Chapter 5, Problem 5G.3E

(a)

Interpretation Introduction

Interpretation:

The solubility in water of sparingly soluble silver bromide in $\text{1}{\text{.4×10}}^{\text{-3}}{\text{ M NaBr}}_{\text{(aq)}}$ solution has to be estimated.

Answer to Problem 5G.3E

The solubility in water of sparingly soluble silver bromide in $\text{1}{\text{.4×10}}^{\text{-3}}{\text{ M NaBr}}_{\text{(aq)}}$ solution is $\text{5}{\text{.5×10}}^{\text{-10}}{\text{ mol dm}}^{\text{-3}}$.

Explanation of Solution

The solubility of silver bromide in $\text{1}{\text{.4×10}}^{\text{-3}}{\text{ M NaBr}}_{\text{(aq)}}$ in water can be estimated by the dissolution of $\text{AgBr}$.

  ${\text{AgBr}}_{\text{(s)}}\rightleftharpoons {\text{Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+ Br}}^{\text{-}}{}_{\text{(aq)}}$

Assuming the solution is sufficiently dilute, the solubility constant can be written as

  ${\text{K}}_{\text{s}}\text{ = }\frac{{\text{a}}_{{\text{Ag}}^{\text{+}}{}_{\text{(aq)}}}{\text{a}}_{{\text{Br}}^{\text{-}}{}_{\text{(aq)}}}}{{\text{a}}_{{\text{AgBr}}_{\text{(s)}}}}\approx {\text{([Ag}}^{\text{+}}{\text{]/c}}^{\sout{\text{ο}}}{\text{)([Br}}^{\text{-}}{\text{]/c}}^{\sout{\text{ο}}}\text{)}$

The concentration of ${\text{Ag}}^{\text{+}}$ ions is equal to the solubility, ${\text{[Ag}}^{\text{+}}\text{] = s}$.

The ${\text{Br}}^{\text{-}}$ ions forms from the $\text{NaBr}$ solution and from the dissolution of $\text{AgBr}$.  So, ${\text{[Br}}^{\text{-}}\text{] = C + s}$.

As $\text{AgBr}$ is only sparingly soluble, $\text{C >> s}$.  Hence, ${\text{[Br}}^{\text{-}}\text{] }\approx \text{ C}$.

  ${\text{K}}_{\text{s}}\approx {\text{ sC/c}}^{\sout{\text{ο}}\text{2}}$

On rearrangement of the above expression,

  $\text{s = }\frac{{\text{K}}_{\text{s}}{\text{c}}^{\sout{\text{ο}}\text{2}}}{\text{C}}$

The solubility constant of $\text{AgBr}$ is $\text{7}{\text{.7×10}}^{\text{-13}}$.  Substitution of the value in the above expression

  $\begin{array}{l}\text{s = }\frac{\text{7}{\text{.7×10}}^{\text{-13}}{\text{ × (1 mol dm}}^{\text{-3}}{\text{)}}^{\text{2}}}{\text{1}{\text{.4×10}}^{\text{-3}}{\text{ mol dm}}^{\text{-3}}}\\ \\ \text{s = 5}{\text{.5×10}}^{\text{-10}}{\text{ mol dm}}^{\text{-3}}\end{array}$

The solubility in water of sparingly soluble silver bromide in $\text{1}{\text{.4×10}}^{\text{-3}}{\text{ M NaBr}}_{\text{(aq)}}$ solution is $\text{5}{\text{.5×10}}^{\text{-10}}{\text{ mol dm}}^{\text{-3}}$.

(b)

Interpretation Introduction

Interpretation:

The solubility in water of sparingly soluble magnesium carbonate in $\text{1}{\text{.1×10}}^{\text{-5}}{\text{ M Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}$ solution has to be estimated.

Answer to Problem 5G.3E

The solubility in water of sparingly soluble magnesium carbonate in $\text{1}{\text{.1×10}}^{\text{-5}}{\text{ M Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}$ solution is $\text{3}{\text{.2 × 10}}^{\text{-3}}{\text{ mol dm}}^{\text{-3}}$.

Explanation of Solution

The solubility of magnesium carbonate in $\text{1}{\text{.1×10}}^{\text{-5}}{\text{ M Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}$ in water can be estimated by the dissolution of ${\text{MgCO}}_{\text{3}}$.

