Question

The \[pH\] of the resultant solution of $20mL$ of $0.1M$ ${H_3}P{O_4}$ and $20mL$ of $0.1M$ $N{a_3}P{O_4}$ solution is
(A) $p{K_{a1}} + \log 2$
(B) $p{K_{a1}}$
(C) $p{K_{a2}}$
(D) $\dfrac{{p{K_{a1}} + p{K_{a2}}}}{2}$

Answer 1

Hint: The above is the solution of a weak acid and a salt of a strong base. ${H_3}P{O_4}$ is a weak tribasic acid which breaks into 3 steps with $K{a_1},K{a_2},K{a_3}$ being the dissociation constant for each step of the disassociation. Whereas $N{a_3}P{O_4}$ is a salt of a strong base.

Complete step by step answer:
${H_3}P{O_4}$ is a tribasic acid which breaks in three steps
${H_3}P{O_4} \rightleftharpoons {H_2}PO_4^ - + {H^ + }$ with $K{a_1}$
${H_2}PO_4^ - \rightleftharpoons HPO_4^{2 - } + {H^ + }$ with $K{a_2}$
$HPO_4^{2 - } \rightleftharpoons PO_4^ - + {H^ + }$ with $K{a_3}$
steps with $K{a_1},K{a_2},K{a_3}$ being the dissociation constant for each step of the disassociation.
The reaction in the above solution of ${H_3}P{O_4}$ and $N{a_3}P{O_4}$ can be represented as
${H_3}P{O_4} + PO_4^{3 - } \to {H_2}PO_4^ - + HPO_4^{2 - }$ , where $PO_4^{3 - }$ comes from full dissociation of $N{a_3}P{O_4}$
The molarity of the solution is given so the millimoles of the components before and after the reaction can be calculated by multiplying the molarity with the capacity in a milliliter.
So, the initial millimoles present in the reaction and final millimoles present in the reaction can be given as
Initial millimoles of ${H_3}P{O_4}$ is $2$ and initial millimoles of $PO_4^{3 - }$ are also $2$ millimoles.
Initial millimoles of ${H_2}PO_4^ - $ and $HPO_4^{2 - }$ being $0$ before the reaction.
After reaction the reactants completely and the final concentration can be seen as
Final millimoles of ${H_3}P{O_4}$ and $PO_4^{3 - }$ is $0$ since they fully reacted
Final millimoles of ${H_2}PO_4^ - $ and $HPO_4^{2 - }$ being $2$ millimoles after the reaction.
This reaction leaves the buffer solution of ${H_2}PO_4^ - $ and $HPO_4^ - $ former being the Lewis acid and the latter being the corresponding Lewis base.
The pH of buffer solution is calculated by the formula as stated below
$pH = p{K_a} + \log \dfrac{{\text{[conjugate base]}}}{{[acid]}}$ , where $\text{[conjugate base]}$ represent the concentration of the conjugate base in the solution and $[acid]$ represent the concentration of Lewis acid in the final solution.
Thus, putting the values in the above formula we get,
$pH = p{K_{a2}} + \log \dfrac{2}{2}$
$\Rightarrow pH = p{K_{a2}} + \log 1$
$\Rightarrow pH = p{K_{a2}}$
Here, we have taken $K{a_2}$ because it is the dissociation constant of the step in which ${H_2}PO_4^ - $ and $HPO_4^{2 - }$ exists in equilibrium.

So, the correct answer is Option C

Note: The solution forms a buffer of acid and its corresponding conjugate base.
The dissociation constant of that reaction in which the final products are in equilibrium is taken into account for the final calculation.
There are different ways of measuring pH. They are indicator methods, metal- electrode methods, glass- electrode methods, semiconductor sensor methods.