Question

A mixture contains 1.0 x 10^-3 M Cu⁺² and Fe⁺² cations, and is saturated with .10 M H₂S. Determine the pH where CuS precipitates, but FeS does not precipitate. K_sp for CuS = 4.0 x 10^-36 and K_sp for FeS = 4.0 x 10^-17

Answer 1

CuS    ⇌   Cu ²⁺ +S ^2-

K_sp for CuS= 4*10^-36

FeS      ⇌          Fe ²⁺ +S ^2-

K_sp for FeS=4*10^-17  

H₂S    ⇌            2H⁺ +S ^2-

The sulfide concentrations needed to cause CuS and FeS to precipitate are

Ksp=[Cu²⁺] [S²⁻]

4*10^-36 = (0.10M) [S²⁻]

[S²⁻]=4*10^-36/1*10 ^-1=4*10 ^-35 M       (1)

and

Ksp=[Fe ²⁺][S^2-]

4*10^-17   =(0.10M) [S2−]

[S²⁻]=4*10 ^-16 M      (2)

sulfide concentrations between 4*10 ^-35 M and 4*10 ^-16 M      will precipitate CuS and not FeS.

H₂S       ⇌     2H⁺ +S ^2-

_pKa1=6.9=-logKa1

Ka1=10^-pKa = 1.1*10^-7   (3)

         

HS    ⇌    H⁺+S^2-

_pKa2=12.9

Ka2=1.3×10⁻¹³       (4)

The pK_a1for H₂S is 6.97, and pK_a2 corresponding to the formation of [S²⁻] is 12.90

Ka1=1.1*10^-7   = [H⁺][HS⁻] / [H2S]

= x² / 0.10 M

x =1.1×10⁻⁴ M

= [H⁺]=[HS⁻]

Substituting this value for [H⁺] and [HS⁻] into (4)

Ka2=1.3×10⁻¹³= [H+][S2−] / [HS−]=(1.1×10⁻⁴ ) x /1.1×10⁻⁴

x=[S²⁻]

equilibrium constant K for the overall reaction, which is the product of K_a1 and K_a2, and the concentration of H₂S in a saturated solution to calculate the H⁺ concentration needed to produce [S²⁻] of

K=Ka1 Ka2

= (1.1×10⁻⁷)(1.3×10⁻¹³)=1.4×10⁻²⁰   = [H⁺][HS⁻] / [H2S]* [H+][S2−] / [HS−]

=[H⁺][H⁺][S^2-] / [H2S]

= [H⁺]² [S^2-] / [H2S]

[H⁺]²=K[H2S][S^2-]=(1.4×10⁻²⁰)(0.10 M) / 4*10 ^-35 M

[H⁺] = 0.35*10 ¹⁴

pH = -log10[H+]

       =-log₁₀ 0.35*10 ¹⁴

= -log3.5*10¹³

^=12.45