Question

What concentration of SO32– is in equilibrium with Ag2SO3(s) and 5.50 × 10-3 M Ag ? The Ksp of Ag2SO3 can be found here.

Answer 1

Given equation is

Ag2SO3(s) = 2 Ag+ + SO32-

Ksp = [Ag^+]^2 x[SO3^2-]

If x moles of Ag2SO3(s) dissolve we have 2x moles of Ag^+ and x moles of SO3^2-
Given [Ag^+] at equilibrium=5.50x10^-3 M and

Ksp =1.5x10^-14
In 5.50 x 10 ^-3 M Ag + we have [Ag+ ] = 2x + 5.50 x 10 ^-3 moles / liter

Therefore
Ksp = (2x + 5.50 x 10 ^-3)^2 * x = 1.50×10^–14
Let us assume x <<< 5.5 x 10 ^-3
we then have 1.50 x 10 ^-14 = x *(5.50 x 10 ^-3)^2
x = 1.50 x 10 ^-14 / (5.50 x 10 ^-3)^2
= 0.049 x 10 ^-08 moles/L=0.49x10^-9 M.

Therefore the concentration of SO3^2-=0.49x10^-9 M.