NEETChemistryIonic EquilibrumMCQ+4 / −1
pH of a saturated solution of Ca(OH)2 is 9. The solubility product (Ksp) of Ca(OH)2 is :
- A0.125 × 10–15
- B0.5 × 10–10
- C0.5 × 10–15
- D0.25 × 10–10
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Correct answer: C
pH = 9 ; pOH = 5 ; [OH–] = 10–5 = 2S
$ \therefore $ S = $${{{{10}^{ - 5}}} \over 2}$$
Ksp = [Ca+2] [OH–]2
$ \Rightarrow $ Ksp = S × (2S)2
$ \Rightarrow $ Ksp = 4S3
$ \Rightarrow $ Ksp = $$4 \times {\left( {{{{{10}^{ - 5}}} \over 2}} \right)^3}$$ = 0.5 × 10–15
$ \therefore $ S = $${{{{10}^{ - 5}}} \over 2}$$
Ksp = [Ca+2] [OH–]2
$ \Rightarrow $ Ksp = S × (2S)2
$ \Rightarrow $ Ksp = 4S3
$ \Rightarrow $ Ksp = $$4 \times {\left( {{{{{10}^{ - 5}}} \over 2}} \right)^3}$$ = 0.5 × 10–15
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