Question

The solubility of BaSO₄ is 1.1 × 10^-5 mol/L. What is its K_sp?

• A. 1.21 x 10$^{-9}$
• B. 1.21 x 10$^{-10}$
• C. 2.42 x 10$^{-10}$
• D. 2.1 x 10$^{-11}$

Hint:

The Ksp of a sparingly soluble salt is calculated by squaring its solubility.

Answer 1

The Correct Option is B

The solubility product constant, \(K_{\text{sp}}\), is defined by the solubility \(S\) using the equation: \[ K_{\text{sp}} = S^2 \] With a solubility \(S = 1.1 \times 10^{-5}\) mol/L, the solubility product constant is calculated as: \[ K_{\text{sp}} = (1.1 \times 10^{-5})^2 = 1.21 \times 10^{-10} \]