  ${\text{MgCO}}_{\text{3}}{}_{\text{(s)}}\rightleftharpoons {\text{Mg}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}$

The magnitude of the solubility constant is similar to the concentration of the solution that contains common ion, the solubility constant is written as

  $\begin{array}{l}{\text{K}}_{\text{s}}\text{ = }\frac{{\text{a}}_{{\text{Mg}}^{\text{2+}}{}_{\text{(aq)}}}{\text{a}}_{{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}}}{{\text{a}}_{{\text{MgCO}}_{\text{3}}{}_{\text{(s)}}}}\approx {\text{([Mg}}^{\text{2+}}{\text{]/c}}^{\sout{\text{ο}}}{\text{)([CO}}_{\text{3}}{}^{\text{2-}}{\text{]/c}}^{\sout{\text{ο}}}\text{)}\\ \\ {\text{K}}_{\text{s}}{\text{ = s (C + s) / c}}^{\sout{\text{ο}}\text{2}}\\ {\text{s = Mg}}^{\text{2+}}{\text{ and (C + s) = CO}}_{\text{3}}{}^{\text{2-}}\end{array}$

On rearrangement of the above expression, a quadratic equation is obtained.

  ${\text{s}}^{\text{2}}{\text{ + Cs - K}}_{\text{s}}{\text{c}}^{\sout{\text{ο}}\text{2}}\text{ = 0}$

It is in the form of ${\text{ax}}^{\text{2}}\text{ + bx + c = 0}$.  On simplification,

  $\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$

On simplification of ${\text{s}}^{\text{2}}{\text{ + Cs - K}}_{\text{s}}{\text{c}}^{\sout{\text{ο}}\text{2}}\text{ = 0}$,

  $\text{s = }\frac{\text{-C ± }\sqrt{{\text{C}}^{\text{2}}{\text{- 4 × 1 × -K}}_{\text{s}}{\text{c}}^{\sout{\text{ο}}\text{2}}}}{\text{2}}$

The solubility constant of ${\text{MgCO}}_{\text{3}}$ is $\text{1}{\text{.0×10}}^{\text{-5}}$.  Substitution of the value in the above expression

  $\begin{array}{l}\text{s = }\frac{\text{-1}{\text{.1×10}}^{\text{-5}}\text{ ± }\sqrt{{\text{(1}{\text{.1×10}}^{\text{-5}}\text{)}}^{\text{2}}\text{+( 4 × 1}{\text{.0×10}}^{\text{-5}}\text{)}}}{\text{2}}{\text{ mol dm}}^{\text{-3}}\\ \\ \text{s = 3}{\text{.2 × 10}}^{\text{-3}}{\text{ mol dm}}^{\text{-3}}\end{array}$

The solubility in water of sparingly soluble magnesium carbonate in $\text{1}{\text{.1×10}}^{\text{-5}}{\text{ M Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}$ solution is $\text{3}{\text{.2 × 10}}^{\text{-3}}{\text{ mol dm}}^{\text{-3}}$.

(c)

Interpretation Introduction

Interpretation:

The solubility in water of sparingly soluble lead (II) sulphate in $\text{0}{\text{.10 M CaSO}}_{\text{4}}{}_{\text{(aq)}}$ solution has to be estimated.

Answer to Problem 5G.3E

The solubility in water of sparingly soluble lead (II) sulphate in $\text{0}{\text{.10 M CaSO}}_{\text{4}}{}_{\text{(aq)}}$ solution is $\text{1}{\text{.6×10}}^{\text{-7}}{\text{ mol dm}}^{\text{-3}}$.

Explanation of Solution

The solubility of lead (II) sulphate in $\text{0}{\text{.10 M CaSO}}_{\text{4}}{}_{\text{(aq)}}$ in water can be estimated by the dissolution of lead sulfate.

  ${\text{PbSO}}_4{}_{\text{(s)}}\rightleftharpoons {\text{Pb}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ SO}}_4{{}^{\text{2-}}}_{\text{(aq)}}$

Assuming the solution is sufficiently dilute, the solubility constant can be written as

  ${\text{K}}_{\text{s}}\text{ = }\frac{{\text{a}}_{{\text{Pb}}^{\text{2+}}{}_{\text{(aq)}}}{\text{a}}_{{\text{SO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}}}{{\text{a}}_{{\text{PbSO}}_{\text{4}}{}_{\text{(s)}}}}{\text{ = ([Pb}}^{\text{2+}}{\text{]/c}}^{\sout{\text{ο}}}{\text{)([SO}}_{\text{4}}{}^{\text{2-}}{\text{]/c}}^{\sout{\text{ο}}}\text{) }\approx \frac{\text{sC}}{{\text{c}}^{\sout{\text{ο}}\text{2}}}$

