃)₂O-BF₃(g) <=> BF₃(g) + (CH₃)₂O(g)

is 0.17.
(a) Tell which product is the Lewis acid and which is the Lewis base.
(b) What is the F-B-F angle in BF₃? In (CH₃)₂O-BF₃?
(c) What is the hybridization of O in (CH₃)₂O-BF₃? Of the boron atom?
(d) If you place 1.00 g of the complex in a 565 mL flask at 25°C, what is the total pressure in the flask? What are the partial pressures of the Lewis acid, the Lewis base, and the complex?

[from Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing), problem 17.110]

SOLUTION

(a) The Lewis structures for the reactant and products are

           H                    H
    ..     |            ..      |
   :F:  H--C--H        :F:   H--C--H
 .. |      |         .. |       |
:F--B------O:   <=> :F--B  +   :O:
 .. |      |         .. |       |
   :F:  H--C--H        :F:   H--C--H
    ..     |            ..      |
           H                    H

Since the oxygen atom in (CH₃)₂O has two lone electron pairs which can bond with the boron atom in BF₃, BF₃ is the Lewis acid and (CH₃)₂O is the Lewis base.

(b) The boron atom in BF₃ has three bonding electron pairs and no lone pairs surrounding it. Therefore the shape of BF₃ is trigonal planar, and the F-B-F angle is 120°. The boron atom in (CH₃)₂O-BF₃ has three bonding electron pairs and one lone pair surrounding it. Therefore the overall arrangement of the four electron pairs is tetrahedral, and the F-B-F angle is 109.5°.

(c) There are four electron pairs surrounding either the oxygen or the boron atom in (CH₃)₂O-BF₃. Therefore, the hybridization is sp³ in both cases.

(d) The molecular mass of (CH₃)₂O-BF₃ is

M = (2)[12.01 g/mole + (3)(1.008 g/mole)] + 16.00 g/mole + 10.81 g/mole + (3)(19.00 g/mole)
= 113.878 g/mole

Therefore, 1.00 g of (CH₃)₂O-BF₃ equals 8.781 x 10^-3 mole of (CH₃)₂O-BF₃. If none of this dissociated, the pressure would be

P₀ = nRT/V = (8.781 x 10^-3 mole)(0.0821 L atm / mole K)(298 K) / (565 x 10^-3 L) = 0.3803 atm

Let x be the number of atm of (CH₃)₂O-BF₃ which dissociate. Then at equilibrium we have

P[(CH₃)₂O-BF₃] = P₀ - x
P[(CH₃)₂O] = P[BF₃] = x

The equilibrium constant is

K_P = P[(CH₃)₂O]P[BF₃] / P[(CH₃)₂O-BF₃] = x² / (P₀ - x) => K_PP₀ - K_Px = x²
=> x² + K_Px - K_PP₀ = 0
=> x = [- K_P + sqrt(K_P² + 4K_PP₀)] / 2 = {- 0.17 + sqrt[0.17² + (4)(0.17)(0.3803)]} / 2 = 0.1831 atm

The final partial pressures are

P[(CH₃)₂O-BF₃] = P₀ - x = 0.3803 atm - 0.1831 atm = 0.1972 atm
P[(CH₃)₂O] = P[BF₃] = x = 0.1831 atm

The final total pressure is

P = P[(CH₃)₂O-BF₃] + P[(CH₃)₂O] + P[BF₃] = 0.1972 atm + 0.1831 atm + 0.1831 atm
= 0.5633 atm

Alternate Method:

We could solve part (d) using K_C instead of K_P. The equilibrium constant K_C for the above reaction is related to the equilibrium concentrations of the reactant and products according to

K_C = [(CH₃)₂O][BF₃] / [(CH₃)₂O-BF₃]
= {P[(CH₃)₂O] / RT}{P[BF₃] / RT} / {P[(CH₃)₂O-BF₃] / RT}
= K_P / RT = (0.17) / [(0.0821 L atm / mole K)(298 K)] = 6.948 x 10^-3

This follows from the general fact that, for an ideal gas, the pressure P and concentration n/V are related by

PV = nRT => n/V = P/RT

If none of the initial (CH₃)₂O-BF₃ dissociated, its concentration would be

[(CH₃)₂O-BF₃]₀ = (8.781 x 10^-3 mole) / (565 x 10^-3 L) = 1.554 x 10^-2 mole/L = 1.554 x 10^-2 M = C₀

Let x be the number of moles per liter of (CH₃)₂O-BF₃ which dissociate. Then the equilibrium concentrations of (CH₃)₂O-BF₃, (CH₃)₂O, and BF₃ are

[(CH₃)₂O-BF₃] = C₀ - x
[(CH₃)₂O] = [BF₃] = x

We have

K_C = [(CH₃)₂O][BF₃] / [(CH₃)₂O-BF₃] = x² / (C₀ - x) => K_CC₀ - K_Cx = x²
=> x² + K_Cx - K_CC₀ = 0
=> x = {- K_C + sqrt(K_C² + 4K_CC₀)} / 2
= {- 6.948 x 10^-3 + sqrt[(6.948 x 10^-3)² + (4)(6.948 x 10^-3)(1.554 x 10^-2)]} / 2 = 7.483 x 10^-3 M

The equilibrium concentrations of (CH₃)₂O-BF₃, (CH₃)₂O, and BF₃ are

[(CH₃)₂O-BF₃] = C₀ - x = 1.554 x 10^-2 M - 7.483 x 10^-3 M = 8.059 x 10^-3 M
[(CH₃)₂O] = [BF₃] = x = 7.483 x 10^-3 M

The final partial pressures are

P[(CH₃)₂O-BF₃] = RT[(CH₃)₂O-BF₃] = (0.0821 L atm / mole K)(298 K)(8.059 x 10^-3 M)
= 0.1972 atm
P[(CH₃)₂O] = RT[(CH₃)₂O] = (0.0821 L atm / mole K)(298 K)(7.483 x 10^-3 M) = 0.1831 atm
P[BF₃] = RT[BF₃] = (0.0821 L atm / mole K)(298 K)(7.483 x 10^-3 M) = 0.1831 atm

The final total pressure is

P = P[(CH₃)₂O-BF₃] + P[(CH₃)₂O] + P[BF₃] = 0.1972 atm + 0.1831 atm + 0.1831 atm
= 0.5633 atm

PROBLEM

Draw electron dot diagrams to show the possible structure of H₃O⁺, H₅O₂⁺, H₇O₃⁺, and H₉O₄⁺.

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.1]

SOLUTION

H₃O⁺:

   ..
H--O--H
   |
   H

H₅O₂⁺:

               H
   ..          |
H--O--H-hhhhh-:O:     hhhhh = hydrogen bond
   |           |
   H           H

H₇O₃⁺:

 H                       H
 |           ..          |
:O:-hhhhh-H--O--H-hhhhh-:O:     hhhhh = hydrogen bond
 |           |           |
 H           H           H

H₉O₄⁺:

 H                       H
 |           ..          |
:O:-hhhhh-H--O--H-hhhhh-:O:     hhhhh = hydrogen bond
 |           |           |
 H           H           H
             |
             h
             h
             h
             |
             ..
          H--O--H
             ..

PROBLEM

What is the OH ^- concentration of a solution which is (a) 0.001 M HCl, (b) 0.001 M H₂SO₄?

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.5]

SOLUTION

(a) Hydrochloric acid (HCl) is a strong monoprotic acid which ionizes completely. Therefore, [H⁺] = 10^-3 M and

[OH ^-] = K_w / [H⁺] = 10^-14 / 10^-3 = 10^-11 M

(b) Sulfuric acid (H₂SO₄) is a strong acid which ionizes once according to
H₂SO₄ => H⁺ + HSO₄^-

HSO₄^- is a weak acid which ionizes according to

HSO₄^- => H⁺ + SO₄^2-

with ionization constant K_a2 = 1.3 x 10^-2. Let x be the number of moles per liter of H₂SO₄ which ionize only once to form HSO₄^- and let y be the number of moles per liter of H₂SO₄ which ionize twice to form SO₄^2-. At equilibrium, we have

[H⁺] = x + 2y
[HSO₄^-] = x
[SO₄^2-] = y

Let f = 0.001 M, the concentration of H₂SO₄. Since x + y = f, y = f - x, and

[H⁺] = x + 2(f - x) = 2f - x
[HSO₄^-] = x
[SO₄^2-] = f - x

The ionization constant

K_a2 = [H⁺][SO₄^2-]/[HSO₄^-] = (2f - x)(f - x)/x => K_a2x = 2f² - 3fx + x²
=> x² - (3f + K_a2)x + 2f² = 0
=> x = {(3f + K_a2) ± sqrt[(3f + K_a2)² - 8f²]} / 2 = (1.587 x 10^-2, 1.26 x 10^-4)
=> y = f - x = (- 1.487 x 10^-2, 8.74 x 10^-4)

The "+" solution for x yields a negative value for y, so we reject it and get

x = 1.26 x 10^-4
y = 8.74 x 10^-4
[H⁺] = x + 2y = 1.874 x 10^-3
[OH ^-] = K_w/[H⁺] = (10^-14) / (1.874 x 10^-3) = 5.336 x 10^-12

PROBLEM

In a 0.050 M solution of hydrofluoric acid, HF, in water at 25°C, only 11% of the molecules dissociate to form ions. Calculate the ionization constant for hydrofluoric acid.

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.13]

SOLUTION

The ionization constant is

K_a = [H⁺][F ^-]/[HF]

Since 11% of the HF dissociate, we have

[H⁺] = [F ^-] = (0.11)(0.050 M)
[HF] = (0.050 M)(1 - 0.11)

and

K_a = (0.11)²(0.050)²/[(0.050)(1 - 0.11)] = 6.8 x 10^-4

PROBLEM

Identify each of the following as an acid, base, salt, or amphoteric substance:
(a) sodium phosphate (Na₃PO₄)
(b) hydrogen chloride (HCl)
(c) sodium hydroxide (NaOH)
(d) hydrogen sulfate or bisulfate (HSO₄^-)
(e) oxalate (C₂O₄^2-)

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 52]

SOLUTION

(a) Sodium phosphate is the salt of a strong base (sodium hydroxide, NaOH) and a weak acid (phosphoric acid, H₃PO₄).

(b) Hydrochloric acid is a strong acid.

(c) Sodium hydroxide is a strong base.

(d) Bisulfate can either donate a proton or accept a proton and is therefore both a Bronsted acid and a Bronsted base. It is therefore also amphoteric.

(e) Oxalate can accept protons and is therefore a Bronsted base.

