Question

Including activity coefficient, find $\left[{{\mathrm{Hg}}_2}^{2+}\right]$in saturated ${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{Br}}_{\mathbf{2}}$in localid="1654854633986" $\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{}\mathbf{}\mathbf{}\mathbf{00}\mathbf{}\mathbf{}\mathbf{}\mathbf{M}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{KBr}$ .

Short answer

The concentration of $\left[{{\mathrm{Hg}}_2}^{2+}\right]$in saturated ${\mathrm{Hg}}_2{\mathrm{Br}}_2$in $0.00100\mathrm{M}\mathrm{KBr}$ is$7\times {10}^{-17}\mathrm{M}$

Step-by-step solution

Step 1:Finding the activity coefficient of aqueous solution (Hg22+) and (Br-).

In the given problem, we need to find the activity coefficient along with concentration of $\left[{{\mathrm{Hg}}_2}^{2+}\right]$in saturated in ${\mathrm{Hg}}_2{\mathrm{Br}}_2\mathrm{in}0.00100\mathrm{M}\mathrm{KBr}$.

The solubility of is not quite expressed. So we will assume that,

$\mathrm{\mu }=0.001\mathrm{M}=\mathrm{c}\left({\mathrm{Br}}^{-}\right)$

From the Table $8-1$, we can know that the values are,

$$\begin{gathered} \mathrm{\mu }\left({\mathrm{Hg}}^{2+}\right)=0.867 \\ \mathrm{\mu }\left({\mathrm{Br}}^{-}\right)=0.964 \end{gathered}$$

Finding the concentration of (Hg22+) using equilibrium constant equation.

Now, we will use equilibrium constant equation,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=5.6\times {10}^{-23} \\ {\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{c}\left({{\mathrm{Hg}}_2}^{2+}\right)\mathrm{\mu }\left({\mathrm{Hg}}^{2+}\right)\right)\times \left({\mathrm{c}}^2\left(\mathrm{Br}-\right)\right) \\ 5.6\times {10}^{-23}=\left(\mathrm{c}\left({{\mathrm{Hg}}_2}^{2+}\right)\times 0.867\right)\times \left(0.{001}^2\times 0.{964}^2\right) \\ \mathrm{c}\left({{\mathrm{Hg}}_2}^{2+}\right)=7\times {10}^{-17}\mathrm{M} \end{gathered}$$

Thus, the concentration of$\left({{\mathrm{Hg}}_2}^{2+}\right)$in saturated${\mathrm{Hg}}_2{\mathrm{Br}}_2$in localid="1654855580714" $0.00100\mathrm{M}\mathrm{KBr}$is$7\times {10}^{-17}\mathrm{M}$