Chapter 19.3, Problem 19.7BFP

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ in water has to be calculated.

Concept Introduction:

Solubility product constant:

The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.

The equilibrium constant of more soluble ionic compound is given by ${\text{K}}_{\text{SP}}$ and it is expressed by product of each ion present in the compound which are raised to the power of number respective ion present in the compound to give a maximum solubility of the compound.

$\begin{array}{l}{\text{M}}_{\text{m}}{\text{X}}_{\text{x}}{}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ mM}}^{\text{n+}}{}_{\text{(aq)}}{\text{ + xX}}^{\text{y-}}{}_{\text{(aq)}}\\ \\ {\text{K}}_{\text{sp}}{\text{ = [M}}^{\text{n+}}{\text{]}}^{\text{m}}{{\text{[X}}^{\text{y-}}\text{]}}^{\text{x}}\end{array}$

Molar solubility:

Molar solubility (S) is given as solubility in moles per litre.

$\text{S =}\frac{\text{Solubility}\text{in}\text{gram}}{\text{1L}}\text{×}\frac{\text{1}\text{mol}}{\text{Molarmass}\text{(g)}}$