The Kb for NH3 Is Kw/Ka for the Ammonium Ion

Chapter 6 - Equilibrium Chemistry

Chapter 6

1. (a) = 1.75109

NH3(aq) + H2O(l) / / OH-(aq) + NH4+(aq) (1) / Kb = 1.7510-5
HCl(aq) + OH-(aq) / / H2O(l) + Cl-(aq) (2) / Kw-1 = 1.001014
NH3(aq) + HCl(aq) / / NH4+(aq) + Cl-(aq) (1+2) / K = 1.75109

The Kb for NH3 is Kw/Ka for the ammonium ion.

(b) = 31019

(1) PbI2(s) / / Pb2+(aq) + 2I-(aq) / Ksp = 7.910-9
(2) Pb2+(aq) + S2-(aq) / / PbS(s) / Ksp-1 = 3.31027
(1+2) PbI2(s) + S2-(aq) / / PbS(s) + 2I-(aq) / K = 2.61019

(1) CdY2-(aq) / / Cd2+(aq) +Y4-(aq) / Kf-1 = 3.4710-17
(2) Cd2+(aq) + 4CN-(aq) / / Cd(CN)42-(aq) / b4 = 8.321017
(1+2) CdY2-(aq) + 4CN-(aq) / / Cd(CN)42-(aq) + Y4-(aq) / K = 28.9

The overall formation constant, b4, is obtained by multiplying together the first four stepwise formation constants (b4 = K1K2K3K4) for Cd(CN)42-.

(d) = 3.010-3

(1) AgCl(s) / / Ag+(aq) + Cl-(aq) / Ksp =1.810-10
(2) Ag+(aq) + 2NH3(aq) / / Ag(NH3)2+ (aq) / b2 = 1.66107
(1+2) AgCl(s) + 2NH3(aq) / / Ag(NH3)2+ (aq) / K = 3.010-3

The overall formation constant, b2, is obtained by multiplying together the two stepwise formation constants (b2 = K1K2) for Ag(NH)3+.

(e) = 2.4108

(1) BaCO3(s) / / Ba2+(aq) + CO32-(aq) / Ksp = 5.010-9
(2) CO32-(aq) + H2O(l) / / OH-(aq) + HCO3-(aq) / Kb1 = 2.1310-4
(3) HCO3-(aq) + H2O(l) / / OH-(aq) + H2CO3(aq) / Kb2 = 2.2510-8
(4) H3O+(aq) + OH-(aq) / / 2H2O(l) / Kw-1 = 1.001014
(5) H3O+(aq) + OH-(aq) / / 2H2O(l) / Kw-1 = 1.001014
(1+2+3+4+5) BaCO3(s) + 2H3O+(aq) / / Ba2+(aq) + H2CO3(aq) + 2H2O(l) / K = 2.4108

The Kb1 for CO32- is Kw/Ka2 for HCO3- and Kb2 for HCO3- is Kw/Ka1 for H2CO3.

2. The ladder diagram for H3PO4 and HF are shown below.

A reaction between H3PO4 and F- is favorable because their respective areas of predominance do not overlap. On the other hand, a reaction between H2PO4- and F-, which is necessary if the final product is to include HPO42-, is unfavorable because the areas of predominance for H2PO4- and F- overlap. The equilibrium constant for the first reaction is

(1) H3PO4(aq) + H2O(l) / / H2PO4-(aq) + H3O+(aq) / Ka1 = 7.1110-3
(2) F-(aq) + H2O(l) / / HF(aq) + OH-(aq) / Kb = 1.510-11
(3) H3O+(aq) + OH-(aq) / / 2H2O(l) / Kw-1 = 1.001014
(1+2+3) H3PO4(aq) + F-(aq) / / HF(aq) + H2PO4-(aq) / K = 10.7

