Question

pH of a 0.01M solution $({{K}_{a}}=6.6\times {{10}^{-4}})$ is:
A. 7.6
B. 8
C. 2.6
D. 5

Answer 1

Hint: Consider all the information that we are given and look for the formulae connecting those aspects. The formula for $pH$ and the formula for ${{K}_{a}}$
\[pH=-\log [{{H}^{+}}]\]
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}\]
Where,
$[{{H}^{+}}]=$ concentration of protons
$[{{A}^{-}}]=$ concentration of anions
$[HA]=$ concentration of solution

Complete step by step solution:
To find the $pH$ of the acid with the given information, we will consider that it is a monoprotic acid and modify the equation for ${{K}_{a}}$ accordingly.
Consider,
\[{{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}\]
Since it is a monoprotic acid, the concentration of the cations and anions will be the same so we can write the equation as
\[{{K}_{a}}=\dfrac{{{[{{H}^{+}}]}^{2}}}{[HA]}\]
Solving the equation for $[{{H}^{+}}]$ , we get,
\[[{{H}^{+}}]=\sqrt{{{K}_{a}}\cdot [HA]}\]
Now putting the given values in this equation, we get $[{{H}^{+}}]$
\[[{{H}^{+}}]=\sqrt{(6.6\times {{10}^{-4}})\cdot (0.01)}\]
\[[{{H}^{+}}]=\sqrt{6.6\times {{10}^{-6}}}\]
\[[{{H}^{+}}]=2.57\times {{10}^{-3}}\]
Now putting this value in the formula,
\[pH=-\log [{{H}^{+}}]\]
\[pH=-\log (2.7\times {{10}^{-3}})\]
\[pH=2.57\]
Therefore, the answer to this question is C. 2.6, by rounding off.

Additional Information:
We are considering that the acid is a monoprotic acid since most acids show such behavior. Any diprotic or polyprotic behaviour is rarely shown since the energy required for that to happen is a lot. Hence, it is safe to assume that it is a monoprotic acid. If it is specifically mentioned in the problem that the given acid is not a monoprotic acid, then the formula for ${{K}_{a}}$ can be modified according to the requirements. The concentration of the $[{{H}^{+}}]$ ions can be twice or thrice than that of the anions.

Note: The key step here is to link both the formulae together and to assume that it is a monoprotic acid. We will not be able to solve the problem if we do not assume this.
Finding the $pH$ is possible even when the ${{K}_{b}}$ of the compound is given. Just use the formulae
\[{{K}_{b}}=\dfrac{[{{B}^{+}}][O{{H}^{-}}}{[BOH]}\],
\[pOH=-\log [O{{H}^{-}}]\], and
\[pH=14-pOH\]
Rearrange these formulae as required and the $pH$ can be calculated.