CHEMISTRY 115 HOMEWORK 4

CHEMISTRY 115 HOMEWORK 2 NAME ______________

April 18, 2001 COMP.HWK:Sol.Eq.: all

1. A solution is 0.15 M in both Ca⁺² and Sr⁺². If solid Na₂SO₄ is slowly added to 1.00 L solution, which salt will precipitate first? Ksp CaSO₄ = 2.4 x 10^-5; Ksp SrSO₄ = 3.2 x 10^-7

Same type chemical formula so smaller will precipitate first

a. Will Ca⁺² or Sr⁺². precipitate first? Sr²⁺

b. Calculate the concentration of this ion that is left in solution when the second ion just begins to precipitate. Work with second ion first because it is what you are trying to keep in solution. Ksp CaSO₄ = 2.4 x 10^-5 = [Ca²⁺][SO₄^2-] = [0.15][x]

x = 2.4 x 10^-5 / 0.15 = 1.6 x 10^-4 M maximum sulfate concentration before Ca²⁺ precipitates

Ksp SrSO₄ = 3.2 x 10^-7 = [Sr²⁺][SO₄^2-] = [Sr²⁺][1.6 x 10^-4]

[Sr²⁺] = 3.2 x 10^-7 / 1.6 x 10^-4 = 2.0 x 10^-3, which means that 1.3% of Sr²⁺ is still in soln.

Conc. = |_2.0 x 10^-3M______

2. Can selective separation of Mn⁺² and Sn⁺² be accomplished by suitable adjustment of the pH of a solution that is 0.010 M in Mn⁺² and 0.010 M in Sn⁺²with saturated with H2S?
Ksp MnS = 3 x 10^-11 (pink) = 3 x 10^-14 (green); Ksp SnS = 1 x 10^-26; sat’d H₂S soln = 0.10 M

Calculate the pH of the solution that would lead to the maximum precipitation of one ion but not permit the precipitation of the other ion
Because SnS has smaller Ksp, it will precipitate at the lower pH. Work with the more soluble species to keep it in soln.

Because this will be acidic solution, use Kspa = Ksp/KwxKa₁ = Ksp/1 x 10^-21

MnS(s) + 2 H⁺ = Mn²⁺ + H₂S; Kspa = [Mn²⁺][H₂S] / [H⁺]² = [0.010][0.10] / [H⁺]² = 3 x 10¹⁰

[H⁺] = (1.0 x 10^-3 / 3 x 10¹⁰)^1/2 = (3.33 x 10^-14)^1/2 = 1.826 x 10^-7 M; pH = 6.74 (pink)

[H⁺] = (1.0 x 10^-3 / 3 x 10⁷)^1/2 = (3.33 x 10^-11)^1/2 = 5.77 x 10^-6 M; pH = 5.24 (green)

pH = |_6.74 pink; 5.24 green_

3. Calculate the molar solubility of CoCO3 in a strongly buffered solution of pH = 9.00.
Ksp CoCO₃ = 1.0 x 10^-10; Ka₁H₂CO₃ = 4.5 x 10^-7; Ka₂ = 4.7 x 10^-11 ; [H⁺] = 1.0 x 10^-9

CoCO₃(s) = Co²⁺ + CO₃^2- ; CO₃^2- + H⁺ = HCO₃^-; HCO₃^- + H⁺ =H₂CO₃; soln is not so acid that it likely would be saturated with CO₂(g) – so can’t use acid method in text.

Ka₁ = [H⁺][HCO₃^-] / [H₂CO₃]; Ka₂ = [H⁺][CO₃^2-] / [HCO₃^-]

[Co²⁺] = [CO₃^2-] + [HCO₃^-] +[H₂CO₃] = solubility since it is not changed

[HCO₃^-] = [H⁺][CO₃^2-] / Ka₂ = [1.0 x 10^-9] [CO₃^2-] /[ 4.7 x 10^-11] = 21.28 [CO₃^2-]

[H₂CO₃] = [H⁺][HCO₃^-] / Ka₁ = [1.0 x 10^-9] [21.28CO₃^2-] /[ 4.5 x 10^-7] = .0473 [CO₃^2-]

[Co²⁺] = [CO₃^2-] + [HCO₃^-] +[H₂CO₃] = [1 + 21.28 + 0.0473][CO₃^2-] = 22.32[CO₃^2-]

