Chapter 17, Problem 16QP

Calculate the $\text{pH}$ of 1.00 L of the buffer $\text{1}\text{.00 }M{\text{CH}}_{\text{3}}\text{COONa/1}\text{.00 }M{\text{ CH}}_{\text{3}}\text{COOH}$ before and after the addition of (a) 0.080 mol $\text{NaOH}$ and (b) 0.12 mol $\text{HCl}$. (Assume that there is no change in volume.)

Interpretation Introduction

Interpretation:

The $\text{pH}$

of the buffer solution ${\text{CH}}_3{\text{COONa/CH}}_3\text{COOH}$, and that after the addition of $\text{HCL}$

and $\text{NaOH}$

in the solution are to be calculated.

Concept introduction:

Acid–base titration is a technique to analyse the unknown concentration of the acid or base through a known concentration of the acid and base.

The buffers are those solutions that resist a change in pH on the addition or dilution of a small amount of an acid or alkali, and they consist of a mixture of the weak acid with its salt of the strong base or the weak base with its salt of the strong acid.

Answer to Problem 16QP

Solution:

a)

$\text{pH}=4.74$

The $\text{pH}$

of the buffer solution is $4.82$

after the addition of $\text{NaOH}$ in the buffer solution. The $\text{pH}$

of the buffer solution is increased from $4.74\text{to}4.82$

b)

$\text{pH}=4.74$

The $\text{pH}$

of the buffer solution is $4.64$

after the addition of $\text{HCl}$ in the buffer solution. The $\text{pH}$

of the buffer solution is decreased from $4.74\text{to}4.64$.

Explanation of Solution

Given information: The concentration of acetic acid and sodium acetate is $1.0\text{ M}$

each.

The volume of the buffer solution is $1\text{ L}$.

a) $0.08\text{ mol}$

NaOH

As it is a buffer solution, it is a mixture of weak base with its conjugate acid.

The $K_{\text{a}}$

value of ${\text{CH}}_3\text{COOH is 1}\text{.8}\times {\text{10}}^{-5}$.

The pH of the buffer solution is given by the expression as follows:

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{[\text{Conjugate}\text{Base}]}{[\text{Weak Acid}]}$

Here, $\left[{\text{CH}}_3\text{COOH}\right]$

is the concentration of acetic acid and $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of acetate ion.

Substitute the values of the concentrations of acetic acid and acetate ion and the dissociation constant of acetic acid in the above expression,

$\begin{array}{c}\text{pH}={\text{pK}}_{\text{a}}+\log \frac{[{\text{CH}}_3{\text{COO}}^{-}]}{[{\text{CH}}_3\text{COOH}]}\\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{[1\text{M}]}{[1\text{M}]}\\ =4.74\end{array}$

Hence, $\text{pH}$

of the buffer solution is $4.74$.

$\text{NaOH}$ will completely react with ${\text{CH}}_3\text{COOH}$

in the buffer solution. $\text{NaOH}$

is a strong base. It completely dissociates into its ions in the aqueous solution.

The equation for the dissociation of $\text{NaOH}$ in the aqueous solution is as follows:

$\text{NaOH}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{Na}}^{+}\left(aq\right)+{\text{ OH}}^{-}\left(aq\right)$

The Initial Change Equilibrium table for dissociation of $\text{NaOH}$ is as follows:

$\begin{array}{l}\text{NaOH}\left(s\right)\text{+}{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{Na}}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mol}\right)& 0.080& -& 0& 0\\ \text{Change}\left(\text{mol}\right)& -0.080& -& +0.080& +0.080\\ \text{Equilibrium}\left(\text{mol}\right)& 0& -& 0.080& 0.080\end{array}\end{array}$

The Initial Change Equilibrium table for neutralization reaction of $\text{NaOH}$ with ${\text{CH}}_3\text{COOH}$

is as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mol}\right)& 1& 0.080& 1& 0\\ \text{Change}\left(\text{mol}\right)& -0.080& -0.080& +0.080& +0.080\\ \text{Equilibrium}\left(\text{mol}\right)& 0.92& 0& 1.080& 0.080\end{array}\end{array}$

Consider $x$

to be a degree of dissociation.

The Initial Change Equilibrium table for ionisation of ${\text{CH}}_3\text{COOH}$

is as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& 0.92& 1.08& 0\\ \text{Change}\left(\text{M}\right)& -x& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.92-x& 1.08+x& x\end{array}\end{array}$

The equilibrium expression for a reaction is represented below.

$K_{\text{a}}=\frac{\left[{\text{CHCOO}}^{-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{CH}}_3\text{COOH}\left(aq\right)\right]}$

Here, $\left[{\text{CH}}_3\text{COOH}\right]$

is the concentration of acetic acid, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of acetate ion, $\left[{\text{H}}^{+}\right]$ is the concentration of hydronium ion and $K_a$ is the acid dissociation constant.

