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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...

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

Explanation

Solution

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

Complete step by step solution:
To find the pHpH of the acid with the given information, we will consider that it is a monoprotic acid and modify the equation for Ka{{K}_{a}} accordingly.
Consider,
Ka=[H+][A−][HA]{{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
Ka=[H+]2[HA]{{K}_{a}}=\dfrac{{{[{{H}^{+}}]}^{2}}}{[HA]}
Solving the equation for [H+][{{H}^{+}}] , we get,
[H+]=Ka⋅[HA][{{H}^{+}}]=\sqrt{{{K}_{a}}\cdot [HA]}
Now putting the given values in this equation, we get [H+][{{H}^{+}}]
[H+]=(6.6×10−4)⋅(0.01)[{{H}^{+}}]=\sqrt{(6.6\times {{10}^{-4}})\cdot (0.01)}
[H+]=6.6×10−6[{{H}^{+}}]=\sqrt{6.6\times {{10}^{-6}}}
[H+]=2.57×10−3[{{H}^{+}}]=2.57\times {{10}^{-3}}
Now putting this value in the formula,
pH=−log⁡[H+]pH=-\log [{{H}^{+}}]
pH=−log⁡(2.7×10−3)pH=-\log (2.7\times {{10}^{-3}})
pH=2.57pH=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 Ka{{K}_{a}} can be modified according to the requirements. The concentration of the [H+][{{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 pHpH is possible even when the Kb{{K}_{b}} of the compound is given. Just use the formulae
Kb=[B+][OH−[BOH]{{K}_{b}}=\dfrac{[{{B}^{+}}][O{{H}^{-}}}{[BOH]},
pOH=−log⁡[OH−]pOH=-\log [O{{H}^{-}}], and
pH=14−pOHpH=14-pOH
Rearrange these formulae as required and the pHpH can be calculated.