Question

Find $\left[\mathrm{Cu}*\right]$ in equilibrium with CuBr(s) and 0.10MB${\mathbf{r}}^{\mathbf{-}}$.

Short answer

The $\left[{\mathrm{Cu}}^{+}\right]$was calculated as role="math" localid="1663330959468" $5\times {10}^{-8}\mathrm{M}$.

Step-by-step solution

The concept used.

Solubility Product:

The solubility product is a kind of equilibrium constant whose value depends on temperature.

The concentration

For given data, the equilibrium reaction is:

$\mathrm{CuBr}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Cu}}^{+}+{\mathrm{Br}}^{-}$

The

$\begin{array}{l}\left[C{u}^{+}\right]\left[B{r}^{-}\right]=K_{sp}\\ \left[C{u}^{+}\right]\left[0.10\right]=5\times {10}^{-9}M\end{array}$

Therefore,

$\begin{array}{l}\left[{\mathrm{Cu}}^{+}\right]=\frac{5\times {10}^{-8}\mathrm{M}}{0.10}\\ \left[{\mathrm{Cu}}^{+}\right]=5\times {10}^{-8}\mathrm{M}\end{array}$