Question

What is the pH of the solution when 0.2mole of hydrochloric acid is added to one litre of a solution containing 1M acetic acid and acetate ion? Assume that the total volume is one litre.(\[{K_a}\] for \[C{H_3}COOH\]\[ = 1.8 \times {10^{ - 5}}\])

Answer 1

Hint: Mixture of acetic acid and acetate ions is called mixture of acid and its conjugate base and to find pH of these mixtures, Henderson-Hasselbalch equation is used which is as below.
\[pH = p{K_a} + \log \left( {\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}} \right)\]

Complete answer:
Remember that in the presence of strong acid, the weak acid does not dissociate into ions.
So, we have 1M acetic acid present in the solution but as we will add hydrochloric acid, the respective amount of acetate ions will turn into acetic acid molecules. This reaction can be given as,
\[\mathop {C{H_3}CO{O^ - }}\limits_{0.2mol} + \mathop {HCl}\limits_{0.2mol} \to \mathop {C{H_3}COOH}\limits_{0.2mol} + C{l^ - }\]
So, Concentration of acetate ions present in the solution after the addition of hydrochloric acid can be give by
Concentration of acetate ions = Initial concentration – Moles of acetate ions converted to acetic acid
Concentration of acetate ions = 1M - 0.2M
Concentration of acetate ions = 0.8M
Same way, concentration will increase because of the addition of acetic acid molecules into the solution.
Concentration of acetic acid = 1.0M + 0.2M
Concentration of acetic acid =1.2M
We know that the pH of a weak acid and its conjugate base mixture can be obtained by the Henderson-Hasselbalch equation which is written as below.
\[pH = p{K_a} + \log \left( {\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}} \right)\]..................(1)
Now we will put all the available values into equation (1) where concentration of salt is equal to concentration of acetate ions and concentration of acid is equal to concentration of acetic acid.
\[pH = - \log (1.8 \times {10^{ - 5}}) + \log \left( {\dfrac{{\left[ {0.8} \right]}}{{\left[ {1.2} \right]}}} \right)\]
\[pH = - \log (1.8 - 5\log 10) + \log \dfrac{{0.8}}{{1.2}}\]
\[pH = - (0.2552 - 5) + \log 8 - log12\]
\[pH = - 0.2552 + 5 + 0.9031 - 1.0791\]
\[pH = 4.5688\]
So, we can say that the pH of the resulting solution will be equal to 4.5688.

Additional Information:
-Note that Hydrochloric acid and Sulfuric acid are known as strong acids. Acetic acid is a type of a weak acid.
- Weak acid always dissociates into its salt forms according to its dissociation constant. Dissociation constant change with changes in temperature.

Note:
Do not just add up the protons produced by the given amount of acetic acid and hydrochloric acid in order to find the pH. Remember that the weak acid always gets protonated when any strong acid gets dissociated in the solution.