Chapter 16, Problem 16.45QA

Interpretation Introduction

To sketch:

A titration curve for the titration of $50.0 mL$ of $0.200 MHN{O}_2$ with $1.00 MNaOH$. Find the $pH$ of the sample after the addition of the titrant?

a) $2.50 mL$

b) $5.00 mL$

c) $7.50 mL$

d) $10.00 mL$

Answer to Problem 16.45QA

Solution:

a) $pH$ of solution after addition of $2.50 mL$ of $1.00 MNaOH$ is $2.92.$

b) $pH$ of solution after addition of $5.00 mL$ of $1.00 MNaOH$ is $3.40$.

c) $pH$ of solution after addition of $7.50 mL$ of $1.00 MNaOH$ is $3.88$.

d) $pH$ of solution after addition of $10.00 mL$ of $1.00 MNaOH$ is $8.31$.

Explanation of Solution

1) Concept:

The given titration is the titration of a weak acid and a strong base. The volume of the sample of the acid and its molarity is given. We can find the initial moles of the acid present using these values. When $HN{O}_2$ reacts with N $aOH$, it is an acid base reaction to form $NaN{O}_2$. Calculating moles of $NaOH$ at each addition and the moles of the acid remaining in the solution, we can find $pH$ ofthe solution for each step. The acid base reaction between $HN{O}_2$ and $NaOH$ is written as:

$HN{O}_2\left(aq\right)+{OH}^{-}\left(aq\right)\to {NO}_2^{-}\left(aq\right)+H_{2}O\left(l\right)$

The initial moles of $HN{O}_2$ and $NaOH$, $\left[HN{O}_2\right]$ and [${NO}_2^{-}\left(aq\right)]$ can be determined in the final solution and pH can be found by using the 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 $HN{O}_2$ 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) $Molarity=\frac{molesofsolute}{volumeinL}$

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 $HN{O}_2=0.200 M$

ii) Volume of $HN{O}_2=50.0 mL=0.050L$

iii) Molarity of $NaOH$= $1.00 M$

iv) $K_a=4.0\times {10}^{-4}$ (Appendix 5 Table A5.1)

4) Calculation:

Calculating the initial moles of $HN{O}_2$

$MolofHN{O}_2=\frac{0.200 mol}{L}\times 0.050 L=0.010 mol$

a) Calculating the pH of solution when $2.50 mL \left(0.0025 L\right)of 1.00 MNaOH is added$

Calculating the moles of $NaOH$

$MolesofNaOH=\frac{1.00 mol}{L}\times 0.00250 L=0.00250 mol$

Set up the RICE table to determine the reacted moles of $HN{O}_2$ and how many moles of ${CH}_{3}CO{O}^{-}$ are formed

Reaction	$HN{O}_2\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${NO}_2^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HN{O}_2 \left(mol\right)$		${OH}^{-}\left(mol\right)$		${NO}_2^{-}\left(mol\right)$
Initial	$0.010$		$0.00250$		0
Change	$-0.00250$		$-0.00250$		$+0.00250$
Final	$0.00750$		0		$0.00250$

$Totalvolume = 0.050 L+0.00250 L=0.0525 L$

Calculating the concentration of $HN{O}_2$ and ${NO}_2^{-}$

$\left[HN{O}_2\right]=\frac{0.00750 mol}{0.0525 L}=0.142857 M$

$\left[{NO}_2^{-}\right]=\frac{0.00250 mol}{0.0525 L}=0.047619 M$

${pK}_a=-\log (4.0\times {10}^{-4})=3.40$

Plug in the values in the Henderson–Hasselbalch equation,

$pH=3.40+\log \frac{\left[0.047619\right]}{\left[0.142857\right]}$

$pH=3.40+\log \left(0.33333\right)$

$pH=3.40-0.4771$

$pH=2.92$

b) Calculating the pH of solution when $5.0 mL \left(0.005 L\right)of 1.00 MNaOH is added$

Calculating the moles of $NaOH$

$MolofNaOH=\frac{1.00 mol}{L}\times 0.0050 L=0.0050 mol$

Set up the RICE table to determine the reacted moles of $HN{O}_2$ and how many moles of ${NO}_2^{-}$ are formed,

Reaction	$HN{O}_2\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${NO}_2^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HN{O}_2 \left(mol\right)$		${OH}^{-}\left(mol\right)$		${NO}_2^{-}\left(mol\right)$
Initial	$0.010$		$0.0050$		0
Change	$-0.0050$		$-0.0050$		$+0.0050$
Final	$0.0050$		0		$0.0050$

