Question

What is the silver ion concentration in a solution prepared by mixing 431 mL of 0.398 M silver nitrate with 445 mL of 0.422 M sodium chromate? The ?sp of silver chromate is 1.2×10^−12 .

Answer 1

AgNO3 Molarity initially M1= 0.398 M

initial volume V1 = 431 ml

final solution volume after mixing V2 = 431 + 445 = 876 ml

using M1V1 = M2V2 we find final molarity of AgNO3

0.398 M x 431 ml = M2 x 876 ml

M2 = 0.196

[Ag+] = 0.196 M ( since AgNO3 is strong electrolyte and dissociates nearly to complete extent to give Ag+ and NO3-)

similarly for Sodium chromate Na2CrO4  

0.422M x 445 ml = M2 x 876 ml

M2 = 0.214

[CrO4^2- ] = 0.214 M   

Now 2Ag+ combines with 1 CrO4^2- to form Ag2CrO4 complex , hence [CrO4^2-] reacted = ( 1/2) [Ag+]

= ( 1/2) ( 0.196) = 0.098

[CrO4^2- left = 0.214-0.098 = 0.116 M

now we have dissociation eq Ag2CrO4 (s) <---> 2Ag+ (aq) + CrO4^2- (aq)

Ksp = [Ag+]^2 [ CrO4^2-]

1.2 x 10^ -12 = [Ag+]^2 ( 0.116)

[Ag+] = 3.2 x 10^ -6 M