Question

For $0.1\;{\text{M}}$ weak acid $HA\;(pH = 3)$ is titrated with $0.05\;{\text{M }}NaOH$ solution. Calculate the pH (approx.) when $25\% $ of acid has been neutralized. ($\log 3 = 0.48$)
A. $4.52$
B. $5.41$
C. $4$
D. $3.52$

Answer 1

Hint: A chemical reaction in which an acid quantitatively reacts with a base without any presence of external hydrogen and hydroxide ions, then the formation of respective salt and water takes place and the reaction is known as neutralization reaction.

Complete answer:
Acid-base titrations are usually used to calculate the amount of a known acid or base with the help of neutralization reactions.
As per question, the given data is as follows:
Initial concentration of weak acid $HA = 0.1{\text{ M}}$
Initial concentration of $NaOH = 0.05{\text{ M}}$
Dissociation of acid after neutralization $ = 25\% $
Calculation of the value of dissociation constant at initial conditions:
The weak acid dissociated as follows:
\[HA \rightleftharpoons {H^ + } + {A^ - }\]
As the value of pH $ = 3$, so the concentration of the hydrogen ion for the given weak acid can be determined as follows-
$pH = - \log \left[ {{H^ + }} \right]$
$ \Rightarrow - \log \left[ {{H^ + }} \right] = 3$
Taking antilog on both sides of the equation.
$ \Rightarrow {H^ + } = {10^{ - 3}}$
Therefore, the value of acid dissociation constant will be as follows:
${K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
Substituting values:
${K_a} = \dfrac{{{{10}^{ - 3}} \times {{10}^{ - 3}}}}{{0.1}}$
$ \Rightarrow {K_a} = {10^{ - 5}}$
Now, it is given that $25\% $ acid has been neutralized. That means $\dfrac{1}{4}th$ of the acid dissociates on reacting it with sodium hydroxide. The equation for neutralization reaction is given as follows:
$HA + NaOH \rightleftharpoons NaA + {H_2}O$
The initial concentrations and change in concentration after neutralization reaction of the weak acid and its salt can be represented as per following table:

	$\left[ {HA} \right]$	$\left[ {NaA} \right]$
Initial concentration	$0.1$	$0$
Change	$ - 0.1 \times \dfrac{1}{4}$	$0.1 \times \dfrac{1}{4}$
Final concentration	$0.1 - 0.1 \times \dfrac{1}{4} = 0.075$	$0.025$

Therefore, the pH of the solution after the neutralization reaction can be determined as following expression:
$pH = p{K_a} + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$
$ \Rightarrow pH = - \log ({K_a}) + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$
Substituting values:
$ \Rightarrow pH = - \log \left( {{{10}^{ - 5}}} \right) + \log \dfrac{{0.025}}{{0.075}}$
$ \Rightarrow pH = 5 - 0.477$
$ \Rightarrow pH = 4.523$
Hence, the pH of the solution after neutralization of weak acid with $NaOH = 4.52$.
So, option (A) is the correct answer.

Note:
It is important to note that the pH of the acid-base titrations can alternatively be calculated with the help of base dissociation constant if the pOH of the base is provided. The expression for the calculation is $pOH = p{K_b} + \log \dfrac{{[salt]}}{{[base]}}$. Further, the pH is calculated from the value of pOH with the help of an expression that is $pH = 14 - pOH$.