Chapter 15, Problem 15.96SP

(a)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{BaCrO}}_{\text{4}}$ should 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 multiplication of number of each ion present in the compound rise 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:

The solubility of gram mole of compound is given by the molar solubility it is,

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

(b)

Interpretation Introduction

Interpretation:

The molar solubility of $\text{Mg(OH)}$ should 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 multiplication of number of each ion present in the compound rise 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:

The solubility of gram mole of compound is given by the molar solubility it is,

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

(c)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}$ solution should 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 multiplication of number of each ion present in the compound rise 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:

The solubility of gram mole of compound is given by the molar solubility it is,

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