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Calculate pH of a solution containing 0.1M $C{{H}_{3}}COOH$ and 0.1M benzoic acid. ${{K}_{a}}$ for $C{{H}_{3}}COOH$ and benzoic acid are $1.8\times {{10}^{-5}}$and $6.5\times {{10}^{-5}}$ respectively.

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Last updated date: 11th Jun 2024
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Answer
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Hint: pH is used to measure hydrogen ion concentration in an aqueous solution. The pH scale ranges from 0 to 14. Whereas if a compound having pH value less than 7 they are acidic in nature and higher than 7 indicates basic compounds whereas pH exactly 7 is neutral in nature.

Complete Solution :
The pH value helps us to determine whether it is an acid or a base. Whereas ${{K}_{a}}$ and ${{K}_{b}}$ values helps us to determine whether the given species will donate or accept protons at a specific pH value. ${{K}_{a}}$ is acid dissociation constant while ${{K}_{b}}$ is base dissociation constant.
- Both the compounds given in the question are acidic in nature. Acetic acid dissociates into:
$C{{H}_{3}}COOH\rightleftarrows C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$

- In the similar manner benzoic acid dissociates into:
${{C}_{6}}{{H}_{5}}COOH\rightleftarrows {{C}_{6}}{{H}_{5}}CO{{O}^{-}}+{{H}^{+}}$

Now we know the formula of calculating ${{H}^{+}}$which is as shown below:
$[{{H}^{+}}]=\sqrt{{{K}_{1}}{{C}_{1}}+{{K}_{2}}{{C}_{2}}}$
Here value of ${{C}_{1}}=0.1,{{C}_{2}}=0.1,{{K}_{1}}=1.8\times {{10}^{-5}},{{K}_{2}}=6.5\times {{10}^{-5}}$

- By putting all the values in the formula
$[{{H}^{+}}]=\sqrt{1.8\times {{10}^{-5}}\times 0.1+6.5\times {{10}^{-5}}\times 0.1}$
$[{{H}^{+}}]=\sqrt{1.8\times {{10}^{-6}}+6.5\times {{10}^{-6}}}$
$[{{H}^{+}}]=\sqrt{8.3\times {{10}^{-6}}}$
\[[{{H}^{+}}]=2.88\times {{10}^{-3}}\]
\[pH=-\log [H+]=-\log (2.88\times {{10}^{-3}})\]
 \[=3-\log 2.88=2.540\]
Hence the pH of a solution containing 0.1M $C{{H}_{3}}COOH$ and 0.1M benzoic acid. ${{K}_{a}}$ for $C{{H}_{3}}COOH$ and benzoic acid are $1.8\times {{10}^{-5}}$ and $6.5\times {{10}^{-5}}$ respectively is 2.540.

Note: ${{K}_{a}}$ and ${{K}_{b}}$ describe the degree of ionization of an acid or base and are true indicators of acid or base strength because adding water to a solution will not change the equilibrium constant. ${{K}_{a}}$ is related to acids while ${{K}_{b}}$ represents the base.