Question

What is the value of \( K_{sp} \) for saturated solution of \( \text{Ba(OH)_2 \) having pH 12 ?

• A. \( 4 \times 10^{-4} \)
• B. \( 4 \times 10^{-6} \)
• C. \( 5 \times 10^{-6} \)
• D. \( 5 \times 10^{-7} \)

Hint:

For a $AB_2$ type electrolyte, $K_{sp} = s(2s)^2 = 4s^3$, where $s$ is the solubility.

Answer 1

The Correct Option is D

Step 1: Concept
The solubility product constant ($K_{sp}$) is calculated from the molar concentrations of ions in a saturated solution.

Step 2: Meaning
pH 12 means pOH = $14 - 12 = 2$. Therefore, $[\text{OH}^-] = 10^{-2} \text{ M}$.

Step 3: Analysis
For $\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-$, if $[\text{OH}^-] = 10^{-2} \text{ M}$, then $[\text{Ba}^{2+}] = \frac{1}{2} \times 10^{-2} = 5 \times 10^{-3} \text{ M}$.
$K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 = (5 \times 10^{-3})(10^{-2})^2$.

Step 4: Conclusion
$K_{sp} = 5 \times 10^{-3} \times 10^{-4} = 5 \times 10^{-7}$. Final Answer: (D)