Difficult chemistry, crystallisation of oxalic acid??

Question

A titration was completed by placing 0.075g of oxalic acid with 25ml of water & 10ml of 05M H2SO4, this was then heated to 40 0c

After this a burette was filled with KMnO4 and this was added until the solution was neutralised, the average volume of KMnO4 added to achieve this was 23.3cm3

How would I find the number of water molecules of crystallisation in oxalic acid using the average volume of KMnO4 and the weight of the oxalic acid? I think the molar mass is also used.

Also what is the purpose of adding H2SO4?

please help i'm very confused, and it would be much appreciated.

Answer 1

we need to find the moles of H2C2O4 & the moles of H2O

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the molarity of your KMnO4 is missing,

we need it,

I have worked your problem in reverse I believe that your KMnO4 was likely a 0.01021 Molar solution

you need to change these calculations, however using your KMnO4 Molarity:

find moles of KMnO4, from the volume used ...by using its Molarity:

0.0233 litres @ 0.01021 mol/litre = 2.3766 e-4 moles KMnO4

using the equation that 5 H2C2O4 & 2 KMnO4 --> , find the moles of oxalic acid:

2.3766 e-4 mol KMnO4 @ 5 mol H2C2O4 / 2 mol KMnO4 = 5.942e-4 moles of H2C2O4

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now find the grams of H2C2O4 using its molar mass:

5.942e-4 mol H2C2O4 @ 90.03 g/mol = 0.05349 grams of H2C2O4

find the grams of water in the sample:

0.075 grams total - 0.535 grams acid = 0.0215 grams of H2O

find the moles of water  using its molar mass:

0.0215 moles of H2O @ 18.02 g / mol = 0.00119 moles of water

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now look @ the moles:

H2C2O4 = 5.94e-4 moles

H2O = 0.00119 moles

ratio the moles:

H2C2O4 = 5.94e-4 moles / 5.94e-4 = 1 mole H2C2O4

H2O = 0.00119 moles / 5.94e-4 = 2 moles H2O

your answer is: H2C2O4 dot 2 H2O

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an edit: the rest of your question:

KMnO4 is an oxidizer that goes from purple KMnO4 to colorless Mn+2 , when the reaction is done in an acid environment.

Otherwise the reaction usually produces the brown precipitate MnO2, which is a bit of a pain to work with.

When doing a titration, it is so much easier to have Mn+2 as your product.

to get a balanced equation:

KMnO4 takes 5 electrons --> Mn+2   (when Mn+7 --> Mn+2)

H2C2O4 --> 2 CO2 loses 2 electrons  (when 2 C+3 --> 2 C+4)

to balance the electrons lost & taken:

KMnO4    &   5 e- --> Mn+2 .....becomes:

2 KMnO4    &   10 e- --> 2 Mn+2

H2C2O4 --> 2 CO2   & 2 e-   ... becomes:

5 H2C2O4 --> 10  CO2   & 10 e-

combining gives:

2KMnO4 & 5 H2C2O4 --> 2Mn+2 & 2K+   & 10 CO2

adding H2SO4 provides sulfate for Mn+2 & K+, as well as makes water out of  KMnO4 's oxygens:

2 KMnO4 & 5 H2C2O4  & 3 H2SO4 --> 2 MnSO4 & K2SO4   & 10 CO2 & 8 H2O

Answer 2

Potassium permanganate reacts with oxalic acid in a redox reaction - the reaction is complicated and I don't know the stoichiometry - I guess you need to assume a 1:1 reaction. It does follow first order kinetics. You also need to know the concentration of the potassium permanganate solution - you do not give this. I don't know the purpose of the sulfuric acid - probably to prevent any acid-base reactions with the oxalate.

From the volume and concentration of the potassium permanganate find the mol of oxalic acid. If it is a 1:1 reaction mol oxalic acid = mol potassium permanganate = M x V(in L)

From mol oxalic acid find the mass of oxalic acid. Then the initial mass minus the calculated mass gives you the mass of water in your original sample. Calculate the # mol water then find the ratio mol oxalic acid/mol water.