NEET Solved Paper 2019 Question 19
Question: pH of a saturated solution of $ Ca{{(OH)}_2} $ is 9. The solubility product $ (K _{sp}) $ of $ Ca{{(OH)}_2} $ is- [NEET 5-5-2019]
Options:
A) $ 0.125\times 10^{15} $
B) $ 0.5\times 10^{10} $
C) $ 0.5\times 10^{15} $
D) $ 0.25\times 10^{10} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ Ca{{(OH)}_2}\rightarrow {{\underset{s}{\mathop{Ca}},}^{+2}}+{{\underset{2s}{\mathop{OH}},}^{-}} $ $ pH=9pOH=5 $ $ [O{H^{}}]=10^{5}=2s $ $ s=\frac{{10^{-5}}}{2} $ $ K _{sp}=\text{(s)}O{H^{}}^{2} $ $ =\frac{{10^{-5}}}{2}\times {{({10^{-5}})}^{2}} $ $ \frac{1}{2}\times {10^{-15}} $ $ =0.5\times 10^{15} $
