Question

pH of a strong diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ at
concentrations:
(i) ${10^{ - 4}}{\text{ M}}$
(ii) ${10^{ - 4}}{\text{ N}}$
are respectively:
(A) ${\text{3}}{\text{.7}}$ and ${\text{4}}{\text{.0}}$
(B) ${\text{4}}$ and ${\text{3}}{\text{.7}}$
(C) ${\text{4}}$ and ${\text{4}}$
(D) ${\text{3}}{\text{.7}}$ and ${\text{3}}{\text{.7}}$

Answer 1

Hint: We know that pH measures the acidity of an acid or it measures how easily an acid gives away a proton. In the normality, the number of moles of the protons provided in the reaction are considered for calculating the pH.

Complete step by step solution:
We are given a diprotic acid. Diprotic acid means there are two ionisable hydrogen atoms and the acid donates two protons when dissolved in water. It is also known as diprotic acid.
We know that pH is the negative logarithm of the hydrogen ion concentration. Thus,
${\text{pH}} = - \log [{{\text{H}}^ + }]$
The concentration of diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ is ${10^{ - 4}}{\text{ M}}$:
The given diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ on dissociation produces two moles of hydrogen ion. Thus, the concentration of hydrogen ions is $2 \times {10^{ - 4}}{\text{ M}}$.
Thus, the pH of ${10^{ - 4}}{\text{ M}}$ diprotic acid is,
${\text{pH}} = - \log [2 \times {10^{ - 4}}]$
${\Rightarrow \text{pH}} = 3.7$
Thus, pH of a strong diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ at concentration ${10^{ - 4}}{\text{ M}}$ is ${\text{3}}{\text{.7}}$.
The concentration of diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ is ${10^{ - 4}}{\text{ N}}$:
The relation between normality and molarity is as follows:
\[{\text{Normality}} = n \times {\text{Molarity}}\]
Where \[n\] is the number of moles of hydrogen ions formed.
Thus,
\[{\text{Molarity}} = \dfrac{{{\text{Normality}}}}{n}\]
\[\Rightarrow {\text{Molarity}} = \dfrac{{{\text{1}}{{\text{0}}^{ - 4}}{\text{ N}}}}{2}\]
\[\Rightarrow {\text{Molarity}} = 5 \times {10^{ - 5}}{\text{ M}}\]
The given diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ on dissociation produces two moles of hydrogen ion. Thus, the concentration of hydrogen ions is \[5 \times {10^{ - 5}}{\text{ M}}\].
Thus, the pH of \[5 \times {10^{ - 5}}{\text{ M}}\] diprotic acid is,
${\text{pH}} = - \log [5 \times {10^{ - 5}}]$
$\Rightarrow {\text{pH}} = 4$
Thus, pH of a strong diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ at concentration ${10^{ - 4}}{\text{ N}}$ is ${\text{4}}{\text{.0}}$.

Thus, the pH of a strong diprotic acid $\left( {{{\text{H}}_{\text{2}}}{\text{A}}} \right)$ at concentrations ${10^{ - 4}}{\text{ M}}$ and ${10^{ - 4}}{\text{ N}}$ is ${\text{3}}{\text{.7}}$ and ${\text{4}}{\text{.0}}$ respectively.

Thus the correct option is (A).

Note: The diprotic acid loses its first proton more easily than it loses its second proton. With each ionization step, the removal of protons becomes difficult. This is because the negative charge increases.