Question

A 0.2121 g sample of pure Na2C2O4 was titrated with 43.31mL of KMnO4. What is the normality of KMnO4 solution?

Answer 1

To find the normality of the KMnO4 solution used to titrate the Na2C2O4, we need to follow these steps:

1. **Determine moles of Na2C2O4 in the sample:**
- The molar mass of Na2C2O4 (sodium oxalate) can be calculated as follows:
- Na: \( 22.99 \, \text{g/mol} \times 2 = 45.98 \, \text{g/mol} \)
- C: \( 12.01 \, \text{g/mol} \times 2 = 24.02 \, \text{g/mol} \)
- O: \( 16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol} \)
- Total molar mass of Na2C2O4:
\[
45.98 + 24.02 + 64.00 = 134.00 \, \text{g/mol}
\]

- Now, calculate the moles of Na2C2O4 in the sample:
\[
\text{Moles of Na2C2O4} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.2121 \, \text{g}}{134.00 \, \text{g/mol}} \approx 0.00158 \, \text{mol}
\]

2. **Determine the reaction between Na2C2O4 and KMnO4:**
- The reaction between Na2C2O4 and KMnO4 in acidic medium follows this balanced equation:
\[
5 \, \text{C}_2\text{O}_4^{2-} + \text{MnO}_4^{-} + 8 \, \text{H}^+ \rightarrow 5 \, \text{CO}_2 + \text{Mn}^{2+} + 4 \, \text{H}_2\text{O}
\]

- From the reaction, we see that 1 mole of KMnO4 reacts with 5 moles of C2O4^2- (from Na2C2O4). Therefore, for every mole of Na2C2O4, there are 5 equivalents.

3. **Calculate the equivalents of Na2C2O4:**
- The equivalents of Na2C2O4:
\[
\text{Equivalents of Na2C2O4} = \text{moles} \times 5 = 0.00158 \, \text{mol} \times 5 = 0.00790 \, \text{eq}
\]

4. **Calculate the normality of KMnO4:**
- Normality (N) is defined as equivalents per liter of solution. We have the equivalents of Na2C2O4 and the volume of KMnO4 solution.
- The volume of KMnO4 solution = 43.31 mL = 0.04331 L.
- Thus,
\[
N = \frac{\text{equivalents of KMnO4}}{\text{volume of solution in L}} = \frac{0.00790 \, \text{eq}}{0.04331 \, \text{L}} \approx 0.182 \, \text{N}
\]

Therefore, the normality of the KMnO4 solution is approximately **0.182 N**.