Chapter 16, Problem 16.53QA

Interpretation Introduction

To find:

When $100 mL$ of $0.0125 M$ ascorbic acid is titrated with $0.010 M NaOH$, how many equivalence points will the titration curve have, and what pH indicator(s) could be used? Refer to Figure 16.5 for the colors of indicators.

Answer to Problem 16.53QA

Solution:

When $100 mL$ of $0.0125 M$ ascorbic acid is titrated with $0.010 M NaOH$, the titration curve will have two equivalence points, and the indicator bromthymol blue will be used for the first equivalent point and Alizarin yellow R for the second.

Explanation of Solution

1) Concept:

The number of equivalence points depends on the number of protons that can dissociate from the molecule in an aqueous form.

Ascorbic acid is a weak diprotic acid, i.e., there are two protons that can dissociate from one molecule of ascorbic acid. 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

$H_2{C}_6{H}_6{O}_6\left(aq\right)+{OH}^{-}\left(aq\right)\to H{C}_6{H}_6{O}_6^{-}\left(aq\right)+H_{2}O\left(aq\right)$

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

$H{C}_6{H}_6{O}_6^{-}\left(aq\right)+{OH}^{-}\left(aq\right)\to C_6{H}_6{O}_6^{2-}\left(aq\right)+H_{2}O\left(aq\right)$

Equation for the complete titration is

$H_2{C}_6{H}_6{O}_6\left(aq\right)+2{OH}^{-}\left(aq\right)\to C_6{H}_6{O}_6^{2-}\left(aq\right)+2H_{2}O\left(aq\right)$

Therefore, there are two equivalence points in the titration of ascorbic acid with $NaOH$.

The volume of NaOH required to reach the second equivalence point is double the required volume of NaOH for the first equivalence point.

At equivalence point, the equal moles of weak acid and strong base react and undergo neutralization.

2) Formula:

i) $Molarity=\frac{moles}{volume in L}$

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

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

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

3) Given:

i) $\left[H_2{C}_6{H}_6{O}_6\right]=0.0125 M$

ii) Volume of $H_2{C}_6{H}_6{O}_6=100 mL$

iii) $\left[NaOH\right]=0.010 M$

iv) $K_{a1}=9.1\times {10}^{-5}$ (Appendix 5 Table A5.1)

v) $K_{a2}=5\times {10}^{-12}$ (Appendix 5 Table A5.1)

vi) ${pK}_{a1}=4.04$(Appendix 5 Table A5.1)

vii) ${pK}_{a2}=11.3$(Appendix 5 Table A5.1)

4) Calculation:

Converting the volume in mL to L,

$100 mL H_2{C}_6{H}_6{O}_6\times \frac{1 L}{1000 mL}=0.100 L H_2{C}_6{H}_6{O}_6$

Calculating the moles of $H_2{C}_6{H}_6{O}_6$,

$\frac{0.0125 mol}{ L}\times 0.100 L=0.00125 mol H_2{C}_6{H}_6{O}_6$

At first equivalence point, moles of $H_2{C}_6{H}_6{O}_6$ equal the moles of NaOH; therefore,

$moles of NaOH=0.00125 mol$ and

$Volume of NaOH=\frac{1 L}{0.010 mol}\times 0.00125 mol=0.125 L$

Set up the RICE table to determine the reacted moles of $H_2{C}_6{H}_6{O}_6$ and to determine how many moles of $H{C}_6{H}_6{O}_6^{-}$ are formed.

Reaction	$H_2{C}_6{H}_6{O}_6\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	$H{C}_6{H}_6{O}_6^{-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$H_2{C}_6{H}_6{O}_6 \left(mol\right)$		${OH}^{-}\left(mol\right)$		$H{C}_6{H}_6{O}_6^{-}\left(mol\right)$
Initial	$0.00125$		$0.00125$		0
Change	$-0.00125$		$-0.00125$		$+0.00125$
Final	$0$		0		$0.00125$

Calculation of $pH$ at first equivalence point:

After first equivalence point, total volume of solution is $0.100+0.125=0.225 L$.

