Chapter 12, Problem 12.47PAE

Interpretation Introduction

To determine:

The equilibrium equation and $K_{sp}$expression for dissolution of silver bromide in water

Explanation of Solution

When the salt is dissolved in water, it forms the ions. When AgBr is dissolved in water, it dissociates as;

$\text{AgBr(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Ag}}^{\text{+}}{\text{(aq)+Br}}^{\text{-}}\text{(aq)}$

Applying Law of Chemical equilibrium, $\text{K=}\frac{{\text{[Ag}}^{\text{+}}{\text{(aq)][Br}}^{\text{-}}\text{(aq)]}}{\text{[AgBr(s)]}}$

Since dissociation of undissociated solid remains constant, so [AgBr(s)] = 1 and K here represents solubility for the salt, it is expressed as ${\text{K}}_{\text{sp}}$

So, ${\text{K}}_{\text{sp}}{\text{=[Ag}}^{\text{+}}{\text{][Br}}^{\text{-}}\text{]}$

Interpretation Introduction

To determine:

The equilibrium equation and $K_{sp}$expression for dissolution of copper hydroxide in water

Explanation of Solution

${\text{Cu(OH)}}_{\text{2}}$ dissociates as :

$\begin{array}{l}{\text{Cu(OH)}}_{\text{2}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Cu}}^{\text{2+}}{\text{(aq)+2OH}}^{\text{-}}\text{(aq)}\\ \end{array}$

Applying Law of Chemical equilibrium,

$\text{K=}\frac{{\text{[Cu}}^{\text{2+}}{\text{(aq)][OH}}^{\text{-}}{\text{(aq)]}}^{\text{2}}}{\left[{\text{Cu(OH)}}_{\text{2}}\text{(s)}\right]}$

Since dissociation of undissociated solid remains constant, so ${\text{Cu(OH)}}_{\text{2}}$ (s)= 1 and K here represents solubility for the salt, it is expressed as ${\text{K}}_{\text{sp}}$

So, ${\text{Ksp=[Cu}}^{\text{2+}}{\text{][OH}}^{\text{-}}{\text{]}}^{\text{2}}$

Interpretation Introduction

To determine:

The equilibrium equation and $K_{sp}$expression for dissolution of barium sulfate in water

Explanation of Solution

${\text{BaSO}}_{\text{4}}$ dissociates as;

${\text{BaSO}}_{\text{4}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Ba}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Applying Law of Chemical equilibrium, $K=\frac{[B{a}^{2+}(aq)][S{O}_4{}^{2-}(aq)]}{[BaS{O}_4(s)]}$

Since dissociation of undissociated solid remains constant, so ${\text{BaSO}}_{\text{4}}$ (s)= 1 and K here represents solubility for the salt, it is expressed as $K_{sp}$

So, ${\text{Ksp=[Ba}}^{\text{2+}}{\text{][SO}}_{\text{4}}{}^{\text{2-}}\text{]}$

Interpretation Introduction

To determine:

To write equilibrium equation and $K_{sp}$expression for dissolution of lead chromate in water

Explanation of Solution

${\text{PbCrO}}_{\text{4}}$ dissociates as;

${\text{PbCrO}}_{\text{4}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Pb}}^{\text{2+}}{\text{(aq)+CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Applying Law of Chemical equilibrium,

$\text{K=}\frac{{\text{[Pb}}^{\text{2+}}{\text{(aq)][CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)]}}{{\text{[PbCrO}}_{\text{4}}\text{(s)]}}$

Since dissociation of undissociated solid remains constant, so ${\text{PbCrO}}_{\text{4}}$ (s)= 1 and K here represents solubility for the salt, it is expressed as ${\text{K}}_{\text{sp}}$

So, ${\text{K}}_{\text{sp}}$ = ${\text{[Pb}}^{\text{2+}}{\text{][CrO}}_{\text{4}}{}^{\text{2-}}\text{]}$

Interpretation Introduction

To determine:

To write equilibrium equation and $K_{sp}$expression for dissolution of magnesium phosphate in water

Explanation of Solution

${\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ dissociates as;

${\text{Mg}}_{\text{3}}{\text{(PO4)}}_{\text{2}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{3Mg}}^{\text{2+}}{\text{(aq)+2PO}}_{\text{4}}{}^{\text{3-}}\text{(aq)}$

Applying Law of Chemical equilibrium, $K=\frac{{[M{g}^{2+}(aq)]}^3{[P{O}_4{}^{3-}(aq)]}^2}{[M{g}_3{(P{O}_4)}_2(s)]}$

Since dissociation of undissociated solid remains constant, so ${\text{Mg}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ (s)= 1 and K here represents solubility for the salt, it is expressed as ${\text{K}}_{\text{sp}}$

So, ${\text{Ksp=[Mg}}^{\text{2+}}{\text{]}}^{\text{3}}{{\text{[PO}}_{\text{4}}{}^{\text{3-}}\text{]}}^{\text{2}}$

Conclusion

${\text{K}}_{\text{sp}}$ expression for each of the following equilibrium for dissolution of salts in water is written as;$\text{AgBr(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Ag}}^{\text{+}}{\text{(aq)+Br}}^{\text{-}}\text{(aq)}$

${\text{K}}_{\text{sp}}{\text{=[Ag}}^{\text{+}}{\text{][Br}}^{\text{-}}\text{]}$

$\begin{array}{l}{\text{Cu(OH)}}_{\text{2}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Cu}}^{\text{2+}}{\text{(aq)+2OH}}^{\text{-}}\text{(aq)}\\ \end{array}$

${\text{Ksp=[Cu}}^{\text{2+}}{\text{][OH}}^{\text{-}}{\text{]}}^{\text{2}}$

${\text{BaSO}}_{\text{4}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Ba}}^{\text{2+}}{\text{(aq)+SO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

${\text{Ksp=[Ba}}^{\text{2+}}{\text{][SO}}_{\text{4}}{}^{\text{2-}}\text{]}$

${\text{PbCrO}}_{\text{4}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{Pb}}^{\text{2+}}{\text{(aq)+CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq})$

${\text{K}}_{\text{sp}}$ = ${\text{[Pb}}^{\text{2+}}{\text{][CrO}}_{\text{4}}{}^{\text{2-}}\text{]}$

${\text{Mg}}_{\text{3}}{\text{(PO4)}}_{\text{2}}\text{(s)}\underset{}{\overset{}{\rightleftarrows }}{\text{3Mg}}^{\text{2+}}{\text{(aq)+2PO}}_{\text{4}}{}^{\text{3-}}\text{(aq)}$

${\text{Ksp=[Mg}}^{\text{2+}}{\text{]}}^{\text{3}}{{\text{[PO}}_{\text{4}}{}^{\text{3-}}\text{]}}^{\text{2}}$