how do i calculate the Ka value for acetic acid through titration...?

Question

A titration was done, 50mL of 0.1 mol/L NaOH added into 25mL of CH3COOH (initial pH = 3.5)

this is what i've worked out so far:
chem equation: CH3COOH(aq) + NaOH(aq) <--> NaCH3COO(aq) + H2O(l)

dissociation of CH3COOH = CH3COOH(aq) + H2O(l) <--> CH3COOˉ(aq) + H3O + (l)

inital pH of CH3COOH(aq) = 3.5
initial [CH3COOH(aq)] = 0.2 mol/L (n = 5mmol, V = 25mL)
inital [NaOH] = 0.1mol/L (n = 5mmol, V = 50mL)

this is what i'm asked to do:

Use the initial pH of the CH3COOH (before you added any based (3.5)) to find the initial [H3O]. What was the initial [CH3COO-]

Assume the amount of CH3COOH that dissociates is small compared with the initial concentration of the acid. If this is true, the equilibrium valye of [CH3COOH] is equal to the initial concentration of the acid. Use your valuse of [H3O+],[CH3COO-] and [CH3COOH] to calculate Ka for acetic acid.

i worked it out to be 5 x 10^-7, but i'm pretty sure that is wrong.

Answer 1

3.16 x 10^-4