Question

The solubility of $CuBr$ is $2\times {10}^{-4}$ $mol{L}^{-1}$. at ${25}^{o}C.$ The $K_{sp}$ value for $CuBr$ is:

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

Answer 1

The correct option is A $4\times {10}^{-8}mo{l}^2{L}^{-2}$

Given, $s=2\times {10}^{-4}$ $mol{L}^{-1}$
the dissociation of $CuBr$ is given by
$CuBr\left(s\right)\rightleftharpoons C{u}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)t=0100ateqb1-sss$
solubility product: $K_{sp}=\left[C{u}^{+}\right]\left[B{r}^{-}\right]=s\times s$
$s^2=(2\times {10}^{-4}{)}^2=4\times {10}^{-8}mo{l}^2{L}^{-2}$