# Etha

Solution X contains a mixture of a dibasic acid, ethanedioic acid H2C2O4, and a salt, sodium ethanedioate Na2C2O4. A student performed two titrations to determine the concentrations of ethanedioic acid and sodium ethanedioate in solution X.

In the first titration, 25.0 cm3 of solution X required 14.75 cm3 of 0.1 mol/dm3 sodium hydroxide solution for neutralisation.

In the second titration, 25.0 cm3 of solution X required 32 cm3 of 0.0205 mol/dm3 potassium manganate (VII) solution for complete oxidation. In the oxidation reaction, all the (bivalent ethanedioate) C2O4 ions are oxidised to CO2 in acidic conditions.

(a) Write the equation for the reaction between ethanedioic acid H2C2O4 abd sodium hydroxide solution to form sodium ethanedioate Na2C2O4 and water. Use this equation and the results if the first titration to calculate tge concentration of ethanedioic acid in solution X.

(b) (i) Construct a balanced ionic equation for the reaction between (bivalent ethanedioate) ions C2O4 and (univalent) manganese (VII) ions MnO4-.

(b) (ii) Using the balanced equation in (b)(i) abd the result of the second titration, calculate the total amount of (bivalent ethanedioate) C2O4 ions present in 25 cm3 of solution X.

(c) Hence, calculate the concentration of sodium ethanedioate in solution X.
Original post by Dennis 106
Solution X contains a mixture of a dibasic acid, ethanedioic acid H2C2O4, and a salt, sodium ethanedioate Na2C2O4. A student performed two titrations to determine the concentrations of ethanedioic acid and sodium ethanedioate in solution X.

In the first titration, 25.0 cm3 of solution X required 14.75 cm3 of 0.1 mol/dm3 sodium hydroxide solution for neutralisation.

In the second titration, 25.0 cm3 of solution X required 32 cm3 of 0.0205 mol/dm3 potassium manganate (VII) solution for complete oxidation. In the oxidation reaction, all the (bivalent ethanedioate) C2O4 ions are oxidised to CO2 in acidic conditions.

(a) Write the equation for the reaction between ethanedioic acid H2C2O4 abd sodium hydroxide solution to form sodium ethanedioate Na2C2O4 and water. Use this equation and the results if the first titration to calculate tge concentration of ethanedioic acid in solution X.

(b) (i) Construct a balanced ionic equation for the reaction between (bivalent ethanedioate) ions C2O4 and (univalent) manganese (VII) ions MnO4-.

(b) (ii) Using the balanced equation in (b)(i) abd the result of the second titration, calculate the total amount of (bivalent ethanedioate) C2O4 ions present in 25 cm3 of solution X.

(c) Hence, calculate the concentration of sodium ethanedioate in solution X.
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Original post by charco
You appear to have typed out a question in full.
Which part are you having difficulties with?
Part (c). Thanks
Original post by Dennis 106
Part (c). Thanks
You need these to work out c)
The equation for the reaction is:
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
Using this equation and the results of the first titration:
Volume of H2C2O4 solution × Molarity of H2C2O4
= Volume of NaOH solution × Molarity of NaOH
25 cm3 x Molarity of H2C2O4
= 14.75 cm3 x 0.1 mol/dm3
Molarity of H2C2O4
= 14.75 cm3 x 0.1 mol/dm3 divided by 25 cm3
= 0.059 mol/dm3.
==
The balanced ionic equation for the reaction between C₂O₄{2-} and MnO₄{-}is:
5C₂O₄{2-} + 2MnO₄{-} + 8H{+} 10CO₂ + 2Mn{2+} + 4H₂O

From the second titration,
Volume of KMnO₄ solution used = 32 cm3 = 0.032 dm3.
Concentration of KMnO₄ solution = 0.0205 mol/dm³
Number of moles of MnO₄ ions used
= 0.032 dm3 x 0.0205 mol/dm³)
= 0.000656 mol

