Question

The concentration and percentage purity of oxalic acid can be determined by titration with \(KMnO_4\) in presence of dil. \(H_2SO_4\). Instead of dil. \(H_2SO_4\), dil. HCl cannot be used because

• A. HCl can also reduce \(MnO_4^-\) to \(Mn^{2+}\)
• B. HCl can also provide \(H^+\) ions in addition to \(H^+\) ions from oxalic acid
• C. HCl can also oxidise oxalic acid to \(CO_2\) and \(H_2O\)
• D. Oxalic acid oxidises HCl to \(Cl_2\)

Hint:

Always use acids like \(H_2SO_4\) in redox titrations with \(KMnO_4\). Avoid HCl or \(HNO_3\) as they may participate in side reactions.

Answer 1

The Correct Option is A

Step 1: Understand the role of \(KMnO_4\) in titration.
In acidic medium, \(KMnO_4\) acts as a strong oxidising agent.
It oxidises oxalic acid to \(CO_2\) while itself gets reduced to \(Mn^{2+}\).
\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]

Step 2: Role of dilute \(H_2SO_4\).
Dilute sulphuric acid provides the required acidic medium without interfering in the reaction.
It does not react with \(KMnO_4\), hence it is suitable for titration.

Step 3: Behaviour of HCl in this reaction.
Hydrochloric acid contains \(Cl^-\) ions, which can act as reducing agents.
These chloride ions can be oxidised by \(KMnO_4\).
\[ Cl^- \rightarrow Cl_2 \]

Step 4: Side reaction with \(KMnO_4\).
\(KMnO_4\) oxidises \(Cl^-\) ions to \(Cl_2\) gas while getting reduced to \(Mn^{2+}\).
\[ MnO_4^- + Cl^- \rightarrow Mn^{2+} + Cl_2 \] This causes consumption of \(KMnO_4\) apart from the main reaction.

Step 5: Effect on titration results.
Due to this side reaction, extra \(KMnO_4\) is consumed, leading to incorrect results.
Thus, HCl interferes with the titration process.

Step 6: Final conclusion.
Since HCl reduces \(MnO_4^-\) by oxidising chloride ions to chlorine gas, it cannot be used in place of dilute \(H_2SO_4\).
Final Answer:
The correct reason is:
\[ \boxed{\text{(A) HCl can also reduce } MnO_4^- \text{ to } Mn^{2+}} \]