What mass of solid potassium oxalate (K2C2O4), when dissolved in an acidic solution, would be titrated by exactly 50.0 mL of 0.0358 M potassium permanganate (KMnO4) solution?
The correct answer and explanation is:
To solve this problem, we need to understand the redox titration between potassium permanganate (KMnO₄) and potassium oxalate (K₂C₂O₄) in an acidic solution.
Balanced Redox Reaction in Acidic Medium
The redox reaction between MnO₄⁻ and C₂O₄²⁻ in acidic medium is: 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2 \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}
This tells us:
- 2 moles of MnO₄⁻ react with 5 moles of C₂O₄²⁻.
Step-by-step Calculation
Step 1: Moles of KMnO₄ used Moles of KMnO₄=M×V=0.0358 mol/L×0.0500 L=0.00179 mol\text{Moles of KMnO₄} = M \times V = 0.0358 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.00179 \, \text{mol}
Each mole of KMnO₄ provides 1 mole of MnO₄⁻.
Step 2: Use mole ratio to find moles of oxalate (C₂O₄²⁻)
From the balanced equation: 2mol MnO4−:5mol C2O42−2 \text{mol MnO}_4^- : 5 \text{mol C}_2\text{O}_4^{2-}
So: Moles of C2O42−=0.00179×52=0.004475 mol\text{Moles of C}_2\text{O}_4^{2-} = 0.00179 \times \frac{5}{2} = 0.004475 \, \text{mol}
Step 3: Find molar mass of K₂C₂O₄ K₂C₂O₄=2(39.10)+2(12.01)+4(16.00)=78.20+24.02+64.00=166.22 g/mol\text{K₂C₂O₄} = 2(39.10) + 2(12.01) + 4(16.00) = 78.20 + 24.02 + 64.00 = 166.22 \, \text{g/mol}
Step 4: Calculate mass of K₂C₂O₄ Mass=moles×molar mass=0.004475×166.22=0.744 g\text{Mass} = \text{moles} \times \text{molar mass} = 0.004475 \times 166.22 = \boxed{0.744} \, \text{g}
✅ Final Answer: 0.744 grams of potassium oxalate
🧪 Explanation Summary (300 Words)
In this problem, potassium permanganate (KMnO₄) is used as a titrant to determine the amount of potassium oxalate (K₂C₂O₄) in an acidic medium. KMnO₄ is a strong oxidizing agent, and oxalate ions (C₂O₄²⁻) are reducing agents. In an acidic environment, the permanganate ion (MnO₄⁻) is reduced to Mn²⁺, and oxalate is oxidized to carbon dioxide (CO₂). The stoichiometry of this redox reaction shows that 2 moles of permanganate react with 5 moles of oxalate.
First, we calculate how many moles of KMnO₄ are in 50.0 mL of 0.0358 M solution. This comes out to 0.00179 moles. Using the balanced chemical equation, we apply the 2:5 mole ratio to find the moles of oxalate required, which is 0.004475 moles.
Potassium oxalate (K₂C₂O₄) contains one mole of oxalate ion per mole of compound, so we assume a 1:1 ratio between K₂C₂O₄ and C₂O₄²⁻. We then multiply the moles of potassium oxalate by its molar mass (166.22 g/mol) to get the mass required: 0.744 grams.
Thus, 0.744 grams of solid K₂C₂O₄ would exactly react with 50.0 mL of 0.0358 M KMnO₄ solution in acidic conditions.