To solve the problem step by step, we will first analyze the reaction between sodium hydrogen oxalate (NaHC₂O₄) and potassium permanganate (KMnO₄) in acidic medium, and then determine the volume of sodium hydroxide (NaOH) required to neutralize the same amount of NaHC₂O₄.
### Step 1: Determine the reaction between NaHC₂O₄ and KMnO₄
In acidic medium, the balanced reaction between NaHC₂O₄ and KMnO₄ can be represented as follows:
\[ \text{5 NaHC}_2\text{O}_4 + \text{2 KMnO}_4 + \text{6 H}_2\text{SO}_4 \rightarrow \text{5 CO}_2 + \text{2 MnSO}_4 + \text{K}_2\text{SO}_4 + \text{6 H}_2\text{O} + \text{5 Na}_2\text{SO}_4 \]
From the balanced equation, we can see that 2 moles of KMnO₄ react with 5 moles of NaHC₂O₄.
### Step 2: Calculate the equivalents of KMnO₄ used
Given that 50 mL of 0.1 M KMnO₄ is used, we can calculate the number of equivalents of KMnO₄:
\[
\text{Equivalents of KMnO}_4 = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor}
\]
For KMnO₄ in acidic medium, the n-factor is 5 (as it changes from MnO₄⁻ to Mn²⁺).
\[
\text{Equivalents of KMnO}_4 = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} \times 5 = 0.025 \, \text{equivalents}
\]
### Step 3: Calculate the equivalents of NaHC₂O₄
From the stoichiometry of the reaction, we know that:
\[
\frac{5 \, \text{equivalents of NaHC}_2\text{O}_4}{2 \, \text{equivalents of KMnO}_4} = \frac{5}{2}
\]
Thus, the equivalents of NaHC₂O₄ can be calculated as:
\[
\text{Equivalents of NaHC}_2\text{O}_4 = \frac{5}{2} \times \text{Equivalents of KMnO}_4 = \frac{5}{2} \times 0.025 = 0.0625 \, \text{equivalents}
\]
### Step 4: Calculate the molarity of NaHC₂O₄
We know that the volume of NaHC₂O₄ solution is 100 mL (or 0.1 L). The molarity (M) can be calculated as:
\[
\text{Molarity} = \frac{\text{Equivalents}}{\text{Volume (L)}}
\]
\[
\text{Molarity of NaHC}_2\text{O}_4 = \frac{0.0625 \, \text{equivalents}}{0.1 \, \text{L}} = 0.625 \, \text{M}
\]
### Step 5: Determine the volume of NaOH required
The reaction between NaHC₂O₄ and NaOH can be written as:
\[
\text{NaHC}_2\text{O}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{O}
\]
In this case, the n-factor for NaHC₂O₄ is 1 because it has one acidic hydrogen that reacts with NaOH.
Using the formula:
\[
N_1V_1 = N_2V_2
\]
Where:
- \(N_1\) = Normality of NaHC₂O₄ = 0.625 M (since n-factor = 1)
- \(V_1\) = Volume of NaHC₂O₄ = 100 mL
- \(N_2\) = Normality of NaOH = 0.1 M
- \(V_2\) = Volume of NaOH to be calculated
Substituting the values:
\[
0.625 \times 100 = 0.1 \times V_2
\]
\[
62.5 = 0.1 \times V_2
\]
\[
V_2 = \frac{62.5}{0.1} = 625 \, \text{mL}
\]
### Final Answer
The volume of 0.1 M NaOH required by 100 mL of NaHC₂O₄ is **625 mL**.
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