equilibrium

Firstly, consider the titration (35 cm3 of NaOH added):

HA + NaOH → NaA + H2O
Mole ratio HA : NaOH = 1 : 1
No. of moles NaOH used = MV = 0.1 x (35/1000) = 0.0035 mol
No. of moles of HA in 40 cm3 of HA solution = 0.0035 mol

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Consider the point when 20 cm3 of NaOH is added, and neglect the dissociation of HA.

HA + NaOH → NaA + H2O
Mole ratio HA : NaOH = 1 : 1
No. of moles of HA added = 0.0035 mol
No. of moles of NaOH added = MV = 0.1 x (20/1000) = 0.002 mol

HA is in excess. All NaOH is reacted with part of HA to give NaA.
No. of moles of HA left = 0.0035 - 0.002 = 0.0015 mol
No. of moles of NaA formed = 0.002 mol

Final volume of the solution = 40 + 20 = 60 cm3 = 0.06 dm3
Molarity of HA left = mol/V = 0.0015/0.06 = 0.025 M
Molarity of A- = Molarity of NaA = mol/V = 0.002/0.06 = 0.0333 M

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Consider the dissociation of HA in the presence of A-.

[HA]o = 0.025 M
[A-]o = 0.0333 M

HA + H2O ≒ H3O+ + A-
pH = [H3O+]
Hence, [H3O+] at equilibrium = 10-pH = 10-5.75 M
The dissociation of HA give H3O+ and A-.
[A-] at equilibrium = (0.0333 + 10-5.75) M
[HA] at equilibrium = (0.025 - 10-5.75) M

Ka of HA
= [H3O+] [A-] / [HA]
= 10-5.75 x (0.0333 + 10-5.75) / (0.025 - 10-5.75)
= 2.37 x 10-6 M