is 10.7, which, as expected, is a favorable value. For the second reaction

(1)H3PO4(aq) + H2O(l) / / H2PO4-(aq) + H3O+(aq) / Ka1 = 7.1110-3
(2)H2PO4-(aq) + H2O(l) / / HPO42-(aq) + H3O+(aq) / Ka2 = 6.3210-8
(3)F-(aq) + H2O(l) / / HF(aq) + OH-(aq) / Kb = 1.510-11
(4)F-(aq) + H2O(l) / / HF(aq) + OH-(aq) / Kb = 1.510-11
(5)H3O+(aq) + OH-(aq) / / 2H2O(l) / Kw-1 = 1.001014
(6)H3O+(aq) + OH-(aq) / / 2H2O(l) / Kw-1 = 1.001014
(1+2+3+4+5+6) H3PO4(aq) + 2F-(aq) / / 2HF(aq) + HPO42-(aq) / K = 1.010-3

the equilibrium constant is 0.0010, which, as expected, is an unfavorable value.

N.B. Note that there was an error in the textbook; the phosphoric acid species product of the second reaction is HPO42-

3. The potential is calculated using the Nernst equation; thus

4. To balance these redox reaction use the half-reaction method outlined in class or in Appendix 4 in which we balance, in order, (i) all elements but oxygen and hydrogen, (ii) oxygen using H2O, (iii)hydrogen using H3O+/H2O or H2O/OH-, and (iv) charge.

(a) The reduction half-reaction is

MnO4- Mn2+

MnO4- Mn2+ + 4H2O

MnO4- + 8H3O+ Mn2+ + 12H2O

MnO4- + 8H3O+ + 5e- Mn2+ + 12H2O

The oxidation half-reaction is

H2SO3 SO42-

H2SO3 + H2O SO42-

H2SO3 + 5H2O SO42- + 4H3O+

H2SO3 + 5H2O SO42- + 4H3O+ + 2e-

Combining the two half-reactions gives

2MnO4- + 5H2SO3 + H2O 2Mn2+ + 5SO42- + 4H3O+

The standard state potential for the reaction is

0.172 = 1.338 V

for which the equilibrium constant is

(b) The reduction half-reaction is

IO3- I2

2IO3- I2

2IO3- I2 + 6H2O

2IO3- + 12H3O+ I2 + 18H2O

2IO3- + 12H3O+ + 10e- I2 + 18H2O

The oxidation half-reaction is

I- I2

2I- I2

2I- I2 + 2e-

Combining the two half-reactions gives, after simplifying the stoichiometric coefficients,

IO3- + 5I- + 6H3O+ 3I2 + 9H2O

The standard state potential for the reaction is

for which the equilibrium constant is

ClO- Cl-

ClO- Cl- + H2O

ClO- + H2O Cl- + 2OH-

ClO- + H2O + 2e- Cl- + 2OH-

The oxidation half-reaction is

I- IO3-

I- + 3H2O IO3-

I- + 6OH- IO3- + 3H2O

I- + 6OH- IO3- + 3H2O + 6e-

Combining the two half-reactions gives

3ClO- + I- 3Cl- + IO3-

The standard state potential for the reaction is

for which the equilibrium constant is

5. (a) As the pH of the solution decreases (becomes more acidic), SO42- will be converted to HSO4-. With a decreasing concentration of SO42-, the solubility reaction will shift to the right, increasing the solubility of BaSO4.

(b) Adding BaCl2, which is a soluble salt of Ba2+, increases the concentration of Ba2+ in solution, pushing the solubility reaction to the left and decreasing the solubility of BaSO4.

(c) Decreasing the volume of solution (by evaporation) causes the concentrations of both Ba2+ and SO42- to increase, pushing the solubility reaction to the left and decreasing the solubility of BaSO4.

