A 0.2121 g sample of pure Na2C2O4 was treated with 43.3 ml of KMnO4.What is the normality of KMnO4 solution? The chemical reaction is:

Question

2MnO4^- + 5C2O4^-2 + 16H^+ = 2Mn^+2 + 10CO2 + 8H2O
Give the answer to 4 significant figures .

Answer 1

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

1. Calculate the moles of Na2C2O4 in the sample.
2. Use the stoichiometry of the reaction to find moles of KMnO4.
3. Calculate the normality of the KMnO4 solution.

Step 1: Calculate 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: \( 16.00 , \text{g/mol} \times 4 = 64.00 , \text{g/mol} \)

Adding these gives us the total molar mass: \[ 45.98 + 24.02 + 64.00 = 134.00 , \text{g/mol} \]

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

Step 2: Use stoichiometry of the reaction

From the balanced reaction: \[ 2 , \text{MnO}_4^{-} + 5 , \text{C}_2\text{O}_4^{2-} + 16H^{+} = 2 , \text{Mn}^{2+} + 10 , \text{CO}_2 + 8H_2O \]

We can see that:

• 5 moles of \( \text{C}_2\text{O}_4^{2-} \) react with 2 moles of \( \text{MnO}_4^{-} \).
• Therefore, the mole ratio of \( \text{MnO}_4^{-} \) to \( \text{C}_2\text{O}_4^{2-} \) is \( \frac{2}{5} \).

Thus, the moles of \( \text{MnO}_4^{-} \) that reacted will be: \[ \text{Moles of MnO}_4^{-} = \frac{2}{5} \times \text{Moles of Na}_2\text{C}_2\text{O}_4 = \frac{2}{5} \times 0.00158 \approx 0.000632 , \text{moles} \]

Step 3: Calculate normality of KMnO4 solution

Normality (N) is defined as the number of equivalents per liter of solution. Since \( \text{MnO}_4^{-} \) has a change of 5 for each mole of \( \text{C}_2\text{O}_4^{2-} \), the number of equivalents of KMnO4 is: \[ \text{Equivalents of KMnO}_4 = 0.000632 \times 2 = 0.001264 , \text{equivalents} \]

Normality can be calculated as: \[ \text{Normality (N)} = \frac{\text{equivalents}}{\text{liters of solution}} \] Given that we have \( 43.3 , \text{ml} \) of KMnO4 solution, converting this to liters: \[ \text{Volume in liters} = \frac{43.3}{1000} = 0.0433 , \text{L} \]

Now we can calculate the normality: \[ N = \frac{0.001264 , \text{equivalents}}{0.0433 , \text{L}} \approx 0.0292 , \text{N} \]

Final Answer

The normality of the KMnO4 solution is: \[ \boxed{0.02920 , \text{N}} \] (to four significant figures).