Question

Titration of $0.1467\text{ g}$ of primary standard $\text{Na}_2\text{C}_2\text{O}_4$ required $28.85\text{ mL}$ of $\text{KMnO}_4$ solution. Calculate the molar concentration of $\text{KMnO}_4$ solution.

• A. \( 0.01518\text{ M} \)
• B. \( 0.001518\text{ M} \)
• C. \( 0.15180\text{ M} \)
• D. \( 1.5180\text{ M} \)

Hint:

To solve analytical titrations significantly faster without worrying about full balanced equations, use the Normality Law of Equivalents: \(N_1 V_1 = N_2 V_2\), which expands to \((M \times \text{n-factor})_1 \times V_1 = \left(\frac{\text{mass}}{\text{equivalent mass}}\right)_2\). For this setup, remember that the n-factor of \(\text{KMnO}_4\) in an acidic medium is *always* 5 and the n-factor of \(\text{Na}_2\text{C}_2\text{O}_4\) is 2. Plugging this in directly gives: \(M_{\text{KMnO}_4} \times 5 \times 0.02885 = \frac{0.1467}{134.02} \times 2\), isolating your answer in a single step!

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
This problem presents a quantitative redox titration scenario where a primary standard substance, sodium oxalate ($\text{Na}_2\text{C}_2\text{O}_4$), is used to standardize a solution of potassium permanganate ($\text{KMnO}_4$). We are given the mass of sodium oxalate and the volume of permanganate solution required to reach the end point. Our goal is to determine the molarity of the potassium permanganate solution by using the stoichiometric relationships of the chemical reaction.
Step 2 : Key Formulas and Approach:
The reaction occurs in an acidic medium, where permanganate ions oxidize oxalate ions. The balanced chemical equation for this redox reaction is:
\[ 2\text{MnO}_4^-(aq) + 5\text{C}_2\text{O}_4^{2-}(aq) + 16\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l) \]
From this balanced equation, we establish the stoichiometric ratio between the reacting species:
\[ \text{Moles of }\text{KMnO}_4 = \frac{2}{5} \times \text{Moles of }\text{Na}_2\text{C}_2\text{O}_4 \]
We can calculate the moles of sodium oxalate using its mass and molar mass:
\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]
Once the moles of permanganate are found, the molarity of the permanganate solution is calculated using the volume in liters:
\[ \text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}} \]
Step 3 : Detailed Explanation:

First, we calculate the molar mass of sodium oxalate ($\text{Na}_2\text{C}_2\text{O}_4$). Using the atomic weights of sodium ($23.00\text{ g/mol}$), carbon ($12.01\text{ g/mol}$), and oxygen ($16.00\text{ g/mol}$), we get: $\text{Molar Mass} = 2(23.00) + 2(12.01) + 4(16.00) = 46.00 + 24.02 + 64.00 = 134.02\text{ g/mol}$.

Next, we find the number of moles of sodium oxalate used in the titration by dividing the given mass by its molar mass: $\text{Moles of }\text{Na}_2\text{C}_2\text{O}_4 = \frac{0.1467\text{ g}}{134.02\text{ g/mol}} \approx 1.0946 \times 10^{-3}\text{ moles}$.

Using the stoichiometric ratio from our balanced chemical equation, we find the moles of potassium permanganate that reacted: $\text{Moles of }\text{KMnO}_4 = \frac{2}{5} \times (1.0946 \times 10^{-3}\text{ moles}) \approx 4.3784 \times 10^{-4}\text{ moles}$.

We then convert the volume of the potassium permanganate solution from milliliters to liters: $\text{Volume} = 28.85\text{ mL} = \frac{28.85}{1000}\text{ L} = 0.02885\text{ L}$.

Finally, we calculate the molarity of the potassium permanganate solution by dividing the moles of permanganate by the volume in liters: $\text{Molarity} = \frac{4.3784 \times 10^{-4}\text{ moles}}{0.02885\text{ L}} \approx 0.015176\text{ M}$.

Rounding this result to four significant figures gives the final molar concentration of $0.01518\text{ M}$.

Step 4 : Final Answer:
The molar concentration of the $\text{KMnO}_4$ solution is $0.01518\text{ M}$, which matches Option (A).