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

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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Alcohols, Phenols and Ethers - Classification

Classification of Alcohols Phenols and Ethers are as follows:

On the basis of dependence on the attachment of hydroxyl groups, alcohol can be classified into three different types:

  • 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
  • Tertiary alcohols

On the basis of the attachment of the different number of hydroxyl groups, the phenols can be classified further into three types:

  • Monohydric phenols
  • Dihydric phenols
  • Trihydric phenols

The Classification is dependant on the type of the alkyl or aryl groups attached to the oxygen atom in ether group, thus it can be classified into two types:

  • Symmetrical ether
  • Unsymmetrical ether