Question

At 25 °C only 0.0770 mol of the generic salt AB2 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?

Answer 1

When dissolved in water, your salt will dissociate into cations and anions according to the following equilibrium

AB₂ (s)<=====> A²⁺ (aq) + 2B^- (aq)

Since we're dealing with a 1.00-L solution, the molar solubility of the salt will be

C=n/V = 0.0770 moles/ 1.00 L=0.0770 mol/L

see the mole ratio that exists between the species involved in the equilibrium. You have 1 mole of AB₂ dissociating to produce 1 mole of A²⁺ and 2moles of B−.

Make an ICE table to determine the value of the Ksp.

                 AB₂ (s)<=====> A²⁺ (aq) + 2B^- (aq)

I                   _                           0               0

C                 _                            +X           +2X

E                 _                              X              2X

By definition, Ksp will be

ksp=[A²⁺]⋅[B^-]2

          =x⋅(2x)²

       = 4X³

But x is actually the molar solubility of the salt, 0.0770 mol/L, which means that the value of the solubility product constant will be

Ksp=4⋅(0.0770)³

Ksp   = 0.001826132

Ksp = 1.82 x 10^-3