Chapter 22, Problem 22.95P

Interpretation Introduction

Interpretation:

The solubility of ${\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)$ in solution buffered at $pH=2,4,6\text{,8 and10}$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

  ${\text{M}}_{\text{n}}{\text{X}}_{\text{m}}\text{(s)}\stackrel{}{\rightleftharpoons }\text{n}{\text{M}}^{\text{m+}}\text{(aq)+m}{\text{X}}^{\text{n+}}\text{(aq)}$

The relation between solubility product and molar solubility is as follows:

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{M}}^{\text{m+}}\right]}^{\text{n}}{\left[{\text{X}}^{\text{n-}}\right]}^{\text{m}}$

Here

The solubility product of salt is ${\text{K}}_{\text{sp}}$.

The molar solubility of ${\text{M}}^{\text{m+}}$ ion is $\left[{\text{M}}^{\text{m+}}\right]$

The molar solubility of ${\text{X}}^{\text{n-}}$ ion is $\left[{\text{X}}^{\text{n-}}\right]$

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

Answer to Problem 22.95P

The solubility of ${\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)$ in solution buffered at $pH=2,4,6\text{,8 and10}$ are given as $1.02\text{M}$, $1.02\times 0^{-3}M$, $1.02\times {10}^{-5}\text{M}$, $1.02\times {10}^{-7}\text{M}$ and $1.02\times {10}^{-9}\text{M}$ respectively.

Explanation of Solution

Given data is as follows:

  $\text{pH }=2.00$

The concentration of hydronium ion can be calculated as follows:

  $\begin{array}{c}\text{pH }=2.00\\ \\ \text{pH}\text{=}-{\text{log[H}}^{+}\text{]}\\ \\ {\text{[H}}^{+}\text{]}={10}^{\left(-2\right)}\\ \\ {\text{[H}}^{+}\text{]}=0.01\text{M}\end{array}$

The solubility can be calculated as follows:

  $\begin{array}{c}{\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}{\text{COO}}_{}^{-}\left(aq\right)\left(1\right)\\ \\ K_{sp}=\left[{\text{Ag}}^{+}\right]\left[{\text{CH}}_{\text{3}}{\text{COO}}_{}^{-}\right]=1.9\times {10}^{-3}{M}^2\\ \\ {\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)+{\text{H}}_3{O}^{+}\left(aq\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{H}}_{2}O\left(l\right)\left(2\right)\\ \\ K_c=K_{sp}\frac{1}{K_a}=\left(1.9\times {10}^{-3}{M}^2\right)\left(5.5\times {10}^4{M}^{-1}\right)\text{= 104}\text{.5M}\end{array}$

One $A{g}^{+}$ ion will be there for every formula unit of silver acetate that dissolves. Hence, the solubility will be equal to the concentration of $A{g}^{+}$ ion. Comparing the $K_{sp}$ and $K_c$ values, the value of $K_c$ is much greater than $K_{sp}$ and thus the additional contribution to solubility from (1) can be ignored.

  $s=\left[{\text{Ag}}^{+}\right]=\left[{\text{CH}}_{\text{3}}\text{COOH}\right]$

The solubility at $\text{pH }=2.00$ can be calculated as follows:

  $\begin{array}{l}{\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)+{\text{H}}_3{O}^{+}\left(aq\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{H}}_{2}O\left(l\right)\\ \\ Initial1.0\times {10}^{-2}M0M0.200M\\ \\ Change+s+s\\ \\ Equilibrium1.0\times {10}^{-2}Mss\end{array}$

Substituting the values,

  $\begin{array}{l}{K}_c=\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\left[{\text{H}}_3{O}^{+}\right]}=\text{ 104}\text{.5M}\\ \text{104}\text{.5M}=\frac{s\times \text{s}}{{\left[1.0\times {10}^{-2}M\right]}^2}\\ 10.2=\frac{s}{0.1}\\ \\ s=1.02\text{M}\end{array}$

The solubility is calculated to be $1.02\text{M}$.

