Question

The solubility product constant of mercury(I) iodide is 25^oc 1.2x10^-28 at 25^oc . Estimate the concentration of and in equilibrium with solid.

Short answer

Solubility product

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

Step-by-step solution

Step(1):- Answer of the question

Given:- Solubility product constant of mercury(I) iodide is 1.2X10^-28 at 25⁰c.

The concentration of Hg₂²⁺ and I^- in equilibrium with solid Hg₂I₂ are 3.1X10^-10Mand 6.2X10^-10M .

Step(2):- Calculation of molar solubility

Calculation of molar solubility of Hg₂I₂.

The solubility equilibrium is-

$H{g}_2{I}_2\left(s\right)\rightleftarrows \left[H{g_2}^{2+}\right]X{\left[I^{-}\right]}^2$

The solubility product is-

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

If S mole of is Hg₂I₂ dissolved in water then the molar concentration of [Hg₂²⁺] and [I^-] will be-

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

$$\begin{gathered} K_{SP}=SX{\left(2S\right)}^2 \\ {K}_{SP}=SX4S^2 \\ {K}_{SP}=4S^3 \\ {K}_{SP}=1.2X{10}^{-28} \\ {S}^3=\frac{1.2X{10}^{-28}}{4} \\ {S}^3=0.3X{10}^{-28} \\ S=3.1X{10}^{10} \end{gathered}$$

So the molar solubility of Hg₂I₂ is 3.1X10^-10 M.

Step(3):- Calculation of concentration

Calculation of concentration of [Hg₂²⁺] and [I^-]

The molar solubility of Hg₂I₂ is 3.1X10^-10 M. So, the equilibrium concentrations of the ions are follows-

$\begin{array}{rcl}\left[H{g_2}^{2+}\right]& =& s\\ & =& 3.1X{10}^{-10}M\\ \left[I^{-}\right]& =& 2S\\ & =& 6.2X{10}^{-10}M\end{array}$

Equilibrium concentration means the amount of solute of each ions that is soluble in water.

Hence, [Hg₂²⁺] is 3.1X10^-10 M and [I^-] is 6.2X10^-10M .