Question

The dissociation constant of a weak acid is ${{10}^{-6}}$ . Then the pH of 0.01 N of that acid is
A. 2
B. 7
C. 8
D. 4

Answer 1

The dissociation constant of an acid is going to represent with a symbol ${{K}_{a}}$ . By using acid dissociation constant we can measure the strength of the acid in a solution. There is a formula to calculate the pH of an acid by using the below formula
$pH=\dfrac{1}{2}[{{p}^{{{K}_{a}}}}-\log ({{H}^{+}})]$
pH = pH of the solution
${{p}^{{{K}_{a}}}}$ = negative logarithm of acid dissociation constant.

Complete answer:
- In the question it is asked to calculate the pH of the 0.01 N of an acid whose dissociation constant is ${{10}^{-6}}$ .
- There is a formula to calculate the pH of the solution with the given data.
$pH=\dfrac{1}{2}[{{p}^{{{K}_{a}}}}-\log ({{H}^{+}})]$
pH = pH of the solution
${{p}^{{{K}_{a}}}}$ = negative logarithm of acid dissociation constant.
- In the question the given data is ${{K}_{a}}={{10}^{-6}}$ , concentration of the acid is 0.01 N.
- Substitute all the known values in the above formula to get the pH of the given acid and it is as follows.

& {{p}^{{{K}_{a}}}}=-\log ({{k}_{a}}) \\\ & {{p}^{{{K}_{a}}}}=-\log ({{10}^{-6}}) \\\ & {{p}^{{{K}_{a}}}}=6 \\\ \end{aligned}$$ \- Substitute the above ${{p}^{{{K}_{a}}}}$ value in the above formula to get the pH of the acid. $$\begin{aligned} & pH=\dfrac{1}{2}[{{p}^{{{K}_{a}}}}-\log ({{H}^{+}})] \\\ & pH=\dfrac{1}{2}[6-\log (0.01)] \\\ & pH=4 \\\ \end{aligned}$$ \- Therefore the pH of the acid is 4. **So, the correct option is D.** **Note:** By using ${{p}^{{{K}_{a}}}}$ value we can determine the strength of the acid. If ${{p}^{{{K}_{a}}}}$ value of an acid is too high then the acid is very weak and the ${{p}^{{{K}_{a}}}}$ value of any acid is very less then the acid is a strong acid. Means ${{p}^{{{K}_{a}}}}$ value and strength are inversely proportional to each other.