Chapter 15, Problem 15.70QA

Interpretation Introduction

To find:

The $pH$ of $0.0288M$ Oxalic acid.

Answer to Problem 15.70QA

Solution

$pH=1.67$

Explanation of Solution

1) Concept:

We are asked to determine $pH$ of a weak diprotic acid. There are two ionizable H atoms in Oxalic acid. The $K_{a_1}$ and $K_{a_2}$ values are given in appendix 5. $\left[C_2{H}_2{O}_4\right]$ value in the denominator of the $K_{a_1}$ expression will be constant at $0.0288M.$

The large difference between $K_{a1} and K_{a2}$ values means that the $pH$ of the solution is controlled by the first ionization equilibrium.

$C_2{H}_2{O}_{4\left(aq\right)}+H_2{O}_{\left(l\right)} ⥨{HC}_2{O}_{4\left(aq\right)}^{-}+H_3{O}_{\left(aq\right)}^{+}{K}_{a_1}=5.9\times {10}^{-2}$

Because of the small value of $K_{a1}$ and the high concentration of Oxalic acid, we should obtain a $pH$ greater than 1.

${HC}_2{O}_{4\left(aq\right)}^{-}+H_2{O}_{\left(l\right)}⥨ H_3{O}_{\left(aq\right)}^{+}+C_2{O}_{4\left(aq\right)}^{2-}{K}_{a_2}=6.4\times {10}^{-5}$

First we set up a RICE table for first ionization equilibrium in which $x=\left[H_{3}O\right]=\left[{HC}_2{O}_4^{-}\right]$ at equilibrium and the value of the $\left[C_2{H}_2{O}_4\right]$ is a constant $0.0288M$.

2) Formula:

i) $pH= -\log \left[{H_{3}O}^{+}\right]$

ii) $x=\frac{\left(-b\pm \sqrt{b^2-4ac }\right)}{2a}$

3) Given:

Concentration of $Oxalic acid is 0.0288M.$

4) Calculations:

Reaction	$C_2{H}_2{O}_{4\left(aq\right)}+H_2{O}_{\left(l\right)} ⥨{HC}_2{O}_{4\left(aq\right)}^{-}+H_3{O}_{\left(aq\right)}^{+}$
	$C_2{H}_2{O}_4\left] \right(M)$	${HC}_2{O}_4^{-}\left] \right(M)$	$H_3{O}^{+}\left]\right(M)$
Initial	$0.0288$	$0$	$0$
Change	$-x$	$+x$	$+x$
Equilibrium	$(0.0288-x)$	$x$	$x$

Inserting the equilibrium concentrations in the equilibrium constant expression for $K_{a1}$

$K_{a1}= \frac{{[H}_3{O}^{+}\left] {[HC}_2{O}_4^{-}\right]}{{[C}_2{H}_2{O}_4] }$

$5.9\times {10}^{-2}= \frac{\left(x\right)\left(x\right)}{(0.0288-x)}$

$x^2+5.9\times {10}^{-2}x-{1.6992\times 10}^{-3}=0$

Here, $a=1$, $b= 5.9\times {10}^{-2}$ and $c=-{1.6992\times 10}^{-3}$.

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

$x=\frac{\left(-(5.9\times {10}^{-2})\pm \sqrt{{(5.9\times {10}^{-2})}^2-(4\times 1\times (-{1.6992\times 10}^{-3}) }\right)}{2\times 1}$

$x=\frac{\left(-5.9\times {10}^{-2}\pm \sqrt{0.010277 }\right)}{2}$

Solving this equation for $x$ yields a positive value and a negative value

$x=0.021189 and x= -0.0801897$

The negative value for $x$ has no physical meaning because it gives us a negative ${[HC}_2{O}_4^{-}]$ value. Therefore,

${[H}_3{O}^{+}]=x=0.021189$

The corresponding $pH$ is

$pH= -\log \left[{H_{3}O}^{+}\right]= -\log \left(0.021189 \right)=1.67$

Therefore, the  $pH$ of $0.0288M$ Oxalic acid is $1.67$.

Conclusion

The Oxalic acid is a weak acid and its concentration here is large, so we obtain a $pH$ value that is greater than 1.