5 mol C2O4 react with 2 mol MnO4.
In other words,
2.5 mol C2O4 react with 1 mol MnO4.
Number of moles of bivalent C2O4 ions
= 0.000656 mol MnO4 x 2.5 mol C2O4
= 0.00164 mol.
Original post by Dennis 106
The equation for the reaction is:
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
Using this equation and the results of the first titration:
Volume of H2C2O4 solution × Molarity of H2C2O4
= Volume of NaOH solution × Molarity of NaOH
25 cm3 x Molarity of H2C2O4
= 14.75 cm3 x 0.1 mol/dm3
Molarity of H2C2O4
= 14.75 cm3 x 0.1 mol/dm3 divided by 25 cm3
= 0.059 mol/dm3.
==
The balanced ionic equation for the reaction between C₂O₄{2-} and MnO₄{-}is:
5C₂O₄{2-} + 2MnO₄{-} + 8H{+} 10CO₂ + 2Mn{2+} + 4H₂O

From the second titration,
Volume of KMnO₄ solution used = 32 cm3 = 0.032 dm3.
Concentration of KMnO₄ solution = 0.0205 mol/dm³
Number of moles of MnO₄ ions used
= 0.032 dm3 x 0.0205 mol/dm³)
= 0.000656 mol

5 mol C2O4 react with 2 mol MnO4.
In other words,
2.5 mol C2O4 react with 1 mol MnO4.
Number of moles of bivalent C2O4 ions
= 0.000656 mol MnO4 x 2.5 mol C2O4
= 0.00164 mol.
Your ionic equation is not balanced.

MnO4- + 8H+ + 5e ===> Mn2+ + 4H2O

C2O42- ==> 2CO2 + 2e
----------------------------------------------------------------------------------------- combine
2MnO4- + 16H+ + 10e ===> 2Mn2+ + 8H2O
5C2O42- ==> 10CO2 + 10e
-----------------------------------------------------------------------------------------
2MnO4- + 16H+ + 5C2O42- ==> 2Mn2+ + 8H2O + 10CO2

However the ratio is still 2 x manganate to 5 x ethandioate

First titration gives mol of acid only:
Mol NaOH = 0.001475
Therefore mol of acid = 0.001475/2 = 7.375 x 10-4 mol in 25ml

Second titration gives total moles of ethandioate
In the second titration, 25.0 cm3 of solution X required 32 cm3 of 0.0205 mol/dm3 potassium manganate (VII) solution for complete oxidation.

In the oxidation reaction, all the (bivalent ethanedioate) C2O4 ions are oxidised to CO2 in acidic conditions.

mol manganate = 0.032 x 0.0205 = 6.56 x 10-4
Hence total mol ethandioate = 5/2 x 6.56 x 10-4 = 1.64 x 10-3 mol in 25ml

Hence mol of sodium ethandioate in 25ml = total ethandioate ethandioic acid
= 1.64 x 10-3 mol - 7.375 x 10-4 mol = 9.025 x 10-4

Hence concentration = 9.025 x 10-4/0.025 = 0.0361 mol dm-3
Original post by charco
Your ionic equation is not balanced.

MnO4- + 8H+ + 5e ===> Mn2+ + 4H2O

C2O42- ==> 2CO2 + 2e
----------------------------------------------------------------------------------------- combine
2MnO4- + 16H+ + 10e ===> 2Mn2+ + 8H2O
5C2O42- ==> 10CO2 + 10e
-----------------------------------------------------------------------------------------
2MnO4- + 16H+ + 5C2O42- ==> 2Mn2+ + 8H2O + 10CO2

However the ratio is still 2 x manganate to 5 x ethandioate

First titration gives mol of acid only:
Mol NaOH = 0.001475
Therefore mol of acid = 0.001475/2 = 7.375 x 10-4 mol in 25ml

Second titration gives total moles of ethandioate
In the second titration, 25.0 cm3 of solution X required 32 cm3 of 0.0205 mol/dm3 potassium manganate (VII) solution for complete oxidation.

In the oxidation reaction, all the (bivalent ethanedioate) C2O4 ions are oxidised to CO2 in acidic conditions.

mol manganate = 0.032 x 0.0205 = 6.56 x 10-4
Hence total mol ethandioate = 5/2 x 6.56 x 10-4 = 1.64 x 10-3 mol in 25ml

Hence mol of sodium ethandioate in 25ml = total ethandioate ethandioic acid
= 1.64 x 10-3 mol - 7.375 x 10-4 mol = 9.025 x 10-4

Hence concentration = 9.025 x 10-4/0.025 = 0.0361 mol dm-3
Thank you very much, Charco, for your help