Question:
$50 \, cm^3$ of $0.2 \, N \, HCl$ is titrated against $0.1 \, N \, NaOH$ solution. The titration was 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
$50 \, cm^3$ of $0.2 \, N \, HCl$ is titrated against $0.1 \, N \, NaOH$ solution. The titration was 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
Updated On: Apr 19, 2024
- $10\, cm^3$
- $12 \,cm^3$
- $16.2 \, cm^3$
- $21.0\, cm^3$
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The Correct Option is A
Solution and Explanation
No. of equivalent of $HCl$ remaining after adding
$50\, cm ^{3}$ of $0.1\, N\, NaOH =\frac{0.2 \times 50-0.1 \times 50}{100}$
$=\frac{0.5}{100}$
$\therefore$ Volume of $0.5\, N$ KOH required $\frac{0.5}{100}$ eq
$\equiv \frac{V \times 0.5}{1000}$
$V =\frac{0.5}{100} \times \frac{1000}{0.5}$
$=10\, cm ^{3}$
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Concepts Used:
Alcohols, Phenols and Ethers - Classification
Classification of Alcohols Phenols and Ethers are as follows:
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- Monohydric alcohols
- Dihydric alcohols
- Trihydric alcohols
On the basis of dependence on the different number of carbon atoms in a molecule that are bonded with the -OH group, alcohol can be classified into three different types:
- Primary alcohols
- Secondary alcohols
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