Ksp CoCO₃ = 1.0 x 10^-10 = [Co²⁺][CO₃^2-] = 22.32[CO₃^2-][CO₃^2-] = 22.32[CO₃^2-]²

[CO₃^2-] = (1.0 x 10^-10/22.32)^1/2 = (4.49 x 10^-12)^1/2 = 2.116 x 10^-6M

[Co²⁺] = 22.32[CO₃^2-] = 22.32 x 2.12 x 10^-6 = 4.72 x 10^-5 M

Sol. = |_4.7 x 10^-5 M______

Will CuS dissolve in 8 M HCl solution? Perform calculations that prove your answer.
CuS(s) + 2 H⁺ = Cu²⁺ + H₂S(aq) since this is a very acid solution

Ksp CuS = 6 x 10^{-37 à} Kspa = 6 x 10^-16 = [Cu²⁺][H₂S] / [H⁺]² = [x][x] /[8- 2x]²

[x] ~ (6 x 10^-16 x 64)^1/2 = (3.84 x 10^-14 )^1/2= 1.96 x 10^-7 M so doesn’t dissolve

Soluble/ insoluble

A student performing the chloride ion test obtained a white precipitate on addition of silver nitrate to the acidified solution. This precipitate disappeared when ammonia was added. Write the chemical reactions and the equilibrium expressions associated with this process, the overall reaction and calculate the overall equilibrium constant.
Ag⁺ + Cl^- = AgCl(s); Ksp = 1.8 x 10^-10 = [Ag⁺][Cl^-]

Ag⁺ + 2 NH₃(aq) = Ag(NH₃)₂⁺; K_f = 1.6 x 10⁷ = [Ag(NH₃)₂⁺] / [Ag⁺] [NH₃]²

AgCl(s) + 2 NH₃(aq) = Ag(NH₃)₂⁺ + Cl^-; K = [Ag(NH₃)₂⁺][Cl^-] / [NH₃]² = Ksp x K_f

K = 1.8 x 10^-10 x 1.6 x 10⁷ = 2.88 x 10^-3

K = |_2.9 x 10^-3_______

Silver ion forms a complex with thiosulfite ion, S₂O₃^2-, that has the formula Ag(S₂O₃)₂^3-. This complex has a Kf = 2.0 x 10¹³. How many grams of AgBr (Ksp = 5.0 x 10^-13) will dissolve in 125 mL 1.20 M Na₂S₂O₃ solution?
AgBr(s) + 2 S₂O₃^2- = Ag(S₂O₃)₂^3- + Br^-; K = Ksp x K_f = 5.0 x 10^-13 x 2.0 x 10¹³ = 10

K = [Ag(S₂O₃)₂^3-][Br^-] / [S₂O₃^2-]² = [x][x] / [1.20-2x]² = 10

(10)^1/2 = x / [1.20 – 2x] = 3.16;

3.79 – 6.32x = x

7.32 x = 3.79; x = 0.518 M

MW AgBr = 187.8 g/mol;

0.518 mol/L x 0.125 L x 187.8 g/mol = 12.16 g

Mass = |_12 g_________

7. A total of 0.010 mol of Cd(NO3)2 is dissolved in 1.00 liter of solution that is 1.0 M in NaSCN. Kf for [Cd(SCN)₄]^2-. Calculate the concentrations of Cd²⁺and [Cd(SCN)₄]^2- at equilibrium.
Cd²⁺ + 4 SCN^- = Cd(SCN)₄^2-; K_f = 1 x 10³ = [Cd(SCN)₄^2-] / [Cd²⁺][SCN^-]⁴

K_f is rather large, so mainly products, i.e. 0.010 M [Cd(SCN)₄^2-], [SCN^-] = [1.00 – 0.040] M

Thus [Cd²⁺] = x formed from dissociation of complex ion, etc.

K_f = 1 x 10³ = [Cd(SCN)₄^2-] / [Cd²⁺][SCN^-]⁴ = [0.010 –x] / [x][0.96 + 4x]⁴ ~ [0.010] /[x][0.96]⁴

x ~ [0.010] / ([0.849] * 1 x 10³) = 1.18 x 10^-5 ~ [Cd²⁺]; approx ok

[Cd^-2] = |_1.2 x 10^-5 M_____
[Cd(SCN)₄^2-] = 0.010 – 1.2 x 10^-5 = 0.010 M

[Cd(SCN)₄]^2- = |_0.010 M__