Substitute the value of $\left[{\text{CH}}_3\text{COOH}\right]$, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$, $\left[{\text{H}}^{+}\right]$, and $K_{\text{a}}$ in the above expression,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left[1.08+x\right]\left[x\right]}{\left[0.92-x\right]}\\ 1.8\times {10}^{-5}=\frac{\left[1.08\right]\left[x\right]}{\left[0.92\right]}\\ \left[x\right]=\frac{\left(1.8\times {10}^{-5}\right)\times \left(0.92\right)}{\left(1.08\right)}\end{array}$

Solve further,

The value $x$ is very small as compared to $1.08\text{and}0.92$. Therefore, it can be neglected.

$x=1.5\times {10}^{-5}\text{M}$

The concentration of $\left[{\text{H}}^{+}\right]$ is calculated as follows:

$\begin{array}{l}\left[{\text{H}}^{+}\right]=x\\ =1.5\times {10}^{-5}\text{M}\end{array}$

The value of $\text{pH}$

in the solution is calculated by the expression as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression,

$\begin{array}{l}\text{pH}=-\log \left[1.5\times {10}^{-5}\right]\\ =4.82\end{array}$

Therefore, $\text{pH}$

of the buffer solution is $4.82$ after the addition of $\text{NaOH}$ in the buffer solution. The $\text{pH}$

of the buffer solution is increased from $4.74\text{to}4.82$.

b) $0.12\text{ mol}$

$\text{HCl}$.

$\text{HCl}$ will completely react with ${\text{CH}}_3{\text{COO}}^{-}$

in the buffer solution. $\text{HCl}$ is a strong acid. It completely dissociates into its ions in the aqueous solution.

The equation for the dissociation of $\text{HCl}$ in the aqueous solution as follows:

$\text{HCl}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}^{+}\left(aq\right)+{\text{ Cl}}^{-}\left(aq\right)$

The Initial Change Equilibrium table for dissociation of $\text{HCl}$

is as follows:

$\begin{array}{l}\text{HCl}\left(aq\right)\text{+}{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{Na}}^{\text{+}}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mol}\right)& 0.12& -& 0& 0\\ \text{Change}\left(\text{mol}\right)& -0.12& -& +0.12& +0.12\\ \text{Equilibrium}\left(\text{mol}\right)& 0& -& 0.12& 0.12\end{array}\end{array}$

The Initial Change Equilibrium table for neutralization reaction of $\text{HCl}$

with ${\text{CH}}_3{\text{COO}}^{-}$

is as follows:

$\begin{array}{l}{\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\rightleftharpoons {\text{CH}}_3\text{COOH}\left(\text{aq}\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mol}\right)& 1& 0.12& 1& -\\ \text{Change}\left(\text{mol}\right)& -0.12& -0.12& +0.12& -\\ \text{Equilibrium}\left(\text{mol}\right)& 0.88& 0& 1.12& -\end{array}\end{array}$

Consider $x$

to be a degree of dissociation.

The Initial Change Equilibrium table for ionisation of ${\text{CH}}_3\text{COOH}$

is as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{M}\right)& 1.12& 0.88& 0\\ \text{Change}\left(\text{M}\right)& -x& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 1.12-x& 0.88+x& x\end{array}\end{array}$

The equilibrium expression for a reaction is represented by the expression as follows:

$K_{\text{a}}=\frac{\left[{\text{CHCOO}}^{-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{CH}}_3\text{COOH}\left(aq\right)\right]}$

Here, $\left[{\text{CH}}_3\text{COOH}\right]$

is the concentration of acetic acid, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of acetate ion, $\left[{\text{H}}^{+}\right]$

is the concentration of hydronium ion, and $K_{\text{a}}$ is the acid dissociation constant.

Substitute the values of $\left[{\text{CH}}_3\text{COOH}\right]$, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$, $\left[{\text{H}}^{+}\right]$, and $K_{\text{a}}$

in the above expression,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left[0.08+x\right]\left[x\right]}{\left[1.12-x\right]}\\ 1.8\times {10}^{-5}=\frac{\left[0.08\right]\left[x\right]}{\left[1.12\right]}\\ \left[x\right]=\frac{\left(1.8\times {10}^{-5}\right)\times \left(1.12\right)}{\left(0.08\right)}\end{array}$

The value $x$

is very small as compared to $1.08\text{ and}0.92$. Therefore, it can be neglected.

$x=2.3\times {10}^{-5}\text{M}$

The concentration of $\left[{\text{H}}^{+}\right]$ is calculated as follows:

$\begin{array}{l}\left[{\text{H}}^{+}\right]=x\\ =2.5\times {10}^{-5}\text{M}\end{array}$

The value of $\text{pH}$

in the solution is calculated by using expression as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression,

$\begin{array}{l}\text{pH}=-\log \left[2.3\times {10}^{-5}\right]\\ =4.64\end{array}$

Therefore, $\text{pH}$

of the buffer solution is $4.64$ after addition of $\text{HCl}$ in the buffer solution. The $\text{pH}$

of the buffer solution is decreased from $4.74\text{to}4.64$.