Question

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.

Answer 1

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
`K_(a)` for the acid HX is
`6.902 * 10^(-6)`.