Question

Excess amount of solid calcium sulphate (CaSO\(_4\)) was added to a pure water sample (pH = 7) so that some solids remain undissolved at the equilibrium. The solubility product of CaSO\(_4\) is \( 2 \times 10^{-5} \, \text{mol}^2/\text{L}^2 \). The molar concentration of SO\(_4^{2-}\) in this water sample at equilibrium will be _______ mol/L (rounded off to three decimal places).

Hint:

The solubility product can be used to calculate the concentration of ions at equilibrium in cases of sparingly soluble salts.

Answer 1

Correct Answer: 0.004

The solubility product \( K_{sp} \) for calcium sulphate is given by: \[ K_{sp} = [\text{Ca}^{2+}] [\text{SO}_4^{2-}]. \] Let the concentration of \( \text{SO}_4^{2-} \) at equilibrium be \( x \). Since the dissociation of calcium sulphate produces one mole of \( \text{Ca}^{2+} \) and one mole of \( \text{SO}_4^{2-} \) per mole of calcium sulphate, the concentration of \( \text{Ca}^{2+} \) is also \( x \). Thus, the solubility product becomes: \[ K_{sp} = x \cdot x = x^2 = 2 \times 10^{-5}. \] Solving for \( x \): \[ x = \sqrt{2 \times 10^{-5}} \approx 0.0045 \, \text{mol/L}. \] Thus, the molar concentration of \( \text{SO}_4^{2-} \) at equilibrium is \( 0.0045 \, \text{mol/L} \).