$Totalvolume = 0.050 L+0.0050 L=0.055 L$

Calculating the concentration of $HN{O}_2$ and ${NO}_2^{-}$

$\left[HN{O}_2\right]=\left[{NO}_2^{-}\right]=\frac{0.0050 mol}{0.055 L}=0.09091 M$

Plug in the values in the Henderson–Hasselbalch equation,

$pH=3.40+\log \frac{\left[0.09091 \right]}{\left[0.09091 \right]}$

$pH=3.40+\log \left(1\right)$

$pH=3.40$

c) Calculating the pH of solution when $7.50 mL \left(0.0075 L\right)of 1.00 MNaOH is added$:

Calculating the moles of $NaOH$

$MolesofNaOH=\frac{1.00 mol}{L}\times 0.00750 L=0.00750 mol$

Set up the RICE table to determine the reacted moles of $HN{O}_2$ and how many moles of ${NO}_2^{-}$ are formed,

Reaction	$HN{O}_2\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${NO}_2^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HN{O}_2 \left(moles\right)$		${OH}^{-}\left(moles\right)$		${NO}_2^{-}\left(moles\right)$
Initial	$0.010$		$0.00750$		0
Change	$-0.00750$		$-0.00750$		$+0.00750$
Final	$0.00250$		0		$0.00750$

$Totalvolume = 0.050 L+0.00750 L=0.0575 L$

Calculating the concentration of $HN{O}_2$ and ${NO}_2^{-}$

$\left[HN{O}_2\right]=\frac{0.00250 mol}{0.0575 L}=0.043478 M$

$\left[{NO}_2^{-}\right]=\frac{0.00750 mol}{0.0575 L}= 0.1304349 M$

${pK}_a=-\log (4.0\times {10}^{-4})=3.40$

Plug in the values in the Henderson–Hasselbalch equation,

$pH=3.40+\log \frac{\left[0.1304349\right]}{\left[0.043478\right]}$

$pH=3.40+\log \left(3.000\right)$

$pH=3.40+0.4771$

$pH=3.88$

d) Calculating the pH of solution when $10 mL \left(0.010 L\right)of 1.00 MNaOH is added$

Calculating the moles of $NaOH$

$Mol NaOH=\frac{1.00 mol}{L}\times 0.0100 L=0.010 mol$

Set up the RICE table to determine the reacted and remaining moles of $HN{O}_2$ and moles of $N{O}_2^{-}$ that are produced:

Reaction	$HN{O}_2\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	${NO}_2^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HN{O}_2 \left(mol\right)$		${OH}^{-}\left(mol\right)$		${NO}_2^{-}\left(mol\right)$
Initial	$0.010$		$0.010$		0
Change	$-0.010$		$-0.010$		$+0.010$
Final	$0$		0		$0.010$

Since the acid and added base have reacted completely, and an equivalent amount of conjugate base has formed, this represents the equivalence point.

$Totalvolume = 0.050 L+0.010 L=0.060 L$

Calculating the concentration of ${NO}_2^{-}$ formed

$\left[{NO}_2^{-}\right]=\frac{0.010 mol}{0.060 L}=0.166667 M$

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

Reaction	${NO}_2^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$	$\rightleftharpoons$	$HN{O}_2\left(aq\right)$	+	${OH}^{-}\left(aq\right)$
Initial	$0.166667 M$				0		0
Change	$-x$				$+x$		$+x$
Equilibrium	$(0.166667 –x)$				$x$		$x$

$K_b=\frac{\left[HN{O}_2\right]\left[{OH}^{-}\right]}{\left[{NO}_2^{-}\right]}$

Calculating $K_b$

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

$K_b=\frac{1.0\times {10}^{-14}}{4.0\times {10}^{-4}}$

$K_b=2.50\times {10}^{-11}$

$2.50\times {10}^{-11}=\frac{x^2}{(0.166667 –x)}$

$2.50\times {10}^{-11}=\frac{x^2}{\left(0.166667 \right)}$

$x=2.0412\times {10}^{-6}M=\left[{OH}^{-}\right]$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{2.0412\times {10}^{-6}}$

$\left[H_3{O}^{+}\right]=4.8991\times {10}^{-9}\mathrm{ }\mathrm{M}$

Calculating the $pH$

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

$pH=-\log (4.8991\times {10}^{-9})$

$pH=8.31$

A titration curve for the titration of $50.0 mL$ of $0.200 MHN{O}_2$ vs.$mLof 1.00 MNaOH$

Conclusion:

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