Question

A weak acid HA has pH = 4. This can confirm: (i) $C = {{10}^{-3}},\alpha $ = 10 %, (ii) $C={{10}^{-3}},{{K}_{a}}={{10}^{-6}}$ (iii) $\left| {{A}^{-}} \right|={{10}^{-4}}$
A. Only (i) is correct
B. both (i) and (ii) are correct
C. (i),(ii),(iii) are correct
D. All are correct

Answer 1

Hint:. The concept of dissociation of weak acids is to be used in this question. By knowing the value of concentration and dissociation constant of the acid, we can find out the concentration of ${{H}^{+}}$

Complete step by step answer:
In order to answer this question, we need to learn about the acids and their dissociation. According to Arrhenius concept they are strong acids and bases as they are able to completely dissociate and produce ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ ions respectively in the medium. Alternatively, the strength of an acid or base may also be gauged in terms of Bronsted-Lowry concept of acids and bases, where in a strong acid means a good proton donor and a strong base implies a good proton acceptor. Consider the acid-base dissociation equilibrium of a weak acid, HA.
\[HA+{{H}_{2}}O(l)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+{{A}^{-}}(aq)\]
- The acid or base dissociation equilibrium is dynamic involving a transfer of protons in forward and reverse directions. Now, we have been given that we have a weak acid whose pH = 4. Now, $pH = -\log [{{H}^{+}}]$. So, we can write
\[\begin{align}
 & \log [{{H}^{+}}]=-4 \\
 & [{{H}^{+}}]={{10}^{-4}}M \\
\end{align}\]
Now, let us see the dissociation of the weak acid:
\[HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}\]
- Now, at t = o, the concentration of HA is c and both ${{H}^{+}}$ and ${{A}^{-}}$ have 0 concentration. As the equilibrium is reached, we have at the time of equilibrium , concentration of HA as $c-c\alpha $ and concentrations of ${{H}^{+}}$ and ${{A}^{-}}$ both as $c\alpha $.
- Now, $[{{H}^{+}}]=c\alpha ={{10}^{-4}}$. In option (i), we have $C={{10}^{-3}},\alpha = $10 %, as their product gives the answer as ${{10}^{-4}}$, hence the option is correct. Also, ${{K}_{a}} = c{{\alpha }^{2}}$, so $\alpha =\sqrt{\dfrac{{{K}_{a}}}{C}}$. So, $[{{H}^{+}}] = c\alpha = c\sqrt{\dfrac{{{K}_{a}}}{c}}=\sqrt{{{K}_{a}}\times c}$, so from the data given in the question, we have $[{{H}^{+}}]=\sqrt{{{10}^{-6}}\times {{10}^{-3}}}=\sqrt{{{10}^{-9}}}$, but it is not the same concentration as LHS, so it is incorrect option.
- Now, we can see from the dissociation equation of the acid that both $[{{H}^{+}}]$ and $[{{A}^{-}}]$ have concentration as $c\alpha $. As their concentrations are same and $[{{H}^{+}}]={{10}^{-4}}$, so accordingly, we also can write that $[{{A}^{-}}]={{10}^{-4}}$.
Hence, options (i) and (iii) are the correct options in the question.
So, the correct answer is “Option i and iii”.

Note: In case of a weak electrolyte, at a given temperature, the degree of ionisation is inversely proportional to the square root of the molar concentration or directly proportional to square root of the volume of the solution which contains one mole electrolyte. This is called Ostwald's dilution law.