Given the molar solubility of BaCrO₄ in pure water is 1.08×10⁻⁵ M, what is the solubility product constant (Ksp) for BaCrO₄?

• A. 1.08×10⁻¹⁰
• B. 1.08×10⁻⁵
• C. 1.17×10⁻¹⁰
• D. 2.16×10⁻⁵

Answer: (C)

Verified step by step guidance

1. Understand the concept of solubility product constant (Ksp). It is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For BaCrO₄, the dissolution can be represented as: BaCrO₄(s) ⇌ Ba²⁺(aq) + CrO₄²⁻(aq).
2. Write the expression for the solubility product constant (Ksp) based on the dissolution equation. The Ksp expression is: $K_{\mathrm{sp}}=\; [Ba²⁺][CrO₄²⁻]$.
3. Recognize that the molar solubility of BaCrO₄ in water is given as 1.08×10⁻⁵ M. This means that at equilibrium, the concentration of Ba²⁺ ions and CrO₄²⁻ ions in the solution will both be 1.08×10⁻⁵ M.
4. Substitute the molar solubility values into the Ksp expression. Since both ions have the same concentration, the expression becomes: $K_{\mathrm{sp}}=\; (1.08\times 10⁻⁵)(1.08\times 10⁻⁵)$.
5. Calculate the product of the concentrations to find the Ksp value. This involves multiplying the molar solubility of Ba²⁺ and CrO₄²⁻ ions to get the solubility product constant.