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.



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