PROBLEM

Determine whether each of the following oxides dissolve in water to form acidic or basic solutions:
(a) sodium oxide (Na₂O)
(b) calcium oxide (CaO)
(c) aluminum oxide (Al₂O₃)
(d) sulfur trioxide (SO₃)

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Diagnostic Test problem 68]

SOLUTION

(a) All alkali metal oxides and all alkaline earth metal oxides except beryllium oxide (BeO) dissolve in water to form basic solutions. Sodium oxide dissolves in water to form sodium hydroxide, a base:

Na₂O(s) + H₂O(l) => 2NaOH(aq)

(b) Calcium oxide dissolves in water to form calcium hydroxide, which is insoluble in aqueous solution:

CaO(s) + H₂O(l) => Ca(OH)₂(s)

(c) Aluminum oxide is amphoteric.

(d) Most nonmetallic oxides dissolve in water to form acidic solutions. Sulfur trioxide reacts with water to form sulfuric acid:

SO₃(g) + H₂O(l) => H₂SO₄(aq)

PROBLEM

If each of the following salts undergoes hydrolysis, indicate whether the resulting solution will be acidic or basic:
(a) Na₂SO₄
(b) K₂SO₄
(c) NaNO₃
(d) Cu(NO₃)₂

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), problem 8.5]

SOLUTION

(a) Na₂SO₄ + 2H₂O => 2NaOH + H₂SO₄

Sodium hydroxide (NaOH) is a strong base and sulfuric acid (H₂SO₄) is a strong acid, but since there are twice as many moles of NaOH as H₂SO₄, the solution will be basic.

(b) K₂SO₄ + 2H₂O => 2KOH + H₂SO₄

Potassium hydroxide (KOH) is a strong base and sulfuric acid (H₂SO₄) is a strong acid, but since there are twice as many moles of KOH as H₂SO₄, the solution will be basic.

(c) NaNO₃ + H₂O => NaOH + HNO₃

Sodium hydroxide (NaOH) is a strong base and HNO₃ is a strong acid. There will be equal numbers of H⁺ and OH^- ions, so the solution will be neutral.

(d) Cu(NO₃)₂ + 2H₂O => Cu(OH)₂ + 2HNO₃

Cuprous hydroxide (Cu(OH)₂) is a weak base and nitric acid (HNO₃) is a strong acid, so the solution will be acidic.

PROBLEM

The following three mixtures have been prepared: CaO plus water, SiO₂ plus water, and CO₂ plus water. For each mixture, predict whether the pH is less than 7, equal to 7, or greater than 7. Justify your answer.

[from 2005 AP Chemistry Free-Response Question 5]

SOLUTION

CaO: Calcium oxide is basic. It reacts with water to form calcium hydroxide according to

CaO + H₂O => Ca(OH)₂

Most metallic oxides are basic. The pH will be greater than 7.

SiO₂: SiO₂ does not react with water, so the pH will be equal to 7.

CO₂: When CO₂ is added to water, carbonic acid is formed according to

CO₂ + H₂O => H₂CO₃

The pH will be less than 7.

PROBLEM

Five acids are listed below in order of decreasing acid strength:

HCl > HC₂H₃O₂ > HCN > H₂O > NH₃

Determine whether the equilibrium constant for each of the following reactions is less than, equal to, or greater than one:
(a) HCl(aq) + CN ^-(aq) <=> HCN(aq) + Cl ^-(aq)
(b) HCl(aq) + H₂O(l) <=> H₃O⁺(aq) + Cl ^-
(c) HC₂H₃O₂(aq) + OH ^-(aq) <=> C₂H₃O₂^-(aq) + H₂O(l)
(d) H₂O(aq) + NH₂^-(aq) <=> NH₃(aq) + OH ^-(aq)
(e) HCN(aq) + C₂H₃O₂^- <=> HC₂H₃O₂(aq) + CN ^-(aq)

[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 21]

SOLUTION

The five acids listed are hydrochloric acid (HCl), acetic acid (HC₂H₃O₂), hydrocyanic acid (HCN), water (H₂O), and ammonia (NH₃). The ionization constants for the above acids are

K_a1 > K_a2 > K_a3 > K_a4 > K_a5

where

K_a1 = [H₃O⁺][Cl ^-] / [HCl]
K_a2 = [H₃O⁺][C₂H₃O₂^-] / [HC₂H₃O₂]
K_a3 = [H₃O⁺][CN ^-] / [HCN]
K_a4 = [H₃O⁺][OH ^-]
K_a5 = [H₃O⁺][NH₂^-] / [NH₃]

(a) HCl(aq) + CN ^-(aq) <=> HCN(aq) + Cl ^-(aq)

K_eq = [HCN][Cl ^-] / [HCl][CN ^-] = [HCN][Cl ^-][H₃O⁺] / [HCl][CN ^-][H₃O⁺] = K_a1/K_a3 > 1

(b) HCl(aq) + H₂O(l) <=> H₃O⁺(aq) + Cl ^-

K_eq = [H₃O⁺][Cl ^-] / [HCl] > 1

(c) HC₂H₃O₂(aq) + OH ^-(aq) <=> C₂H₃O₂^-(aq) + H₂O(l)

K_eq = [C₂H₃O₂^-] / [HC₂H₃O₂][OH ^-] = [C₂H₃O₂^-][H₃O⁺] / [HC₂H₃O₂][OH ^-][H₃O⁺]
= K_a2/K_a4 > 1

(d) H₂O(aq) + NH₂^-(aq) <=> NH₃(aq) + OH ^-(aq)

K_eq = [NH₃][OH ^-] / [NH₂^-] = [NH₃][OH ^-][H₃O⁺] / [NH₂^-][H₃O⁺] = K_a4/K_a5 > 1

(e) HCN(aq) + C₂H₃O₂^- <=> HC₂H₃O₂(aq) + CN ^-(aq)

K_eq = [HC₂H₃O₂][CN ^-] / [HCN][C₂H₃O₂^-] = [HC₂H₃O₂][CN ^-][H₃O⁺] / [HCN][C₂H₃O₂^-][H₃O⁺]
= K_a3/K_a2 < 1

PROBLEM

Write the molecular and ionic equations for the following neutralizations and predict the pH (< 7, > 7, or ≈ 7) of a solution of the salt formed:
(a) phosphoric acid and calcium hydroxide
(b) acetic acid and magnesium hydroxide
(c) carbonic acid and potassium hydroxide

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercises 14.3, 14.4]

SOLUTION

(a) The molecular equation is

2H₃PO₄ + 3Ca(OH)₂ => 6H₂O + Ca₃(PO₄)₂

The ionic equation is

6H⁺ + 2PO₄^3- + 3Ca²⁺ + 6OH^- => 6H₂O + 3Ca²⁺ + 2PO₄^3-

The net ionic equation is

6H⁺ + 6OH^- => 6H₂O

Phosphoric acid is a weak acid with K_a = 7.5 x 10^-3. Calcium hydroxide is a weak base with K_b = 5.5 x 10^-6. Since K_a > K_b, the solution of the salt Ca₃(PO₄)₂ is acidic with pH < 7.

(b) The molecular equation is

2HCH₃COO + Mg(OH)₂ => 2H₂O + Mg(CH₃COO)₂

The ionic equation is

2H⁺ + 2CH₃COO^- + Mg²⁺ + 2OH^- => 2H₂O + Mg²⁺ + 2CH₃COO^-

The net ionic equation is

2H⁺ + 2OH^- => 2H₂O

Acetic acid is a weak acid with K_a = 1.8 x 10^-5. Magnesium hydroxide is a weak base with K_b = 1.8 x 10^-11. Since K_a > K_b, the solution of the salt Mg(CH₃COO)₂ is acidic with pH < 7.

(c) The molecular equation is

H₂CO₃ + 2KOH => 2H₂O + K₂CO₃

The ionic equation is

2H⁺ + CO₃^2- + 2K⁺ + 2OH^- => 2H₂O + 2K⁺ + CO₃^2-

The net ionic equation is

2H⁺ + 2OH^- => 2H₂O

Since carbonic acid is a weak acid and potassium hydroxide is a strong base, the solution of the salt K₂CO₃ is basic with pH > 7.

PROBLEM

In a 0.1 N solution of carbonic acid, H₂CO₃, in water at 25°C, only 0.29% ionization takes place. Calculate the ionization constant for carbonic acid.

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 15.14]

SOLUTION

A 0.1 N solution of carbonic acid contains 0.1 equivalents of H₂CO₃ per liter of solution. Each mole of H₂CO₃ is equal to two equivalents of H₂CO₃. Therefore, a 0.1 N solution of carbonic acid contains 0.05 moles of H₂CO₃ per liter of solution, so it is a 0.05 M solution of H₂CO₃. Carbonic acid ionizes according to

H₂CO₃ + H₂O <=> H₃O⁺ + HCO₃^-

At equilibrium, we have

[H₂CO₃] = (0.05)(1 - 0.0029) = 0.049855
[HCO₃^-] = [H₃O⁺] = (0.05)(0.0029) = 1.45 x 10^-4

The ionization constant of carbonic acid is

K_a = [H₃O⁺][HCO₃^-]/[H₂CO₃] = (1.45 x 10^-4)²/(0.049855) = 4.217 x 10^-7

PROBLEM

The pH of a water solution of sodium carbonate may be (A) 1.5, (B) 3.5, (C) 5.5, (D) 8.5.

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), multiple choice question 15.5]

SOLUTION

Sodium carbonate, Na₂CO₃, is soluble in aqueous solution:

Na₂CO₃ => 2Na⁺ + CO₃^2-

The carbonate can hydrolyze with H₂O to form the hydrogen carbonate ion HCO₃^-,

CO₃^2- + H₂O <=> HCO₃^- + OH^-

for which the hydrolysis constant is K_h = K_w/K_a2 = (10^-14) / (4.8 x 10^-11) = 2.08 x 10^-4, where K_a2 = 4.8 x 10^-11 is the ionization constant of the hydrogen carbonate ion. The carbonate can also react with H₃O⁺ to form the hydrogen carbonate ion,

CO₃^2- + H₃O⁺ <=> HCO₃^- + H₂O

for which the equilibrium constant is 1/K_a2 = 1/(4.8 x 10^-11) = 2.083 x 10¹⁰. Whichever species, H₂O or H₃O⁺, the carbonate reacts with, the result will be more OH ^- and less H₃O⁺, resulting in a pH which is greater than 7 or basic.

In general, the solution of a salt derived from a strong base and a weak acid, or a strong base and a neutral entity, is basic. The carbonate ion, CO₃^2- is the conjugate base of a weak acid, HCO₃^-, and is therefore a strong base. The sodium ion, Na⁺, is neither acidic nor basic. The correct answer is (D).

PROBLEM

Calculate the hydrolysis constant for the reaction

HCO₂^- + H₂O <=> HCO₂H + OH^-

and find the concentrations of H₃O⁺, OH^-, HCO₂^-, and HCO₂H in a solution of 0.15 M HCO₂Na. The dissociation constant of formic acid (HCO₂H) is K_a = 1.8 x 10^-4.