The solubility at $\text{pH }=4.00$ can be calculated as follows:

  $\begin{array}{l}{\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)+{\text{H}}_3{O}^{+}\left(aq\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{H}}_{2}O\left(l\right)\\ \\ Initial1.0\times {10}^{-4}M0M0.200M\\ \\ Change+s+s\\ \\ Equilibrium1.0\times {10}^{-4}Mss\end{array}$

Substituting the values,

  $\begin{array}{l}{K}_c=\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\left[{\text{H}}_3{O}^{+}\right]}=\text{ 104}\text{.5M}\\ \text{104}\text{.5M}=\frac{s\times \text{s}}{{\left[1.0\times {10}^{-4}M\right]}^2}\\ 10.2=\frac{s}{1.0\times {10}^{-4}M}\\ \\ s=1.02\times 0^{-3}M\end{array}$

The solubility is calculated to be $1.02\times 0^{-3}M$.

The solubility at $\text{pH }=6.00$ can be calculated as follows:

  $\begin{array}{l}{\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)+{\text{H}}_3{O}^{+}\left(aq\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{H}}_{2}O\left(l\right)\\ \\ Initial1.0\times {10}^{-6}M0M0.200M\\ \\ Change+s+s\\ \\ Equilibrium1.0\times {10}^{-6}Mss\end{array}$

Substituting the values,

  $\begin{array}{l}{K}_c=\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\left[{\text{H}}_3{O}^{+}\right]}=\text{ 104}\text{.5M}\\ \text{104}\text{.5M}=\frac{s\times \text{s}}{{\left[1.0\times {10}^{-6}M\right]}^2}\\ 10.2=\frac{s}{1.0\times {10}^{-6}M}\\ \\ s=1.02\times {10}^{-5}\text{M}\end{array}$

The solubility is calculated to be $1.02\times {10}^{-5}\text{M}$.

The solubility at $\text{pH }=8.00$ can be calculated as follows:

  $\begin{array}{l}{\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)+{\text{H}}_3{O}^{+}\left(aq\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{H}}_{2}O\left(l\right)\\ \\ Initial1.0\times {10}^{-8}M0M0.200M\\ \\ Change+s+s\\ \\ Equilibrium1.0\times {10}^{-8}Mss\end{array}$

Substituting the values,

  $\begin{array}{l}{K}_c=\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\left[{\text{H}}_3{O}^{+}\right]}=\text{ 104}\text{.5M}\\ \text{104}\text{.5M}=\frac{s\times \text{s}}{{\left[1.0\times {10}^{-8}M\right]}^2}\\ 10.2=\frac{s}{1.0\times {10}^{-8}M}\\ \\ s=1.02\times {10}^{-7}\text{M}\end{array}$

The solubility is calculated to be $1.02\times {10}^{-7}\text{M}$.

The solubility at $\text{pH }=10.00$ can be calculated as follows:

  $\begin{array}{l}{\text{AgCH}}_{\text{3}}\text{COO}\left(\text{s}\right)+{\text{H}}_3{O}^{+}\left(aq\right)\rightleftharpoons {\text{Ag}}^{+}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{H}}_{2}O\left(l\right)\\ \\ Initial1.0\times {10}^{-10}M0M0.200M\\ \\ Change+s+s\\ \\ Equilibrium1.0\times {10}^{-10}Mss\end{array}$

Substituting the values,

  $\begin{array}{l}{K}_c=\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}{\left[{\text{H}}_3{O}^{+}\right]}=\text{ 104}\text{.5M}\\ \text{104}\text{.5M}=\frac{s\times \text{s}}{{\left[1.0\times {10}^{-10}M\right]}^2}\\ 10.2=\frac{s}{1.0\times {10}^{-10}M}\\ \\ s=1.02\times {10}^{-9}\text{M}\end{array}$

The solubility is calculated to be $1.02\times {10}^{-9}\text{M}$.