Question

The solubility of $CuBr$ is $2\times {10}^{-4}$ $mol/L$ at ${25}^{o}C$. The $K_{sp}$ value of $CuBr$ is:

• A. $4\times {10}^{-8}$ ${mol}^2{l}^{-2}$
• B. $4\times {10}^{-4}$ ${mol}^2{l}^{-2}$
• C. $4\times {10}^{-11}$ ${mol}^2{L}^{-1}$
• D. $4\times {10}^{-15}$ ${mol}^2{l}^{-2}$

Answer 1

Answer: (A)

$4\times {10}^{-8}$ ${mol}^2{l}^{-2}$

$CuBr\rightleftharpoons {Cu}^{+}+{Br}^{-}$

[$CuBr$ will be fully dissociated]

$\therefore$ $\left[{Cu}^{+}\right]=2\times {10}^{-4}$ and $\left[{Br}^{-}\right]=2\times {10}^{-4}$

Therefore, $\left[{Cu}^{+}\right]\left[{Br}^{-}\right]=2\times {10}^{-4}\times 2\times {10}^{-4}$ $=4\times {10}^{-8}$ ${mol}^2{L}^{-2}$