Chapter 16, Problem 85QP

Calculate the pH at $\text{25°C}$ of a $0.25-M$ aqueous solution of oxalic acid $\left({\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right)$. ( $K_{\text{a}}$, and $K_{\text{a}}$, for oxalic acid are $6.5 \times {10}^{-2}\text{ and }6.1 \times {10}^{-5}$, respectively.)

Interpretation Introduction

Interpretation:

The pH of the given concentration of aqueous solution that contains oxalic acid is to be determined.

Concept introduction:

The first ionization of the diprotic acid takes place as:

${\text{H}}_2\text{A}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HA}}^{-}\left(aq\right)$

$K_{\text{a1}}$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific at a particular temperature.

$K_{\text{a1}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$ …… (1)

The second ionization of the diprotic acid takes place as:

${\text{HA}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

$K_{\text{a2}}$ is the measure of the dissociation of the second proton of an acid and is known as the second acid-ionization constant, which is specific at a particular temperature.

$K_{\text{a2}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$ …… (2)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$\%\text{ dissociation}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[{\text{H}}_2\text{A}\right]}_{\text{o}}}\times 100\%$ …… (3)

Here, ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}$ is the hydronium ion concentration at equilibrium and ${\left[{\text{H}}_2\text{A}\right]}_{\text{o}}$ is the original acid concentration.

The pH of the solution is calculated as:

$\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$ …… (4)

Answer to Problem 85QP

Solution:

The pH of the oxalic acid solution is $1.00$.

Given information:

The concentration of oxalic acid $\left({\text{H}}_2{\text{C}}_2{\text{O}}_4\right)$ at $25{}^{\circ }\text{C}$ is $0.25M$. The ${\text{K}}_{\text{a1}}{\text{ and K}}_{\text{a2}}$ values for oxalic acid are ${\text{K}}_{\text{a1}}=6.5\times {10}^{-2}{\text{ and K}}_{\text{a2}}=6.1\times {10}^{-5}$.

Explanation of Solution

When the oxalic acid is dissolved in water, the dissociation takes place in two steps as it is a diprotic acid. First, one proton is partially dissociated since oxalic acid is a weak acid. Thus, the concentration of ${\text{HC}}_2{\text{O}}_4{}^{-}$ is determined by the ${\text{K}}_{\text{a1}}$ of the acid. The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is determined by the contribution made by both the proton dissociations.

The reaction of the first proton dissociation of oxalic acid is depicted as:

${\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{HC}}_2{\text{O}}_4{}^{-}\left(aq\right)$

First, prepare an equilibrium table and represent each of the species in terms of $x$ as:

$\begin{array}{ccccc}& {\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{HC}}_2{\text{O}}_4{}^{-}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.25& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.25-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as:

${\text{K}}_{\text{a1}}=\frac{\left(x\right)\left(x\right)}{\left(0.25-x\right)}$

Since the value of ${\text{K}}_{\text{a1}}$ is small, the amount of acid dissociated is less. Therefore, $\left(0.25-x\right)$ can be approximated as $0.25$. Now, substitute the value of ${\text{K}}_{\text{a1}}$ in the above equation as:

$\begin{array}{c}6.5\times {10}^{-2}=\frac{\left(x\right)\left(x\right)}{0.25}\\ x^2=\left(6.5\times {10}^{-2}\right)0.25\\ x=\sqrt{1.625\times {10}^{-2}}\\ x=0.13\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1\left(aq\right)=0.13M$, ${\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\right]}_{\left(\text{O}\right)}=0.25M$

Calculate the percent dissociation from equation (3) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{0.13}{\text{0}\text{.25}}\times 100\%\\ =50.99\%\end{array}$

Since, the percent dissociation is much more than $5\%$, the approximation taken is not valid. Thus, again solve for the value of $x$ as:

$\begin{array}{c}6.5\times {10}^{-2}=\frac{\left(x\right)\left(x\right)}{\left(0.25-x\right)}\\ x^2+\left(6.5\times {10}^{-2}\right)x=\left(1.625\times {10}^{-2}\right)\\ x=-0.164,\text{ 0}\text{.099}\end{array}$

Since, concentration cannot be negative so, $x=0.099$.

Thus,

$\begin{array}{l}\left[{\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}{}^{-}\right]=\text{0}\text{.099 }M\\ {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{\text{1}}=0.099M\end{array}$

Now, the reaction of the second proton dissociation of oxalic acid is depicted as:

${\text{HC}}_2{\text{O}}_4{}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{C}}_2{\text{O}}_4{}^{-}\left(aq\right)$

First, prepare an equilibrium table and represent each of the species in terms of $y$ as:

$\begin{array}{ccccc}& {\text{HC}}_2{\text{O}}_4{}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{C}}_2{\text{O}}_4{}^{-}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.099& -& 0.099& 0\\ \text{Change in concentration}\left(M\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(M\right)& 0.099-y& -& 0.099+y& y\end{array}$

Now, substitute these concentrations in equation (2) as:

${\text{K}}_{\text{a2}}=\frac{\left(0.099+y\right)\left(y\right)}{\left(0.099-y\right)}$

Since the value of ${\text{K}}_{\text{a2}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(0.099-y\right)$ and $\left(0.099+y\right)$ can be approximated as $0.099$. Now, substitute the value of ${\text{K}}_{\text{a2}}$ in the above equation as:

$\begin{array}{c}6.1\times {10}^{-5}=\frac{\left(\cancel{0.099}\right)\left(y\right)}{\cancel{0.099}}\\ y=6.1\times {10}^{-5}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2(aq)=6.1\times {10}^{-5}M$, ${\left[{\text{HC}}_{\text{2}}{\text{O}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=0.086M$

Calculate the percent dissociation from equation (3) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{6.1\times {10}^{-5}}{\text{0}\text{.099}}\times 100\%\\ =0.07\%\end{array}$

Since, the percent dissociation is much less than $5\%$, the approximation taken is valid.

Also,

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_1+{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_2\\ =0.099+\left(6.1\times {10}^{-5}\right)\\ \simeq 0.099M\end{array}$

Now, pH is calculated using equation (4) as:

$\begin{array}{c}\text{pH}=-\log \left[0.099\right]\\ =1.00\end{array}$

Therefore, the pH is $1.00$.

Conclusion

The pH of $0.25M$ aqueous solution of oxalic acid is $1.00$.