Question

The solubility product of BaSO₄ is \(1 \times 10^{-10}\) at 298 K. The solubility of BaSO₄ in 0.1 M K₂SO₄(aq) is \(x \times 10^{-9}\) g L⁻¹. Calculate the value of \(x\).

Answer 1

Correct Answer: 8

To find the solubility of BaSO₄ in a 0.1 M K₂SO₄ solution, we first need to understand the equilibrium involved. BaSO₄ dissociates in water as:

\( \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \)

The solubility product (\(K_{sp}\)) of BaSO₄ is given by:

\( K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \)

Given \(K_{sp} = 1 \times 10^{-10}\), we know that the solution already contains \([\text{SO}_4^{2-}] = 0.1\) M from K₂SO₄(aq) dissociation. Assume the solubility of BaSO₄ to be 's' M. So:

\( [\text{Ba}^{2+}] = s \)

\( [\text{SO}_4^{2-}] = 0.1 + s \approx 0.1 \)

because s is much smaller than 0.1. Substitute back into \(K_{sp}\):

\( 1 \times 10^{-10} = s \times 0.1 \)

Solve for 's':

\( s = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \) M

Now convert molarity (mol L⁻¹) to g L⁻¹. Molar mass of BaSO₄ = 137.33 (Ba) + 32.07 (S) + 4×16 (O) = 233.39 g/mol.

\( \text{Solubility in g L}^{-1} = s \times 233.39 = 1 \times 10^{-9} \times 233.39 = 2.3339 \times 10^{-7} \text{ g L}^{-1} \)

Since solubility is given as \(x \times 10^{-9}\), equate:

\( 2.3339 \times 10^{-7} = x \times 10^{-9} \)

Solve for \(x\):

\( x = \frac{2.3339 \times 10^{-7}}{10^{-9}} = 233.39 \)