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.16]

SOLUTION

The hydrolysis constant for the hydrolysis of HCO₂^- is

K_h = K_w/K_a = (10^-14) / (1.8 x 10^-4) = 5.556 x 10^-11

HCO₂Na is soluble in aqueous solution. Let x be the number of moles per liter of HCO₂^- which hydrolyze. At equilibrium,

[HCO₂^-] = 0.15 - x
[HCO₂H] = x
[OH^-] = x

The hydrolysis constant is

K_h = [HCO₂H][OH^-] / [HCO₂^-] = x² / (0.15 - x)

Solve for x.

(0.15 - x)K_h = x²
x² + K_hx - 0.15K_h = 0
x = {- K_h + sqrt[K_h² + 4(0.15)K_h]} / 2
= {- 5.556 x 10^-11 + sqrt[(5.556 x 10^-11)² + 4(0.15)(5.556 x 10^-11)]} / 2
= 2.887 x 10^-6 M

Thus,

[OH^-] = [HCO₂H] = x = 2.887 x 10^-6 M
[HCO₂^-] = 0.15 - x = 0.15 - 2.887 x 10^-6 = 0.15 M
[H₃O⁺] = K_w / [OH^-] = (10^-14) / (2.887 x 10^-6) = 3.464 x 10^-9 M

PROBLEM

A solution of equal concentrations of lactic acid and sodium acetate was found to have pH = 3.08. (a) What are the values of pK_a and K_a of lactic acid? (b) What would the pH be if the acid had twice the concentration of the salt?

SOLUTION

(a) Let L^- and Ac^- represent the lactate and acetate ions. Lactic and acetic acid ionize according to

HL + H₂O <=> L^- + H₃O⁺, K_a1 = ?
HAc + H₂O <=> Ac^- + H₃O⁺, K_a2 = 1.8 x 10^-5

Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac^- which combine with H₃O⁺ to form HAc. Assume that the initial concentrations of HL and NaAc are both 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are

[HL] = 1 - x = a - x
[L^-] = x
[HAc] = y
[Ac^-] = 1 - y = a - y
[H₃O⁺] = x - y

where a = 1. Define

z = x - y => x = y + z

Then the equilibrium concentrations can be written as

[HL] = a - y - z
[L^-] = y + z
[HAc] = y
[Ac^-] = a - y
[H₃O⁺] = z = 10^-pH = 10^-3.08 = 8.318 x 10^-4 M

The equilibrium constants are

K_a1 = [L^-][H₃O⁺] / [HL] = (y + z)z / (a - y - z) (1)
K_a2 = [Ac^-][H₃O⁺] / [HAc] = (a - y)z / y => K_a2y = az - yz => y = az / (K_a2 + z) (2)

From (2), we get

y = (1)(8.318 x 10^-4) / (1.8 x 10^-5 + 8.318 x 10^-4) = 0.9788

From (1), we get

K_a1 = (0.9788 + 8.318 x 10^-4)(8.318 x 10^-4) / (1 - 0.9788 - 8.318 x 10^-4) = 4.004 x 10^-2

Note: This value for K_a1 differs from the accepted value of 1.4 x 10^-4 under standard conditions (Kotz and Treichel 1996, p. 820; Nims and Smith 1936, p. 151, Fig. 3), but this is what results from assuming pH = 3.08.

Kotz, John C., and Treichel, Paul Jr. 1996, Chemistry & Chemical Reactivity, Third Edition (New York: Saunders College Publishing).

Nims, Leslie Frederick, and Smith, Paul K. 1936, "The Ionization of Lactic Acid from Zero to Fifty Degrees," J. Biol. Chem., 113, 145.

(b) Use the result from part (a), K_a1 = 4.004 x 10^-2. Let L^- and Ac^- represent the lactate and acetate ions. Lactic and acetic acid ionize according to

HL + H₂O <=> L^- + H₃O⁺, K_a1 = 4.004 x 10^-2
HAc + H₂O <=> Ac^- + H₃O⁺, K_a2 = 1.8 x 10^-5

Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac^- which combine with H₃O⁺ to form HAc. Assume that the initial concentration of HL is 2 M and the initial concentration of NaAc is 1 M. NaAc is soluble in aqueous solution. The equilibrium concentrations are

[HL] = 2 - x = a - x
[L^-] = x
[HAc] = y
[Ac^-] = 1 - y = b - y
[H₃O⁺] = x - y

where a = 2 and b = 1. The ionization constants are

K_a1 = [L^-][H₃O⁺] / [HL] = x(x - y) / (a - x) => y = x - K_a1(a - x) / x (1)
K_a2 = [Ac^-][H₃O⁺] / [HAc] = (b - y)(x - y) / y => x = y + K_a2y / (b - y) (2)

Substitute (2) into (1).

y = y + K_a2y / (b - y) - K_a1{a - [y + K_a2y / (b - y)]} / [y + K_a2y / (b - y)]

Subtract y from both sides.

0 = K_a2y / (b - y) - K_a1{a - [y + K_a2y / (b - y)]} / [y + K_a2y / (b - y)] = f(y)

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.999 and 0.9999. We can narrow down the solution to y = 0.99951641. Using (2), we get x = 1.036719985 => x - y = 0.037203575 = [H₃O⁺] => pH = - log [H₃O⁺] = 1.429. See LacticAcetic070421.xls.

Note: If we use the correct value for the ionization constant of lactic acid, K_a1 = 1.4 x 10^-4, we get y = 0.904076789, x = 0.90424644, x - y = 0.00016965 = [H₃O⁺], and pH = 3.770. See LacticAcetic070421.2.xls.

PROBLEM

One liter of a 2 M solution of lactic acid (CH₃CHOHCO₂H) is combined with one liter of a 2 M solution of sodium acetate (NaCH₃CO₂). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.

SOLUTION

(a) Let "L" and "Ac" represent the lactate (CH₃CHOHCO₂^-) and acetate (CH₃CO₂^-) ions, respectively. Then lactic and acetic acids ionize according to

HL + H₂O <=> L^- + H₃O⁺, K_a1 = 1.4 x 10^-4
HAc + H₂O <=> Ac^- + H₃O⁺, K_a2 = 1.8 x 10^-5

After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H₃O⁺ will combine with Ac^- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac^- which combine with H₃O⁺ to form HAc. At equilibrium, we have

[HL] = 1 - x = a - x
[L^-] = x
[HAc] = y
[Ac^-] = 1 - y = a - y
[H₃O⁺] = x - y

where a = 1. The ionization constants are

K_a1 = [L^-][H₃O⁺] / [HL] = x(x - y) / (a - x) => y = x - K_a1(a - x) / x (1)
K_a2 = [Ac^-][H₃O⁺] / [HAc] = (a - y)(x - y) / y => x = y + K_a2y / (a - y) (2)

Substitute (2) into (1).

y = y + K_a2y / (a - y) - K_a1{a - [y + K_a2y / (a - y)]} / [y + K_a2y / (a - y)]

Subtract y from both sides.

0 = K_a2y / (a - y) - K_a1{a - [y + K_a2y / (a - y)]} / [y + K_a2y / (a - y)] = f(y)

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10^-5 = [H₃O⁺] => pH = - log [H₃O⁺] = 4.299. See LacticAcetic070407.xls.

(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10^-5 = [H₃O⁺] => pH = - log [H₃O⁺] = 4.299. See LacticAcetic070407.2.xls.

(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10^-5 = [H₃O⁺] => pH = - log [H₃O⁺] = 4.300. See LacticAcetic070407.3.xls.

PROBLEM

The dissociation constant of HCN, hydrocyanic acid, is K_a = 4.8 x 10^-10. What is the concentration of H₃O⁺, OH^-, and HCN in a solution prepared by dissolving 0.160 mole of NaCN in 450 mL of water?

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.17]

SOLUTION

NaCN is soluble in aqueous solution. Let x be the number of moles per liter of CN^- which hydrolyze to form HCN according to

CN^- + H₂O <=> HCN + OH^-

for which the hydrolysis constant is

K_h = K_w/K_a = (10^-14) / (4.8 x 10^-10) = 2.083 x 10^-5

At equilibrium,

[CN^-] = (0.160 mole) / (0.450 L) - x = 0.3556 - x
[OH^-] = x
[HCN] = x

We have

K_h = [HCN][OH^-] / [CN^-] = x² / (0.3556 - x)
=> K_h(0.3556 - x) = x²
=> x² + K_hx - 0.3556K_h = 0
=> x = {- K_h + sqrt[K_h² + 4(0.3556)K_h]} / 2
= {- 2.083 x 10^-5 + sqrt[(2.083 x 10^-5)² + 4(0.3556)(2.083 x 10^-5)]} / 2 = 2.711 x 10^-3 M
=> [OH^-] = [HCN] = 2.711 x 10^-3 M
=> [H₃O⁺] = K_w / [OH^-] = (10^-14) / (2.711 x 10^-3) = 3.688 x 10^-12 M

PROBLEM

Calculate the solubility of lead sulfate in a solution of 0.100 M H₃O⁺ by taking account of the reaction

HSO₄^- + H₂O => H₃O⁺ + SO₄^2-

The solubility product for PbSO₄ is K_sp = 1.8 x 10^-8. Be sure to justify all approximations.

[from Mahan, Bruce A. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.19]

SOLUTION

The hydrogen sulfate or bisulfate ion, HSO₄^-, is a weak acid with ionization constant K_a = 1.2 x 10^-2. The addition of PbSO₄ to the H₃O⁺ solution causes the H₃O⁺ to combine with SO₄^2-. Let x be the amount of PbSO₄ which dissolves per liter of solution and let y be the amount of HSO₄^- produced per liter. At equilibrium, we have

[Pb²⁺] = x
[SO₄^2-] = x - y
[H₃O⁺] = 0.1 - y
[HSO₄^-] = y

The solubility product for PbSO₄ is

K_sp = [Pb²⁺][SO₄^2-] = x(x - y) (1)

The ionization constant of hydrogen sulfate is

K_a = [H₃O⁺][SO₄^2-] / [HSO₄^-] = (0.1 - y)(x - y) / y (2)

(1) and (2) are two nonlinear equations in two unknowns x and y. We can assume a value for y, solve (1) for x, substitute the result into (2) and solve for a new value of y, and keep repeating this until the values of x and y obtained converge. From (1), we get

K_spy = x² - xy => x² - xy - K_spy = 0
=> x = {y ± sqrt[y² + 4K_spy]} / 2 => {y + sqrt[y² + 4K_spy]} / 2 (3)

since x cannot be negative. From (2) we get

K_ay = 0.1x - 0.1y - xy + y² => y² - (0.1 + x + K_a)y + 0.1x = 0
=> y = {(0.1 + x + K_a) ± sqrt[(0.1 + x + K_a)² - 4(0.1x)]} / 2 (4)

For a trial value of y, we note that y cannot be greater than 0.1, the initial amount of H₃O⁺ per liter of solution. So start with y = 0.05.

Substituting y = 0.05 into (3), we get x = 5 x 10^-2. Substituting this into (4), we get y = (0.121, 4.15 x 10^-2) from the (+, -) solution, so we reject the + solution.

