Answer:
The answer is 2.2
Explanation:
Since the above substance is a weak acid it's pH can be found by using the formula
[tex]pH = \frac{1}{2} ( - log(Ka ) - log(c) )[/tex]
where
Ka is the acid dissociation constant
c is the concentration of the acid
From the question
Ka = 4.0 × 10-⁴
c = 0.1 M
So we have
[tex]pH = \frac{1}{2} ( - log(4.0 \times {10}^{ - 4} ) - log(0.1) ) \\ = \frac{1}{2} (3.4 + 1) \\ = \frac{1}{2}(4.4) \: \: \: \: \: \: \: \: [/tex]
We have the final answer as
2.2
Hope this helps you
