Question

pH of a strong diprotic acid $({{H}_{2}}A)$ at concentration are, respectively:
(i) ${{10}^{-4\,}}M$ (ii) ${{10}^{-4}}N$
(A) $3.7\,and\,4$
(B) $4\,\,and\,3.7$
(C) $4\,and\,4$
(D) $3.7\,and\,3.7$

Answer 1

Hint: The pH which measures the extent of acidity of acid, that is, its ease to give away its protons in water. Also, in the normality of acid in relation to the molarity, the number of moles of protons provided in the reaction is considered before calculating the pH.

Complete step by step solution:
The diprotic acid is an acid which has two ionizable hydrogen atoms. It donates its two hydrogen ions when dissolved in water. Thus, it is also called dibasic acid. All diprotic acids have two protons per molecule.
In order to calculate the pH of protic acid, we use the formula:
\[pH=-\log \left[ {{H}^{+}} \right]\]
(i) Given the concentration of acid, $\left[ {{H}_{2}}A \right]$ as\[\text{1}{{\text{0}}^{\text{-4}}}\text{ M}\],where it produces two moles of hydrogen ions on dissolution. The concentration of $\left[ {{H}^{+}} \right]$ becomes $2\times {{10}^{-4}}M$.
Then, putting the value of $\left[ {{H}^{+}} \right]$ in the formula for pH, we get,
\[pH=-\log \left[ \text{2}\times \text{1}{{\text{0}}^{\text{-4}}} \right]\]
\[=-\log 2-(-4\log 10)\]
\[=-\log 2+(4\log 10)\]
\[=-0.301+4\,\,=3.699\sim \,3.7\,\]

(ii) Also, for the concentration of the acid, $\left[ {{H}_{2}}A \right]$ given as ${{10}^{-4}}\,N$, the relation between normality and molarity is as follows: $Normality\,=\,n\,\,\times \,Molarity$, where n is the number of moles of hydrogen ions formed in the reaction. Therefore, n = 2.
Then, the concentration of acid is,
\[\text{Molarity= }\dfrac{\text{Normality}}{\text{n}}\text{=}\dfrac{\text{1}{{\text{0}}^{\text{-4}}}}{\text{2}}\text{M}\]
We can then, get the concentration of $\left[ {{H}^{+}} \right]$ to be simply ${{10}^{-4}}\,M$and
\[pH=-\log \left[ {{10}^{-4}} \right]\]
\[pH=\,\,4\log 10\,\,=4\]
Therefore, we get the pH of a strong diprotic acid for concentration ${{10}^{-4}}\,M$ and ${{10}^{-4}}\,N$ as 3.7 and 4 respectively.

That is, option (A).

Note: The ease with which these acids lose their first and second (or third and so on) protons has a large difference. For example, in the diprotic acid the first dissociation constant, ${{K}_{a1}}$ is larger than 1, whereas, for the second proton loss given by second dissociation constant ${{K}_{a2}}$ is minimal.
Therefore, ${{K}_{a1}}>\,{{K}_{a2}}>\,{{K}_{a3}}$ is found true for polyprotic acids, wherewith each ionization step the removal of $({{H}^{+}})$ from the molecule becomes difficult as the negative charge increases.
But, in case of ionic salts of these acids, they dissociate $100%$ in a single step.