We substitute y = 4.15 x 10^-2 into (3) and get x = 4.15 x 10^-2. We substitute this into (4) and get y = 3.50 x 10^-2. If we repeat this process a number of times, the results converge to x = 4.10 x 10^-4 and y = 3.66 x 10^-4. Thus, the solubility of PbSO₄ is 4.10 x 10^-4 M.

Note: In order for this method to work, we must choose a way to solve for x and y iteratively in such a way that the solutions converge.

PROBLEM

A solution is prepared by dissolving 0.200 mole of sodium formate, HCO₂Na, and 0.250 moles of formic acid, HCO₂H, in approximately 200(±50) mL of water. Calculate the concentrations of H₃O⁺ and OH^-. The dissociation constant of formic acid is K_a = 1.8 x 10^-4.

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.21]

SOLUTION

Sodium formate is soluble in aqueous solution. Let x be the number of moles per liter of HCO₂H at equilibrium. Then at equilibrium we have

[HCO₂H] = x
[HCO₂^-] = [(0.250 mole) / (0.200 L)] - x + [(0.200 mole) / (0.200 L)] = 2.25 - x
[H₃O⁺] = [(0.250 mole) / (0.200 L)] - x = 1.25 - x

The dissociation constant of formic acid, which dissociates according to

HCO₂H + H₂O <=> H₃O⁺ + HCO₂^-

is

K_a = [H₃O⁺][HCO₂^-] / [HCO₂H] = (1.25 - x)(2.25 - x) / x
=> K_ax = 2.8125 - 3.5x + x²
=> x² - (K_a + 3.5)x + 2.8125 = 0
=> x = {K_a + 3.5 ± sqrt[(K_a + 3.5)² - 4(2.8125)]} / 2 = (2.2504 M, 1.2498 M)

Reject the + solution because this would yield a negative value for [HCO₂^-]. Thus, at equilbrium we have

[HCO₂H] = x = 1.2498 M
[HCO₂^-] = 2.25 - x = 1.000225 M
[H₃O⁺] = 1.25 - x = 2.25 x 10^-4 M
[OH^-] = K_w / [H₃O⁺] = (10^-14) / (2.25 x 10^-4) = 4.446 x 10^-11 M

PROBLEM

A carbonate buffer solution is prepared by dissolving 30.0 g of Na₂CO₃ in 350 mL of water and adding 150 mL of 1.00 M HCl. Calculate the pH of the solution.

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.27]

SOLUTION

The molecular weight of Na₂CO₃ is

(2)(22.99 g/mole) + 12.01 g/mole + (3)(16.00 g/mole) = 105.99 g/mole
=> 30.0 g Na₂CO₃ = [(30.0 g) / (105.99 g/mole)] Na₂CO₃ = 0.2830 mole Na₂CO₃

Na₂CO₃ is soluble in aqueous solution:

Na₂CO₃ => 2Na⁺ + CO₃^2-

The initial concentrations of Na⁺ and CO₃^2- are

[Na⁺] = [(2)(0.2830 mole)] / (0.500 L) = 1.132 M
[CO₃^2-] = (0.2830 mole) / (0.500 L) = 0.5661 M

The initial concentration of H₃O⁺ is

[H₃O⁺] = (1.00 M)(0.150 L) / (0.500 L) = 0.3 M

Now consider what could happen. The CO₃^2- could hydrolyze according to

CO₃^2- + H₂O <=> HCO₃^- + OH^-

The hydrolysis constant for this reaction is

K_h = K_w/K_a2 = (10^-14) / (4.8 x 10^-11) = 2.08 x 10^-4

where K_a2 = 4.8 x 10^-11 is the ionization constant for the reaction

HCO₃^- + H₂O <=> H₃O⁺ + CO₃^2-

The CO₃^2- could also be consumed by reacting with H₃O⁺, for which the equilibrium constant is

1/K_a2 = 1 / (4.8 x 10^-11) = 2.083 x 10¹⁰

Because 1/K_a2 >> 1, all of the excess H₃O⁺ will react with CO₃^2- and be converted into H₂O. At equilibrium, we have

[CO₃^2-] = 0.5661 M - 0.3 M = 0.2661 M
[HCO₃^-] = 0.3 M

The ionization constant

K_a2 = [H₃O⁺][CO₃^2-] / [HCO₃^-]
=> [H₃O⁺] = K_a2[HCO₃^-] / [CO₃^2-] = (4.8 x 10^-11)(0.3) / (0.2661) = 5.412 x 10^-11 M
=> pH = - log [H₃O⁺] = - log(5.412 x 10^-11) = 10.27

PROBLEM

One liter of a 2 M solution of lactic acid (CH₃CHOHCO₂H) is combined with one liter of a 2 M solution of sodium acetate (NaCH₃CO₂). (a) What is the pH of the resulting solution? (b) Repeat if the initial concentrations are 4 M instead of 2 M. (c) Repeat if the initial concentrations are 0.1 M instead of 2 M.

SOLUTION

(a) Let "L" and "Ac" represent the lactate (CH₃CHOHCO₂^-) and acetate (CH₃CO₂^-) ions, respectively. Then lactic and acetic acids ionize according to

HL + H₂O <=> L^- + H₃O⁺, K_a1 = 1.4 x 10^-4
HAc + H₂O <=> Ac^- + H₃O⁺, K_a2 = 1.8 x 10^-5

After the two solutions are mixed, we have a solution containing 1 M lactic acid and 1 M sodium acetate. Sodium acetate is soluble in aqueous solution. The lactic acid will ionize to some extent and some of the resulting H₃O⁺ will combine with Ac^- to form HAc. Let x be the moles per liter of HL which ionize and let y be the moles per liter of Ac^- which combine with H₃O⁺ to form HAc. At equilibrium, we have

[HL] = 1 - x = a - x
[L^-] = x
[HAc] = y
[Ac^-] = 1 - y = a - y
[H₃O⁺] = x - y

where a = 1. The ionization constants are

K_a1 = [L^-][H₃O⁺] / [HL] = x(x - y) / (a - x) => y = x - K_a1(a - x) / x (1)
K_a2 = [Ac^-][H₃O⁺] / [HAc] = (a - y)(x - y) / y => x = y + K_a2y / (a - y) (2)

Substitute (2) into (1).

y = y + K_a2y / (a - y) - K_a1{a - [y + K_a2y / (a - y)]} / [y + K_a2y / (a - y)]

Subtract y from both sides.

0 = K_a2y / (a - y) - K_a1{a - [y + K_a2y / (a - y)]} / [y + K_a2y / (a - y)] = f(y)

We can solve this graphically for y. Plot f(y) as a function of y for 0 < y < 1. We see that the value of y that makes f(y) zero is between 0.73 and 0.74. We can narrow down the solution to y = 0.736043742. Using (2), we get x = 0.736093935 => x - y = 5.01931 x 10^-5 = [H₃O⁺] => pH = - log [H₃O⁺] = 4.299. See LacticAcetic070407.xls.

(b) If the initial concentrations are 4 M, the solution to the problem is the same except that a = 2. x and y must both be between 0 and 2. Plot f(y) as a function of y for 0 < y < 2. The value of y that makes f(y) zero is between 1.45 and 1.50. We can narrow down the solution to y = 1.47211258. Using (2), we get x = 1.472162777 => x - y = 5.01964 x 10^-5 = [H₃O⁺] => pH = - log [H₃O⁺] = 4.299. See LacticAcetic070407.2.xls.

(c) If the initial concentrations are 0.1 M, the solution to the problem is the same except that a = 0.05. x and y must both be between 0 and 0.05. Plot f(y) as a function of y for 0 < y < 0.05. The value of y that makes f(y) zero is between 0.0365 and 0.0370. We can narrow down the solution to y = 0.036778392. Using (2), we get x = 0.036828462 => x - y = 5.00704 x 10^-5 = [H₃O⁺] => pH = - log [H₃O⁺] = 4.300. See LacticAcetic070407.3.xls.

PROBLEM

A solution is prepared by adding 2.05 g of sodium acetate, NaCH₃COO, to 100 mL of 0.100 M HCl solution. What is the resulting concentration of H₃O⁺? A subsequent addition of 6.00 mL of 0.100 M HCl is made. What is the new concentration of H₃O⁺?

[from Mahan, Bruce A. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 6.29]

SOLUTION

The molecular weight of NaCH₃COO is [22.99 + (2)(12.01) + (3)(1.008) + (2)(16.00)] g/mole = 82.034 g/mole.

2.05 g NaCH₃COO = (2.05 g)/(82.034 g/mole) NaCH₃COO = 2.499 x 10^-2 mole NaCH₃COO

The initial concentration of NaCH₃COO is (2.499 x 10^-2 mole) / (0.100 L) = 0.2499 M. Since NaCH₃COO is an ionic compound containing an alkali metal ion (Na⁺), it is soluble in aqueous solution, and the initial concentrations of Na⁺ and CH₃COO^- are both 0.2499 M. Since HCl is a strong acid, the initial concentrations of H₃O⁺ and Cl^- are both 0.100 M. Acetic acid, CH₃COOH, is a weak acid with ionization constant K_a = 1.85 x 10^-5 (Mahan 1966, p. 193), which ionizes in water according to

CH₃COOH + H₂O <=> H₃O⁺ + CH₃COO^-

The addition of the sodium acetate causes this reaction to proceed to the left. Let x be the number of moles per liter of H₃O⁺ which combine with CH₃COO^- to form CH₃COOH. At equilibrium,

[CH₃COOH] = x
[CH₃COO^-] = 0.2499 - x
[H₃O⁺] = 0.100 - x

The ionization constant of acetic acid is

K_a = [H₃O⁺][CH₃COO^-] / [CH₃COOH] = (0.100 - x)(0.2499 - x) / x
=> K_ax = 0.02499 - 0.100x - 0.2499x + x² => x² - (K_a + 0.3499)x + 0.02499 = 0
=> x = {(K_a + 0.3499) ± sqrt[(K_a + 0.3499)² - 4(0.02499)]} / 2 = (0.24993, 0.099988) M

Reject the + solution because 0.24993 M is greater than the initial concentration of CH₃COO^-. We therefore have

x = 0.099988 M
[CH₃COOH] = x = 0.099988 M
[CH₃COO^-] = 0.2499 - x = 0.1499 M
[H₃O⁺] = 0.100 - x = 1.234 x 10^-5 M

This is in 100 mL of solution. Now if 6.00 mL of 0.100 M HCl is added, we have a total of 106 mL of solution. The starting concentrations of H₃O⁺, CH₃COOH, and CH₃COO^- are

[H₃O⁺] = [(0.100 L)(1.234 x 10^-5 M) + (0.006 L)(0.100 M)] / (0.106 L)
= 5.672 x 10^-3 M
[CH₃COOH] = (0.100 L)(0.099988 M) / (0.106 L) = 9.433 x 10^-2 M
[CH₃COO^-] = (0.100 L)(0.1499 M) / (0.106 L) = 0.1414 M

The additional HCl causes more of the CH₃COO^- to combine with H₃O⁺ to form CH₃COOH. Let x be the number of moles per liter of CH₃COO^- which combine with H₃O⁺. At equilibrium, the new concentrations are

[H₃O⁺] = 5.672 x 10^-3 - x = A - x
[CH₃COOH] = 9.433 x 10^-2 + x = B + x
[CH₃COO^-] = 0.1414 - x = C - x

where A = 5.672 x 10^-3, B = 9.433 x 10^-2, and C = 0.1414. We have

K_a = [H₃O⁺][CH₃COO^-] / [CH₃COOH]
= (A - x)(C - x) / (B + x)
=> K_a(B + x) = (A - x)(C - x) => K_aB + K_ax = AC - (A + C)x + x²
=> x² - (K_a + A + C)x + (AC - K_aB) = 0
=> x = {(K_a + A + C) ± sqrt[(K_a + A + C)² - 4(AC - K_aB)]} / 2 = (0.1415, 5.658 x 10^-3) M

Reject the + solution because 0.1415 M is greater than the initial concentration of H₃O⁺. So x = 5.658 x 10^-3 M and the final concentration of H₃O⁺ is

[H₃O⁺] = 5.672 x 10^-3 M - 5.658 x 10^-3 M = 1.362 x 10^-5 M

An electrochemical cell generates electricity through the use of a spontaneous redox reaction.

