Question

A $50{\text{mL}}$ of $0.1{\text{M}}$ ${\text{HCl}}$ is titrated with $0.1{\text{M}}$ ${\text{NaOH}}$ , at ${\text{pH}} = 3$ . Volume of ${\text{NaOH}}$ used is approximately:
A.$49{\text{mL}}$
B.$50{\text{mL}}$
C.$45{\text{mL}}$
D.$41{\text{mL}}$

Answer 1

Here we need to first calculate the equivalent of acid and base by using the formula given. By comparing equivalents, we can predict the nature of the mixture.
Formula Used: ${\text{no of equivalent}} = {\text{normality}} \times \dfrac{{{\text{volume}}\left( {{\text{inmL}}} \right)}}{{1000}}$

Complete step by step answer:
To determine the pH of mixture of strong acid and strong base, some step should be followed to easily determine the pH of mixture:
First, calculate the number of equivalents or milli equivalent of strong acid and strong base given. Now compare the number of equivalents of acid and base to decide the nature of the resultant solution. If equivalents of acids are greater than equivalents of base then the solution will be acidic and the extra equivalent will be of hydrogen ion. If equivalents of base are more than equivalents of acid then the solution will be basic and the extra equivalent will be of hydroxide ion.
The number of equivalents present in $50{\text{mL}}$ of $0.1{\text{M}}$ ${\text{HCl}}$ can be given as: ${\text{no of equivalent}} = 0.1 \times \dfrac{{50}}{{1000}} = 0.005$ and on the other hand, number of equivalent in V $\left( {{\text{assumed}}} \right)$ volume of $0.1{\text{M}}$ ${\text{NaOH}}$ can be given as: ${\text{no of equivalent}} = 0.1 \times \dfrac{{\text{V}}}{{1000}} = 0.000{\text{V}}$ . As it is given that pH of mixture is 3 means acidic, therefore equivalents of acid is more than equivalent of base. The $0.005$ equivalent of acid will completely neutralize the $0.000{\text{V}}$ equivalent of base and the remaining equivalent of acid will correspond to hydrogen ion leftover in the reaction mixture. Due to this leftover hydrogen ion, pH of mixture is acidic that is 3.
As pH is equal to 3 and as we know it is negative logarithm of hydrogen ion. Therefore, final hydrogen ion in mixture can be given as:
${\text{pH}} = - {\text{log}}\left( {{{\text{H}}^ + }} \right)$
$\left[ {{{\text{H}}^ + }} \right] = {10^{ - {\text{pH}}}} = {10^{ - 3}}$ .
Equivalent of hydrogen ion can be given as:
${\text{N}} \times {\text{total volume}} = 0.005 - 0.000{\text{V}}$
${10^{ - 3}} \times \left( {\dfrac{{50 + {\text{V}}}}{{1000}}} \right) = 0.005 - 0.000{\text{V}}$
On calculating, V is equal to $49{\text{mL}}$ .

Thus, the correct option is A.

Note:
Here we calculate the hydrogen ion present in a solution which has a pH equals to 3 and then we compare it with the hydrogen which will be leftover by the mixing of more number of equivalents of acid then equivalents of base. We are considering equivalence because it accounts for the real hydrogen ion or hydroxide ion produced by the acid and base respectively.