Chapter 17.3, Problem 28SSC

Interpretation Introduction

Interpretation:

The use of solubility product in calculating the solubility of an ionic compound needs to be explained.

Concept introduction:

The solubility product constant is the product of the concentration of its dissolved ions raised to the power of their respective stoichiometric coefficients that calculated from the equilibrium reaction.

Answer to Problem 28SSC

The solubility product is the product of the concentration of its constituent ions with each ion concentration term raised to the power equal to the number of ions produced or dissociated

Consider the reaction of a sparingly soluble salt.

${\text{AB}}_{\text{2}}\rightleftharpoons {\text{A}}^{\text{+}}\text{+}{\text{2B}}^{\text{-}}$

${\text{K}}_{\text{sp}}\text{=}\left[{\text{A}}^{\text{+}}\right]{\left[{\text{2B}}^{\text{-}}\right]}^{\text{2}}$

Explanation of Solution

Consider a reversible reaction example of a sparingly soluble salt

${\text{PbCl}}_{\text{2}}\rightleftharpoons {\text{Pb}}^{\text{+2}}\text{+}{\text{2Cl}}^{\text{-}}$

So, the solubility product expression is given as

$\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{}^{}\frac{\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{Cl}}^{\text{-}}\right]}^{\text{2}}}{\left[{\text{PbCl}}_{{}_{\text{2}}}\right]}\\ {\text{K}}_{\text{sp}}\left[{\text{PbCl}}_{{}_{\text{2}}}\right]\text{=}\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{Cl}}^{\text{-}}\right]}^{\text{2}}\\ {\text{K}}_{\text{sp}}\text{=}\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{Cl}}^{\text{-}}\right]}^{\text{2}}\end{array}$

The concentration of pure solids and liquid is constant.

The above equation can be represented as follows:

$Ksp=s\left({\left(2s\right)}^2\right)=s\left(4s^2\right)=4s^3$

Thus,

$s=\sqrt[3]{\frac{Ksp}{4}}$

Here, s is solubility.

Conclusion

The solubility product is given as

${\text{K}}_{\text{sp}}\text{=}\left[{\text{A}}^{\text{+}}\right]{\left[{\text{2B}}^{\text{-}}\right]}^{\text{2}}$