In an electrolytic cell, electrical energy causes a nonspontaneous chemical reaction to occur. The process is called electrolysis.

An example of an electrochemical cell is the following.

             +-+
   +---------|V|---------+
   |         +-+         |
   |                     |
   |     salt bridge     |
  +-+  +-------------+  +-+
| | |  | +---------+ |  | | |
|-| |--| |-|     |-| |--| |-|
| | |  | | |     | | |  | | |
| +-+  | | |     | | |  +-+ |
| Zn       |     |       Cu |
|          |     |          |
+----------+     +----------+
   ZnSO₄             CuSO₄
 solution           solution

On the (left, right) side we have a (zinc, copper) electrode immersed in a solution of (ZnSO₄, CuSO₄). Both ZnSO₄ and CuSO₄ are soluble in aqueous solution. The standard reduction potentials (25°C and 1 M concentrations of Zn²⁺ and Cu²⁺) for the reduction of Zn²⁺ and Cu²⁺ are

Zn²⁺ + 2e^- => Zn, E⁰ = - 0.76 V
Cu²⁺ + 2e^- => Cu, E⁰ = + 0.34 V

Cu²⁺ is easier to reduce than Zn²⁺ because its standard reduction potential is more positive, so in the above electrochemical cell the Cu²⁺ will be reduced and the Zn will be oxidized. The actual half reactions that occur are

Zn => Zn²⁺ + 2e^- => Zn, E⁰ = + 0.76 V
Cu²⁺ + 2e^- => Cu, E⁰ = + 0.34 V

A voltmeter would measure an electromotive force (emf) of

E_cell⁰ = + 0.76 V + 0.34 V = + 1.10 V

The (zinc, copper) electrode is the (anode, cathode). Electrons flow through the wire from the zinc electrode to the copper electrode, from the anode to the cathode. The (anode, cathode) is the (-, +) terminal. A battery is often represented schematically by the following symbol

    - |+
-----||-----
      |

where the (shorter, longer) line represents the (anode, cathode). In physics, the charge carriers are thought of as positive charges which flow into the - terminal and out of the + terminal, even though physically the charge carriers are electrons which are negatively charged and flow in the opposite direction.

PROBLEM

Oxalic acid (H₂C₂O₄) is present in many plants and vegetables.
(a) Balance the following equation in acid solution:

MnO₄^- + C₂O₄^2- => Mn²⁺ + CO₂

(b) If a 1.00 g sample of H₂C₂O₄ requires 24.0 mL of 0.0100 M KMnO₄ solution to reach the equivalence point, what is the percent by mass of H₂C₂O₄ in the sample?

[from Chang, Raymond 2002, Chemistry, Seventh Edition (New York: McGraw-Hill), problem 19.66]

SOLUTION

(a) Write down the oxidation numbers of each element.

+7-2    +3-2      +2    +4-2
MnO₄^- + C₂O₄^2- => Mn²⁺ + CO₂

Carbon is oxidized from +3 to +4. Manganese is reduced from +7 to +2. The oxidation-reduction half reactions are

C₂O₄^2- => CO₂ (oxidation)
MnO₄^- => Mn²⁺ (reduction)

We balance these two half reactions by first balancing the non-oxygen non-hydrogen elements, then balancing oxygen by adding H₂O, then balancing hydrogen by adding H⁺, then balancing charge by adding e^-. The balanced half reactions are

C₂O₄^2- => 2CO₂ + 2e^- (oxidation)
MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O (reduction)

Add 5 times the first equation to 2 times the second equation.

5C₂O₄^2- +2MnO₄^- + 16H⁺ => 10CO₂ + 2Mn²⁺ + 8H₂O

(b) The amount of KMnO₄ is (0.024 L)(0.0100 M) = 2.4 x 10^-4 mole. The amount of C₂O₄^2- is (5/2)(2.4 x 10^-4 mole) = 6 x 10^-4 mole => 6 x 10^-4 mole of H₂C₂O₄. The molar mass of H₂C₂O₄ is (2)(1.008 g/mole) + (2)(12.01 g/mole) + (4)(16.00 g/mole) = 90.036 g/mole. The mass of H₂C₂O₄ is (6 x 10^-4)(90.036 g/mole) = 5.402 x 10^-2 g. The sample is (5.402 x 10^-2 g) / (1.00 g) = 5.402% H₂C₂O₄ by mass.

PROBLEM

Write balanced equations for the following reactions and determine the maximum potential voltage:
(a) Zn + Ag⁺ =>
(b) Sn + Cu²⁺ =>
(c) Zn + Fe²⁺ =>
(d) Pb + MnO₄^- + H⁺ =>
(e) Fe + NO₃^- + H⁺ =>

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), problem 16.1]

SOLUTION

(a) Referring to the table of half-reactions, the complete unbalanced equation is

Zn + Ag⁺ => Zn²⁺ + Ag

The half-reactions are

Zn => Zn²⁺ + 2e^-, E⁰ = 0.76 V (oxidation)
Ag⁺ + e^- => Ag, E⁰ = 0.80 V (reduction)

Add the first equation to two times the second equation.

Zn + 2Ag⁺ => Zn²⁺ + 2Ag, E⁰ = 1.56 V

(b) Referring to the table of half-reactions, the complete unbalanced equation is

Sn + Cu²⁺ => Sn²⁺ + Cu

The half-reactions are

Sn => Sn²⁺ + 2e^-, E⁰ = 0.14 V (oxidation)
Cu²⁺ + 2e^- => Cu, E⁰ = 0.34 V (reduction)

Add the two equations.

Sn + Cu²⁺ => Sn²⁺ + Cu, E⁰ = 0.48 V

(c) Referring to the table of half-reactions, the complete unbalanced equation is

Zn + Fe²⁺ => Zn²⁺ + Fe

The half-reactions are

Zn => Zn²⁺ + 2e^-, E⁰ = 0.76 V (oxidation)
Fe²⁺ + 2e^- => Fe, E⁰ = - 0.44 V (reduction)

Add the two equations.

Zn + Fe²⁺ => Zn²⁺ + Fe, E⁰ = 0.32 V

(d) Referring to the table of half-reactions, the complete unbalanced equation is

Pb + MnO₄^- + H⁺ => Pb²⁺ + Mn²⁺

The half-reactions are

Pb => Pb²⁺ (oxidation)
MnO₄^- => Mn²⁺ (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The standard half-cell potentials are as indicated.

Pb => Pb²⁺ + 2e^-, E⁰ = 0.13 V (oxidation)
MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O, E⁰ = 1.52 V (reduction)

Add five times the first equation to two times the second equation. The maximum potential voltage is obtained by adding the two half-cell potentials.

5Pb + 2MnO₄^- + 16H⁺ => 5Pb²⁺ + 2Mn²⁺ + 8H₂O, E⁰ = 1.65 V

(e) Referring to the table of half-reactions, the complete unbalanced equation is

Fe + NO₃^- + H⁺ => Fe²⁺ + NO

Iron is oxidized from 0 to +2 while nitrogen is reduced from +5 to +2. The half-reactions are

Fe => Fe²⁺ (oxidation)
NO₃^- => NO (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The standard half-cell potentials are as indicated.

Fe => Fe²⁺ + 2e^-, E⁰ = 0.44 V (oxidation)
NO₃^- + 4H⁺ + 3e^- => NO + 2H₂O, E⁰ = 0.96 V (reduction)

Add three times the first equation to two times the second equation.

3Fe + 2NO₃^- + 8H⁺ => 3Fe²⁺ + 2NO + 4H₂O, E⁰ = 1.40 V

PROBLEM

The following are some substances that contain the element nitrogen: KNO₃, KNO₂, NH₄Cl, N₂H₅, N₂H₄, NO₂, N₂, NH₃, N₂O₄, HNO₃.
(a) What is the oxidation state of nitrogen in each substance?
(b) In relation to nitrogen, which substances can act as oxidizing agents?
(c) In relation to nitrogen, which substances can act as reducing agents?
(d) In which of the substances is nitrogen completely oxidized?
(e) In which substance does nitrogen have the lowest oxidation number?
(f) Name the substances that can, in relation to nitrogen, act both as an oxidizing agent and as a reducing agent.

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 16.6, p. 389]

SOLUTION

(a) KNO₃ (potassium nitrate):

+1+5-2
  KNO₃

KNO₂ (potassium nitrite):

+1+3-2
  KNO₂

NH₄Cl (ammonium chloride):

+5-1-1
 NH₄Cl

N₂H₅ (dinitrogen pentahydride):

+5/2-1
  N₂H₅

N₂H₄ (dinitrogen tetrahydride):

+2-1
N₂H₄

NO₂ (nitrogen dioxide):

+4-2
 NO₂

N₂ (nitrogen):

0
N₂

NH₃ (ammonia):

+3-1
 NH₃

N₂O₄ (dinitrogen tetraoxide):

+4-2
N₂O₄

HNO₃ (nitric acid):

+1+5-2
  HNO₃

(b) Since the electronic structure of nitrogen is 1s²2s²2p³, it can acquire up to three additional valence shell electrons, resulting in an oxidation state of -3. Thus, all of the above substances can be reduced and all can act as oxidizing agents.

(c) Nitrogen can lose up to five valence shell electrons, resulting in an oxidation state of +5. All of the substances can be oxidized and act as reducing agents except those which have oxidation states of +5: KNO₃, NH₄Cl, HNO₃.

(d) Nitrogen is completely oxidized in KNO₃, NH₄Cl, and HNO₃.

(e) Nitrogen has the lowest oxidation number in N₂.

(f) All except KNO₃, NH₄Cl, and HNO₃ can act both as an oxidizing agent and as a reducing agent.

