Chapter 16, Problem 16.43QA

Interpretation Introduction

Interpretation & Concept Introduction:

A $25.0 mL$ sample of $0.100 M$ acetic acid is titrated with $0.125 M NaOH$ at ${25}^{o}C$. What is the pH of the solution after the following volume of base have been added?

a) $10.0 mL$

b) $20.0 mL$

c) $30.0 mL$

Answer to Problem 16.43QA

Solution:

a) $pH$ of solution after addition of $10.0 mL$ of $0.125 M NaOH$ is $4.75.$

b) $pH$ of solution after addition of $20.0 mL$ of $0.125 M NaOH$ is $8.75$.

c) $pH$ of solution after addition of $30.0 mL$ of $0.125 M NaOH$ is $12.36$.

Explanation of Solution

1) Concept:

The known concentration and volume of a weak acid and molarity of a strong base $\left(NaOH\right)$ are given. We are asked to find the pH of solutions upon the addition of different volumes of $NaOH$ to the acid.

We can calculate the initial moles of acetic acid using the given molarity and volume. The reaction for the titration of acetic acid and $NaOH$ is:

${CH}_{3}COOH\left(aq\right)+{OH}^{-}\left(aq\right)\to {CH}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)$

From the stoichiometry of the above reaction and using initial moles of acetic acid and $NaOH$, we can find moles of acetic acid remaining in the solution and amount of conjugate base that is formed. We then use the Henderson–Hasselbalch equation to calculate the $pH$ of the buffer.

$pH=p{K}_a+\log \frac{\left[Base\right]}{\left[Acid\right]}$

Where, $p{K}_a=-\log K_a$

The $K_b$ value for ${CH}_{3}CO{O}^{-}$ can be calculated using the relation between $K_a$ and $K_b$

$K_a\times K_b=K_w$(where,$K_w=1.0\times {10}^{-14}$)

The concentration of $H_3{O}^{+}$ can be calculated using the following relation

$\left[H_3{O}^{+}\right]\left[{OH}^{-}\right]=K_w$

The relation for $pH$ calculation is:

$pH=-\log \left[H_3{O}^{+}\right]$

2) Formula:

i) $pH=p{K}_a+\log \frac{\left[Base\right]}{\left[Acid\right]}$

ii) $M=\frac{n}{V}$

iii) $K_a\times K_b=K_w$

iv) $\left[H_3{O}^{+}\right]\left[{OH}^{-}\right]=K_w$

v) $pH=-\log \left[H_3{O}^{+}\right]$

vi) $p{K}_a=-\log K_a$

3) Given:

i) Molarity of acetic acid = $0.100 M$

ii) Volume of acetic acid = $25.0 mL=0.025 L$

iii) Molarity of $NaOH$= $0.125 M$

iv) $K_a=1.75\times {10}^{-5}$ (Appendix 5 Table A5.1)

4) Calculation:

Calculating the initial moles of ${CH}_{3}COOH$

$Moles of {CH}_{3}COOH=\frac{0.100 mol}{L}\times 0.0250 L=0.00250 mol$

a) Calculating $pH when 0.010L of 0.125 M NaOH is added$ as:

Calculating the moles of $NaOH$,

$Mol of NaOH=\frac{0.125 mol}{L}\times 0.0100 L=0.00125 mol$

Set up the RICE table to determine the moles of ${ CH}_{3}COOH$ remaining after the reaction and moles of ${CH}_{3}CO{O}^{-}$ that is formed:

Reaction	${ CH}_{3}COOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${CH}_{3}CO{O}^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	${ CH}_{3}COOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		${CH}_{3}CO{O}^{-}\left(mol\right)$
Initial	$0.00250$		$0.00125$		0
Change	$-0.00125$		$-0.00125$		$+0.00125$
Final	$0.00125$		0		$0.00125$

$Total volume = 0.0250 L+0.0100 L=0.0350 L$

Since the moles of ${CH}_{3}COOH$ and ${CH}_{3}CO{O}^{-}$ are same, the ratio of base: acid will reduce it to $1$, so, this reduces the Henderson-Hasselbalch equation to $pH=p{K}_a$, hence, this is the “half equivalence” point in the titration.

