Equilibrium Question 282
Question: The solubility of $ CuBr $ is $ 2\times {10^{-4}}mol/l $ at $ 25{}^\circ C $ . The $ K _{sp} $ value for $ CuBr $ is [AIIMS 2002]
Options:
A) $ 4\times {10^{-8}}mol^{2}{l^{-2}} $
B) $ 4\times {10^{-11}}mol^{2}{L^{-1}} $
C) $ 4\times {10^{-4}}mol^{2}{l^{-2}} $
D) $ 4\times {10^{-15}}mol^{2}{l^{-2}} $
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Answer:
Correct Answer: A
Solution:
$ \underset{K _{sp}}{\mathop{CuBr}}, $ ⇌ $ \underset{(S)}{\mathop{C{u^{+}}}},+\underset{(S)}{\mathop{B{r^{-}}}}, $
$ K _{sp}=S^{2}={{(2\times {10^{-4}})}^{2}}=4\times {10^{-8}}\frac{mol^{2}}{l^{2}} $
