Chapter 19, Problem 19.89P

Interpretation Introduction

Interpretation:

When $\text{7}\text{.5}\text{mg}\text{of}{\text{BaCl}}_{\text{2}}$ is dissolved in 500.0 mL of $\text{0}\text{.023}\text{M}{\text{NaIO}}_{\text{3}}$, whether the solid $\text{Ba}{\left({\text{IO}}_{\text{3}}\right)}_{\text{2}}$ will form or not this has to be checked.

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.

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

Ion product:

The multiplication of concentrations of ions which are raised to the power of number of respective ions present in the molecular formula.  This gives the ion product.

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

Precipitation occurs when solution is supersaturated, $\text{IP}\text{>}{\text{K}}_{\text{sp}}$.

Precipitation will not occurs when solution is saturated and in equilibrium, $\text{IP}\text{>}{\text{K}}_{\text{sp}}$.

Precipitation will not occurs when solution is unsaturated, $\text{IP}<{\text{K}}_{\text{sp}}$.