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

Concept: This problem involves an analytical redox titration calculation based on the principle of equivalence or stoichiometric molar ratios:
• Potassium permanganate (\(\text{KMnO}_4\)) acts as a powerful oxidizing agent. In its standard lab medium (acidic), the manganese center reduces from an oxidation state of \(+7\) to \(+2\).
• Sodium oxalate (\(\text{Na}_2\text{C}_2\text{O}_4\)) is a reliable primary standard acting as a reducing agent. The oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)) oxidizes into carbon dioxide (\(\text{CO}_2\)), shifting carbon's oxidation state from \(+3\) to \(+4\). The complete net balanced ionic equation for this specific redox tracking 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 the balanced stoichiometry, we establish a direct reacting molar bridge: \[ \text{Moles of }\text{KMnO}_4 = \frac{2}{5} \times \text{Moles of }\text{Na}_2\text{C}_2\text{O}_4 \]

Step 1: Calculating the number of moles of sodium oxalate ($\text{Na}_2\text{C}_2\text{O}_4$).
First, let's compute the formula mass of the primary standard salt \(\text{Na}_2\text{C}_2\text{O}_4\): Molar Mass &= (2 \times 23.00) + (2 \times 12.01) + (4 \times 16.00)
&= 46.00 + 24.02 + 64.00 = 134.02 g/mol Using the given experimental mass \(m = 0.1467\text{ g}\): \[ \text{Moles of }\text{Na}_2\text{C}_2\text{O}_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.1467\text{ g}}{134.02\text{ g/mol}} \approx 0.0010946\text{ moles} \]

Step 2: Finding the required moles of potassium permanganate ($\text{KMnO}_4$).
Using the stoichiometric relationship from our balanced equation: \[ \text{Moles of }\text{KMnO}_4 = \frac{2}{5} \times 0.0010946 = 0.4 \times 0.0010946 \approx 0.00043784\text{ moles} \]

Step 3: Calculating the molarity of the solution.
Molarity (\(M\)) is defined as the total moles of solute divided by the total volume of the solution in liters (\(\text{L}\)). Convert the given volume from milliliters to liters: \[ V = 28.85\text{ mL} = \frac{28.85}{1000}\text{ L} = 0.02885\text{ L} \] Now, compute the final molar concentration: \[ M = \frac{\text{Moles}}{V(\text{L})} = \frac{0.00043784\text{ moles}}{0.02885\text{ L}} \approx 0.015176\text{ M} \] Rounding to matching significant figures gives \(0.01518\text{ M}\).