PROBLEM

Balance the following redox equation:

Cr₂O₇^2-(aq) + HNO₂(aq) + H⁺(aq) => Cr³⁺ + NO₃^- + H₂O(l)

[from Sample Questions for Chemistry Advanced Placement Test, College Board, Question 18]

SOLUTION

Write the oxidation numbers for each element.

+6 -2   +1+3-2   +1    +3    +5-2   +1-2
Cr₂O₇^2- + HNO₂ + H⁺ => Cr³⁺ + NO₃^- + H₂O

Nitrogen is oxidized from +3 to +5. Chromium is reduced from +6 to +3. The half reactions are

HNO₂ => NO₃^- (oxidation)
Cr₂O₇^2- => Cr³⁺ (reduction)

Balance non-oxygen, non-hydrogen elements. Then balance oxygen by adding H₂O, balance hydrogen by adding H⁺, and balance charge by adding e^-. The balanced half reactions are

HNO₂ + H₂O => NO₃^- + 3H⁺ + 2e^- (oxidation)
Cr₂O₇^2- + 14H⁺ + 6e^- => 2Cr³⁺ + 7H₂O (reduction)

Add three times the first equation to the second equation.

3HNO₂ + 3H₂O + Cr₂O₇^2- + 14H⁺ + 6e^- => 3NO₃^- + 9H⁺ + 6e^- + 2Cr³⁺ + 7H₂O

Subtract 6e^-, 9H⁺, and 3H₂O from both sides. The balanced redox equation is

Cr₂O₇^2- + 3HNO₂ + 5H⁺ => 2Cr³⁺ + 3NO₃^- + 4H₂O

PROBLEM

Balance the equations for each of the following reactions using the half-reaction (ion-electron) method. Also indicate which is the oxidant, which is the reductant, and the changes in oxidation numbers of the main elements involved.
(a) Cu + H⁺ + NO₃^- => Cu²⁺ + NO + H₂O
(b) Fe²⁺ + MnO₄^- + H⁺ => Fe³⁺ + Mn²⁺ + H₂O
(c) PbO₂ + Mn²⁺ + H⁺ => Pb²⁺ + MnO₄^- + H₂O
(d) Br₂ + SO₂ + H₂O => H₂SO₄ + HBr
(e) I₂ + S₂O₃^2- => I ^- + S₄O₆^2-
(f) Bi₂O₃ + NaOH + NaOCl => NaBiO₃ + NaCl + H₂O

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), reasoning exercise 16.10, p. 390]

SOLUTION

(a) Write down the oxidation numbers for each element.

0   +1   +5-2     +2    +2-2 +1-2
Cu + H⁺ + NO₃^- => Cu²⁺ + NO + H₂O

Copper is oxidized from 0 to +2 and is the reductant. Nitrogen is reduced from +5 to +2 and is the oxidant. The half-reactions are

Cu => Cu²⁺ (oxidation)
NO₃^- => NO (reduction)

Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

Cu => Cu²⁺ + 2e^- (oxidation)
NO₃^- + 4H⁺ + 3e^- => NO + 2H₂O (reduction)

Add three times the first equation to two times the second equation.

3Cu + 8H⁺ + 2NO₃^- => 3Cu²⁺ + 2NO + 4H₂O

(b) Write down the oxidation numbers for each element.

+2     +7-2    +1    +3     +2    +1-2
Fe²⁺ + MnO₄^- + H⁺ => Fe³⁺ + Mn²⁺ + H₂O

Iron is oxidized from +2 to +3 and is the reductant. Manganese is reduced from +7 to +2 and is the oxidant. The half-reactions are

Fe²⁺ => Fe³⁺ (oxidation)
MnO₄^- => Mn²⁺ (reduction)

Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

Fe²⁺ => Fe³⁺ + e^- (oxidation)
MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O (reduction)

Add five times the first equation to the second equation.

5Fe²⁺ + MnO₄^- + 8H⁺ => 5Fe³⁺ + Mn²⁺ + 4H₂O

(c) Write down the oxidation numbers for each element.

+4-2   +2     +1    +2     +7-2   +1-2
PbO₂ + Mn²⁺ + H⁺ => Pb²⁺ + MnO₄^- + H₂O

Manganese is oxidized from +2 to +7 and is the reductant. Lead is reduced from +4 to +2 and is the oxidant. The half-reactions are

Mn²⁺ => MnO₄^- (oxidation)
PbO₂ => Pb²⁺ (reduction)

Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

Mn²⁺ + 4H₂O => MnO₄^- + 8H⁺ + 5e^- (oxidation)
PbO₂ + 4H⁺ + 2e^- => Pb²⁺ + 2H₂O (reduction)

Add two times the first equation to five times the second equation.

2Mn²⁺ + 8H₂O + 5PbO₂ + 20H⁺ => 2MnO₄^- + 16H⁺ + 5Pb²⁺ + 10H₂O

Subtract 16H⁺ and 8H₂O from both sides.

5PbO₂ + 2Mn²⁺ + 4H⁺ => 5Pb²⁺ + 2MnO₄^- + 2H₂O

(d) Write down the oxidation numbers for each element.

 0   +4-2  +1-2   +1+6-2  +1-1
Br₂ + SO₂ + H₂O => H₂SO₄ + HBr

Sulphur is oxidized from +4 to +6 and is the reductant. Bromine is reduced from 0 to -1 and is the oxidant. The half-reactions are

SO₂ => H₂SO₄ (oxidation)
Br₂ => HBr (reduction)

Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

SO₂ + 2H₂O => H₂SO₄ + 2H⁺ + 2e^- (oxidation)
Br₂ + 2H⁺ + 2e^- => 2HBr (reduction)

Add the two equations.

Br₂ + SO₂ + 2H₂O => H₂SO₄ + 2HBr

(e) Write down the oxidation numbers for each element.

0   +2-2       -1 +5/2-2
I₂ + S₂O₃^2- => I^- + S₄O₆^2-

Sulfur is oxidized from +2 to +5/2 and is the reductant. Iodine is reduced from 0 to -1 and is the oxidant. The half-reactions are

S₂O₃^2- => S₄O₆^2- (oxidation)
I₂ => I ^- (reduction)

Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

2S₂O₃^2- => S₄O₆^2- + 2e^- (oxidation)
I₂ + 2e^- => 2I ^- (reduction)

Add the two equations.

I₂ + 2S₂O₃^2- => 2I ^- + S₄O₆^2-

(f) Write down the oxidation numbers for each element.

+3-2   +1-2+1 +1-2+1    +1+5-2   +1-1   +1-2
Bi₂O₃ + NaOH + NaOCl => NaBiO₃ + NaCl + H₂O

Bismuth is oxidized from +3 to +5 and is the reductant. Chlorine is reduced from +1 to -1 and is the oxidant. Write the oxidation and reduction half-reactions.

Bi₂O₃ => NaBiO₃ (oxidation)
NaOCl => NaCl (reduction)

Balance non-hydrogen, non-oxygen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

Bi₂O₃ + 2Na⁺ + 3H₂O => 2NaBiO₃ + 6H⁺ + 4e^- (oxidation)
NaOCl + 2H⁺ + 2e^- => NaCl + H₂O (reduction)

Add the oxidation half-reaction to twice the reduction half-reaction.

Bi₂O₃ + 2Na⁺ + 3H₂O + 2NaOCl + 4H⁺ => 2NaBiO₃ + 6H⁺ + 2NaCl + 2H₂O

Subtract 4H⁺ from each side.

Bi₂O₃ + 2Na⁺ + 3H₂O + 2NaOCl => 2NaBiO₃ + 2H⁺ + 2NaCl + 2H₂O

Add 2OH^- to both sides.

Bi₂O₃ + 2Na⁺ + 3H₂O + 2NaOCl + 2OH^- => 2NaBiO₃ + 2H⁺ + 2NaCl + 2H₂O + 2OH^-

Combine the Na⁺ and the OH^- on the left side, and combine the H⁺ and the OH^- on the right side.

Bi₂O₃ + 2NaOH + 3H₂O + 2NaOCl => 2NaBiO₃ + 2NaCl + 4H₂O

Subtract 3H₂O from both sides. The balanced net equation is

Bi₂O₃ + 2NaOH + 2NaOCl => 2NaBiO₃ + 2NaCl + H₂O

PROBLEM

Balance the equations for each of the following reactions, indicating what is oxidized and what is reduced. Indicate the oxidizing and reducing agents and the changes in oxidation numbers of the main elements involved.
(a) H₂S(g) + H₂SO₃(aq) => S(s) + H₂O
(b) KI(aq) + O₂(g) => KI₃(aq) + H₂O
(c) Fe(NO₃)₂(aq) + HNO₃(aq) => Fe(NO₃)₃(aq) + NO(g) + H₂O
(d) K₂Cr₂O₇ + HCl => KCl + CrCl₃ + Cl₂ + H₂O
(e) CuS + HNO₃ => Cu(NO₃)₂ + SO₂ + NO₂ +H₂O
(f) KMnO₄ + HCl => KCl + MnCl₂ + Cl₂ + H₂O
(g) H₂S + HNO₃ => NO₂ + S + H₂O
(h) Cl₂ + H₂O + SO₂ => SO₄^2- + Cl^- + H⁺
(i) HI + H₂SO₄ => H₂S + H₂O + I₂
(j) Cu + HNO₃ => Cu(NO₃)₂ + NO + H₂O

[from Kavanah, Patrick, and Robbins, Jack 1971, Concepts in Modern Chemistry (New York: Cambridge Book Company), problem 16.2]

SOLUTION

(a) Write down the oxidation numbers for each element

+1-2     +1+4-2        0     +1-2
 H₂S(g) + H₂SO₃(aq) => S(s) + H₂O

Sulfur is oxidized from -2 to 0 and reduced from +4 to 0. Sulfur is both an oxidizing and a reducing agent. The oxidation and reduction half-reactions are

H₂S => S (oxidation)
H₂SO₃ => S (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

H₂S => S + 2H⁺ + 2e^- (oxidation)
H₂SO₃ + 4H⁺ + 4e^- => S + 3H₂O (reduction)

Add two times the first equation to the second equation.

2H₂S + H₂SO₃ => 3S + 3H₂O

(b) Write down the oxidation numbers for each element.

+1-1      0       +1-1/3    +1-2
 KI(aq) + O₂(g) => KI₃(aq) + H₂O

Iodine is oxidized from -1 to -1/3 and is the reducing agent. Oxygen is reduced from 0 to -2 and is the oxidizing agent. The oxidation and reduction half-reactions are

KI => KI₃ (oxidation)
O₂ => H₂O (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

3KI => KI₃ + 2K⁺ + 2e^- (oxidation)
O₂ + 4H⁺ + 4e^- => 2H₂O (reduction)

Add two times the first equation to the second equation.

6KI + O₂ + 4H⁺ => 2KI₃ + 4K⁺ + 2H₂O

Add 4OH^- to both sides.

