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The concentration of H2C2O4 in the rhubarb extract is closest to A - VCE - SSCE Chemistry - Question 17 - 2018 - Paper 1

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Question 17

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The concentration of H2C2O4 in the rhubarb extract is closest to A. 5.43 x 10^-3 M B. 5.00 x 10^-2 M C. 2.17 x 10^-2 M D. 7.40 x 10^-4 M

Worked Solution & Example Answer:The concentration of H2C2O4 in the rhubarb extract is closest to A - VCE - SSCE Chemistry - Question 17 - 2018 - Paper 1

Step 1

Step 1 - Analyzing the Titration Results

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Answer

To find the concentration of oxalic acid (H₂C₂O₄) in the rhubarb extract, we first consider the balanced chemical equation for the reaction between oxalic acid and potassium permanganate (KMnO₄):

2MnO4−(aq)+5C2O42−(aq)+16H+(aq)→2Mn2+(aq)+10CO2(g)+8H2O(l)2MnO_4^-(aq) + 5C_2O_4^{2-}(aq) + 16H^+(aq) \rightarrow 2Mn^{2+}(aq) + 10CO_2(g) + 8H_2O(l)

This shows that 2 moles of KMnO₄ react with 5 moles of H₂C₂O₄, establishing a stoichiometric ratio.

Step 2

Step 2 - Relating Volume and Concentration

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Answer

From the titration data, we know:

  • Volume of KMnO₄ used = volume of the titration (average of concordant titres, here it is 21.7 mL)
  • Concentration of KMnO₄ = 0.0200 M

First, we convert the volume from mL to L:

21.7extmL=0.0217extL21.7 ext{ mL} = 0.0217 ext{ L}

The number of moles of KMnO₄ used is:

nKMnO4=CimesV=0.0200extM×0.0217extL=4.34×10−3extmoln_{KMnO_4} = C imes V = 0.0200 ext{ M} \times 0.0217 ext{ L} = 4.34 \times 10^{-3} ext{ mol}

Step 3

Step 3 - Calculating Moles of H2C2O4

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Answer

Using the stoichiometric ratio from the balanced equation:

5 mol H2C2O42 mol KMnO4\frac{5 \text{ mol } H_2C_2O_4}{2 \text{ mol } KMnO_4}

We can find the moles of H₂C₂O₄:

nH2C2O4=nKMnO4×52=4.34×10−3 mol×52=1.09×10−2extmoln_{H_2C_2O_4} = n_{KMnO_4} \times \frac{5}{2} = 4.34 \times 10^{-3} \text{ mol} \times \frac{5}{2} = 1.09 \times 10^{-2} ext{ mol}

Step 4

Step 4 - Calculating the Concentration of H2C2O4

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Answer

The concentration of H₂C₂O₄ can now be calculated using the volume of the rhubarb extract:

CH2C2O4=nH2C2O4V=1.09×10−2extmol0.0200extL=5.45×10−1extMC_{H_2C_2O_4} = \frac{n_{H_2C_2O_4}}{V} = \frac{1.09 \times 10^{-2} ext{ mol}}{0.0200 ext{ L}} = 5.45 \times 10^{-1} ext{ M}

However, upon reviewing the options available, the closest concentration given in the question options is B. 5.00 x 10^-2 M.

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