equilibrium2

Firstly, consider the titration to the equivalence point (30.40 cm3 of NaOH added):

HX + NaOH → NaX + H2O
Mole ratio HX : NaOH = 1 : 1
No. of moles NaOH used = MV = 0.25 x (30.4/1000) = 0.0076 mol
No. of moles of HX in the 25 cm3 of HX solution = 0.0076 mol

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Consider the point when 5.0 cm3 of NaOH is added, and neglect the dissociation of HX.

HX + NaOH → NaX + H2O
Mole ratio HX : NaOH : NaX= 1 : 1 : 1
No. of moles of HX added = 0.0076 mol
No. of moles of NaOH added = MV = 0.25 x (5/1000) = 0.00125 mol

HX is in excess. All NaOH is reacted with part of HX to give NaX.
No. of moles of HX left = 0.0076 - 0.00125 = 0.00635 mol
No. of moles of NaX formed = 0.00125 mol

Final volume of the solution = 25 + 5 = 30 cm3 = 0.03 dm3
Molarity of HX left = mol/V = 0.00635/0.03 = 0.2117 M
Molarity of X- = Molarity of NaX = mol/V = 0.00125/0.03 = 0.04167 M

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Consider the dissociation of HX in the presence of X-.

[HX]o = 0.2117 M
[X-]o = 0.04167 M

HX + H2O ≒ H3O+ + X-
pH = [H3O+]
Hence, [H3O+] at equilibrium = 10-4.5 M
The dissociation of HX give H3O+ and X-.
[X-] at equilibrium = (0.04167 + 10-4.5) M
[HX] at equilibrium = (0.2117 - 10-4.5) M

Ka of HX
= [H3O+] [X-] / [HX]
= 10-4.5 x (0.04167 + 10-4.5) / (0.2117 - 10-4.5)
= 6.23 x 10-6 M