Chapter 16, Problem 16.46QA

Interpretation Introduction

To sketch:

A titration curve for the titration of $40.0 mL$ of $0.100 M HOOCCOOH$(oxalic acid) with $0.100 M NaOH$. What is the $pH$ of the sample after the addition of the titrant?

a. $10.0 mL$

b. $20.0 mL$

c. $30.0 mL$

d. $40.0 mL$

Answer to Problem 16.46QA

Solution:

a) $pH$ of solution after titration of $40.0 mL$ of $0.100 M HOOCCOOH$ with $10.0 mL$ of $0.100 M NaOH$ is $0.75.$

b) $pH$ of solution after addition of $40.0 mL$ of $0.100 M HOOCCOOH$ with $20.0 mL$ of $0.100 M NaOH$ is $1.23.$

c) $pH$ of solution after addition of $40.0 mL$ of $0.100 M HOOCCOOH$ with $30.0 mL$ of $0.100 M NaOH$ is $1.71.$

d) $pH$ of solution after addition of $40.0 mL$ of $0.100 M HOOCCOOH$ with $40.0 mL$ of $0.100 M NaOH$ is $2.72.$

Explanation of Solution

1) Concept:

Oxalic acid is a weak diprotic acid, i.e., there are two protons that can dissociate from one molecule of oxalic acid in aqueous solution. The dissociation of protons occurs stepwise.

The first equivalence point occurs where the first dissociated proton is completely neutralized when titrated with a strong base $NaOH$, and the equation is

$HOOCCOOH\left(aq\right)+{OH}^{-}\left(aq\right)\to HOOCCO{O}^{-}\left(aq\right)+H_{2}O\left(aq\right)$

The second equivalence point occurs where both dissociated protons fully neutralized, and the equation is

$HOOCCO{O}^{-}\left(aq\right)+{OH}^{-}\left(aq\right)\to C_2{O}_4^{2-}\left(aq\right)+H_{2}O\left(aq\right)$

Equation for complete titration is

$HOOCCOOH\left(aq\right)+2{OH}^{-}\left(aq\right)\to C_2{O}_4^{2-}\left(aq\right)+2H_{2}O\left(aq\right)$

At equivalence point, the equal moles of weak acid and strong base react and undergo neutralization. The second $p{K}_a$ of oxalic acid is $4.71$, so there is not much difference between $p{K}_{a1}$ and $p{K}_{a2}$.We need to consider the second dissociation of oxalic acid to determine the $pH$ at the first equivalence point.

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

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

3) Given:

i) Molarity of oxalic acid, $HOOCCOOH=0.100 M$

ii) Volume of oxalic acid,$HOOCCOOH=40.0 mL=0.040L$

iii) Molarity of $NaOH=0.100 M$

iv) $K_{a1}=5.9\times {10}^{-2}$(Appendix 5 Table A5.1)

v) $K_{a2}=6.4\times {10}^{-5}$(Appendix 5 Table A5.1)

vi) pK_a1 = 1.23 (Appendix 5 Table A5.1)

vii) pK_a2 = 4.19 (Appendix 5 Table A5.1)

4) Calculation:

Converting volume in $mL to L$

$40.0 mL oxalic acid\times \frac{1 L}{1000 mL}=0.040 L oxalic acid$

Calculating moles of oxalic acid

$\frac{0.100 mol}{1 L}\times 0.040 L HOOCCOOH=0.0040 mol oxalic acid$

These are the initial moles of oxalic acid present before any $NaOH$ is added to the acid.

Now, calculating the $pH$ for each of the following additions of $NaOH$ solution:

a) $10.0 mL$ of $0.100 M NaOH$

Converting volume in ml to L,

$10.0 mL NaOH\times \frac{1 L}{1000 mL}=0.010 L NaOH$

Calculating the moles of $NaOH,$

$\frac{0.100 mol}{1 L}\times 0.010 L NaOH=0.0010 mol NaOH$

Set up the RICE table that is used to calculate the moles of acid reacted and moles of conjugate base formed.

$HOOCCOOH\left(aq\right)+{OH}^{-}\left(aq\right)\to HOOCCO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)$

Reaction	$HOOCCOOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	$HOOCCO{O}^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HOOCCOOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		$HOOCCO{O}^{-}\left(mol\right)$
Initial	$0.004$ 0		$0.0010$		0
Change	$-0.0010$		$-0.0010$		$+0.0010$
Final	$0.0030$		0		$0.0010$

$Total volume =0.040 L+0.010 L=0.050 L$

Calculating the concentration of $HOOCCO{O}^{-}$ and $HOOCCO{O}^{-}$,

$\left[HOOCCOOH\right]=\frac{0.0030 mol}{0.050 L}=0.060 M$

$\left[HOOCCO{O}^{-}\right]=\frac{0.0010 mol}{0.050 L}=0.020 M$

$pH=1.23+\log \frac{\left(0.020\right)}{\left(0.060\right)}$

$pH=0.75$

b) $20.0 mL$ of $0.100 M NaOH$

Converting volume in ml to L,

$20.0 mL NaOH\times \frac{1 L}{1000 mL}=0.020 L NaOH$

Calculating the moles of $NaOH$,

$\frac{0.100 mol}{1 L}\times 0.020 L NaOH=0.0020 mol NaOH$

Setting up the RICE table to calculate the moles of acid reacted and moles of conjugate base formed,

