Question

What concentrations of SO3^2- is in equilibrium with Ag2SO3 and 8.7x10^-3 M of Ag+?

Answer 1

Ag2SO3(s) = 2 Ag+ + SO3^2-
The Ksp of Ag2SO3 is 1.50×10^–14
Ksp = [Ag+]^2 * [SO3 ^2- ]
if x moles of Ag2SO3(s) dissolve we have 2x moles of Ag+ and x moles of SO3 2-
in .8.7 x 10 ^-3 M Ag + we have [Ag+ ] = 2x + 8.7 x 10 ^-3 moles / liter

Ksp = (2x + 8.7 x 10 ^-3)^2 * x = 1.50×10^–14
we can solve the quadratic equation or assume (x <<< 8.7x 10 ^-3, use aprox)
we then have 1.50 x 10 ^-14 = x * (8.70 x 10 ^-3)^2
x = 1.50 x 10 ^-14 / (8.7 x 10 ^-3)^2
[SO3^2-]= 1.98 x 10 ^-10 moles per liter