6KI + O₂ + 4H₂O => 2KI₃ + 4KOH + 2H₂O

Subtract 2H₂O from both sides.

6KI + O₂ + 2H₂O => 2KI₃ + 4KOH

(c) Write down the oxidation numbers for each element.

+2+5-2    +1+5-2   +3+5-2    +2-2 +1-2
Fe(NO₃)₂ + HNO₃ => Fe(NO₃)₃ + NO + H₂O

Iron is oxidized from +2 to +3 and is the reducing agent. Nitrogen is reduced from +5 to +2 and is the oxidizing agent. Rewrite the above equation with the terms separated into ions.

Fe²⁺ + 2NO₃^- + H⁺ + NO₃^- => Fe³⁺ + 3NO₃^- + NO + H₂O

The oxidation and reduction half-reactions are

Fe²⁺ => Fe³⁺ (oxidation)
NO₃^- => NO (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

Fe²⁺ => Fe³⁺ + e^- (oxidation)
NO₃^- + 4H⁺ + 3e^- => NO + 2H₂O (reduction)

Add three times the first equation to the second equation.

3Fe²⁺ + NO₃^- + 4H⁺ => 3Fe³⁺ + NO + 2H₂O

Add 9NO₃^- to both sides.

3Fe²⁺ + 10NO₃^- + 4H⁺ => 3Fe³⁺ + NO + 2H₂O + 9NO₃^-

Combine 3Fe²⁺ with 6NO₃^- and 4H⁺ with 4NO₃^- on the left side, and 3Fe³⁺ with 9NO₃^- on the right side.

3Fe(NO₃)₂ + 4HNO₃ => 3Fe(NO₃)₃ + NO + 2H₂O

(d) Write down the oxidation numbers for each element.

+1+6-2   +1-1   +1-1   +3-1     0   +1-2
K₂Cr₂O₇ + HCl => KCl + CrCl₃ + Cl₂ + H₂O

Chlorine is oxidized from -1 to 0 and is the reducing agent. Chromium is reduced from +6 to +3 and is the oxidizing agent. Rewrite the above equation with the terms written as separate ions.

2K⁺ + Cr₂O₇^2- + H⁺ + Cl^- => K⁺ + Cl^- + Cr³⁺ + 3Cl^- + Cl₂ + H₂O

The oxidation and reduction half-reactions are

Cl^- => Cl₂ (oxidation)
Cr₂O₇^2- => Cr³⁺ (reduction)

Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

2Cl^- => Cl₂ + 2e^- (oxidation)
Cr₂O₇^2- + 14H⁺ + 6e^- => 2Cr³⁺ + 7H₂O (reduction)

Add three times the first equation to the second equation.

6Cl^- + Cr₂O₇^2- + 14H⁺ => 3Cl₂ + 2Cr³⁺ + 7H₂O

Add 8Cl^- and 2K⁺ to both sides.

14Cl^- + Cr₂O₇^2- + 14H⁺ + 2K⁺ => 3Cl₂ + 2Cr³⁺ + 7H₂O + 8Cl^- + 2K⁺

Combine 2K⁺ with Cr₂O₇^2- and 14H⁺ with 14Cl^- on the left side and combine 2K⁺ with 2Cl^- and 2Cr³⁺ with 6Cl^- on the right side.

K₂Cr₂O₇ + 14HCl => 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O

(e) Write down the oxidation numbers for each element.

+2-2 +1+5-2    +2+5-2    +4-2  +4-2  +1-2
 CuS + HNO₃ => Cu(NO₃)₂ + SO₂ + NO₂ + H₂O

Sulfur is oxidized from -2 to +4 and is the reducing agent. Nitrogen is reduced from +5 to +4 and is the oxidizing agent. The half-reactions are

CuS => SO₂ (oxidation)
HNO₃ => NO₂ (reduction)

Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

CuS + 2H₂O => SO₂ + Cu²⁺ + 4H⁺ + 6e^- (oxidation)
HNO₃ + H⁺ + e^- => NO₂ + H₂O (reduction)

Add the first equation to six times the second equation.

CuS + 2H₂O + 6HNO₃ + 6H⁺ => SO₂ + Cu²⁺ + 4H⁺ + 6NO₂ + 6H₂O

Add 2NO₃^- to both sides.

CuS + 2H₂O + 6HNO₃ + 6H⁺ + 2NO₃^- => SO₂ + Cu(NO₃)₂ + 4H⁺ + 6NO₂ + 6H₂O

Combine 2H⁺ and 2NO₃^- on the left side and subtract 4H⁺ and 2H₂O from both sides.

CuS + 8HNO₃ => Cu(NO₃)₂ + SO₂ + 6NO₂ + 4H₂O

(f) Write down the oxidation numbers for each element.

+1+7-2  +1-1   +1-1   +2-1     0   +1-2
 KMnO₄ + HCl => KCl + MnCl₂ + Cl₂ + H₂O

Chlorine is oxidized from -1 to 0 and is the reducing agent. Manganese is reduced from +7 to +2 and is the oxidizing agent. The half-reactions are

HCl => Cl₂ (oxidation)
KMnO₄ => MnCl₂ (reduction)

Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

2HCl => Cl₂ + 2H⁺ + 2e^- (oxidation)
KMnO₄ + 2Cl^- + 8H⁺ + 5e^- => MnCl₂ + K⁺ + 4H₂O (reduction)

Add five times the first equation to two times the second equation.

10HCl + 2KMnO₄ + 4Cl^- + 16H⁺ => 5Cl₂ + 10H⁺ + 2MnCl₂ + 2K⁺ + 8H₂O

Add 2Cl^- to both sides and subtract 10H⁺ from both sides.

10HCl + 2KMnO₄ + 6Cl^- + 6H⁺ => 5Cl₂ + 2MnCl₂ + 2K⁺ + 8H₂O + 2Cl^-

Combine the 6H⁺ and 6Cl^- on the left side and the 2K⁺ and 2Cl^- on the right side.

2KMnO₄ + 16HCl => 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

(g) Write down the oxidation numbers for each element.

+1-2 +1+5-2   +4-2   0  +1-2
 H₂S + HNO₃ => NO₂ + S + H₂O

Sulfur is oxidized from -2 to 0 and is the reducing agent. Nitrogen is reduced from +5 to +4 and is the oxidizing agent. The half-reactions are

H₂S => S (oxidation)
HNO₃ => NO₂ (reduction)

Balance the non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

H₂S => S + 2H⁺ + 2e^- (oxidation)
HNO₃ + H⁺ + e^- => NO₂ + H₂O (reduction)

Add the first equation to two times the second equation.

H₂S + 2HNO₃ => 2NO₂ + S + 2H₂O

(h) Write down the oxidation numbers for each element.

 0   +1-2  +4-2   +6-2     -1   +1
Cl₂ + H₂O + SO₂ => SO₄^2- + Cl^- + H⁺

Sulfur is oxidized from +4 to +6 and is the reducing agent. Chlorine is reduced from 0 to -1 and is the oxidizing agent. The half-reactions are

SO₂ => SO₄^2- (oxidation)
Cl₂ => Cl^- (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

SO₂ + 2H₂O => SO₄^2- + 4H⁺ + 2e^- (oxidation)
Cl₂ + 2e^- => 2Cl^- (reduction)

Add the two half-reactions. The balanced net equation is

Cl₂ + SO₂ + 2H₂O => SO₄^2- + 2Cl^- + 4H⁺

(i) Write down the oxidation numbers for each element.

+1-1 +1+6-2   +1-2  +1-2   0
 HI + H₂SO₄ => H₂S + H₂O + I₂

Iodine is oxidized from -1 to 0 and is the reducing agent. Sulfer is reduced from +6 to -2 and is the oxidizing agent. The half-reactions are

HI => I₂ (oxidation)
H₂SO₄ => H₂S (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-.

2HI => I₂ + 2H⁺ + 2e^- (oxidation)
H₂SO₄ + 8H⁺ + 8e^- => H₂S + 4H₂O (reduction)

Add four times the first equation to the second equation.

8HI + H₂SO₄ => H₂S + 4H₂O + 4I₂

(j) Write down the oxidation numbers for each element.

0  +1+5-2    +2+5-2    +2-2 +1-2
Cu + HNO₃ => Cu(NO₃)₂ + NO + H₂O

Copper is oxidized from 0 to +2 and is the reducing agent. Nitrogen is reduced from +5 to +2 and is the oxidizing agent. The half-reactions are

Cu => Cu(NO₃)₂ (oxidation)
HNO₃ => NO (reduction)

Balance non-oxygen, non-hydrogen elements, then balance oxygen by adding H₂O, then balance hydrogen by adding H⁺, then balance charge by adding e^-. The balanced half-reactions are

Cu + 2NO₃^- => Cu(NO₃)₂ + 2e^- (oxidation)
HNO₃ + 3H⁺ + 3e^- => NO + 2H₂O (reduction)

Add three times the first equation to two times the second equation.

3Cu + 6NO₃^- + 2HNO₃ + 6H⁺ => 3Cu(NO₃)₂ + 2NO + 4H₂O

Combine the 6NO₃^- and 6H⁺ on the left side.

3Cu + 8HNO₃ => 3Cu(NO₃)₂ + 2NO + 4H₂O

PROBLEM

Consider the following oxidation-reduction reaction:

KMnO₄ + H₂SO₃ => K₂SO₄ + MnSO₄ + H₂SO₄ + H₂O

(a) Which species is/are oxidized and reduced?
(b) Write the balanced oxidation and reduction half-reactions.
(c) Write the balanced equation for the full reaction.

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), Practice Test 1, problem 51]

SOLUTION

(a) Write the oxidation numbers for each species.

+1+7-2  +1+4-2   +1+6-2  +2+6-2  +1+6-2  +1-2
 KMnO₄ + H₂SO₃ => K₂SO₄ + MnSO₄ + H₂SO₄ + H₂O

We see that Mn is reduced from +7 to +2 and S is oxidized from +4 to +6.

(b) Rewrite the unbalanced equation, separating each component, except water, into ions.

K⁺ + MnO₄^- + 2H⁺ + SO₃^2- => 2K⁺ + SO₄^2- + Mn²⁺ + SO₄^2- + 2H⁺ + SO₄^2- + H₂O

The unbalanced oxidation and reduction half-reactions are

SO₃^2- => SO₄^2- (oxidation)
MnO₄^- => Mn²⁺ (reduction)

Add one H₂O to the left side of the oxidation half-reaction and 4H₂O to the right side of the reduction half-reaction to balance the oxygens.

SO₃^2- + H₂O => SO₄^2- (oxidation)
MnO₄^- => Mn²⁺ + 4H₂O (reduction)

Add 2H⁺ to the right side of the oxidation half-reaction and 8H⁺ to the left side of the reduction half-reaction to balance the hydrogens.

