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Titration calculations

Can anyone help with a method of how to complete this question:

A 0.1575g sample of ethanedioic acid crystals, H2C2O4.nH20 was dissolved in water. In a titration, 25.0 cm3 of this solution of acid reacted with exactly 15.6 cm3 of 0.16 mol/dm3 NaOH. Calculate the value of Mr of the acid and n.

I've worked out the moles of NaOH but not sure where to go from there.

Any help appreciated :smile:
Original post by Revise_
Can anyone help with a method of how to complete this question:

A 0.1575g sample of ethanedioic acid crystals, H2C2O4.nH20 was dissolved in water. In a titration, 25.0 cm3 of this solution of acid reacted with exactly 15.6 cm3 of 0.16 mol/dm3 NaOH. Calculate the value of Mr of the acid and n.

I've worked out the moles of NaOH but not sure where to go from there.

Any help appreciated :smile:


Go to the moles of acid in the 25ml aliquot, remembering that there are two acidic protons in the acid...
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Original post by charco
Go to the moles of acid in the 25ml aliquot, remembering that there are two acidic protons in the acid...



I worked out the number of moles of the acid by dividing the moles of NaOH by 2, which gave me 1.248 x 10-3, because the equation has 2 moles of NaOH to one of ethanedioc acid. To work out the Mr of the acid do I have to divide the 0.1575 by the moles of acid? The value I get is too small to be the Mr.

Thanks for the help
Original post by Revise_
I worked out the number of moles of the acid by dividing the moles of NaOH by 2, which gave me 1.248 x 10-3, because the equation has 2 moles of NaOH to one of ethanedioc acid. To work out the Mr of the acid do I have to divide the 0.1575 by the moles of acid? The value I get is too small to be the Mr.

Thanks for the help


I presume that the original mass was dissolved in water and MADE UP to 250 ml (the usual procedure)

Now you have the moles of acid in 25ml

You must now multiply to find the moles of acid in 250ml

THEN you divide the mass by the moles
(edited 6 years ago)
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Revise_
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Original post by charco
I presume that the original mass was dissolved in water and MADE UP to 250 ml (the usual procedure)

Now you have the moles of acid in 25ml

You must now multiply to find the moles of acid in 250ml

THEN you divide the mass by the moles


Thank you this really helped :smile:

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