Here is a titration curve recorded using a pH meter::
Determine the volume at the equivalence point and calculate the initial molarity of $NaOH$ !
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$V_e$ $=$ $40\;mL$ $c_{NaOH}$ $=$ $\frac{40\cdot 0.05}{20}$ $=$ $0.10M$
Recalculate the pH at point 1
Strong base: $pH$ $=$ $14+log0.1$ $=$ $13$.
Recalculate the pH at point 2
Rest of strong base: $pH$ $=$ $14+log\frac{0.064\cdot 0.05-0.020\cdot 0.1}{0.084}$ $=$ $11.9$.
Recalculate the pH at point 3
Excess of strong acid: $pH$ $=$ $-log\frac{0.020\cdot 0.1-0.032\cdot 0.05}{0.052}$ $=$ $1.8$.
