Question

Find the pH of 0.1M ${{\text{K}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ solution:
The third dissociation constant of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is $1.3 \times {10^{ - 12}}$ . Assume that the hydrolysis proceeds only in the first step.
(A) pH=12.44
(B) pH=12.04
(C) pH=12.64
(D) pH=12.94

Answer 1

pH can be considered as a number that indicates the acidic or basic nature of a solution. It is the negative logarithm of the hydrogen ion concentration. The pH of a solution can be calculated from the pOH of the solution which in turn can be calculated from the hydroxide ion concentration.

Complete step by step solution:
Given that the third dissociation constant of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is equal to $1.3 \times {10^{ - 12}}$ .
Also, given that the hydrolysis of ${{\text{K}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ proceeds only in the first step.
We are to calculate the pH of 0.1 M ${{\text{K}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ solution using this given data.
${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ or phosphoric acid or orthophosphoric acid is a weak acid and contains three removable hydrogens. All of them are acidic and when all three are removed, the phosphate ion ${\left( {{\text{P}}{{\text{O}}_{\text{4}}}} \right)^{{\text{3 - }}}}$ is formed. Removal of one hydrogen leads to the formation of the dihydrogen phosphate ion ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ -$ and the removal of two hydrogens leads to the formation of the hydrogen phosphate ion ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$ .
The dissociation of phosphoric acid can be shown as follows:
${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\overset {{\text{Ka1}}} \leftrightarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}^{\text{ - }}$
${{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\overset {{{\text{K}}_{{\text{a2}}}}} \leftrightarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ + HP}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$
${\text{HP}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\overset {{{\text{K}}_{{\text{a3}}}}} \leftrightarrows {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ + P}}{{\text{O}}_{\text{4}}}^{{\text{3 - }}}$
Here, ${{\text{K}}_{{\text{a1}}}}$ , ${{\text{K}}_{{\text{a2}}}}$ and ${{\text{K}}_{{\text{a3}}}}$ are the respective dissociation constants of the first, second and third dissociation steps.
Now, we need to have an idea about hydrolysis. The general hydrolysis reaction for a salt A- of a weak acid HA and strong base is given by:
${{\text{A}}^{\text{ - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{HA + O}}{{\text{H}}^{\text{ - }}}$
The hydrolysis constant is given by ${{\text{K}}_{\text{h}}}{\text{ = }}\dfrac{{\left[ {{\text{HA}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}{{\left[ {{{\text{A}}^{\text{ - }}}} \right]}}$
It is related to the dissociation constant of the acid as ${{\text{K}}_{\text{h}}}{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}$ where ${{\text{K}}_{\text{w}}}$ is the ionic product of water and is equal to ${10^{ - 14}}$ .
Here, the salt is ${{\text{K}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ , the weak acid is ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ and so ${{\text{K}}_{\text{a}}}$ is equal to ${{\text{K}}_{{\text{a3}}}}$ .
So, ${K_h} = \dfrac{{{K_w}}}{{{K_{a3}}}}$
By question, K_{a3}= 1.3\times 10^{-12} .
So, K_{h}= \dfrac{10^{-14}}{1.3\times 10^{-12}} .
If h is the degree of hydrolysis and c is the concentration of the salt, then:
${{\text{A}}^{\text{ - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{HA + O}}{{\text{H}}^{\text{ - }}}$

Before Hydrolysis	c	0	0	0
After Hydrolysis	c-ch	c-ch	ch	ch

And we will have:
$h= \sqrt{\dfrac{K_{h}}{c}} \Rightarrow h= \sqrt{\dfrac{K_{w}}{K_{a}\times c}}$
Also,
$
\left [ OH^{-} \right ]= ch
\Rightarrow \left [ OH^{-} \right ]= c\sqrt{\dfrac{K_{h}}{c}}
\Rightarrow \left [ OH^{-} \right ]= c\sqrt{\dfrac{K_{w}}{K_{a3}\times c}}
\Rightarrow \left [ OH^{-} \right ]= \sqrt{\dfrac{c^{2}\times K_{w}}{K_{a3}\times c}}
\Rightarrow \left [ OH^{-} \right ]= \sqrt{\dfrac{c\times K_{w}}{K_{a3}}}

$Since c is equal to 0.1 M, therefore,$
\left [ OH^{-} \right ]= \sqrt{\dfrac{10^{-14}\times 0.1}{1.3\times 10^{-12}}}
\Rightarrow \left [ OH^{-} \right ]= 0.02773
\Rightarrow pOH=-log\left ( 0.02773 \right )
\Rightarrow pOH=1.56
$Since$
{\text{pH + pOH = 14}}
\Rightarrow {\text{pH = 14 - pOH}}
\Rightarrow {\text{pH = 14 - 1}}{\text{.56}}
\Rightarrow {\text{pH = 12}}{\text{.44}}
$
So the correct option is A.

Note:
Another way to calculate pH is by calculating the hydrogen ion concentration directly from the ionic product of water and the hydroxide ion concentration. This alternative way to calculate pH is shown below:
$
\left [ H^{+} \right ]= \dfrac{K_{w}}{\left [ OH^{-} \right ]}= \dfrac{K_{w}}{ch}
\Rightarrow \left [ H^{+} \right ]= \sqrt{\dfrac{K_{w}\times K_{a3}}{c}}
\Rightarrow pH= -log\sqrt{\dfrac{K_{w}\times K_{a3}}{c}}

$