Chapter 16.7, Problem 5CYU

Calculate the pH after mixing 15 mL of 0.12 M acetic acid with 15 ml of 0.12 M NaOH. What are the major species in solution at equilibrium (besides water), and what are their concentrations?

Interpretation Introduction

Interpretation:

The $\text{pH}$ of a solution has to be calculated when mixing $15\text{ mL}$ of $0.12\text{ M}$ $\text{NaOH}$ with $15\text{ mL}$ of $0.12\text{ M}$ ${\text{CH}}_{\text{3}}\text{COOH}$ and The major species besides water in the solution at equilibrium and their concentration is to be stated.

Concept introduction:

The $\text{pH}$ of a solution of acetic acid can be calculated by using the hydronium ion concentration by using the expression, $\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$.

Hydronium ion concentration is calculated by considering the equilibrium conditions and from the value of acid dissociation constant.

The concentration of acetic acid and acetate ion is determined by using equilibrium condition The concentration of acetate ion and hydronium ion is equal at equilibrium according to the reaction stoichiometry.

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

$\begin{array}{l}{\text{HA + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ A}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

${\text{K}}_{\text{b}}$ is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

The $\text{pOH}$ is a measure of hydroxide ion concentration. The expression for $\text{pOH}$ is,

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ (1)

The sum of $\text{pH}$ and $\text{pOH}$ is equal to $14$ at $25{}^{°}\text{C}$.

$\text{pH}+\text{pOH}=14$ (2)

The $\text{pH}$ varies from $0$ to $14$ for an aqueous solution.

$\begin{array}{l}\text{pH}=7\left(\text{Neutral}\right)\\ \text{pH}>7\left(\text{Basic}\right)\\ \text{pH}<7\left(\text{Acidic}\right)\end{array}$

Answer to Problem 5CYU

The $\text{pH}$ of a solution after mixing $15\text{ mL}$ of $0.12\text{ M}$ $\text{NaOH}$ with $15\text{ mL}$ of $0.12\text{ M}$ ${\text{CH}}_{\text{3}}\text{COOH}$ is $8.76$.

The major species except water in the solution at equilibrium are ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ion and ${\text{Na}}^{+}$ ion. The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ion is $0.06\text{M}$ and the concentration of ${\text{Na}}^{\text{+}}$ ion is $0.06\text{M}$.

Explanation of Solution

An equilibrium constant $\left(K\right)$ is the ratio of the concentration of products and reactants raised to appropriate stoichiometric coefficient at equilibrium.

For any base B, the chemical reaction is written as,

  $\text{B}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons \text{BH}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

$K_{\text{b}}=\frac{\left[\text{BH}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$ (3)

An equilibrium constant $\left(K\right)$ with subscript $\text{b}$ indicates that it is an equilibrium constant of the base in water.

Given:

Refer to the table $\text{16}\text{.2}$ for the value of dissociation constant.

Dissociation constant $\left(K_{\text{b}}\right)$ of acetate ion is $5.6\times {10}^{-10}$.

The concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.12\text{ mol}\cdot {\text{L}}^{-1}$.

The concentration of $\text{NaOH}$ is $0.12\text{ mol}\cdot {\text{L}}^{-1}$.

The volume of ${\text{CH}}_{\text{3}}\text{COOH}$ is $0.015\text{}\text{L}$.

The volume of $\text{NaOH}$ is $0.015\text{}\text{L}$.

The balanced chemical equation involving acetic acid and sodium hydroxide is written as,

  ${\text{CH}}_{\text{3}}\text{COOH}\left(\text{aq}\right)+\text{NaOH}\left(\text{aq}\right)\rightleftharpoons {\text{CH}}_{\text{3}}\text{COONa}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

$\begin{array}{c}{\text{Amount of CH}}_3\text{COOH consumed}=\left(\text{0}\text{.015 L}\right)\left(\text{0}\text{.12 mol}\cdot {\text{L}}^{-\text{1}}\right)\\ =18\times {\text{10}}^{-\text{4}}\text{mol}\end{array}$

