The equivalent point in a titration of `40.0mL` of a sodium of a weak monoprotic acid occurs when `35.0mL` of a `0.10M NaOH` solutio has been added. The `pH` of the solution is 5.5 after the addition of `20.0mL` of `NaOH` solution. What is the dissociation constant of the acid ?

Explanation

Let `x` mmol of `HA` is taken initially. Find `x`.
Consider the equation of neutralisation:
`{:(HA+,NaOHrarr,NaA +,H_(2)O,),(x,y,-,,),(x-y,-,y,,):}`
At equivalent point, millimoles of acid `=` millimoles of `NaOH` [i.e., `x = y]`
`rArr` millimoles `HA = x = 3.5 [35 mL of 0.1M NaOH]`
At` pH = 5.5, mmol` of `NaOH` added `= 2 = y [20mL` of `0.1M NaOH]`
`rArrr mmol` of `HA` left `= x - y = 1.5`
and mmol of `NaA` formed `= 2`
Now such a solution will behave as an acidic buffer, whose `pH` si given as:
`[H^(o+)] = (K_(a)["acid"])/(["salt"])rArr 10^(-5.5) = K_(a) (1.5)/(2) [10^(-5.5) = sqrt(10) xx 10^(-6)]`
`rArr K_(a) = 4.22 xx 10^(-6)`