Question

What is the pH of 0.2 M aqueous solution of sodium butyrate$\left( {{\text{K}}_{\text{a}}}=2.0\times {{10}^{-5}} \right)$?

Answer 1

This question deals with the hydrolysis of salt of weak acid with strong base. The strong base here is $\text{NaOH}$, the weak acid is but-1-yne and the salt of weak acid is sodium butyrate. The reaction of hydrolysis is:

Complete answer:
Let us solve this question step by step like derivation of hydrolysis of salt of weak acid and strong base.
Step (1) - Write the reaction and introduce degree of hydrolysis and hydrolysis constant:
Reaction is:

| $\text{NaBu}+{{\text{H}}_{2}}\text{O}\rightleftharpoons \text{NaOH}+\text{BuH}$
---|---
Conc. before hydrolysis| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 0 \\\
Conc. after hydrolysis| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1-h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ h\\\

h is the degree of hydrolysis which represents the extent of hydrolysis.
${{\text{K}}_{\text{h}}}$ is like other dissociation or equilibrium constants which defines equilibrium of hydrolysis.
It is expressed as ${{\text{K}}_{\text{h}}}=\dfrac{\left[ \text{NaOH} \right]\left[ \text{BuH} \right]}{\left[ \text{NaBu} \right]}$.
Step (2) - In this question, we have to find pH.
For that find the ${{\text{K}}_{\text{h}}}$ of the reaction using the formula,${{\text{K}}_{\text{h}}}=\dfrac{{{\text{K}}_{\text{w}}}}{{{\text{K}}_{\text{a}}}}$,
Where ${{\text{K}}_{\text{h}}}$ is the hydrolysis constant, ${{\text{K}}_{\text{w}}}$ is the ionic product of water and ${{\text{K}}_{\text{a}}}$ is the dissociation constant of the weak acid.
The value of ${{\text{K}}_{\text{w}}}$ is fixed at standard temperature of ${{25}^{\text{o}}}\text{C}$.
This constant value is ${{10}^{-14}}$.
So, the hydrolysis constant will be
${{\text{K}}_{\text{h}}}=\dfrac{{{10}^{-14}}}{2\times {{10}^{-5}}}$ Or, ${{\text{K}}_{\text{h}}}$ is equal to $5\times {{10}^{-10}}$.
Step (3) - We have calculated the hydrolysis constant or ${{\text{K}}_{\text{h}}}$,
Now find the concentration of $\left[ \text{O}{{\text{H}}^{-}} \right]$ using the formula $\left[ \text{O}{{\text{H}}^{-}} \right]=\sqrt{{{\text{K}}_{\text{h}}}\times \text{c}}$
Where $\left[ \text{O}{{\text{H}}^{-}} \right]$ is the concentration of hydroxide ion, c is the concentration of the salt which is given as 0.2 M and the hydrolysis constant will be ${{\text{K}}_{\text{h}}}=5\times {{10}^{-10}}$.
$\left[ \text{O}{{\text{H}}^{-}} \right]$ = $\sqrt{0.2\times 5\times {{10}^{-10}}}$ = ${{10}^{-5}}\text{ M}$.
Step (4) - Calculate the pOH of the base using the formula,
$\text{pOH=}-\text{log}\left[ \text{O}{{\text{H}}^{-}} \right]$ , and $\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{-5}}\text{M}$,
The pOH will be,
$-\text{log}\left[ {{10}^{-5}} \right]=5$ (as log(10) is 1).
Step (5) - Calculate the concentration of $\left[ {{\text{H}}^{+}} \right]$ ion from the formula
pH= 14-pOH, pOH is 5.
The pH will be 14-5 = 9.
Hence the pH is 9.

Note:
These formulae are to be used when the hydrolysis of salt is considered to be negligible. Here, butyric acid is a weak acid because of the long alkyl chain. So, we can consider the hydrolysis of its salt to be very less. We can consider hydrolysis negligible till the time it is not mentioned to consider it and the compound is acidic.