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

Answer:

C. CuS and PbS  

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

$K_{\text{c}} = \dfrac{\text{[S ^{2-} ][H _{3} O ^{+} ]}^{2}}{\text{[H _{2} S ]}}\\\\\text{[S ^{2-} ]} = \dfrac{K_{\text{a}}\text{[H _{2} S]}}{\text{[H}_{3}\text{O}^{+}]^{2}}= \dfrac{1.1 \times 10^{-20} \times 0.10}{0.100^{2}} = 1.1 \times 10^{-19}$

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.