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: Understanding the Concept:
The solubility product constant (\( K_{sp} \)) represents the equilibrium between a solid solute and its ions in a saturated solution.
For a sparingly soluble salt like \( \text{Ba(OH)}_2 \), the \( K_{sp} \) is determined by the concentrations of \( \text{Ba}^{2+} \) and \( \text{OH}^- \) ions.
The pH provides a direct pathway to find the hydroxide ion concentration.
Step 2: Key Formula or Approach:
The dissociation equation is: \( \text{Ba(OH)}_{2(s)} \rightleftharpoons \text{Ba}^{2+}_{(aq)} + 2\text{OH}^-_{(aq)} \)
The \( K_{sp} \) expression is: \( K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 \)
Formulas relating pH to concentration:
\( \text{pOH} = 14 - \text{pH} \)
\( [\text{OH}^-] = 10^{-\text{pOH}} \)
From stoichiometry, if \( [\text{OH}^-] \) is known, \( [\text{Ba}^{2+}] = \frac{[\text{OH}^-]}{2} \).
Step 3: Detailed Explanation:
Given that the pH of the solution is 12.
First, calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
Now, calculate the hydroxide ion concentration:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \text{ M} = 0.01 \text{ M} \]
Let the molar solubility of \( \text{Ba(OH)}_2 \) be \( s \).
From the balanced dissociation equation, 1 mole of \( \text{Ba(OH)}_2 \) produces 1 mole of \( \text{Ba}^{2+} \) and 2 moles of \( \text{OH}^- \).
Therefore, \( [\text{OH}^-] = 2s \) and \( [\text{Ba}^{2+}] = s \).
We know \( [\text{OH}^-] = 10^{-2} \text{ M} \).
So, \( 2s = 10^{-2} \text{ M} \implies s = \frac{10^{-2}}{2} = 0.5 \times 10^{-2} \text{ M} = 5 \times 10^{-3} \text{ M} \).
Thus, \( [\text{Ba}^{2+}] = 5 \times 10^{-3} \text{ M} \).
Finally, calculate the \( K_{sp} \):
\[ K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 \]
\[ K_{sp} = (5 \times 10^{-3}) \times (10^{-2})^2 \]
\[ K_{sp} = 5 \times 10^{-3} \times 10^{-4} \]
\[ K_{sp} = 5 \times 10^{-7} \]
Step 4: Final Answer:
The value of \( K_{sp} \) is \( 5 \times 10^{-7} \).