Chemistry - Electrochemistry Question with Solution | TestHub

From the given graph 2 ml NaOH solution is used for neutralisation of HCl and 3 ml NaOH solution is used for neutralisation of ${\mathrm{CH}}_3\mathrm{COOH}$.
$\therefore$ Mole of $\mathrm{HCl}=$ Mole of NaOH used
$\begin{array}{rl}& \mathrm{M}\times 40=0.1\times 2\\ & \mathrm{M}=0.005\end{array}$
$\therefore$ Mole of ${\mathrm{CH}}_3\mathrm{COOH}=$ Mole of NaOaH used
$\begin{array}{rl}& \mathrm{M}\times 40=0.1\times 3\\ & \mathrm{M}=0.0075\end{array}$
HCl is strong acid and will be neutralised first.