40 mL of a mixture of CH_3COOH and HCl (aque | vedclass

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(B) In a conductometric titration of a mixture of a strong acid $(HCl)$ and a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the strong acid is ne

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The concentration of $HCl$ in the original mixture is $0.005 \ M$

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(B) In a conductometric titration of a mixture of a strong acid $(HCl)$ and a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the strong acid is neutralized first because it dissociates completely.From the titration curve,the volume of $0.1 \ M$ $NaOH$ required to neutralize $HCl$ is $2 \ mL$ and the volume required to neutralize $CH_3COOH$ is $3 \ mL$.For $HCl$: $M_{HCl} 	imes 40 \ mL = 0.1 \ M 	imes 2 \ mL \implies M_{HCl} = 0.005 \ M$.For $CH_3COOH$: $M_{CH_3COOH} 	imes 40 \ mL = 0.1 \ M 	imes 3 \ mL \implies M_{CH_3COOH} = 0.0075 \ M$.Thus,the concentration of $HCl$ in the original mixture is $0.005 \ M$.

$40 \ mL$ of a mixture of $CH_3COOH$ and $HCl$ (aqueous solution) is titrated against $0.1 \ M$ $NaOH$ solution conductometrically. Which of the following statements is correct?

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