Molarity of

$H{C}_6{H}_6{O}_6^{-}=\frac{0.00125 mol}{0.225 L}=0.005556 M$

Reaction	$H{C}_6{H}_6{O}_6^{-}\left(aq\right)$	+	$H_{2}O \left(l\right)$	$\to$	$C_6{H}_6{O}_6^{2-}\left(aq\right)$	+	$H_3{O}^{+}\left(aq\right)$
	$H{C}_6{H}_6{O}_6^{-}\left(M\right)$				$C_6{H}_6{O}_6^{2-}\left(M\right)$		$H_3{O}^{+}\left(M\right)$
Initial	$0.005556$				0	0 $+x$ $x$
Change	$-x$				$+x$
Final	$0.005556-x$				$x$

${K_a}_2=\frac{\left[C_6{H}_6{O}_6^{2-}\right]\left[H_3{O}^{+}\right]}{\left[H{C}_6{H}_6{O}_6^{-}\right]}$

$5\times {10}^{-12}= \frac{x^2}{(0.005556-x)}$

For simplification, we assume the value of $x$ to be very small and neglect it from the denominator.

$5\times {10}^{-12}= \frac{x^2}{0.005556}$

$x=1.668\times {10}^{-7}=\left[H_3{O}^{+}\right]$

$pH=-\log (1.668\times {10}^{-7})$

$pH=6.78$

Refer Figure 16.5 from the text book for the colors of indicators. Therefore, to indicate first equivalence point, bromthymol blue indicator can be used.

For the second equivalence point,

Reaction	$H{C}_6{H}_6{O}_6^{-}\left(aq\right)$	+	${OH}^{-}\left(aq\right)$	$\to$	$C_6{H}_6{O}_6^{2-}\left(aq\right)$	+	$H_{2}O\left(l\right)$
	$H{C}_6{H}_6{O}_6^{-} \left(mol\right)$		${OH}^{-}\left(mol\right)$		$C_6{H}_6{O}_6^{2-}\left(mol\right)$
Initial	$0.00125$		$0.00125$		0
Change	$-0.00125$		$-0.00125$		$+0.00125$
Final	$0$		0		$0.00125$

$Volume of O{H}^{-}, i.e., NaOH=0.125 L\times 2=0.250 L$

Total volume of solution after the second equivalence point, i.e., complete titration is

$0.100 L+0.250 L=0.350 L$

Calculating the concentration of $C_6{H}_6{O}_6^{2-}$

$\left[C_6{H}_6{O}_6^{2-}\right]=\frac{0.00125 mol}{0.350 L}=0.00357143 M$

The dissociation equation is

$C_6{H}_6{O}_6^{2-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H{C}_6{H}_6{O}_6^{-}\left(aq\right)+{OH}^{-}\left(aq\right)$

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

Reaction	$C_6{H}_6{O}_6^{2-}$	+	$H_{2}O\left(l\right)$	$\rightleftharpoons$	$H{C}_6{H}_6{O}_6^{-}\left(aq\right)\left(aq\right)$	+	${OH}^{-}\left(aq\right)$
Initial	$0.00357143$				0			0
Change	$-x$				$+x$			$+x$
Final	$0.00357143-x$				$x$			$x$

Calculating $K_{b2}$

$K_{b2}=\frac{K_w}{K_{a2}}=\frac{1.0\times {10}^{-14}}{5\times {10}^{-12}}=0.0020$

$K_{b2}=\frac{\left[H{C}_6{H}_6{O}_6^{-}\right]\left[{OH}^{-}\right]}{\left[C_6{H}_6{O}_6^{2-}\right]}$

$0.0020=\frac{x^2}{\left(0.00357143-x\right)}$

$0.00000714286-0.0020x=x^2$

$x^2+0.0020x-0.00000714286=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-0.0020\pm \sqrt{{0.0020}^2-4(-0.00000714286)}}{2\times \left(-0.00000714286\right)}$

$x=0.001651769=\left[{OH}^{-}\right]$

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

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{0.001651769}=6.05411\times {10}^{-12}M$

$pH=-\log (6.05411\times {10}^{-12})$

$pH=11.22$

Refer Figure 16.5 from the text book for the colors of indicators. The ${pK}_a$ value of Alizarin yellow R is near to this $pH,$ i.e., 11; therefore, to indicate the second equivalence point, Alizarin yellow R indicator can be used.

Conclusion:

Calculate the $pH$ at both first and second equivalence point and determine the suitable indicators.