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. Ksp for CuS = 4.0 x 10-36 and Ksp for FeS = 4.0 x 10-17
CuS ⇌ Cu 2+ +S 2-
Ksp for CuS= 4*10-36
FeS ⇌ Fe 2+ +S 2-
Ksp 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+][S2-]
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++S2-
pKa2=12.9
Ka2=1.3×10−13 (4)
The pKa1for H2S is 6.97, and pKa2 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 Ka1 and Ka2, 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+][S2-] / [H2S]
= [H+]2 [S2-] / [H2S]
[H+]2=K[H2S][S2-]=(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
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