Question

The dissociation constant of a substituted benzoic acid at 25⁰C is $1\times {{10}^{-4}}$. The pH of a 0.01 M solution of its sodium salt is :

Answer 1

Hint: Sodium salt is a mixture of strong base and weak acid salt. And it is salt of benzoic acid. In the question the dissociation constant of benzoic acid $({{C}_{6}}{{H}_{5}}COOH)$ is given and here we have solve the pH of its sodium salt $({{C}_{6}}{{H}_{5}}COONa)$.Use the hydrolysis reaction of conjugate base of acid . The formula of pH is given as $pH=-\log [{{H}^{+}}]$.

Complete answer:
From your chemistry lessons you have learned about the hydrolysis of salt.
Hydrolysis of salt is a type of reaction in which one ions from a given salt react with water to form an acidic or basic solution.
In this question we are reacting a weak acid (benzoic acid) with a strong base which yields a solution that is weakly basic. Here a solution of weak acid reacts with a strong base solution which forms the conjugate base of the weak acid and conjugate acid of the strong base. In comparison to water the conjugate acid of a strong base is weaker and thus it shows any effect on the acidity of the solution.
Whereas the conjugate base of weak acid has the tendency to get slightly ionized in water and due to this the amount of hydroxide increases in the solution produced during the reaction and thus makes it slightly basic.
The reaction hydrolysis of a conjugate base of weak acid is given as,
\[{{A}^{-}}(aq)+{{H}_{2}}O\to H{{O}^{-}}+HA\]
From the question the degree of dissociation of weak acid (benzoic acid ) is given as $1\times {{10}^{-4}}$ and we have to find the pH of 0.01 M ${{C}_{6}}{{H}_{5}}COONa$ (conjugate base)
So the hydrolysis reaction of conjugate base (${{C}_{6}}{{H}_{5}}COONa$ )will be,

                   \[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}+{{H}_{2}}O\to {{C}_{6}}{{H}_{5}}COOH+O{{H}^{-}}\]
Initial conc. 0.01 0 0
Final conc. 0.01(1-h) 0.01h 0.01h (h is the dissociation constant)
We know that ${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}=\dfrac{{{10}^{-14}}}{{{10}^{-4}}}={{10}^{-10}}$
Where ${{K}_{h}}$= hydrolysis constant
 ${{K}_{w}}$ = ionic product of water
  ${{K}_{a}}$ =ionization constant of weak acid
Here, $[O{{H}^{-}}]=0.01h=0.01\times {{10}^{-4}}={{10}^{-6}}M$
As we know that $[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}$
So, $[{{H}^{+}}][{{10}^{-6}}]={{10}^{-14}}$
Therefore, $[{{H}^{+}}]={{10}^{-8}}M$
Now we have to find the pH, so from the Arrhenius theorem we know,
\[pH=-\log [{{H}^{+}}]\]
\[pH=-\log [{{10}^{-8}}]=8\]
Thus the pH will be 8.

Note:
As you know that sodium salt is a strong base weak acid salt of benzoic acid so you can also find the pH as, $pH=\dfrac{(p{{K}_{w}}+p{{K}_{a}}+\log C)}{2}$. The ionic product of water which is ${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$. Hydrolysis constant of water is related to the ionic product of water and ionization constant of the acid.