Question

The third dissociation constant of ${H_3}P{O_4}$ is $1.3 \times {10^{ - 12}}$. Assume that the hydrolysis proceeds only in the first step, the $pH$ of $0.1M$ ${K_3}P{O_4}$ solution is:
(A) $pH = 13.12$
(B) $pH = 12.44$
(C) $pH = 12.64$
(D) $pH = 12.49$

Answer 1

Hint: First we have to find how ${K_3}P{O_4}$ dissociates in its solution. Since this compound and ${H_3}P{O_4}$ contains the same phosphate ion, we can find a relation between the dissociation constants of both the compounds. After getting the dissociation constant of the first hydrolysis step, we can substitute it in the formula for $pH$ to get our answer.

Formulas used: $pH = 7 + \dfrac{1}{2}\left[ {p{K_a} + \log c} \right]$
Where $p{K_a}$ denotes the negative logarithm of the acid dissociation constant and $c$ denotes the concentration of the ion given.

Complete step by step answer:
Let us first see how the given compound ionises:
${K_3}P{O_4} \rightleftharpoons 3{K^ + } + PO_4^{3 - }$
Now, this phosphate ion is what will undergo hydrolysis, and the reaction is as follows:
$PO_4^{3 - } + {H_2}O \rightleftharpoons HPO_4^{2 - } + O{H^ - }$
Now let us find what the third dissociation of ${H_3}P{O_4}$ would produce. Each dissociation of this molecule will remove one hydrogen ion. Therefore, the first and second dissociations are given below:
${H_3}P{O_4} \rightleftharpoons {H_2}PO_4^ - + {H^ + }$
${H_2}PO_4^ - \rightleftharpoons HPO_4^{2 - } + {H^ + }$
Therefore, in the third dissociation, we get:
$HPO_4^{2 - } \rightleftharpoons PO_4^{3 - } + {H^ + }$
When we compare this dissociation with the dissociation of the phosphate ion from the ${K_3}P{O_4}$, we find that the same ions are being involved, that is, $PO_4^{3 - }$ and $HPO_4^{2 - }$. Hence, we can take the third dissociation constant of ${H_3}P{O_4}$ to be equal to the dissociation constant of the ${K_3}P{O_4}$ molecule.
Hence, we get: ${K_a} = 1.3 \times {10^{ - 12}}$
Hence, to find $p{K_a}$, we take the negative logarithm of this value.
$ \Rightarrow p{K_a} = - \log {K_a}$
$ \Rightarrow p{K_a} = - \log (1.3 \times {10^{ - 12}}) = 11.88$
As we know, the formula for finding $pH$ is given as:
$pH = 7 + \dfrac{1}{2}\left[ {p{K_a} + \log c} \right]$
Where $p{K_a}$ denotes the negative logarithm of the acid dissociation constant and $c$ denotes the concentration of the ion given.
Here we are given that the concentration is $c = 0.1M$ and as we found out, $p{K_a} = 11.88$. Substituting these values, we get:
$pH = 7 + \dfrac{1}{2}\left[ {11.88 + \log (0.1)} \right]$
$ \Rightarrow pH = 7 + \dfrac{1}{2}\left[ {11.88 - 1} \right]$
On solving this, we get:
$pH = 7 + 5.44 = 12.44$

So, the correct answer is Option B.

Note: The compound given, ${K_3}P{O_4}$ is the salt of a strong base ($KOH$) and a weak acid (${H_3}P{O_4}$). That is why we have used this particular formula for finding the $pH$. Note that the salt of a strong acid and a strong base will produce a neutral solution when dissolved in water, and the $pH$ will be equal to seven for such solutions.