A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility –4 product. If the solubility of BaSO4 in water is 8 × 10 mol dm–3. Calculate its –3 solubility in 0.01 mol dm of H2SO4.
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Ba2+ (aq) + SO 4 (aq) At t = 0 1 0 0 At equilibrium in water 1–S S S At equilibrium in the presence 1–S S (S+0.01) of sulphuric acid 2– 2 K sp for BaSO4 in water = [Ba2+] [ SO 4 ] = (S) (S) = S But S = 8 × 10–4 mol dm–3 –4 2 –8 ∴ Ksp = (8×10 ) = 64 × 10 ... (1) The expression for Ksp in the presence of sulphuric acid will be as follows : Ksp = (S) (S + 0.01) ... (2) Since value of Ksp will not change in the presence of sulphuric acid, therefore from (1) and (2) –8 (S) (S + 0.01) = 64 × 10 2 –8 S + 0.01 S = 64 × 10 2 –8 S + 0.01 S – 64 × 10 = 0 2 –8 – 0.01 ± (0.01) + (4 × 64 × 10 ) S= –4 –8 – 0.01 ± 10 + (256 × 10 ) = –4 –2 – 0.01 ± 10 (1 + 256 × 10 ) = –2 – 0.01 ± 10 1 + 0.256 = –2 – 0.01 ± 10 1.256 = –2 –2 – 10 + (1.12 × 10 ) = –2 (–1+1.12) × 10 0.12 –2 = = × 10 2 2 –4 = 6 × 10 mol dm–3 99 Equilibrium