A 2.35L solution contains 2.20 mol of weak acid HX. If 1.15 mol NaOH is added to this solution, the pH of the resultant solution is 5.20. Calculate Ka for acid HX.

Question

asked Oct 10, 2020 84.3k views

1 Answer

Walv · Answer 1 · 2020-10-14T15:03:08+0000

Step-by-step explanation:

The given reaction will be as follows.

$HX + NaOH \rightarrow NaX + H_(2)O$

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
pK_(a) will be as follows.

pH =
pK_(a) + (log[NaX])/([HX])

pK_(a) = pH - log (log[NaX])/([HX])

=
5.2 - log (log (1.15))/((1.05))

K_(a) =
6.902 * 10^(-6)

Thus, we can conclude that
for the acid HX is
.