Ncert Exercises & Solutions

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 BaSO₄ in water is $8 \times 10^{- 4} m o l d m^{- 3}$ . Calculate its solubility in 0.01 mol dm^-3 of H₂SO₄.

{BaSO}_{4} (s) ⇌ {Ba}^{2 +} (aq) + {SO}_{4}^{2 -} (aq)

K_{sp} for {BaSO}_{4} = [{Ba}^{2 +}] [{SO}_{4}^{2 -}] = s \times s = s^{2}

But s = 8 \times 10^{- 4} mol {dm}^{- 3}

∴ K_{sp} = {(8 \times 10^{- 4})}^{2} = 64 \times 10^{- 8}

In this presence of 0.01 MH₂SO₄, the expression for K_sp will be

K_{sp} = [{Ba}^{2 +}] [{SO}_{4}^{2 -}]

K_{sp} = (s) (s + 0.01) (0.01 M {SO}_{4}^{2 -} ions from 0.01 M H_{2} {SO}_{4})

64 \times 10^{- 8} = s \cdot (s + 0.01)

s^{2} + 0.01 s - 64 \times 10^{- 8} = 0

s = \frac{- 0.01 \pm \sqrt{{(0.01)}^{2} + (4 \times 64 \times 10^{- 8})}}{2}

= \frac{- 0.01 \pm \sqrt{10^{- 4} + (256 \times 10^{- 8})}}{2}

= \frac{- 0.01 \pm \sqrt{10^{- 4} (1 + 256 \times 10^{- 4})}}{2}

= \frac{- 0.01 \pm 10^{- 2} \sqrt{1 + 0.0256}}{2} = \frac{10^{- 2} (- 1 \pm 1.012719)}{2}

= 5 \times 10^{- 3} (- 1 + 1.012719) = 6.4 \times 10^{- 5} mol {dm}^{- 3}

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions