The value of Ksp for BaCrO4(s) in equilibrium with an aqueous solution of its constituent ions at 25°C is 1.2 × 10^−10 M². Write the chemical equation that represents the solubility equilibrium for BaCrO4(s) and calculate its solubility in water in grams per liter at 25°C.

Question

The value of [latex]K_{sp}[/latex] for [latex]BaCrO_{4}(s)[/latex] in equilibrium with an aqueous solution of its constituent ions at 25°C is [latex]1.2 × 10^{−10}\ M^{2}[/latex]. Write the chemical equation that represents the solubility equilibrium for [latex]BaCrO_{4}(s)[/latex] and calculate its solubility in water in grams per liter at 25°C.

Accepted Answer

The chemical equation that describes the solubility equilibrium is

[latex]BaCrO_{4}(s) ⇋ Ba^{2+}(aq) + CrO_{4}^{2−}(aq)[/latex]

The [latex]K_{sp}[/latex] expression for this equation is

[latex]K_{sp} = [Ba^{2+}][CrO_{4}^{2−}] = 1.2 × 10^{–10}\ M^{2}[/latex]

If [latex]BaCrO_{4}(s)[/latex] equilibrated with pure water, then, from the reaction stoichiometry, we have at equilibrium

[latex][Ba^{2+}] = [CrO_{4}^{2−}] = s[/latex]

where s is the solubility of [latex]BaCrO_{4}(s)[/latex] in pure water. Thus,

[latex]K_{sp} = s^{2} = 1.2 × 10^{–10}\ M^{2}[/latex]

and

[latex]s= (1.2 × 10^{–10} M^{2})^{1⁄ 2} = 1.1 × 10^{–5}\ M[/latex]

The solubility in grams per liter is given by

[latex]s = (1.1 × 10^{–5}\ mol·L^{–1}) \left(\frac{253.3\ g\ BaCrO_{4}}{1\ mol\ BaCrO_{4}} \right) = 2.8 × 10^{–3}\ g·L^{–1}[/latex]

Question 22.1: The value of Ksp for BaCrO4(s) in equilibrium with an aqueou......

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