Question

Calculate the molar solubility in pure water for the following compound at 25°C, Ag₂SO₃ K_sp = 1.5x10^-14

What is the molar solubility in 0.0010 M sodium sulfite?  

Answer 1

1)
At equilibrium:
Ag2SO3    <---->    2 Ag+     +          SO32-  

                     2s                 s       

Ksp = [Ag+]^2[SO32-]
1.5*10^-14=(2s)^2*(s)
1.5*10^-14= 4(s)^3
s = 1.554*10^-5 M
Answer: 1.6*10^-5 M

2)
Na2SO3 here is Strong electrolyte
It will dissociate completely to give [SO32-] = 0.001 M
At equilibrium:
Ag2SO3    <---->    2 Ag+     +          SO32-  
                     2s                 1*10^-3 + s
Ksp = [Ag+]^2[SO32-]
1.5*10^-14=(2s)^2*(1*10^-3+ s)
Since Ksp is small, s can be ignored as compared to 1*10^-3
Above expression thus becomes:
1.5*10^-14=(2s)^2*(1*10^-3)
1.5*10^-14= 4(s)^2 * 10^-3
s = 1.936*10^-6 M
Answer: 1.9*10^-6 M