Question

At 25 °C only 0.0200 mol of the generic salt AB is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?

Answer 1

Ans:- K_sp = 4.0 x 10^-4 M²

Explanation :-

Molarity or Solubility of sparingly soluble salt i.e. AB in water is = 0.0200 mol / 1.00 L = 0.0200 mol / L

Now, Partial dissociation of AB in water is :

.........................AB (s) <--------------------> A⁺ (aq) ..................+..........................B^- (aq)

.................................................................0.0200 M............................................0.0200 M

Expression of solubility product i.e. K_sp is :

K_sp = [A⁺(aq)] . [B^- (aq)]

K_sp = (0.0200 M ) .( 0.0200 M)

K_sp = 4.0 x 10^-4 M²