Question

What is the equilibrium concentration of ${C_2}{O_4}^{2 - }$ in a $0.20M$ solution of oxalic acid ?

Answer 1

Hint: To solve the given question, we should have information of oxalic acid, acid dissociation constant and equilibrium constant. Oxalic acid is an organic acid formed from two carboxylic acids. It is a white crystalline solid forming a colourless solution in water. It is used in the cleaning process, extractive metallurgy and in habitat formation.Equilibrium constant is the ratio of concentration of products with power of stoichiometric constant to that of concentration of reactants with power of their stoichiometric constant.
Formula Used :
To solve the given question, we have to use the formula of equilibrium constant :
$K = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
Where ,$[C]$ and $[D]$ are concentrations of products and c and d are the stoichiometric constant.
$[A]$ and $[B]$ are concentrations of reactants and a and b are the stoichiometric constant.
We have to remember that acid dissociation constant ${K_{a1}}$ for oxalic acid is $5.6 \times {10^{ - 2}}$ and ${K_{a2}}$ is $1.5 \times {10^{ - 4}}$ .

Complete answer: Step-1 :
We have given the acid ${H_2}{C_2}{O_4}$ , oxalic acid which will dissociate in water in two steps.
In the first case, only one proton dissociates forming $H{C_2}{O_4}^ - $ ion.
Constructing the table, we get
${H_2}{C_2}{O_4} \Leftrightarrow H{C_2}O_4^ - + {H^ + }$

	${H_2}{C_2}{O_4}$	$H{C_2}O_4^ - $	${H^ + }$
$t = 0$	$0.2$	$0$	$0$
$t = eq$	$0.2 - \alpha $	$\alpha $	$\alpha $

Here, the equilibrium constant will be prepared using the formula as :
${K_{eq}} = \dfrac{{[H{C_2}O_4^ - ][{H^ + }]}}{{[{H_2}{C_2}{O_4}]}}$
Here, ${K_{eq}}$ is equal to first acid dissociation constant ${K_{a1}}$ ,
So,
${K_{a1}} = \dfrac{{{\alpha ^2}}}{{0.2 - \alpha }}$
$5.6 \times {10^{ - 2}} = \dfrac{{{\alpha ^2}}}{{0.2 - \alpha }}$
After solving this we will have a quadratic equation and solving the quadratic equitation we will have
$\therefore \alpha = 8.13 \times {10^{ - 2}}.mol.d{m^{ - 3}}$
So, $[{H^ + }] = [H{C_2}O_4^ - ] = 8.13 \times {10^{ - 2}}$
Step-2 :

Now, we will take $H{C_2}{O_4}^ - $ as a product and perform its dissociation. Now, the second proton dissociates forming ${C_2}{O_4}^{2 - }$ ion.
Constructing the table we get :
${C_2}H{O_4}^ - \Leftrightarrow {C_2}{O_4}^{2 - }_{(aq)} + {H^ + }$

	${C_2}H{O_4}^ - $	${C_2}{O_4}^{2 - }_{(aq)}$	${H^ + }$
$t = 0$	$8.13 \times {10^{ - 2}}$	$0$	$8.13 \times {10^{ - 2}}$
$t = eq$	$8.13 \times {10^{ - 2}} - x$	$x$	$8.13 \times {10^{ - 2}} + x$

Here, equilibrium constant will be prepared using the formula as :
${K_{eq}} = \dfrac{{[{C_2}O_4^ - ][{H^ + }]}}{{[H{C_2}{O_4}^ - ]}}$
Here, ${K_{eq}}$ is equal to second dissociation constant, ${K_{a2}}$ ,
${K_{a2}} = \dfrac{{[{C_2}O_4^ - ][{H^ + }]}}{{[H{C_2}{O_4}^ - ]}}$
To make the calculation easy it is reasonable to consider
 \[8.13 \times {10^{ - 2}} - x \cong 8.13 \times {10^{ - 2}}\]
\[8.13 \times {10^{ - 2}} + x \cong 8.13 \times {10^{ - 2}}\].
Then above formula becomes
So, $1.5 \times {10^{ - 4}} \approx \dfrac{{8.13 \times {{10}^{ - 2}} \times x}}{{8.13 \times {{10}^{ - 2}}}}$
Cancelling the terms we have
$\therefore [{C_2}{O_4}^{2 - }] = x \approx {K_{a2}} = 1.5 \times {10^{ - 4}}.mol.d{m^{ - 3}}$

Note:
While solving any acid dissociation type question, remember to solve the hydrogen dissociation separately in table form. Acid constant or acid dissociation constant, ${K_a}$ is used to quantitatively measure the acidic strength in a solution. The unit of \[{K_a}\] is $mol.d{m^{ - 3}}$.