40 mL of a mixture of CH_3 COOH and HCl (aqueous solution) is titrated against 0.1 M NaOH solution conductometrically. Which of the following statement is correct?

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Correct answer: B. The concentration of HCl in the original mixture is 0.005 M

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{aligned}<br />& \mathrm{M} \times 40=0.1 \times 2 \\<br>& \mathrm{M}=0.005<br />\end{aligned}

\therefore

Mole of

\mathrm{CH}_3 \mathrm{COOH}=

Mole of NaOaH used

\begin{aligned}<br />& \mathrm{M} \times 40=0.1 \times 3 \\<br>& \mathrm{M}=0.0075<br />\end{aligned}

HCl is strong acid and will be neutralised first.

40 mL of a mixture of CH_3 COOH and HCl (aqueous solution) is...

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