Chapter 8, Problem 194MP

(a)

Interpretation Introduction

Interpretation: The equation for the reaction that takes place to increase the solubility of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ in acidic solutions and basic solutions needs to be determined.

Concept Introduction: The solubility of a salt is defined as amount of salt dissolved in per 100 mL of the solvent at a constant temperature.

Explanation of Solution

The reaction of aluminium ions and hydroxide ion forms aluminium hydroxide but it can also form $\text{Al}{\left(\text{OH}\right)}_4^{-}$ .

The precipitate is more soluble in very acidic and basic solutions.

The equations for reaction to increase the solubility of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ .

The reaction in acidic solution is represented as follows:

  $\text{Al}{\left(\text{OH}\right)}_{\text{3}}+3{\text{H}}^{+}\to {\text{Al}}^{3+}+3{\text{H}}_{\text{2}}\text{O}$

The reaction in basic medium is represented as follows:

  $\text{Al}{\left(\text{OH}\right)}_{\text{3}}+3{\text{OH}}^{-}\to \text{Al}{\left(\text{OH}\right)}_4^{-}$

(b)

Interpretation Introduction

Interpretation: This is to be shown that the solubility of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ as a function of concentration of hydrogen ion obeys the following equation:

  $S={\left[H^{+}\right]}^3{K}_{sp}/K_w^2+K{K}_w/\left[H^{+}\right]$

Concept Introduction:

For a salt, the expression for solubility product can be written as follows:

  $A_x{B}_y\rightleftharpoons x{A}^{y+}+y{B}^{x-}$

The solubility product constant can be represented as follows:

  $K_{sp}={\left[A^{y+}\right]}^x{\left[B^{x-}\right]}^y$

Explanation of Solution

The relation between K, $K_{sp}$ and $K_w$ is represented as follows:

  $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

Or,

  $\left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}$

The equilibrium constant expression can be represented as follows:

  $K=\frac{\left[\text{Product}\right]}{\left[\text{Reactant}\right]}$

For the above reaction,

  $K=\frac{\left[\text{Al}{\left(\text{OH}\right)}_4^{-}\right]}{\left[{\text{OH}}^{-}\right]}$

If ionization constant of water is used in this expression, the above expression can be rearranged as follows:

  $K=\frac{\left[\text{Al}{\left(\text{OH}\right)}_4^{-}\right]}{\frac{K_w}{\left[{\text{H}}^{+}\right]}}$

On rearranging,

  $K=\frac{\left[\text{Al}{\left(\text{OH}\right)}_4^{-}\right]}{\frac{K_w}{\left[{\text{H}}^{+}\right]}}$

Or,

  $\left[\text{Al}{\left(\text{OH}\right)}_4^{-}\right]=\frac{K{K}_w}{\left[{\text{H}}^{+}\right]}$   ........ (1)

The solubility product of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ can be written as follows:

  $K_{sp}=\left[{\text{Al}}^{3+}\right]{\left[{\text{OH}}^{-}\right]}^3$

This can be rearranged as follows:

  $\frac{K_{sp}}{{\left[{\text{OH}}^{-}\right]}^3}=\left[{\text{Al}}^{3+}\right]$

Also,

  $\frac{K_{sp}{\left[{\text{H}}^{+}\right]}^3}{K_w^3}=\left[{\text{Al}}^{3+}\right]$   ....... (2)

From equation (1) and (2),

  $S=\frac{K_{sp}{\left[{\text{H}}^{+}\right]}^3}{K_w^3}+\frac{K{K}_w}{\left[{\text{H}}^{+}\right]}$

(b)

Interpretation Introduction

Interpretation: The solubility of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ needs to be plotted in the range of pH 4-12.

Concept Introduction: The relation between solubility and hydrogen ion concentration can be represented as follows:

  $S=\frac{K_{sp}{\left[{\text{H}}^{+}\right]}^3}{K_w^3}+\frac{K{K}_w}{\left[{\text{H}}^{+}\right]}$

The relation between pH and hydrogen ion concentration can be represented as follows:

  $pH=-\log \left[H^{+}\right]$

Explanation of Solution

The given value of K is 40.0 and solubility product constant is $2\times {10}^{-32}$ .

The solubility of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ can be calculated at given pH values using the following equation:

  $S=\frac{K_{sp}{\left[{\text{H}}^{+}\right]}^3}{K_w^3}+\frac{K{K}_w}{\left[{\text{H}}^{+}\right]}$

Putting the value of K, $K_w$ and $K_{sp}$ ,

  $S=\frac{\left(2\times {10}^{-32}\right){\left[{\text{H}}^{+}\right]}^3}{{\left({10}^{-14}\right)}^3}+\frac{\left(40\right)\left({10}^{-14}\right)}{\left[{\text{H}}^{+}\right]}$

Or,

  $S=\left(2\times {10}^{10}\right){\left[{\text{H}}^{+}\right]}^3+\frac{4\times {10}^{-13}}{\left[{\text{H}}^{+}\right]}$

Now, for pH value 4, the hydrogen ion concentration will be ${10}^{-4}$ thus,

  $\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-4}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-4}\right)}\\ =2\times {10}^{-2}+4\times {10}^{-9}\\ =0.02\end{array}$

Similarly, for pH value 5, the hydrogen ion concentration will be ${10}^{-5}$ thus,

  $\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-5}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-5}\right)}\\ =2\times {10}^{-5}+4\times {10}^{-8}\\ =2.004\times {10}^{-5}\end{array}$

Thus, for pH value 6 -12, the solubility of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ will be:

pH	$S=\left(2\times {10}^{10}\right){\left[{\text{H}}^{+}\right]}^3+\frac{4\times {10}^{-13}}{\left[{\text{H}}^{+}\right]}$
6	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-6}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-6}\right)}\\ =2\times {10}^{-8}+4\times {10}^{-7}\\ =4.2\times {10}^{-7}\end{array}$
7	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-7}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-7}\right)}\\ =2\times {10}^{-10}+4\times {10}^{-6}\\ =4.0002\times {10}^{-6}\end{array}$
8	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-8}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-8}\right)}\\ =2\times {10}^{-14}+4\times {10}^{-5}\\ =4\times {10}^{-5}\end{array}$
9	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-9}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-9}\right)}\\ =2\times {10}^{-17}+4\times {10}^{-4}\\ =4\times {10}^{-4}\end{array}$
10	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-10}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-10}\right)}\\ =2\times {10}^{-20}+4\times {10}^{-3}\\ =4\times {10}^{-3}\end{array}$
11	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-11}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-11}\right)}\\ =2\times {10}^{-23}+4\times {10}^{-2}\\ =4\times {10}^{-2}\end{array}$
12	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-12}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-12}\right)}\\ =2\times {10}^{-26}+0.4\\ =0.4\end{array}$
13	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-13}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-13}\right)}\\ =2\times {10}^{-29}+4\\ =4.0\end{array}$
14	$\begin{array}{l}S=\left(2\times {10}^{10}\right){\left({10}^{-14}\right)}^3+\frac{4\times {10}^{-13}}{\left({10}^{-14}\right)}\\ =2\times {10}^{-32}+40\\ =40\end{array}$

The table for pH and solubility can be represented as follows:

pH	S
4	0.02
5	$2.004\times {10}^{-5}$
6	$4.2\times {10}^{-7}$
7	$4.0002\times {10}^{-6}$
8	$4\times {10}^{-5}$
9	$4\times {10}^{-4}$
10	$4\times {10}^{-3}$
11	$4\times {10}^{-2}$
12	0.4
13	4
14	40

The plot can be represented as follows: