91Ó°ÊÓ

The equilibrium between a substance and its ions in an aqueous solution is characterised by the solubility product constant Ksp.

The ratio of moles of solute (in grammes) to the volume of the solution is known as molarity (in litres).

The formula for calculating a solution's molarity is:

$\mathbf{Molarity}\mathbf{}\left(M\right)\mathbf{=}\frac{\mathbf{Moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}\left(\mathrm{in}\mathrm{mol}\right)}{\mathbf{Volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}\left(\mathrm{in}L\right)}$

Given,

Volume of $\mathrm{KI}{\mathrm{O}}_3=34.50\mathrm{mL}$

Volume of iodate at equivalence point =35.69mL

The titration reaction of${\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2\mathrm{VsKI}{\mathrm{O}}_3$ is as follows

${{\mathrm{Hg}}_2}^{2+}+2{\mathrm{IO}}_3^{-}→{\mathrm{Hg}}_2{\left({\mathrm{IO}}_3\right)}_{2\left(\mathrm{s}\right)}$

When the volume of iodate is 34.50mL

the precipitate of ${\mathrm{Hg}}_2^{2+}$is not occurred.

$$\begin{gathered} \left[{\mathrm{Hg}}_2^{2+}\right]=\left(\frac{35.96-34.50}{35.69}\right)\left(0.04132\mathrm{M}\right)\left(\frac{25.00}{25.00+34.50}\right) \\ =5.79×{10}^{-4}\mathrm{M} \end{gathered}$$

When the volume of iodate is $36.50\mathrm{mL},{\mathrm{Hg}}_2{\left({\mathrm{IO}}_3\right)}_2$precipitates out.

Now we are in $\left(36.50\mathrm{mL}-35.69\mathrm{mL}\right)=0.81\mathrm{mL}$

The excess concentration of${{\mathrm{IO}}_3}^{-}$

$$\begin{gathered} \left[{\mathrm{IO}}_3^{-}\right]=\left(0.05789\right)\left(\frac{0.81}{25.00+36.50}\right) \\ =7.6245×{10}^{-4}\mathrm{M} \end{gathered}$$

The concentration of ${\mathrm{Hg}}_2^{2+}$ion which is in equilibrium with ${\mathrm{Hg}}_2{\left({\mathrm{IO}}_3\right)}_2$and iodate ion is

$$\begin{gathered} \left[{{\mathrm{Hg}}_2}^{2+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{[{\mathrm{IO}}_3·{\mathrm{T}}^2} \\ =\frac{1.3×{10}^{-18}}{7.6245×{10}^{-4}\mathrm{M}} \\ =2.2×{10}^{-12}\mathrm{M} \end{gathered}$$