Chapter 16, Problem 16.44QA

Interpretation Introduction

To find:

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

a) $10.0 mL HCl$

b) $20.0 mL HCl$

c) $30.0 mL HCl$

Answer to Problem 16.44QA

Solution:

a) $pH$ of solution after addition of $10.0 mL$ of $0.125 M HCl$ is $9.81$.

b) $pH$ of solution after addition of 2 $0.0 mL$ of $0.125 M HCl$ is $5.53$.

c) $pH$ of solution after addition of $30.0 mL$ of $0.125 M HCl$ is $1.64$.

Explanation of Solution

1) Concept:

We can calculate the initial moles of trimethylamine using the given molarity and volume. The reaction for the titration of trimethylamine and $HCl$ is

${\left({CH}_3\right)}_{3}N\left(aq\right)+HCl\left(aq\right)\to {\left({CH}_3\right)}_3{NH}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

From moles of trimethylamine and $HCl$, ${[\left({CH}_3\right)}_{3}N]$ and [${\left({CH}_3\right)}_3{NH}^{+}]$ can be determined in the final solution, and pH can be found out by using Henderson–Hasselbalch equation

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

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

The $K_a$ value for ${\left({CH}_3\right)}_3{NH}^{+}$ 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 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) $Molarity=\frac{moles of solute}{volume in L}$

iii) $K_a\times K_b=K_w$

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

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

3) Given:

i) Molarity of trimethylamine = $0.100 M$

ii) Volume of trimethylamine = $25.0 mL=0.025L$

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

iv) $K_b=6.46\times {10}^{-5}$ (Appendix 5 Table A5.3)

4) Calculation:

Calculating the initial moles of ${\left({CH}_3\right)}_{3}N$ as:

$Mol of {\left({CH}_3\right)}_{3}N=\frac{0.100 mol}{L}\times 0.0250 L=0.00250 mol$

a) Calculating the $pH$ of solution when $10.0 mL \left(0.010L\right)of 0.125 M HCl are added$

Calculating the moles of $HCl$

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

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	${\left({CH}_3\right)}_{3}N\left(aq\right)$	+	${H^{+}}^{ }\left(aq\right)$	$\to$	${\left({CH}_3\right)}_3{NH}^{+}\left(aq\right)$
	${\left({CH}_3\right)}_{3}N \left(mol\right)$		$H^{+}\left(mol\right)$		${\left({CH}_3\right)}_3{NH}^{+}\left(mol\right)$
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 ${\left({CH}_3\right)}_{3}N$ and ${\left({CH}_3\right)}_3{NH}^{+}$ are same, their concentration is,

$\left[{\left({CH}_3\right)}_{3}N\right]=\left[{\left({CH}_3\right)}_3{NH}^{+}\right]=\frac{0.00125 mol}{0.0350 L}=0.03571 M$

Calculating ${pK}_a$,

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

$K_a=\frac{1.0\times {10}^{-14}}{6.46\times {10}^{-5}}=1.54799\times {10}^{-10}$

${pK}_a=-\log (1.54799\times {10}^{-10})$

${pK}_a=9.810$

Plug in the values in the Henderson–Hasselbalch equation,

$pH=9.810+\log \frac{\left[0.03571\right]}{\left[0.03571\right]}$

$pH=9.810+\log \left(1\right)$

$pH=9.810+0$

$pH=9.810$

This is the $pH$ of the solution at half equivalence point, since $\left[conjugate base\right]$ and $\left[acid\right]$ are equal.

b) Calculating the $pH$ of solution when $20.0 mL \left(0.020L\right)of 0.125 M HCl is added$

Calculating the moles of $NaOH$

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

Set up the RICE table to determine the reacted moles of ${\left({CH}_3\right)}_{3}N$ and how many moles of ${\left({CH}_3\right)}_3{NH}^{+}$ are formed,

Reaction	${\left({CH}_3\right)}_{3}N\left(aq\right)$	+	$H^{+}\left(aq\right)$	$\to$	${\left({CH}_3\right)}_3{NH}^{+}\left(aq\right)$
	${\left({CH}_3\right)}_{3}N \left(moles\right)$		$H^{+}\left(moles\right)$		${\left({CH}_3\right)}_3{NH}^{+}\left(moles\right)$
Initial	$0.00250$		$0.00250$		0
Change	$-0.00250$		$-0.00250$		$+0.00250$
Final	$0$		0		$0.00250$

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

$\left[{\left({CH}_3\right)}_3{NH}^{+}\right]=\frac{0.00250 mol}{0.0450 L}=0.05556 M$

Set up RICE table to calculate the $\left[{H_{3}O}^{+}\right]$

Reaction	${\left({CH}_3\right)}_3{NH}^{+}\left(aq\right)$	+	$H_{2}O\left(l\right)$	$\rightleftharpoons$	${\left({CH}_3\right)}_{3}N\left(aq\right)$	+	${H_{3}O}^{+}\left(aq\right)$
Initial	$0.05556 M$				0		0
Change	$-x$				$+x$		$+x$
Equilibrium	$(0.05556 –x)$				$x$		$x$

$K_a=\frac{\left[{\left({CH}_3\right)}_{3}N\right]\left[{H_{3}O}^{+}\right]}{\left[{\left({CH}_3\right)}_3{NH}^{+}\right]}$

$1.54799\times {10}^{-10}=\frac{x^2}{(0.05556 –x)}$

$1.54799\times {10}^{-10}=\frac{x^2}{\left(0.05556 \right)}$

$x=2.93268\times {10}^{-6} M=\left[H_3{O}^{+}\right]$

Calculating the $pH$

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

$pH=-\log (2.93268\times {10}^{-6})$

$pH=5.53$

When $20.0 mL$$HCl$ is added to the solution of trimethylamine, the initial moles of base are equal to moles of acid added, so, it is the equivalence point of the titration. Thus, the $pH$ at the equivalence point is $5.53$.

c) Calculating the pH of solution when $30.0 mL \left(0.030L\right)of 0.125 M HCl is added$

Calculating the moles of $HCl$

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

Set up the RICE table to determine the reacted moles of ${\left({CH}_3\right)}_{3}N$ and how many moles of ${\left({CH}_3\right)}_3{NH}^{+}$ are formed,

Reaction	${\left({CH}_3\right)}_{3}N\left(aq\right)$	+	$H^{+}\left(aq\right)$	$\to$	${\left({CH}_3\right)}_3{NH}^{+}\left(aq\right)$
	${\left({CH}_3\right)}_{3}N \left(moles\right)$		$H^{+}\left(moles\right)$		${\left({CH}_3\right)}_3{NH}^{+}\left(moles\right)$
Initial	$0.00250$		$0.00375$		0
Change	$-0.00250$		$-0.00250$		$+0.00250$
Equilibrium	$0$		$0.00125$		$0.00250$

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

$\left[H^{+}\right]=\left[H_3{O}^{+}\right]=\frac{0.00125\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}{0.0550\mathrm{ }\mathrm{L}}=0.022727\mathrm{M}$

Dissociation of ${\left({CH}_3\right)}_3{NH}^{+}$ is negligible as compared to $\left[H_3{O}^{+}\right]$

$pH=-\log \left(0.022727\right)$

$pH=1.64$

Conclusion:

In the titration of a weak base and a strong acid, as you increase the volume of the strong acid, the $pH$ of the solution decreases.