$\begin{array}{c}\text{Amount of NaOH consumed}=\left(\text{0}\text{.015 L}\right)\left(\text{0}\text{.12 mol}\cdot {\text{L}}^{-\text{1}}\right)\\ =18\times {\text{10}}^{-\text{4}}\text{ mol }\end{array}$

Amount of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ produced upon completion of the reaction $=18\times {10}^{-4}\text{ mol}$

After completion of the reaction, the total volume of solution $=0.030\text{ L}$

The concentration of acetate ion is,

 $\begin{array}{c}\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]=\frac{18\times {10}^{-4}\text{mol}}{0.030\text{ L}}\\ =0.06\text{ mol}\cdot {\text{L}}^{-1}\\ =0.06\text{M}\end{array}$

Therefore, the concentration of acetate ion is $0.06\text{}\text{M}$.

Amount of ${\text{Na}}^{\text{+}}$ produced upon completion of the reaction $=18\times {10}^{-4}\text{ mol}$

The concentration of sodium ion is,

 $\begin{array}{c}\left[{\text{Na}}^{\text{+}}\right]=\frac{18\times {10}^{-4}\text{mol}}{0.030\text{ L}}\\ =0.06\text{ mol}\cdot {\text{L}}^{-1}\\ =0.06\text{M}\end{array}$

Therefore, the concentration of sodium ion is $0.06\text{}\text{M}$.

With the initial concentration of acetate ion is known, set up an ICE table to calculate the equilibrium concentration of hydroxide ion

$\boxed{\begin{array}{cccccc}\text{Equilibrium}& {\text{CH}}_{\text{3}}{\text{COO}}^{-}+{\text{H}}_{\text{2}}\text{O}& \rightleftharpoons & {\text{CH}}_{\text{3}}\text{COOH}& \text{+}& {\text{OH}}^{-}\\ \text{Initial}\left(\text{M}\right)& \text{0}\text{.06}& & \text{0}& & \text{0}\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.06-x& & x& & x\end{array}}$

Substitute the values at equilibrium from the ICE table into equation (3) to calculate the concentration of hydroxide ion,

$K_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.06-x\right)}$

Substitute the value $5.6\times {10}^{-10}$ for $K_{\text{b}}$ in the above equation to calculate the value of $x$,

$5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.06-x\right)}$

The acetate ion is the very weak base. Therefore, $x$, the concentration of hydroxide ion generated by the reaction of acetate ion with water, is assumed to be very small. So,

$\left(0.06-x\right)\approx 0.06$.

$\begin{array}{c}5.6\times {10}^{-10}=\frac{x^2}{0.06}\\ x=\sqrt{0.336\times {10}^{-10}}\\ =0.58\times {10}^{-5}\end{array}$

Since the concentration of hydroxide ion is equal to $x$. Hence, the concentration of hydroxide ion is $0.58\times {10}^{-5}\text{M}$.

Substitute the concentration of hydroxide ion into equation (1) to calculate $\text{pOH}$ of the solution,

$\begin{array}{c}\text{pOH}=-\log \left[0.58\times {10}^{-5}\right]\\ =5.24\end{array}$

Substitute the value of $\text{pOH}$ in equation (2) to calculate $\text{pH}$ of the solution,

$\begin{array}{c}\text{pH}+5.24=14\\ \text{pH}=14-5.24\\ =8.76\end{array}$

Therefore, the $\text{pH}$ of the solution is $8.76$.

The major species besides water at equilibrium are ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ion and ${\text{Na}}^{+}$ ion. The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ion is $0.06\text{M}$ and the concentration of ${\text{Na}}^{\text{+}}$ ion is $0.06\text{M}$.

Conclusion

The $\text{pH}$ of a solution after mixing $15\text{ mL}$ of $0.12\text{ M}$ $\text{NaOH}$ with $15\text{ mL}$ of $0.12\text{ M}$ ${\text{CH}}_{\text{3}}\text{COOH}$ is $8.76$.

The major species besides water in the solution at equilibrium are ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ion and ${\text{Na}}^{+}$ ion. The concentration of ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ ion is $0.06\text{M}$ and the concentration of ${\text{Na}}^{\text{+}}$ ion is $0.06\text{M}$.