Question

The solubility product constant of Hg₂CI₂ is 2X10^-8 at 25^oc . Estimate the concentration of Hg₂²⁺ and CI^- in equilibrium with solid Hg₂CI₂ at 25^oc .

Short answer

Solubility product

$$\begin{gathered} AB\rightleftarrows A^{+}+B^{-} \\ {K}_{SP}=\left[A^{+}\right]+\left[B^{-}\right] \end{gathered}$$

Step-by-step solution

Step (1):- Answer of the questions

Given:- K_SP of Hg₂CI₂ is 2X10^-8 at 25^oc.

The concentration of Hg₂²⁺ is 7.9X10^-7M and CI^- is 1.6X10^-6M.

Step (2):- Calculation of molar

Calculation of molar solubility ofHg₂CI₂

The solubility equilibrium is-

$H{g}_{2}C{I}_2\rightleftarrows H{g_2}^{2+}+2C{I}^{-}$

So, the equilibrium constant is-

role="math" localid="1663591743145" $K_{SP}=\left[H{g_2}^{2+}\right]+{\left[C{I}^{-}\right]}^2$

If S mole of Hg₂CI₂ dissolves in water then the molar concentration of Hg₂²⁺ and CI^- will be –

$K_{SP}=\left[H{g_2}^{2+}\right]=Sand\left[C{I}^{-}\right]=2S$

$$\begin{gathered} K_{SP}=SX{\left(2S\right)}^2 \\ {K}_{SP}=SX4S^2 \\ {K}_{SP}=4S^3 \\ 4S^3=K_{SP} \\ 4S^3=2X{10}^{-18} \\ {S}^3=0.5X{10}^{-18} \\ S=7.9X{10}^{-17} \end{gathered}$$

So, the molar solubility of Hg₂CI₂ is 7.9X10^-7N.

Step (3):- Calculation of concentration

Calculation of concentration of Hg₂²⁺ and CI^-

The molar solubility of Hg₂CI₂ is 7.9X10^-7N.

So, the equilibrium concentrations of the ions are as follows:-

$\begin{array}{rcl}\left[H{g_2}^2\right]& =& S\\ & =& 7.9X{10}^{-7}N\\ \left[C{I}^{-}\right]& =& 2S\\ & =& 2X7.9X{10}^{-7}N\\ & =& 1.6X{10}^{-6}M\end{array}$

Step (4):- conclusion

Equilibrium concentration means the amount of solute that is dissolved in water.

Hence, the concentration of Hg₂²⁺ is 7.9X10^-7M and CI^- is 1.6X10^-6M.