(B) Initial milliequivalents of $HCl = N \times V = 0.2 \ N \times 50 \ cm^{3} = 10 \ meq$. Milliequivalents of $NaOH$ added $= 0.1 \ N \times 50 \ cm

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(B) Initial milliequivalents of $HCl = N \times V = 0.2 \ N \times 50 \ cm^{3} = 10 \ meq$. Milliequivalents of $NaOH$ added $= 0.1 \ N \times 50 \ cm^{3} = 5 \ meq$. Remaining milliequivalents of $HCl = 10 \ meq - 5 \ meq = 5 \ meq$. To neutralize the remaining $5 \ meq$ of $HCl$,we use $0.5 \ N \ KOH$. Let the volume of $KOH$ be $V_{KOH}$. $N_{KOH} \times V_{KOH} = 5 \ meq$. $0.5 \ N \times V_{KOH} = 5 \ meq$. $V_{KOH} = \frac{5}{0.5} = 10 \ cm^{3}$.

Class 12 Chemistry Principles Related to Practical Chemistry Titrations

Medium

$50 \ cm^{3}$ of $0.2 \ N \ HCl$ is titrated against $0.1 \ N \ NaOH$ solution. The titration is discontinued after adding $50 \ cm^{3}$ of $NaOH$. The remaining titration is completed by adding $0.5 \ N \ KOH$. The volume of $KOH$ required for completing the titration is (in $cm^{3}$)

KCET 2010 All KCET

A
$12$
B
$10$
C
$25$
D
$10.5$

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$50 \ cm^{3}$ of $0.2 \ N \ HCl$ is titrated against $0.1 \ N \ NaOH$ solution. The titration is discontinued after adding $50 \ cm^{3}$ of $NaOH$. The remaining titration is completed by adding $0.5 \ N \ KOH$. The volume of $KOH$ required for completing the titration is (in $cm^{3}$)

Similar Questions

If $25 \ mL$ of $H_2SO_4$ solution is neutralized by $32.63 \ mL$ of $0.164 \ M \ NaOH$ solution,what is the molarity of the $H_2SO_4$ solution?

During the estimation of oxalic acid $V/S$ $KMnO_4$,the self-indicator is:

In the titration of a strong acid and a weak base,the indicator used is:

$3.92 \, g$ of ferrous ammonium sulfate crystals are dissolved in $100 \, mL$ of water. During titration,$20 \, mL$ of this solution is required for complete oxidation by $18 \, mL$ of $KMnO_4$ solution. Calculate the weight of $KMnO_4$ in $1 \, L$ of the solution in grams.

Which of the following reactions occurs during iodometric titration in the laboratory?

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