Explanation:
The given reaction will be as follows.
[tex]HX + NaOH \rightarrow NaX + H_{2}O[/tex]
Initial : 2.2 1.15 0 0
At equili:(2.2 - 1.15) 0 1.15 1.15
As moles of [HX] = (2.2 - 1.15) = 1.05
Moles of [NaX] = 1.15
So, relation between pH and [tex]pK_{a}[/tex] will be as follows.
pH = [tex]pK_{a} + \frac{log[NaX]}{[HX]}[/tex]
[tex]pK_{a} = pH - log \frac{log[NaX]}{[HX]}[/tex]
= [tex]5.2 - log \frac{log (1.15)}{(1.05)}[/tex]
[tex]K_{a}[/tex] = [tex]6.902 \times 10^{-6}[/tex]
Thus, we can conclude that [tex]K_{a}[/tex] for the acid HX is [tex]6.902 \times 10^{-6}[/tex].
