A 100 mL solution of 0.1 N HCl was titrate | vedclass

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(A) For the neutralization of an acid and a base,the total milliequivalents $(N 	imes V)$ must be equal.Total milliequivalents of $HCl = 0.1 \ N 	im

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) For the neutralization of an acid and a base,the total milliequivalents $(N 	imes V)$ must be equal.Total milliequivalents of $HCl = 0.1 \ N 	imes 100 \ mL = 10 \ meq$.Milliequivalents of $NaOH$ added = $0.2 \ N 	imes 30 \ mL = 6 \ meq$.Remaining milliequivalents of $HCl$ to be neutralized = $10 \ meq - 6 \ meq = 4 \ meq$.Let the volume of $KOH$ required be $V \ mL$. Since the normality of $KOH$ is $0.25 \ N$,we have:$0.25 \ N 	imes V \ mL = 4 \ meq$.$V = \frac{4}{0.25} = 16 \ mL$.

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