$HOOCCOOH\left(aq\right)+{OH}^{-}\left(aq\right)\to HOOCCO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)$

Reaction	$HOOCCOOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	$HOOCCO{O}^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HOOCCOOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		$HOOCCO{O}^{-}\left(mol\right)$
Initial	$0.004$ 0		$0.0020$		0
Change	$-0.0020$		$-0.0020$		$+0.0020$
Final	$0.0020$		0		$0.0020$

Moles of the weak acid and conjugate base are equal.Thisis the half equivalence point to the first equivalence point, and the pH at half equivalence point is

$pH=p{K}_{a1}=1.23$

c) $30.0 mL$ of $0.100 M NaOH$

Converting volume in ml to L,

$30.0 mL NaOH\times \frac{1 L}{1000 mL}=0.030 L NaOH$

Calculating the moles of $NaOH$,

$\frac{0.100 mol}{1 L}\times 0.030 L NaOH=0.0030 mol NaOH$

Setting up the RICE that is used to calculate the moles of acid reacted and moles of conjugate base formed,

$HOOCCOOH\left(aq\right)+{OH}^{-}\left(aq\right)\to HOOCCO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)$

Reaction	$HOOCCOOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	$HOOCCO{O}^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HOOCCOOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		$HOOCCO{O}^{-}\left(mol\right)$
Initial	$0.004$ 0		$0.0030$		0
Change	$-0.0030$		$-0.0030$		$+0.0030$
Final	$0.0010$		0		$0.0030$

$Total volume =0.040 L+0.030 L=0.070 L$

Calculating the concentration of $HOOCCO{O}^{-}$ and $HOOCCO{O}^{-}$,

$\left[HOOCCOOH\right]=\frac{0.0010 mol}{0.070 L}=0.014286 M$

$\left[HOOCCO{O}^{-}\right]=\frac{0.0030 mol}{0.050 L}=0.042856 M$

$pH=1.23+\log \frac{\left(0.042856 \right)}{\left(0.014286\right)}$

$pH=1.71$

d) $40.0 mL$ of $0.100 M NaOH$

Converting volume in ml to L,

$40.0 mL NaOH\times \frac{1 L}{1000 mL}=0.040 L NaOH$

Calculating the moles of $NaOH,$

$\frac{0.100 mol}{1 L}\times 0.040 L NaOH=0.0040 mol NaOH$

Set up the RICE table to calculate the moles of acid reacted and moles of conjugate base formed.

$HOOCCOOH\left(aq\right)+{OH}^{-}\left(aq\right)\to HOOCCO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)$

Reaction	$HOOCCOOH\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	$HOOCCO{O}^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$HOOCCOOH \left(mol\right)$		${OH}^{-}\left(mol\right)$		$HOOCCO{O}^{-}\left(mol\right)$
Initial	$0.004$ 0		$0.0040$		0
Change	$-0.0040$		$-0.0040$		$+0.0040$
Final	$0$		0		$0.0040$

This is the first equivalence point since no moles of weak acid or strong base are left.

There is not much difference in the $p{K}_{a1} and p{K}_{a2}$ for the oxalic acid, so we must consider the second dissociation of oxalic acid to determine the pH.

At the equivalence point, the major species present is $HOOCCO{O}^{-}\left(aq\right)$, so by writing the dissociation reaction for this and creating the RICE table, find the $pH$ of solution.

Total volume of solution $=0.080L$

Molarity of $HOOCCO{O}^{-}= \frac{0.0040mol}{0.080 L}=0.05 M$

$HOOCCO{O}^{-}\left(aq\right)+H_{2}O\left(aq\right)\rightleftharpoons OOCCO{O}^{2-}\left(aq\right)+H{H}_3{O}^{+}\left(aq\right) )54ibrium of solution tion reaction for this and creating the RICE table find the nd$
Initial (M)	$0.05$	$0.0$	$0.0$
Change (M)	$-x$	$+x$	$+x$
Equilibrium (M)	$(0.05-x)$	$x$	$x$

Writing the $K_b$ expression and solving for $pH$,

$p{K}_{a2}=4.17$

$p{K}_b= -\log K_b$, $K_b={10}_{}^{-pK}$

$K_{a2}={10}^{-4.17}=6.7608 \times {10}^{-5}$

$K_{a2}= \frac{x^2}{(0.05-x)}= 6.7608 \times {10}^{-5}$

$6.7608 \times {10}^{-5}\times 0.05=x^2$

$x^2=3.3804e-6$

$x=1.8385\times {10}^{-3}=\left[H_3{O}^{+}\right]$

$pH= -\log 1.8385\times {10}^{-3}=2.72$

A titration curve for the titration of $40.0 mL$ of $0.100 M HOOCCOOH$ with $mL of 0.100 M NaO$

Conclusion:

Using the RICE table, we can determinethe $pH$ of solution after the first equivalence pointfor a diprotic acid.