Question

Which of the following precipitate form a solution that has the concentrations indicated? (Show work for each one) a) KClO4: [K+]= 0.01M, [ClO4-]= 0.01 M b) K2PtCl6: [K+]= 0.01 M, [PtCl6^2-]= 0.01 c) PbI2: [Pb2+]= 0.003 M, [I-]= 1.3*10^-3 d) Ag2S: [Ag+]= 1*10^-10 M, [S^2-]= 1*10^-13 M

Answer 1

answer : d) Ag2S: [Ag+]= 1*10^-10 M, [S^2-]= 1*10^-13 M

complete explanation:

a ) ionic product of KClO4 = [K+] [ClO4-] = 0.01 x 0.01 = 10^-4

b) K2PtCl6 ionic product = [K+]^2 [PtCl6-2] = (0.01)^2 (0.01) = 10^-6

c) PbI2 ionic product = [Pb+2] [I-]^2 = (0.003) (1.3*10^-3)^2 = 5.07 x 10^-9

d) Ag2S ionic product = [Ag+]^2 [S-2] = (1 x 10^-10 )^2 (1 x10^-13) = 1.0 x 10^-33

Ksp values for all a,b,c,d

KClO4 highly soluble nomatter of precipitate

K2PtCl6 , Ksp = 7.48×10^–6

PbI2 , Ksp = 9.8×10^–9

Ag2S , Ksp = 3.3×10^–50

for getting precipitate the rule ionic product > Ksp

by comapring ionic product and solubility product of all

only Ag2S ioninc product (1.0 x 10^-33 ) exceeds solubility product ( 3.3×10^–50 )