In permanganate titrations, potassium pemanganate is used an an oxidizing agent in acidic medium.
The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is
`2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O]`
or `MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O`
Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of `KMnO_(4)` solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, `As_(2)O_(3)` etc.
10 ml of `NaHC_(2)O_(4)` solution is neutralized by 10 ml of `0.1` M NaOH . 10 ml of same `NaHC_(2)O_(4)` solution is oxidized by 10 ml of `KMnO_(4)` solution in acidic medium . Hence, molarity of `KMnO_(4)` is

Options

A. `0.1` M
B. `0.2` M
C. `0.04` M
D. `0.02` M

Answer

To find the molarity of the potassium permanganate (KMnO₄) solution used in the titration with sodium hydrogen oxalate (NaHC₂O₄), we can follow these steps:

### Step 1: Determine the molarity of NaHC₂O₄
Given that 10 mL of NaHC₂O₄ solution is neutralized by 10 mL of 0.1 M NaOH, we can use the neutralization reaction to find the molarity of NaHC₂O₄.

- The reaction between NaOH and NaHC₂O₄ is:
\[ \text{NaOH} + \text{NaHC}_2\text{O}_4 \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{O} \]

- Since NaHC₂O₄ is a mono-basic acid and NaOH is a mono-basic base, the stoichiometry is 1:1.

Using the formula for titration:
\[ M_1V_1N_1 = M_2V_2N_2 \]
where:
- \(M_1\) = molarity of NaHC₂O₄
- \(V_1\) = volume of NaHC₂O₄ = 10 mL
- \(N_1\) = number of equivalents of NaHC₂O₄ = 1
- \(M_2\) = 0.1 M (molarity of NaOH)
- \(V_2\) = volume of NaOH = 10 mL
- \(N_2\) = number of equivalents of NaOH = 1

Substituting the values:
\[ M_1 \times 10 \, \text{mL} \times 1 = 0.1 \, \text{M} \times 10 \, \text{mL} \times 1 \]
\[ M_1 \times 10 = 1 \]
\[ M_1 = 0.1 \, \text{M} \]

### Step 2: Write the reaction for KMnO₄ with NaHC₂O₄
The reaction of KMnO₄ in acidic medium can be represented as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

For NaHC₂O₄, the reaction can be simplified as:
\[ \text{NaHC}_2\text{O}_4 \rightarrow \text{CO}_2 + \text{Na}^+ + \text{H}^+ + 2e^- \]

### Step 3: Balance the overall reaction
From the stoichiometry:
- 2 moles of KMnO₄ react with 5 moles of NaHC₂O₄.

Thus, we can write:
\[ 2 \text{MnO}_4^- + 16 \text{H}^+ + 5 \text{NaHC}_2\text{O}_4 \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + 5 \text{CO}_2 + 5 \text{Na}^+ \]

### Step 4: Set up the equivalence relation for KMnO₄
Using the equivalence relation again:
\[ M_1V_1N_1 = M_2V_2N_2 \]
where:
- \(M_1\) = molarity of KMnO₄ (unknown)
- \(V_1\) = volume of KMnO₄ = 10 mL
- \(N_1\) = number of equivalents for KMnO₄ = 5 (since it gains 5 electrons)
- \(M_2\) = 0.1 M (molarity of NaHC₂O₄)
- \(V_2\) = 10 mL
- \(N_2\) = number of equivalents for NaHC₂O₄ = 2 (since it loses 2 electrons)

Substituting the values:
\[ M_1 \times 10 \times 5 = 0.1 \times 10 \times 2 \]
\[ M_1 \times 50 = 2 \]
\[ M_1 = \frac{2}{50} = 0.04 \, \text{M} \]

### Final Answer
The molarity of KMnO₄ is **0.04 M**.

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