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Pre-U chemistry titration question

Hi,

I have attempted this question a number of times and I'm completely stuck - I can see what you're meant to do, but the numbers I get never match up to the question.

1.575 g of hydrated ethanedioic acid, H2C2O4.xH20, was dissolved in 250 cm^3 distilled water. In a titration, 25.0 cm^3 of this acid solution was neutralised by exactly 25.0 cm^3 of dilute sodium hydroxide, concentration 0.100 mol dm^3. The equation is shown below:

H2C2O4 +2NaOH --> Na2C2O4 + 2H20

a) Calculate the number of moles of anhydrous ethanedioic acid in the weighed sample of hydrated ethanedioic acid.

b) Calculate the mass of anhydrous ethanedioic acid in the weighed sample and thus calculate the mass of water of crystallisation in the hydrated acid.

c) Calculate the number of moles of water of crystallisation in the hydrated acid and hence state the formula of the hydrated acid.

Anyone know what to do?

thebrainloves x
Reply 1
Original post by thebrainloves
Hi,

I have attempted this question a number of times and I'm completely stuck - I can see what you're meant to do, but the numbers I get never match up to the question.

1.575 g of hydrated ethanedioic acid, H2C2O4.xH20, was dissolved in 250 cm^3 distilled water. In a titration, 25.0 cm^3 of this acid solution was neutralised by exactly 25.0 cm^3 of dilute sodium hydroxide, concentration 0.100 mol dm^3. The equation is shown below:

H2C2O4 +2NaOH --> Na2C2O4 + 2H20

a) Calculate the number of moles of anhydrous ethanedioic acid in the weighed sample of hydrated ethanedioic acid.

b) Calculate the mass of anhydrous ethanedioic acid in the weighed sample and thus calculate the mass of water of crystallisation in the hydrated acid.

c) Calculate the number of moles of water of crystallisation in the hydrated acid and hence state the formula of the hydrated acid.

Anyone know what to do?

thebrainloves x


Do you get
a) 0.0125

b) 1.125g and 0.45g water

c) 0.025 moles of water
so 0.0125 :0.025 = 1:2
so formula x =2

That's what I got, if it's correct I can show you the steps.
Original post by Mule
Do you get
a) 0.0125

b) 1.125g and 0.45g water

c) 0.025 moles of water
so 0.0125 :0.025 = 1:2
so formula x =2

That's what I got, if it's correct I can show you the steps.


I got the same for a... And that's where I go wrong. The first part of b I was out by a factor of 10 and then I got crazy numbers for the mass of water... I'm lost...
Reply 3
Original post by thebrainloves
I got the same for a... And that's where I go wrong. The first part of b I was out by a factor of 10 and then I got crazy numbers for the mass of water... I'm lost...


So do you have the actual answers? Or are you just saying because it seems like an unreasonable answer you got?
I got 65 for x, first time... Second, I got 3.25.
Reply 5
Original post by thebrainloves
I got 65 for x, first time... Second, I got 3.25.


Do you want me to show you how I did it?

I can't guarantee it's right though.
Reply 6
The factor of 10 comes in when you work out the number of moles of ethanedioic acid in 25 cm3 of solution hence the number of moles in the 250 cm3 made up = 10 x that number
Reply 7
Original post by Mule
Do you want me to show you how I did it?

I can't guarantee it's right though.


You have got it correct Mule!
Reply 8
Original post by GDN
You have got it correct Mule!


Okay cool :smile:

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