Question

Find $\left[{\mathrm{Hg}}_2^{2+}\right]$ in equilibrium with 0.010MKCl saturated with${\mathbf{Hg}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}$.

Short answer

The value of $\left[{\mathrm{Hg}}_2^{2+}\right]$ in equilibrium with 0.010M KCI saturated ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$is$2.2\times {10}^{-14}\mathrm{M}$

Step-by-step solution

Concept used.

Solubility product $\left(K_{sp}\right)$:

The product of a substance's dissolved ion concentrations raised to the power of their stoichiometric coefficients is used to calculate the solubility product.

Step 2: Calculate the value of [Hg22+].

Assume that ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$each contribute a negligible amount of C${\mathrm{l}}^{-}$to 0.010MKCl. This is due to the low solubility of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$.

From the table 8-1 and given information the required values are,

$$\begin{gathered} \mu =0.010M \\ \left[C{l}^{-}\right]=0.010M \\ {\gamma }_{H{g}^{2+}}=0.660and \\ {\gamma }_{Cr}=0.899 \end{gathered}$$

Calculate the value of$\left[{\mathrm{Hg}}_2^{2+}\right]$ using the equation for the solubility product

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=1.2\times {10}^{-18} \\ =\left[{\mathrm{Hg}}_2^{2+}\right]{\mathrm{\gamma }}_{{\lg }^{2+}}{\left[{\mathrm{Cl}}^{-}\right]}^2{\mathrm{\gamma }}^2{\mathrm{Cr}}^2 \\ =\left[{\mathrm{Hg}}_2^{2+}\right]\left(0.660\right){\left(0.01\right)}^2{\left(0.899\right)}^2 \\ \left[{\mathrm{Hg}}_2^{2+}\right]=2.2\times {10}^{-14}\mathrm{M} \end{gathered}$$

The value of $\left[{\mathrm{Hg}}_2^{2+}\right]$in equilibrium with 0.010M KCI saturated ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$is role="math" localid="1663411773731" $2.2\times {10}^{-14}\mathrm{M}$