$\left[{CH}_{3}COOH\right]=\left[{CH}_{3}CO{O}^{-}\right]=\frac{0.00125 mol}{0.0350 L}=0.03571 M$

Plug in the values in the Henderson–Hasselbalch equation,

$pH=4.754+\log \frac{\left[0.03571\right]}{\left[0.03571\right]}$

$pH=4.754+\log \left(1\right)$

$pH=4.754+0$

$pH=4.75$

b) Calculating $pH when 0.020L of 0.125 M NaOH is added$ as:

Calculating the moles of $NaOH$,

$Mol of NaOH=\frac{0.125 mol}{L}\times 0.0200 L=0.00250 mol$

Set up the RICE table to determine the reacted moles of ${ CH}_{3}COOH$ and how many moles of ${CH}_{3}CO{O}^{-}$ are formed,

Reaction	${ CH}_{3}COOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${CH}_{3}CO{O}^{-}(aq$)	+	$H_{2}O\left(l\right)$
	${ CH}_{3}COOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		${CH}_{3}CO{O}^{-}\left(mol\right)$
Initial	$0.00250$		$0.00250$		0
Change	$-0.00250$		$-0.00250$		$+0.00250$
Final	$0$		0		$0.00250$

$Total volume = 0.0250 L+0.0200 L=0.0450 L$

$\left[{CH}_{3}CO{O}^{-}\right]=\frac{0.00250 mol}{0.0450 L}=0.05556 M$

Set up RICE table to calculate the $\left[{OH}^{-}\right]$

Reaction	${CH}_{3}CO{O}^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$	$\rightleftharpoons$	${ CH}_{3}COOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$
Initial	$0.05556 M$				0		0
Change	$-x$				$+x$		$+x$
Equilibrium	$(0.05556 –x)$				$x$		$x$

$K_b=\frac{\left[{CH}_{3}COOH\right]\left[{OH}^{-}\right]}{\left[{CH}_{3}CO{O}^{-}\right]}$

Calculating $K_b$,

$K_b=\frac{K_w}{K_a}$

$K_b=\frac{1.0\times {10}^{-14}}{1.75\times {10}^{-5}}$

$K_b=5.71428\times {10}^{-10}$

$5.71428\times {10}^{-10}=\frac{x^2}{(0.05556 –x)}$

$5.71428\times {10}^{-10}=\frac{x^2}{\left(0.05556 \right)}$

$x=5.6396\times {10}^{-6} M=\left[{OH}^{-}\right]$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{5.6396\times {10}^{-6}}$

$\left[H_3{O}^{+}\right]=1.77315\times {10}^{-9}\mathrm{ }\mathrm{M}$

Calculating the $pH$

$pH=-\log \left[H_3{O}^{+}\right]$

$pH=-\log (1.7798\times {10}^{-9})$

$pH=8.75$

It is noted that initial moles of acetic acid and moles of added base are equal when $20.0 mL$ of $NaOH$ is added, so this is the equivalence point in the titration and the $pH$ at the equivalence point for this titration is $8.75$.

Calculating $pH when 0.030L of 0.125 M NaOH is added$ as:

Calculating the moles of $NaOH$,

$Mol of NaOH=\frac{0.125 mol}{L}\times 0.0300 L=0.00375 mol$

Initial moles of acetic acid $=0.0025 mol$

Moles of $NaOH$ added are more than moles of acetic acid, thus, acetic acid will be the limiting reactant, and hence, the NaOH is the reagent in excess which will decide the $pH$.

Set up the RICE table to determine the reacted moles of ${ CH}_{3}COOH$ and how many moles of ${CH}_{3}CO{O}^{-}$ are formed,

Reaction	${ CH}_{3}COOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${CH}_{3}CO{O}^{-}(aq$)	+	$H_{2}O\left(l\right)$
	${ CH}_{3}COOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		${CH}_{3}CO{O}^{-}\left(mol\right)$
Initial	$0.00250$		$0.00375$		0
Change	$-0.00250$		$-0.00250$		$+0.00250$
Final	$0$		$0.00125$		$0.00250$

$Total volume=0.0250 L+0.0300 L=0.0550 L$

$\left[{OH}^{-}\right]=\frac{0.00125\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}{0.0550\mathrm{ }\mathrm{L}}=0.022727\mathrm{M}$

Dissociation of ${CH}_{3}CO{O}^{-}$ is negligible as compared to $\left[{OH}^{-}\right]$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{0.022727}$

$\left[H_3{O}^{+}\right]=4.400\times {10}^{-13}$

$pH=-\log (4.400\times {10}^{-13})$

$pH=12.36$

Conclusion

In the titration of a weak acid vs. strong base, as you increase the concentration of the strong base, the $pH$ of the solution increases.