Question

A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Q_sp) becomes greater than its solubility product. If the solubility of \[\ce{BaSO4}\] in water is 8 × 10^–4 mol dm^–3. Calculate its solubility in 0.01 mol dm^–3 of \[\ce{H2SO4}\].

Answer 1

	\[\ce{BaSO4 (s)}\]	\[\ce{⇌ Ba^{2+} (aq)}\]	\[\ce{+ SO^{2-}4 (aq)}\]
At t = 0	1	0	0
At equilibrium in water	1 – S	S	S
At equilibrium in the presence of sulphuric acid	1 – S	S	(S + 0.01)

K_sp for \[\ce{BaSO4}\] in water = \[\ce{[Ba^{2+}] [SO^{2-}4] = (S) (S) = S^2}\]

But S = 8 × 10^–4 mol dm^–3

∴ K_sp = (8 × 10^–4)² = 64 × 10^–8  ......(1)

The expression for K_sp in the presence of sulphuric acid will be as follows:

K_sp = (S) (S + 0.01)  ......(2)

Since value of K_sp will not change in the presence of sulphuric acid, therefore from (1) and (2)

(S) (S + 0.01) = 64 × 10^–8

S² + 0.01 S = 64 × 10–8

S² + 0.01 S – 64 × 10^–8 = 0

S = `(- 0.01 +- sqrt((0.01)^2 + (4 xx 64 xx 10^-8)))/2`

= `(- 0.01 +- sqrt(10^-4 + (256 xx 10^-8)))/2`

= `(- 0.01 +- sqrt(10^-4 (1 + 256 xx 10^-2)))/2`

= `(- 0.01 +- 10^-2 sqrt(1 + 0.256))/2`

= `(- 0.01 +- 10^-2 sqrt(1.256))/2`

= `(- 10^2 + (1.12 xx 10^-2))/2`

= `((-1 + 1.12) xx 10^-2)/2`

= `0.12/2 xx 10^-2`

= 6 × 10^–4 mol dm^–3