To determine the normality of the KMnO4 solution used in the titration, we first need to understand the reaction between KMnO4 and oxalic acid (H2C2O4) or its salt (in this case, sodium oxalate, Na2C2O4).
Step 1: Write the balanced equation
The balanced redox reaction between KMnO4 and oxalic acid is:
\[ \text{5 C}_2\text{O}_4^{2-} + \text{2 MnO}_4^{-} + \text{16 H}^+ \rightarrow \text{10 CO}_2 + \text{2 Mn}^{2+} + \text{8 H}_2\text{O} \]
In the titration of Na2C2O4 with KMnO4, each mole of Na2C2O4 provides 1 mole of oxalate ions (C2O4^2-), and 5 moles of oxalate ions react with 2 moles of KMnO4.
Step 2: Determine the moles of Na2C2O4
First, we find the molar mass of Na2C2O4 (sodium oxalate):
- Sodium (Na): 22.99 g/mol × 2 = 45.98 g/mol
- Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol
Total molar mass of Na2C2O4 = 45.98 + 24.02 + 64.00 = 134.00 g/mol
Now calculate the number of moles of Na2C2O4:
\[ \text{Moles of Na}_2\text{C}_2\text{O}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.2121 , \text{g}}{134.00 , \text{g/mol}} \approx 0.00158 , \text{mol} \]
Step 3: Determine the moles of KMnO4 required
From the balanced equation, we know that 5 moles of oxalate react with 2 moles of KMnO4:
- Moles of KMnO4 = \(\frac{2}{5}\) × moles of C2O4^2-
Calculating the moles of KMnO4 required:
\[ \text{Moles of KMnO}_4 = \frac{2}{5} \times 0.00158 \approx 0.000632 , \text{mol} \]
Step 4: Calculate the normality of the KMnO4 solution
Normality (N) is defined as equivalents per liter. In this case, the number of equivalents is equal to the number of moles of KMnO4 because each mole of KMnO4 reacts as one equivalent in the reaction of interest.
To convert moles to normality, we need the volume in liters:
\[ \text{Volume of KMnO}_4 = 43.31 , \text{mL} = 0.04331 , \text{L} \]
Now, calculate the normality:
\[ \text{Normality} (N) = \frac{\text{equivalents}}{\text{volume in L}} = \frac{0.000632, \text{mol}}{0.04331, \text{L}} \approx 0.0146 , \text{N} \]
Conclusion
The normality of the KMnO4 solution is approximately:
\[ \boxed{0.0146 , N} \]
This value is reported to four significant figures.