To determine the normality of the KMnO4 solution, we'll first need to calculate the moles of Na2C2O4 in the sample and then use stoichiometry from the balanced chemical equation to find the equivalent moles of KMnO4 that reacted.
- Calculate the moles of Na2C2O4:
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: \( 16.00 , \text{g/mol} \times 4 = 64.00 , \text{g/mol} \)
Now, add them together: \[ \text{Molar mass of Na2C2O4} = 45.98 + 24.02 + 64.00 = 134.00 , \text{g/mol} \]
Next, we calculate the moles of Na2C2O4 in the 0.2121 g sample: \[ \text{Moles of Na2C2O4} = \frac{0.2121 , \text{g}}{134.00 , \text{g/mol}} \approx 0.00158 , \text{mol} \]
- Determine the moles of C2O4²⁻ (oxalate ions):
According to the reaction, \( 1 , \text{mol} , \text{Na}_2C_2O_4 \) provides \( 1 , \text{mol} , \text{C}_2\text{O}_4^{2-} \). Therefore, moles of \( \text{C}_2\text{O}_4^{2-} \) is equal to the moles of Na2C2O4: \[ \text{Moles of C}_2\text{O}_4^{2-} = 0.00158 , \text{mol} \]
- Use the stoichiometry of the reaction:
From the balanced equation: \[ 2 , \text{MnO}_4^- + 5 , \text{C}_2\text{O}_4^{2-} \rightarrow 2 , \text{Mn}^{2+} + 10 , \text{CO}_2 + 8 , \text{H}_2\text{O} \]
This means 5 moles of \( \text{C}_2\text{O}_4^{2-} \) react with 2 moles of \( \text{MnO}_4^{-} \). Therefore, we can find the moles of \( \text{MnO}_4^{-} \) used: \[ \text{Moles of } \text{MnO}_4^{-} = \frac{2}{5} \times \text{Moles of C}_2\text{O}_4^{2-} = \frac{2}{5} \times 0.00158 , \text{mol} \approx 0.000632 , \text{mol} \]
- Calculate the normality of the KMnO4 solution:
Normality is defined as the number of equivalents per liter of solution. In this case, since \( \text{MnO}_4^{-} \) has a valency of 5 (because it reacts with 5 moles of \( \text{C}_2\text{O}_4^{2-} \)), the normality (N) can be calculated using: \[ \text{Normality} = \frac{\text{Equivalents}}{\text{Volume in liters}} \]
The number of equivalents equals the moles of \( \text{MnO}_4^{-} \) multiplied by the number of electrons transferred (5): \[ \text{Equivalents of } \text{MnO}_4^{-} = 0.000632 , \text{mol} \times 5 = 0.00316 , \text{eq} \]
Next, convert the volume from mL to L: \[ \text{Volume} = 43.31 , \text{mL} = 0.04331 , \text{L} \]
Now, we can find the normality: \[ \text{Normality} = \frac{0.00316 , \text{eq}}{0.04331 , \text{L}} \approx 0.0729 , \text{N} \]
Finally, rounding to four significant figures, the normality of the KMnO4 solution is: \[ \boxed{0.07293 , \text{N}} \]