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Question
pH of a saturated solution of Ca(OH)2 is 9. The Solubility product (Ksp) of Ca(OH)2
Options
0.5 × 10−15
0.25 × 10−10
0.125 × 10−15
0.5 × 10−10
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Solution
0.5 × 10−15
Explanation:
\[\ce{Ca(OH)2 ⇌ Ca^{2+} + 2OH^-}\]
Given that pH = 9
pOH = 14 – 9 = 5
[pOH = – log10 [OH–]]
∴ [OH–] = 10–pOH
[OH–] = 10–5 M
Ksp = [Ca2+] [OH–]2
= `10^-5/2 xx (10^-5)^2`
= 0.5 × 10−15
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