This question is about compounds containing ethanedioate ions. A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4·2H2O) and an inert solid. A volumetric flask contained 1.90 g of this solid mixture in 250 cm³ of aqueous solution. Two different titrations were carried out using this solution. In the first titration 25.0 cm³ of the solution were added to an excess of sulfuric acid in a conical flask. The flask was then correctly titrated with a 0.200 mol dm⁻³ solution of potassium manganate(VII). When 26.50 cm³ of potassium manganate(VII) had been added the solution changed color. The equation for this reaction is MnO4⁻ + 5C2O4²⁻ + 16H⁺ → 2Mn²⁺ + 8H2O + 10CO2 In the second titration 25.0 cm³ of the solution were titrated with a 0.100 mol dm⁻³ solution of sodium hydroxide using phenolphthalein as an indicator. The indicator changed color after the addition of 10.45 cm³ of sodium hydroxide solution. The equation for this reaction is H2C2O4 + 2OH⁻ → C2O4²⁻ + 2H2O Calculate the percentage by mass of sodium ethanedioate in the white solid. Give your answer to the appropriate number of significant figures.

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This question is about compounds containing ethanedioate ions - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

Question 11

This question is about compounds containing ethanedioate ions. A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4·2H2O)... show full transcript

Worked Solution & Example Answer:This question is about compounds containing ethanedioate ions - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

Step 1

Calculate the moles of potassium manganate(VII) used in the first titration

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Answer

To find the moles of potassium manganate(VII) used in the titration, we can use the formula:

ext{Moles} = ext{concentration} imes ext{volume}

Using the given values:

d = 0.200 mol dm⁻³ and V = 26.50 cm³ = 0.0265 dm³,

ext{Moles} = 0.200 imes 0.0265 = 5.30 imes 10^{-3} ext{ mol}

Therefore, the moles of potassium manganate(VII) used are $5.30 imes 10^{-3}$ .

Step 2

Calculate the moles of C2O4²⁻ involved in the reaction

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Answer

From the equation, the mole ratio of MnO4⁻ to C2O4²⁻ is 1:5. Therefore, the moles of C2O4²⁻ reacted can be calculated as:

ext{Moles of C2O4²⁻} = 5 imes ext{moles of MnO4⁻} = 5 imes 5.30 imes 10^{-3} = 2.65 imes 10^{-2} ext{ mol}.

Step 3

Calculate the moles of sodium ethanedioate in the original sample

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Answer

In the second titration, we find the amount of H2C2O4 used. The volume of NaOH used is 10.45 cm³ = 0.01045 dm³, and the concentration is 0.100 mol dm⁻³:

ext{Moles of NaOH} = 0.100 imes 0.01045 = 1.045 imes 10^{-3} ext{ mol}.

From the equation, 1 mole of H2C2O4 reacts with 2 moles of OH⁻:

ext{Moles of H2C2O4} = rac{1.045 imes 10^{-3}}{2} = 5.225 imes 10^{-4} ext{ mol}.

Now, the total moles of oxalate in the original sample is:

ext{Moles of C2O4²⁻} = 5.225 imes 10^{-4} + 2.65 imes 10^{-2} = 2.70225 imes 10^{-2} ext{ mol}.

Step 4

Calculate the mass of sodium ethanedioate in the original sample

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Answer

To find the mass of sodium ethanedioate (Na2C2O4) in the sample, we use its molar mass:

Molar mass of Na2C2O4 = 2(23) + 2(12) + 4(16) = 98 g/mol.

Now, we calculate:

ext{Mass} = ext{moles} imes ext{molar mass} = 2.70225 imes 10^{-2} imes 98 = 2.645215 ext{ g}.

Step 5

Calculate the percentage by mass of sodium ethanedioate

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Answer

Finally, to calculate the percentage of sodium ethanedioate in the original sample:

ext{Percentage} = rac{ ext{mass of Na2C2O4}}{ ext{total mass of sample}} imes 100 = rac{2.645215}{1.90} imes 100 ightarrow 139.18 ext{ %}.

This indicates that there are significant quantities of sodium ethanedioate, suggesting that not all of the mass may be contributing to the sample tested.

Step 6

Write an equation for the ligand substitution reaction

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Answer

The ligand substitution reaction between ethanedioate ions and iron(III) ions can be represented as:

[ ext{Fe(H2O)}_6]^{3+} + 3 ext{C}_2 ext{O}_4^{2-} ightarrow [ ext{Fe(C}_2 ext{O}_4)_3]^{3-} + 6 ext{H}_2 ext{O}.

Step 7

Suggest why the value of the enthalpy change is close to zero

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Answer

The value of the enthalpy change for this ligand substitution reaction is close to zero because the reaction involves the substitution of one ligand for another without altering the overall structure, resulting in minimal change in energy.

Step 8

Draw the displayed formula of the iron complex

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Answer

The displayed formula of the iron complex [Fe(C2O4)3]³⁻ can be illustrated with the iron atom bonded to three ethanedioate ligands:

     O       O
      \     /
       C -- C
        |   |
      O     O
        \   /
         Fe
        /   \
       O     O
      /       \
     C         C
      \       /
       O     O

The iron is at the center with the ethanedioate ligands surrounding it.

Step 9

Indicate the value of the O—Fe—O bond angle

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Answer

The bond angle O—Fe—O in the iron complex is typically around 90° or 180°, depending on the arrangement of the ligands.

Step 10

State the type of isomerism shown by the iron complex

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Answer

The type of isomerism shown by the iron complex is geometrical isomerism, due to the possible different spatial arrangements of the ligands.

This question is about compounds containing ethanedioate ions - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

Worked Solution & Example Answer:This question is about compounds containing ethanedioate ions - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

Calculate the moles of potassium manganate(VII) used in the first titration

Calculate the moles of C2O4²⁻ involved in the reaction

Calculate the moles of sodium ethanedioate in the original sample

Calculate the mass of sodium ethanedioate in the original sample

Calculate the percentage by mass of sodium ethanedioate

Write an equation for the ligand substitution reaction

Suggest why the value of the enthalpy change is close to zero

Draw the displayed formula of the iron complex

Indicate the value of the O—Fe—O bond angle

State the type of isomerism shown by the iron complex

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