50 ~cm^ 3 of 0.2 ~N HCl is titrated against 0.1 ~NNaOH 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

Correct answer: B. 10 ~cm^ 3 When 0.1 ~N NaOH is used, gathered N_ 1 ~V_ 1 =N_ 2 ~V_ 2 (For HCl) (For NaOH) 0.2 ~N × V_ 1 =50 ×0.1 ~N ~V_ 1 =(50 × 0.1)/(0.2)=25 ~cm^ 3 gathered When 0.5 ~N KOH is used, aligned N_ 1 ~V_ 1 =& N_ 3 ~V_ 3 (For remaining HCl) &( For KOH) 0.2 ~N ×25 &=0.5 ~N × V_ 3 ~V_ 3 &=(0.2 × 25)/(0.5) =& 10 ~cm^ 3 aligned

JEE Main

Exam: JEE Main
Subject: Chemistry
Chapter: Redox Reactions

JEE Main · Chemistry · Redox Reactions

50 ~cm^ 3 of 0.2 ~N HCl is titrated against 0.1 ~NNaOH solution....

50 \mathrm{~cm}^{3}

0.2 \mathrm{~N} \mathrm{HCl}

is titrated against

0.1 \mathrm{~N}\mathrm{NaOH}

solution. The titration is discontinued after adding

50 \mathrm{~cm}^{3}

\mathrm{NaOH}

. The remaining titration is completed by adding

0.5 \mathrm{~N} \mathrm{KOH}

. The volume of

\mathrm{KOH}

required for completing the titration is

Options

A $12 \mathrm{~cm}^{3}$
B $10 \mathrm{~cm}^{3}$
C $25 \mathrm{~cm}^{3}$
D $10.5 \mathrm{~cm}^{3}$

Correct answer

B. $10 \mathrm{~cm}^{3}$

Step-by-step solution

When

0.1 \mathrm{~N} \mathrm{NaOH}

is used,

\begin{gathered} \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}\\ \text { (For HCl) }\\ \text { (For NaOH) }\\ 0.2 \mathrm{~N} \times \mathrm{V}_{1}=50 \times0.1 \mathrm{~N}\\ \mathrm{~V}_{1}=\frac{50 \times 0.1}{0.2}=25 \mathrm{~cm}^{3} \end{gathered}

When

0.5 \mathrm{~N} \mathrm{KOH}

is used,

\begin{aligned} \mathrm{N}_{1} \mathrm{~V}_{1}=& \mathrm{N}_{3} \mathrm{~V}_{3}\\ \text { (For remaining HCl) }&(\text { For KOH})\\ 0.2 \mathrm{~N} \times25 &=0.5 \mathrm{~N} \times \mathrm{V}_{3}\\ \mathrm{~V}_{3}&=\frac{0.2 \times 25}{0.5}\\ =& 10 \mathrm{~cm}^{3} \end{aligned}

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