On rearrangement of the above expression,

  $\text{s = }\frac{{\text{K}}_{\text{s}}{\text{c}}^{\sout{\text{ο}}\text{2}}}{\text{C}}$

The solubility constant of ${\text{PbSO}}_{\text{4}}$ is $\text{1}{\text{.6×10}}^{\text{-8}}$.  Substitution of the value in the above expression

  $\begin{array}{l}\text{s = }\frac{\text{1}{\text{.6×10}}^{\text{-8}}{\text{ × (1 mol dm}}^{\text{-3}}{\text{)}}^{\text{2}}}{\text{0}{\text{.10 mol dm}}^{\text{-3}}}\\ \\ \text{s = 1}{\text{.6×10}}^{\text{-7}}{\text{ mol dm}}^{\text{-3}}\end{array}$

The solubility in water of sparingly soluble lead (II) sulphate in $\text{0}{\text{.10 M CaSO}}_{\text{4}}{}_{\text{(aq)}}$ solution is $\text{1}{\text{.6×10}}^{\text{-7}}{\text{ mol dm}}^{\text{-3}}$.

(d)

Interpretation Introduction

Interpretation:

The solubility in water of sparingly soluble nickel (II) hydroxide in $\text{2}{\text{.7×10}}^{\text{-5}}{\text{ M NiSO}}_4{}_{\text{(aq)}}$ solution has to be estimated.

Answer to Problem 5G.3E

The solubility in water of sparingly soluble nickel (II) hydroxide in $\text{2}{\text{.7×10}}^{\text{-5}}{\text{ M NiSO}}_4{}_{\text{(aq)}}$ solution is $\text{2}{\text{.5×10}}^{\text{-7}}{\text{ mol dm}}^{\text{-3}}$.

Explanation of Solution

The solubility of nickel (II) hydroxide in $\text{2}{\text{.7×10}}^{\text{-5}}{\text{ M NiSO}}_4{}_{\text{(aq)}}$ in water can be estimated by the dissolution of nickel hydroxide.

  ${\text{Ni(OH)}}_{\text{2}}{}_{\text{(s)}}\rightleftharpoons {\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}{\text{+ 2 OH}}^{\text{-}}{}_{\text{(aq)}}$

Assuming the solution is sufficiently dilute, the solubility constant can be written as

  ${\text{K}}_{\text{s}}\text{ = }\frac{{\text{a}}_{{\text{Ni}}^{\text{2+}}{}_{\text{(aq)}}}{\text{a}}_{{}_{{\text{OH}}^{\text{-}}{}_{\text{(aq)}}}}^{\text{2}}}{{\text{a}}_{{\text{Ni(OH)}}_2{}_{\text{(s)}}}}{\text{ = ([Ni}}^{\text{2+}}{\text{]/c}}^{\sout{\text{ο}}}{\text{)([OH}}^{\text{-}}{\text{]/c}}^{\sout{\text{ο}}}{\text{)}}^2\approx \frac{{\text{4 s}}^2\text{C}}{{\text{c}}^{\sout{\text{ο}}\text{3}}}$

On rearrangement of the above expression,

  $\text{s = }{\left(\frac{{\text{K}}_{\text{s}}{\text{c}}^{\sout{\text{ο}}\text{3}}}{\text{4C}}\right)}^{\text{1/2}}$

The solubility constant of ${\text{Ni(OH)}}_{\text{2}}$ is $\text{6}{\text{.5×10}}^{\text{-18}}$.  Substitution of the value in the above expression

  $\begin{array}{l}\text{s = }{\left\{\frac{\text{6}{\text{.5 ×10}}^{\text{-18}}{\text{ × (1 mol dm}}^{\text{-3}}{\text{)}}^{\text{3}}}{\text{4 × (2}{\text{.7×10}}^{\text{-5}}{\text{ mol dm}}^{\text{-3}}\text{)}}\right\}}^{\text{1/2}}\\ \\ \text{s = 2}{\text{.5×10}}^{\text{-7}}{\text{ mol dm}}^{\text{-3}}\end{array}$

The solubility in water of sparingly soluble nickel (II) hydroxide in $\text{2}{\text{.7×10}}^{\text{-5}}{\text{ M NiSO}}_4{}_{\text{(aq)}}$ solution is $\text{2}{\text{.5×10}}^{\text{-7}}{\text{ mol dm}}^{\text{-3}}$.