Question

${{\text{H}}_{\text{3}}}{\text{A}}$ is a weak triprotic acid $\left( {{K_{a1}} = {{10}^{ - 5}},\,{K_{a2}} = {{10}^{ - 9}},\,{K_{a3}} = {{10}^{ - 13}}} \right)$. What is the value of $pX$ of $0.1{\text{ M }}{{\text{H}}_3}{\text{A (aq}}{\text{.)}}$ solution where $pX = - \log X$ and $X = \dfrac{{[{A^{3 - }}]}}{{[{\text{H}}{{\text{A}}^{2 - }}]}}$?
A . 7
B. 8
C. 9
D. 10

Answer 1

We know that triprotic acid is the one which donates three protons or three hydrogen ions. From the values of acid dissociation constants we can see that the degree of dissociation for triprotic acid is negligible with respect to 1.

Complete answer:
We are given a weak triprotic acid ${{\text{H}}_{\text{3}}}{\text{A}}$. The acid dissociates into three steps as follows:
${{\text{H}}_{\text{3}}}{\text{A}} \rightleftharpoons {{\text{H}}_{\text{2}}}{{\text{A}}^ - } + {{\text{H}}^ + }{\text{ }}, {K_{a1}} = {10^{ - 5}}$
${{\text{H}}_{\text{2}}}{{\text{A}}^ - } \rightleftharpoons {\text{H}}{{\text{A}}^{2 - }} + {{\text{H}}^ + }{\text{ }}, {K_{a2}} = {10^{ - 9}}$
${\text{H}}{{\text{A}}^{2 - }} \rightleftharpoons {{\text{A}}^{3 - }} + {{\text{H}}^ + }{\text{ }}, {K_{a3}} = {10^{ - 13}}$
Where ${K_{a1}}$, ${K_{a2}}$ and ${K_{a3}}$ are the acid dissociation constants.
The acid dissociation constant for the first reaction ${K_{a1}}$ is very large that the acid dissociation constants for second and third reactions.
Thus, the hydrogen ion concentration is mainly from the first reaction.
Now we can calculate the concentration of hydrogen ion using the equation as follows:
$[{{\text{H}}^ + }] = \sqrt {{K_{a1}}C}$
where
${K_{a1}}$ is the first acid dissociation constant.
$C$ is the molar concentration of the acid.
Substitute ${10^{ - 5}}$ for the first dissociation constant of the acid, $0.1{\text{ M}}$ for the molar concentration of the triprotic acid and solve for the hydrogen ion concentration. Thus,
$[{{\text{H}}^ + }] = \sqrt {{{10}^{ - 5}} \times 0.1{\text{ M}}}$
$\Rightarrow [{{\text{H}}^ + }] = \sqrt {{{10}^{ - 6}}}$
$\Rightarrow [{{\text{H}}^ + }] = {10^{ - 3}}{\text{ M}}$
Thus, the concentration of hydrogen ion is ${10^{ - 3}}{\text{ M}}$.
Now consider the third reaction. For the third reaction, the acid dissociation constant is ${K_{a3}}$. Thus,
${K_{a3}} = \dfrac{{[{{\text{H}}^ + }][{{\text{A}}^{3 - }}]}}{{[{\text{H}}{{\text{A}}^{2 - }}]}}$
Substitute ${10^{ - 13}}$ for the third acid dissociation constant, ${10^{ - 3}}{\text{ M}}$ for the concentration of hydrogen ions. Thus,
$\Rightarrow {10^{ - 13}} = \dfrac{{{{10}^{ - 3}}{\text{ M}} \times [{{\text{A}}^{3 - }}]}}{{[{\text{H}}{{\text{A}}^{2 - }}]}}$
$\Rightarrow \dfrac{{[{A^{3 - }}]}}{{[{\text{H}}{{\text{A}}^{2 - }}]}} = \dfrac{{{{10}^{ - 13}}}}{{{{10}^{ - 3}}}}$
$\Rightarrow \dfrac{{[{A^{3 - }}]}}{{[{\text{H}}{{\text{A}}^{2 - }}]}} = {10^{ - 10}}$
We are given that $X = \dfrac{{[{A^{3 - }}]}}{{[{\text{H}}{{\text{A}}^{2 - }}]}}$. Thus,
$X = {10^{ - 10}}$
Now, we are given that,
$pX = - \log X$
Substitute $X = {10^{ - 10}}$. Thus,
$\Rightarrow pX = - \log \left( {{{10}^{ - 10}}} \right)$
$\Rightarrow pX = 10$
Thus, the value of $pX$ of $0.1{\text{ M }}{{\text{H}}_3}{\text{A (aq}}{\text{.)}}$ solution is 10.

**Thus, the correct option is (D) 10.

Note:**
The triprotic acids have three dissociable protons. It undergoes stepwise ionisation. The example of triprotic acid is phosphoric acid ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$. It dissociates in the following manner.
$\rm{H_3PO_4} \rightleftharpoons H^+\, + \, {H_2PO_4}^-$
$\rm{{H_2PO_4}^- \rightleftharpoons H^+\, + \, {HPO_4}^{2-} }$
$\rm{ {HPO_4}^{2-} \rightleftharpoons H^+\, + \, {PO_4}^{3-}}$