experiment to find the concentration of the ethanedioic acid
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Introduction
A2 Chemistry Experiment 1 - Redox Titration As this is a redox reaction there are two half equations: 5C2O42- -10e- ==> 10CO2 and 2MnO4- + 16H+ + 10e- ==> 2Mn2+ + 8H2O Combining these 2 half equations we get: 5C2O42- + 2MnO4- + 16H+ ==> 10CO2 + 2Mn2+ + 8H2O Therefore to find the concentration of the ethanedioic acid I can titrate with Potassium Permanganate which acts as an oxidising agent, with Ethanedioic acid acting as a reducing agent. The sulphuric acid will act as an acid catalyst, providing extra H+. H2SO4 IN EXCESS The Sulphuric acid is in excess to ensure all the ethanedioic acid will react with the KMnO4 until the equivalence point is reached. I will use a 0.05mol dm-3 concentration of KMnO4. Method: * Set up titration apparatus, and fill the burette with 0.05M Potassium Permanganate, fill up to zero on the burette, use a funnel and place apparatus on a stool if necessary. ...read more.
Middle
And finally to get the concentration of H2C2O4 simply do moles/volume, so moles of H2C2O4 / 0.025dm3 = concentration of H2C2O4 Experiment 2 I can use a gas collection to find the overall concentration of the acid mixture, and as I already have the concentration of H2C2O4 , I can simply minus that from the total concentration to get the concentration of the H2SO4. Both strong and weak acids will react with a metal, therefore I will react the acid mixture with magnesium metal. Quantities of chemicals to use As I have been told that the concentrations are approximately 0.1 mol dm-3 H2C2O4 and 0.2 mol dm-3 H2SO4, I can say that the total concentration is approximately 0.1+0.2=0.3 mol dm-3. I am going to use 25cm3 of the acid mixture, therefore total moles of acid mixture is 0.3 x 0.025 = 0.0075 moles H2SO4 + Mg ==> MgSO4 + H2 and H2C2O4 + Mg ==> MgC2O4 + H2 I need these in one equation so I can get the molar ratios, therefore I formed this ionic equation. ...read more.
Conclusion
Diagram: Safety Ethanedioic acid is a highly poisonous carboxylic acid. It is corrosive and may cause burns. If it comes to contact with skin rinse with plenty of water immediately. Wear a lab coat to prevent exposure to skin. Specimen calculation If 22cm3 of KMnO4 was needed for the redox titration, then 0.022 x 0.05= 1.1x10-3 moles of KMnO4 . Using 5:2 ratio of 5C2O42- 2MnO4- moles of H2C2O4 is (1.1x10-3 /2) x5=2.75x10-3 moles, and finally to get concentration 2.75x10-3 /0.025=0.11 mol dm-3. Which is almost 0.1mol dm-3 as suggested the concentration was approximately. The above was all from experiment 1, and now to get the concentration of the Sulphuric acid was experiment 2. If the volume of gas collected was 80cm3, then moles of Hydrogen would be (80/1000)/24=0.0033moles, however 4:2 ratio with acid so multiply by 2 to get 0.0066moles, and as I used 25cm3 acid, concentration is 0.0066/(25/1000)=0.27mol dm-3. To get the H2SO4 concentration minus the H2C2O4 concentration (0.11), 0.27-0.11=0.16mol dm-3, this is close to 0.20mol dm-3 ?? ?? ?? ?? 1 ...read more.
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