Consider two Group IV metal ions X2+ and Y2+. A solution containing 0.01 M X2+ and 0.01 M Y2+ is saturated with H2S. The pH at which the metal sulphide YS will form as a precipitate is . (Nearest integer) (Given: Ksp(XS) = 1 10-22 at 25 , Ksp(YS) = 4 10-16 at 25 , [H2S] = 0.1 M in solution, Ka1 Ka2(H2S) = 1.0 10-21, 2 = 0.30, 3 = 0.48, 5 = 0.70)
Answer
4
Solution
YS precipitates when [Y2+][S2-] Ksp(YS):
[S2-] = 4 10-160.01 = 4 10-14 M
For H2S 2H+ + S2-:
[S2-] = Ka1Ka2[H2S][H+]2
[H+]2 = 1.0 10-21 0.14 10-14 = 2.5 10-9 = 25 10-10
[H+] = 5 10-5 M
pH = 5 - 5 = 5 - 0.70 = 4.30 4
Answer: 4