SO₃^2- + H₂O => SO₄^2- + 2H⁺ (oxidation)
MnO₄^- + 8H⁺ => Mn²⁺ + 4H₂O (reduction)

Add 2e^- to the right side of the oxidation half-reaction and 5e^- to the left side of the reduction half-reaction to balance the charge.

SO₃^2- + H₂O => SO₄^2- + 2H⁺ + 2e^-(oxidation)
MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O (reduction)

These are the balanced oxidation and reduction half-reactions.

(c) Add five times the oxidation half-reaction to two times the reduction half-reaction.

(5)(SO₃^2- + H₂O => SO₄^2- + 2H⁺ + 2e^-) (oxidation)
(2)(MnO₄^- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O) (reduction)
================================
5SO₃^2- + 5H₂O + 2MnO₄^- + 16H⁺ + 10e^- => 5SO₄^2- + 10H⁺ + 10e^- + 2Mn²⁺ + 8H₂O

Subtract 10e^- from both sides.

5SO₃^2- + 5H₂O + 2MnO₄^- + 16H⁺ => 5SO₄^2- + 10H⁺ + 2Mn²⁺ + 8H₂O

We need to have potassium ions in order to be able to balance the original equation for the reaction, so add 2K⁺ to both sides.

5SO₃^2- + 5H₂O + 2MnO₄^- + 16H⁺ + 2K⁺ => 5SO₄^2- + 10H⁺ + 2Mn²⁺ + 8H₂O + 2K⁺

Combine the 2K⁺ with 2MnO₄^- on the left side and the 2K⁺ with one SO₄^2- on the right side.

5SO₃^2- + 5H₂O + 2KMnO₄ + 16H⁺ => 4SO₄^2- + 10H⁺ + 2Mn²⁺ + 8H₂O + K₂SO₄

Combine 10H⁺ with 5SO₃^2- on the left side and 2Mn²⁺ with 2SO₄^- on the right side.

5H₂SO₃ + 5H₂O + 2KMnO₄ + 6H⁺ => 2SO₄^2- + 10H⁺ + 2MnSO₄ + 8H₂O + K₂SO₄

Combine 4H⁺ with 2SO₄^2- on the right side.

5H₂SO₃ + 5H₂O + 2KMnO₄ + 6H⁺ => 2H₂SO₄ + 6H⁺ + 2MnSO₄ + 8H₂O + K₂SO₄

Subtract 5H₂O and 6H⁺ from both sides.

5H₂SO₃ + 2KMnO₄ => 2H₂SO₄ + 2MnSO₄ + 3H₂O + K₂SO₄

Alternate Method:

Assume that the coefficients in the balanced equation are A, B, C, D, E, and F and solve for them by requiring that the numbers of each type of atom be equal on both sides of the equation.

A · KMnO₄ + B · H₂SO₃ => C · K₂SO₄ + D · MnSO₄ + E · H₂SO₄ + F · H₂O

K: A = 2C
Mn: A = D
O: 4A + 3B = 4C + 4D + 4E + F (1)
H: 2B = 2E + 2F (2)
S: B = C + D + E (3)

We have five equations in six unknowns. Take A = D = 1. Then C = A/2 = 1/2. Substituting into (1), (2), and (3), we get

4 + 3B = 2 + 4 + 4E + F => 3B = 2 + 4E + F (4)
B = E + F (5)
B = 1/2 + 1 + E = 3/2 + E (6)

Use (5) to write B in terms of E and F in (4) and (6).

3E + 3F = 2 + 4E + F => 2F = 2 + E (7)
E + F = 3/2 + E => F = 3/2 (8)

Substitute F = 3/2 into (7) to solve for E.

E = 2F - 2 = 2(3/2) - 2 = 1

Use (6) to solve for B.

B = 3/2 + 1 = 5/2

Thus, we have A = 1, B = 5/2, C = 1/2, D = 1, E = 1, and F = 3/2 or, multiplying all of these by two, A = 2, B = 5, C = 1, D = 2, E = 2, and F = 3. Thus, the balanced equation is

2KMnO₄ + 5H₂SO₃ => K₂SO₄ + 2MnSO₄ + 2H₂SO₄ + 3H₂O

or

5H₂SO₃ + 2KMnO₄ => 2H₂SO₄ + 2MnSO₄ + 3H₂O + K₂SO₄

PROBLEM

If gold foil were placed in a solution containing zinc ions, what would the reaction potential be?

[from Mascetta, Joseph A. 2002, Barron's How to Prepare for the SAT II Chemistry, 7th Edition (Hauppauge, New York: Barron's Educational Series), problem 12.11]

SOLUTION

The half-cell oxidation potentials for zinc and gold are

Zn => Zn²⁺ + 2e^-, E⁰ = 0.76 V
Au => Au³⁺ + 3e^-, E⁰ = - 1.50 V

If gold foil were placed in a solution containing zinc ions, the following combination of reactions could occur:

Zn²⁺ + 2e^- => Zn, E⁰ = - 0.76 V (1)
Au => Au³⁺ + 3e^-, E⁰ = - 1.50 V (2)

for which the balanced equation, obtained by adding three times (1) to two times (2), is

3Zn²⁺ + 2Au => 3Zn + 2Au³⁺

The reaction potential would be E⁰ = - 0.76 V - 1.50 V = - 2.26 V, which indicates that the reaction would not be spontaneous.

PROBLEM

Which of the following oxidizing agents become stronger as the concentration of H⁺ increases? Which are unchanged and which become weaker? (a) Cl₂, (b) Cr₂O₇^2-, (c) Fe³⁺, (d) MnO₄^-.

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.4]

SOLUTION

(a) From the table of standard reduction potentials at 25°C, we see that in an electrochemical cell Cl₂ is reduced according to

Cl₂(g) + 2e^- => 2Cl^-(aq)

The oxidizing strength of Cl₂ is independent of the concentration of H⁺.

(b) Cr₂O₇^2- is reduced according to

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- => 2Cr³⁺(aq) + 7H₂O

The oxidizing strength of Cr₂O₇^2- is increased if the concentration of H⁺ is increased.

(c) Fe³⁺ is reduced according to

Fe³⁺(aq) + e^- => Fe²⁺(aq)

The oxidizing strength of Fe³⁺ is independent of the concentration of H⁺.

(d) MnO₄^- is reduced according to

MnO₄^-(aq) + 2H₂ + 3e^- => MnO₂(s) + 4OH^-(aq)

The oxidizing strength of MnO₄^- is increased if the concentration of OH^- is decreased or if the concentration of H⁺ is increased.

PROBLEM

From the appropriate values of the standard reduction potential E⁰, calculate DE⁰ and the equilibrium constant for the reaction

Hg²⁺ + Hg => Hg₂²⁺

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.6]

SOLUTION

From the table of standard reduction potentials, we have the following half reactions:

2Hg²⁺(aq) + 2e^- => Hg₂²⁺(aq), E⁰ = +0.92 V (1)
Hg₂²⁺(aq) + 2e^- => 2Hg(l), E⁰ = +0.789 V (2)

From (2), we get

2Hg(l) => Hg₂²⁺(aq) + 2e^-, E⁰ = -0.789 V (3)

Adding (1) and (3), we get

2Hg²⁺(aq) + 2Hg(l) => 2Hg₂²⁺, DE⁰ = E⁰(1) + E⁰(3) = +0.92 V - 0.789 V = +0.131 V,
2e^- transferred

or

Hg²⁺(aq) + Hg(l) => Hg₂²⁺, DE⁰ = +0.131 V, 1e^- transferred

The equilibrium constant K of this reaction is given by

log K = nDE⁰/0.059 = (1)(0.131 V)/0.059 = 2.220 => K = 10^2.220 = 166.1

PROBLEM

A half-cell (A) consisting of a strip of nickel dipping into a 1 M solution of Ni²⁺, and a half-cell (B) consisting of a strip of zinc dipping into a 1 M solution of Zn²⁺ were successively connected with a standard hydrogen half-cell. The magnitudes of the individual half-cell potentials were then determined as

(A) Ni²⁺ + 2e^- => Ni, |E⁰| = 0.25 V
(B) Zn²⁺ + 2e^- => Zn, |E⁰| = 0.77 V

(a) When both the half-cells (A) and (B) were connected with the hydrogen half-cell, the metallic electrode (Ni or Zn) was found to be negative. What is the correct sign of the electrode potentials?
(b) Of the substances Ni, Ni²⁺, Zn, Zn²⁺; which is the strongest oxidant? Which is the strongest reductant?
(c) Will a noticeable reaction occur when metallic nickel is placed in a 1 M solution of Zn²⁺? Will a noticeable reaction occur when metallic zinc is placed in a 1 M solution of Ni²⁺?
(d) Zinc forms a complex ion with hydroxide ion, Zn(OH)₄^2-. If hydroxide ion were added to half-cell (B), would its electrode potential become more positive, less positive, or be unaffected?
(e) If half-cells (A) and (B) were connected together, which electrode would be negative? What would the cell voltage be?

[from Mahan, Bruce H. 1966, College Chemistry (Reading, Massachusetts: Addison-Wesley), problem 7.8]

SOLUTION

(a) Since either metallic electrode is found to be negative, and the hydrogen electrode positive, the metallic electrode is being oxidized and is the anode, while hydrogen ions are being reduced at the hydrogen electrode, and the hydrogen electrode is the cathode. Nickel and zinc are both stronger reducing agents than hydrogen. The correct signs of the electrode potentials shown for reactions (A) and (B) are both negative.

(b) Only Ni²⁺ and Zn²⁺ are oxidants, with the following electrode potentials:

(A) Ni²⁺ + 2e^- => Ni, E⁰ = - 0.25 (1) V
(B) Zn²⁺ + 2e^- => Zn, E⁰ = - 0.77 V

Since E⁰ is larger for (A) than for (B), Ni²⁺ is the stronger oxidant.

Only Ni and Zn are reductants, with the following electrode potentials:

(C) Ni => Ni²⁺ + 2e^-, E⁰ = 0.25 V
(D) Zn => Zn²⁺ + 2e^-, E⁰ = 0.77 V

Since E⁰ is larger for (D) than for (C), Zn is the stronger reductant.

(c) Zinc is easier to oxidize than Ni, so if metallic Ni is placed in a 1 M solution of Zn²⁺, no noticeable reaction will occur, but if metallic Zn is placed in a 1 M solution of Ni²⁺, a noticeable reaction will occur.

(d) If the hydroxide ion were added to half cell (B), Zn(OH)₄^2- would be formed, consuming Zn²⁺. The zinc electrode potential would become more negative or less positive.

(e) If half-cells (A) and (B) were connected, Zn would dissolve and Ni would be deposited. Electrons would flow from the Zn electrode to the Ni electrode. The Zn electrode would be negative and the Ni electrode would be positive. The two half-cell reactions would be

Zn => Zn²⁺ + 2e^-, E⁰ = 0.77 V
Ni²⁺ + 2e^- => Ni, E⁰ = - 0.25 V

The cell voltage would be

DE⁰ = 0.77 V - 0.25 V = 0.52 V