Help on A Level Chemistry Question
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 and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.
The equation for this reaction is
2MnO4− + 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 colour 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.
I don’t understand why MnO4- ions react with ethanedioic and ethanedioate.
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 and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.
The equation for this reaction is
2MnO4− + 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 colour 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.
I don’t understand why MnO4- ions react with ethanedioic and ethanedioate.
Original post by Inayawajid123
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 and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.
The equation for this reaction is
2MnO4− + 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 colour 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.
I don’t understand why MnO4- ions react with ethanedioic and ethanedioate.
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 and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.
The equation for this reaction is
2MnO4− + 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 colour 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.
I don’t understand why MnO4- ions react with ethanedioic and ethanedioate.
Check the responses in the thread from five years ago to have some spark.
See here: Help on A Level Chemistry Question
Kind regards,
(edited 1 month ago)
Original post by Inayawajid123
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 and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.
The equation for this reaction is
2MnO4− + 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 colour 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.
I don’t understand why MnO4- ions react with ethanedioic and ethanedioate.
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 and contents were heated to 60 oC and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.
The equation for this reaction is
2MnO4− + 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 colour 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.
I don’t understand why MnO4- ions react with ethanedioic and ethanedioate.
Think of ethanedioate and ethanedioic acid as being equivalents of one another as ethanedioate is just fully dissociated ethanedioic acid.
In the titration, you add a strong acid anyway (H2SO4), which leads to the re-protonation of ethanedioate back to ethanedioic acid. Essentially the ethanedioic acid and ethanedioate from beforehand behave identically under the conditions used in the titration and so both should be capable of undergoing the same reactions.
(edited 1 month ago)
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