For the reaction,

${\mathrm{Hg}}^{2+}+2{\mathrm{Cl}}^{-}\rightleftharpoons {\mathrm{HgCl}}_2,\mathrm{K}=1.65\times {10}^{13}$

concentration of Hg²⁺, at the equivalence point in the titration of 2.0 mmol of Hg²⁺ with CI^- when final volume is 100 mL, is

$$\begin{gathered} \left[{\mathrm{Hg}}^{2+}\right]=2.0\text{ millimoles in }100\mathrm{mL}\text{ solution } \\ =\frac{2\times {10}^{-3}\mathrm{mol}}{0.1\mathrm{L}} \\ =2\times {10}^{-2}\mathrm{M}=0.02\mathrm{M} \end{gathered}$$
Since, stability constant of HgCl₂ is large (1 .65x10¹³), the value of [Hg²⁺] should be small.

At the end point, $\left[{\mathrm{Cl}}^{-}\right]=2\left[{\mathrm{Hg}}^{2+}\right]$

$$\begin{gathered} \left[{\mathrm{HgCl}}_2\right]=0.020-\left[{\mathrm{Hg}}^{2+}\right]\approx 0.02\mathrm{M} \\ \therefore \frac{\left[{\mathrm{HgCl}}_2\right]}{\left[{\mathrm{Hg}}^{2+}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2}=1.65\times {10}^{13} \\ \frac{0.02}{\left[{\mathrm{Hg}}^{2+}\right]{\left(2\left[{\mathrm{Hg}}^{2+}\right]\right)}^2}=1.65\times {10}^{13} \\ \therefore 4{\left[{\mathrm{Hg}}^{2+}\right]}^3=\frac{0.02}{1.65\times {10}^{13}} \\ \therefore \left[{\mathrm{Hg}}^{3+}\right]=6.172\times {10}^6\mathrm{M} \end{gathered}$$