6. (a) The solution contains Na+, Cl-, H3O+, and OH-; thus

charge balance

[H3O+] + [Na+] = [OH-] + [Cl-]

mass balance

0.1 M = [Na+] = [Cl-]

(b) The solution contains, H3O+, Cl-, and OH-; thus

charge balance

[H3O+] = [OH-] + [Cl-]

mass balance

0.1 M = [Cl-]

charge balance

[H3O+] = [OH-] + [F-]

mass balance

0.1 M = [HF] + [F-]

(d) The solution contains Na+, H2PO4-, H3PO4, HPO42-, PO43-, H3O+, and OH-; thus

charge balance

[Na+] + [H3O+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-]

mass balance

0.1 M = [Na+] = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]

(e) The solution contains Mg2+, CO32-, HCO3-, H2CO3, H3O+, and OH-; thus

charge balance

2[Mg2+] + [H3O+] = [OH-] + [HCO3-] + 2[CO32-]

mass balance

[Mg2+] = [H2CO3] + [HCO3-] + [CO32-]

(f) A solution of Ag(CN)2- (prepared from AgNO3 and KCN) contains Ag+, NO3-, K+, CN-, Ag(CN)2-, HCN, H3O+, and OH-; thus

charge balance

[Ag+] + [K+] + [H3O+] = [NO3-] + [CN-] + [Ag(CN)2-] + [OH-]

mass balance

[NO3-] = [Ag+] + [Ag(CN)2-]

[K+] = [CN-] + [HCN] + 2[Ag(CN)2-]

(g) A solution of HCl and NaNO2 contains Cl-, Na+, NO2-, HNO2, H3O+, and OH–; thus

charge balance

[Na+] + [H3O+] = [OH-] + [NO2-] + [Cl-]

mass balance

0.1 M = [Cl-]

0.050 M = [Na+] = [HNO2] + [NO2-]

7. (a) A charge balance equation requires that

[H3O+] = [OH-] + [ClO4-]

Because HClO4 is a strong acid and its concentration is relatively large, we can assume that [OH-] < [ClO4-]; thus

[H3O+] = [ClO4-] = 0.050 M

The pH, therefore, is 1.30.

(b) A charge balance equation requires that

[H3O+] = [OH-] + [Cl-]

Although HCl is a strong acid, its concentration if relatively small and the charge balance equation cannot be simplified. Combining the charge balance equation with a mass balance equation on Cl- gives

[Cl-] = 1.0010-7 M

[H3O+] = [OH-] + 1.0010-7

Rearranging and substituting into Kw gives us

[OH-] = [H3O+] - 1.0010-7

[H3O+]{[H3O+] - (1.0010-7)} = 1.0010-14

[H3O+]2 - (1.0010-7)[H3O+] - 1.0010-14 = 0

Solving this quadratic equation gives [H3O+] = 1.6210-7 or a pH of 6.79.

HOCl(aq) + H2O(aq) H3O+(aq) + OCl-(aq)

for which

The charge balance equation is

[H3O+] = [OH-] + [OCl-]

Because the solution is acidic, we will assume that [OH-] < [OCl-]; thus

[H3O+] = [OCl-]

To find the composition of the solution at equilibrium, we begin with the following table

HOCl(aq) + / H2O(l) / / H3O+(aq) + / OCl-(aq)
Initial / 0.025 / - / 0 / 0
Change / -x / - / +x / +x
Equilibrium / 0.025 - x / - / x / x

Substituting into the Ka expression leaves us with

Because HOCl is a relatively weak acid, we expect x to be significantly smaller than 0.025. We can simplify our calculation by assuming that

0.025 - x ≈ 0.025

Thus,

x = 2.7410-5

The assumption that x is small is reasonable (error of 0.11%). The pH, therefore, is 4.56.

(d) For a solution of HCOOH the pH is determined by the following equilibrium reaction

HCOOH(aq) + H2O(aq) H3O+(aq) + HCOO-(aq)

for which

The charge balance equation is

[H3O+] = [OH-] + [HCOO-]

Because the solution is acidic, we will assume that [OH-] < [HCOO-]; thus

[H3O+] = [HCOO-]

To find the composition of the solution at equilibrium, we begin with the following table

HCOOH(aq) + / H2O(l) / / H3O+(aq) + / HCOO-(aq)
Initial / 0.010 / - / 0 / 0
Change / -x / - / +x / +x
Equilibrium / 0.010 - x / - / x / x

Substituting into the Ka expression leaves us with

In this case an assumption that x < 0.010 is probably not warranted; thus

x2 = 1.810-6 - (1.810-4)x

x2 + (1.810-4)x - 1.810-6 = 0

Solving the quadratic equation gives x = [H3O+] = 1.25 x 10-3 M, or a pH of 2.90.

