Question

Suppose hydrogen sulfide is added to a solution that is 0.10 M in Cu2+, Pb2+, and Ni2+ such that the concentration of H2S is 0.10 M. When the pH of the solution is adjusted to 1.00, a precipitate forms. What is the composition of the precipitate? H2S(aq) + 2H2O(l) 2H3O+(aq) + S2–(aq); Kc = 1.1 × 10–20 Salt Ksp CuS 6.0 × 10–36 PbS 2.5 × 10–27 NiS 3.0 × 10–19

A. NiS only
B. CuS, PbS, and NiS
C. CuS and PbS
D. CuS only
E. PbS and NiS

Answer 1

C. CuS and PbS

Step-by-step explanation:

1. Calculate [H₃O⁺]

pH = 1.00

`\rm [H_(3)O^(+)] = 10^(-pH) = \text{0.100 mol/L}`

2. Calculate the concentration of S²⁻

H₂S + 2H₂O ⇌ S²⁻ + 2H₃O⁺; Kc = 1.1 × 10⁻²⁰

`K_{\text{c}} = \frac{\text{[S$^(2-)$][H$_(3)$O$^(+)$]}^(2)}{\text{[H$_(2)$S ]}}\\\\\text{[S$^(2-)$]} = \frac{K_{\text{a}}\text{[H$_(2)$S]}}{\text{[H}_(3)\text{O}^(+)]^(2)}= (1.1 * 10^(-20) * 0.10)/(0.100^(2)) = 1.1 * 10^(-19)`

3. Calculate Qsp for the sulfides

(a) CuS

CuS ⇌ Cu²⁺ + S²⁻; Ksp = 6.0 × 10⁻³⁶

Qsp = [Cu²⁺][S²⁻] = (0.10)(1.0 × 10⁻¹⁹) = 1.0 × 10⁻²⁰

Qsp > Ksp, so a precipitate of CuS will form.

(b) PbS

PbS ⇌ Pb²⁺ + S²⁻; Ksp = 2.5 × 10⁻²⁷

Qsp = [Pb²⁺][S²⁻] = (0.10)(1.0 × 10⁻¹⁹) = 1.0 × 10⁻²⁰

Qsp > Ksp, so a precipitate of PbS will form.

(iii) NiS

NiS ⇌ Ni²⁺ + S²⁻; Ksp = 3.0 × 10⁻¹⁹

Qsp = [Ni²⁺][S²⁻] = (0.10)(1.0 × 10⁻¹⁹) = 1.0 × 10⁻²⁰

Qsp < Ksp, so a precipitate will not form.