Question

$ pH $ of a saturated solution of $ Ba(OH)_2 $ is $12$ . The value of solubility product $ K_({sp} )$ of $ Ba(OH)_2 $ is

• A. $ 3.3 \times 10^{-7} $
• B. $ 5.0 \times 10^{-7} $
• C. $ 4.0 \times 10^{-6} $
• D. $ 5.0 \times 10^{-6} $

Answer 1

The Correct Option is B

To find the solubility product K_{\text{sp}} of \text{Ba(OH)}_2, we can use the given pH of the solution. Here is the step-by-step explanation:

1. Given that the pH of the saturated solution is 12, we first calculate the pOH.
2. We know that sum of pH and pOH is 14: \text{pH} + \text{pOH} = 14.
3. Given, pH = 12: \text{pOH} = 14 - 12 = 2.
4. The concentration of hydroxide ions, [\text{OH}^-], can be calculated using: [\text{OH}^-] = 10^{-\text{pOH}}.
5. Substituting the value: [\text{OH}^-] = 10^{-2} = 0.01 \text{ M}.
6. The dissociation of \text{Ba(OH)}_2 in water is represented by the equation: \text{Ba(OH)}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{OH}^- (aq).
7. If s is the solubility of \text{Ba(OH)}_2, then:
   • [\text{Ba}^{2+}] = s
   • [\text{OH}^-] = 2s
8. We know [\text{OH}^-] = 0.01 \text{ M}, which gives 2s = 0.01 and hence, s = 0.005 \text{ M}.
9. The expression for the solubility product K_{\text{sp}} is given by: K_{\text{sp}} = [\text{Ba}^{2+}] [\text{OH}^-]^2.
10. Substitute the values: K_{\text{sp}} = (0.005) \times (0.01)^2 = 5 \times 10^{-7}.

Thus, the solubility product K_{\text{sp}} of \text{Ba(OH)}_2 is 5.0 \times 10^{-7}, which matches option 2.