A 0.2121 g sample of pure Na2C2O4 was treated with 43.31 ml of KMnO4.what is the normality of KMnO4 solution? The chemical reaction is:

Question

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

Answer 1

To find the normality of the KMnO4 solution, we need to go through the following steps:

1. Determine the moles of Na2C2O4 in the 0.2121 g sample.
2. Use the stoichiometry of the reaction to find out how many equivalents of KMnO4 were required.
3. Calculate the normality of the KMnO4 solution using the volume of the solution used in the reaction.

Step 1: Moles of Na2C2O4

The molar mass of Na2C2O4 (sodium oxalate) can be calculated as follows:

• \( \text{Na:} , 22.99 , \text{g/mol} \times 2 = 45.98 , \text{g/mol} \)
• \( \text{C:} , 12.01 , \text{g/mol} \times 2 = 24.02 , \text{g/mol} \)
• \( \text{O:} , 16.00 , \text{g/mol} \times 4 = 64.00 , \text{g/mol} \)

Adding these together:

\[ \text{Molar Mass of } \text{Na2C2O4} = 45.98 + 24.02 + 64.00 = 134.00 , \text{g/mol} \]

Now, we calculate the moles of Na2C2O4:

\[ \text{Moles of } \text{Na2C2O4} = \frac{0.2121 , \text{g}}{134.00 , \text{g/mol}} \approx 0.00158 , \text{mol} \]

Step 2: Determine the equivalents of KMnO4

From the chemical reaction:

\[ 2 \text{MnO}_4^- + 5 \text{C}_2\text{O}_4^{2-} + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]

The stoichiometry shows that 5 moles of \(\text{C}_2\text{O}_4^{2-}\) react with 2 moles of \(\text{MnO}_4^-\). This means that:

• 1 mole of \(\text{C}_2\text{O}_4^{2-}\) reacts with \(\frac{2}{5}\) moles of \(\text{MnO}_4^-\).

The number of moles of \(\text{C}_2\text{O}_4^{2-}\) present:

\[ \text{Moles of } \text{C}_2\text{O}_4^{2-} = \text{Moles of Na2C2O4} \times 1 = 0.00158 , \text{mol} \]

Thus the moles of KMnO4 needed:

\[ \text{Moles of } \text{MnO}_4^- = \frac{2}{5} \times \text{Moles of } \text{C}_2\text{O}_4^{2-} = \frac{2}{5} \times 0.00158 \approx 0.000632 , \text{mol} \]

The equivalents of KMnO4 correspond to the moles of \(\text{MnO}_4^-\) since each \(\text{MnO}_4^-\) has one equivalent.

Step 3: Normality of KMnO4

Normality is defined as the number of equivalents per liter of solution. We used 43.31 mL of KMnO4 solution, which is:

\[ 43.31 , \text{mL} = 0.04331 , \text{L} \]

Now, we can calculate the normality of KMnO4:

\[ \text{Normality (N)} = \frac{\text{equivalents of KMnO4}}{\text{volume of solution in L}} = \frac{0.000632 , \text{mol}}{0.04331 , \text{L}} \approx 0.0146 , \text{N} \]

Final Answer

The normality of the KMnO4 solution to four significant figures is:

\[ \boxed{0.01462} \]