Question

${H_3}P{O_4}$ is a weak triprotic acid: approximate $pH$ of $0.1M$ $N{a_2}HP{O_4}(aq)$ is calculated by:
(A) $\dfrac{1}{2}[p{K_{a1}} + p{K_{a2}}]$
(B) $\dfrac{1}{2}[p{K_{a2}} + p{K_{a3}}]$
(C) $\dfrac{1}{2}[p{K_{a1}} + p{K_{a3}}]$
(D) $\dfrac{1}{2}[p{K_{a1}} + p{K_{a2}}]$

Answer 1

Triprotic acid can donate three hydrogen ions per molecule during dissociation. Stepwise dissociation constants are each defined for the loss of a single proton. The constant for dissociation of the first proton may be denoted as ${K_{a1}}$ and the constants for dissociation of successive protons as ${K_{a2}}$ , etc.

Complete answer:
The $p{K_a}$ value is one method used to indicate the strength of an acid. $p{K_a}$ is the negative log of the acid dissociation constant . A lower $p{K_a}$ value indicates a stronger acid. That is, the lower value indicates the acid more fully dissociates in water.
$\Rightarrow {K_a} = \dfrac{{[{A^ - }][{H^ + }]}}{{[HA]}}$
Where, ${K_a} =$ acid dissociation constant
$\Rightarrow [{A^ - }] =$ Concentration of the conjugate base of the acid
$\Rightarrow [{H^ + }] =$ Concentration of hydrogen ions
$\Rightarrow [HA] =$ Concentration of chemical species $HA$
$2HP{O_4}^{2 - } \leftrightarrow P{O_4}^{3 - } + {H_2}P{O_4}^ -$
From this we see that, $[{H_2}P{O_4}^ - ] \cong [P{O_4}^{3 - }]$
$\Rightarrow {K_{a2}} = \dfrac{{[{H_3}{O^ + }][HP{O_4}^{2 - }]}}{{[{H_2}P{O_4}^ - ]}}$
$\Rightarrow {K_{a3}} = \dfrac{{[{H_3}{O^ + }][P{O_4}^{3 - }]}}{{[HP{O_4}^{2 - }]}}$
When, $[{H_2}P{O_4}^ - ] = [P{O_4}^{3 - }]$
$\Rightarrow \dfrac{{[{H_3}{O^ + }][HP{O_4}^{2 - }]}}{{{K_{a2}}}}$ $$
$\Rightarrow {K_{a3}}\dfrac{{[{H_3}{O^ + }][P{O_4}^{3 - }]}}{{[HP{O_4}^{2 - }]}}$
$\Rightarrow {[{H_3}{O^ + }]^2}$
Taking log both sides,
$\Rightarrow \log {K_{a3}} + \log {K_{a2}} = 2\log [{H_3}{O^ + }]$
Or, $pH = \dfrac{1}{2}(p{K_{a2}} + p{K_{a3}})$
So, the correct answer is B) $pH = \dfrac{1}{2}(p{K_{a2}} + p{K_{a3}})$ .

Additional Information:
A knowledge of $p{K_a}$ values is important for the quantitative treatment of systems involving acid–base equilibria in solution. Many applications exist in biochemistry. for example, the $p{K_a}$ values of proteins and amino acid side chains are of major importance for the activity of enzymes and the stability of proteins.

Note:
When the difference between successive $p{K_a}$ values is about four or more, as in this example, each species may be considered as an acid in its own right. In fact salts of ${H_2}P{O_4}^ -$ may be crystallized from solution by adjustment of $pH$ to about $5.5$ and salts of $HP{O_4}^ -$ may be crystallized from solution by adjustment of $pH$ to about $10$ .