Equilibrium Question 284
Question: $ K _{sp} $ for $ Cr{{(OH)}_3} $ is $ 2.7\times {10^{-31}} $ . What is its solubility in moles / litre. [JEE Orissa 2004]
Options:
A) $ 1\times {10^{-8}} $
B) $ 8\times {10^{-8}} $
C) $ 1.1\times {10^{-8}} $
D) $ 0.18\times {10^{-8}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ Cr{{(OH)}_3}\ \to \ \underset{x}{\mathop{C{r^{+3}}}},+\underset{3x}{\mathop{3O{H^{-}}}}, $
$ K _{sp}=x.{{(3x)}^{3}}=27x^{4} $
$ x=\sqrt[4]{\frac{K _{sp}}{27}} $ ; $ x=\sqrt[4]{\frac{2.7\times {10^{-31}}}{27}} $
$ x=1\times {10^{-8}} $ mole/litre.
