Question

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

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

Answer 1

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

The reaction for the dissociation of CuBr is $CuBr\rightleftharpoons \underset{2\times {10}^{-4}}{C{u}^{+}}+\underset{2\times {10}^{-4}}{B{r}^{-}}$

The expression for the solubility product is $K_{sp}=\left[C{u}^{+}\right]\left[B{r}^{-}\right]$

Substitute values in the above expression:

$K_{sp}=(2\times {10}^{-4})(2\times {10}^{-4})=4\times {10}^{-8}mo{l}^{2}l{t}^{-2}$.

Hence, option A is correct.