Question

calculate the molar solubility of Ag2SO4 in each solution below. (the ksp of silver sulfate is 1.5x10^-5)

a.0.34M AgNO3

b.0.34M Na2SO4

Answer 1

a)

[Ag+] from AgNO3 = 0.34 M

Ag2SO4   <—>   2Ag+       +   So42-

   0.34+2s       s

Ksp = [Ag+]^2 [SO42-]

1.5*10^-5 = (0.34+2s)^2 * s

since Ksp is small, s will be small and we can ignore s as compared to 0.34

Above expression thus becomes,

1.5*10^-5 = (0.34)^2 * s

s = 1.3*10^-4 M

Answer: 1.3*10^-4 M

b)

[SO42-] from Na2SO4 = 0.34 M

Ag2SO4   <—>   2Ag+       +   So42-

   2s           0.34+s

Ksp = [Ag+]^2 [SO42-]

1.5*10^-5 = (2s)^2 * (0.34+s)

since Ksp is small, s will be small and we can ignore s as compared to 0.34

Above expression thus becomes,

1.5*10^-5 = (2s)^2 * 0.34

s = 3.3*10^-3 M

Answer: 3.3*10^-3 M