(e) A charge balance equation for a solution of Ba(OH)2 is

2[Ba2+] + [H3O+] = [OH-]

Because Ba(OH)2 is a strong base, we can assume that [H3O+] will be small and

[OH-] = 2[Ba2+] = 2 (0.050 M) = 0.10 M

Solving for the [H3O+] gives

or a pH of 13.00.

(f) For a solution of C5H5N the pH is determined by the following equilibrium reaction

C5H5N(aq) + H2O(aq) OH-(aq) + C5H5NH+(aq)

for which

The charge balance equation is

[C5H5NH+] + [H3O+] = [OH-]

Because the solution is basic, we will assume that [H3O+] < [C5H5NH+]; thus

[C5H5NH+] = [OH-]

To find the composition of the solution at equilibrium, we begin with the following table

C5H5N(aq) + / H2O(l) / / C5H5NH+(aq) + / OH-(aq)
Initial / 0.010 / - / 0 / 0
Change / -x / - / +x / +x
Equilibrium / 0.010 - x / - / x / x

Substituting into the Kb expression gives

Because HOCl is a relatively weak base, we expect x to be significantly smaller than 0.010. We can simplify our calculation by assuming that

0.010 - x ≈ 0.010

Thus,

x = [OH-] = 4.1110-6

The assumption that x is small is reasonable (error of 0.04%). Solving for the [H3O+] gives

or a pH of 8.61.

12. To solve for the solubility of Ag3PO4 we need to account for the form that phosphate will take in solution. A ladder diagram for H3PO4

shows that at a pH of 9.00 the principle form of phosphate is HPO42-. The solubility reaction, therefore, is

Ag3PO4(s) + H3O+(aq) 3Ag+(aq) + HPO42-(aq) + H2O(l)

The equilibrium constant for this reaction is

(1) Ag3PO4(s) / / 3Ag+(aq) + PO43-(aq) / Ksp = 2.810-18
(2) PO43-(aq) + H2O(l) / / HPO4-(l) + OH-(aq) / Kb = 2.210-2
(3) H3O+(aq) + OH-(aq) / / 2H2O(l) / Kw-1 = 1.001014
(4) Ag3PO4(s) + H3O+(aq) / / 3Ag+(aq) + HPO42-(aq) / K = 6.210-6

The solubility of Ag3PO4 is equivalent to the concentration of HPO42- at equilibrium. Because the pH is buffered to 9.00, the concentration of H3O+ is fixed at 1.010-9 M.

Ag3PO4(s) + / H3O+(aq) / / Ag+(aq) + / HPO42-(aq)
Initial / - / 1.0010-9 / 0 / 0
Change / - / - / +3x / +x
Equilibrium / - / 1.0010-9 / 3x / x

Substituting the equilibrium concentrations into the equilibrium constant expression gives

x = [HPO42-] = 1.210-4 M

The solubility of Ag3PO4 at a pH of 9.00, therefore, is 1.210-4 M.

14. (a) A solution of 0.050 M NaCl is 0.050 M Na+ and 0.050 M Cl-; thus

m = 0.5{(0.050)(+1)2 + (0.050)(-1)2} = 0.050 M

(b) A solution of 0.025 M CuCl2 is 0.025 M Cu2+ and 0.050 M Cl-; thus

m = (0.5){(0.025)(+2)2 + (0.050)(-1)2} = 0.075 M

m = (0.5){(0.20)(+1)2 + (0.1)(-2)2} = 0.30 M

19. The pH of an acid-base buffer solution is calculated using the Henderson-Hasselbach equation.

(a)