From the given graph 2 ml NaOH solution is used for neutralisation of HCl and 3 ml NaOH solution is used for neutralisation of CH3COOH.
∴ Mole of HCl= Mole of NaOH used
$\begin{aligned}
& \mathrm{M} \times 40=0.1 \times 2 \
& \mathrm{M}=0.005
\end{aligned}$
∴ Mole of CH3COOH= Mole of NaOaH used
$\begin{aligned}
& \mathrm{M} \times 40=0.1 \times 3 \
& \mathrm{M}=0.0075
\end{aligned}$
HCl is strong acid and will be neutralised first.
