Titration question

cjm1998

Could someone please help me solve this? Thanks! :smile:

Succinicacid has the formula (CH₂)_n(COOH)₂ and reactswith dilute sodium hydroxide as follows: (CH₂)_n(COOH)₂+ 2NaOH à (CH₂)_n(COONa)₂+ 2H₂O2.0 g of succinic acid were dissolved in water and the solution made upto 250 cm³. This solution was placed in a burette and 18.4 cm³was required to neutralise 25 cm³ of 0.1 moldm^-3 NaOH.Deduce the molecular formula of the acid and hence the value of n.

Reply 1

Kvothe the Arcane

Original post by cjm1998

Could someone please help me solve this? Thanks! :smile:

Hi there. What have you tried?

Reply 2

cjm1998

Original post by Kvothe the arcane

Hi there. What have you tried?

I have tried working out the number of moles of the sodium hydroxide (n= v x c / 1000) then dividing the answer by two to find the number of moles of the succinic acid (because of the one to two mole ratio). This gave me 1.25 x 10^-3 as the number of moles of the acid. I then divided 2.0g by this number of moles to find the molecular mass of the acid, but my answer came out as 1600. The answer booklet says n should equal 2, but with this Mr I can't seem to have that as my answer.

Reply 3

Kvothe the Arcane

Original post by cjm1998

I have tried working out the number of moles of the sodium hydroxide (n= v x c / 1000) then dividing the answer by two to find the number of moles of the succinic acid (because of the one to two mole ratio). This gave me as the number of moles of the acid. I then divided 2.0g by this number of moles to find the molecular mass of the acid, but my answer came out as 1600. The answer booklet says n should equal 2, but with this Mr I can't seem to have that as my answer.

The major thing that stands out for me is that you have worked out moles for 18.4 cm3 but you need to scale up for 250cm3

Edit: I have done the question and your thinking is correct. Once you find out the number of moles in the flask after considering that you 1.25 x 10^-3 is only a proportion of the total, the answer follows.

(edited 8 years ago)

Reply 4

cjm1998

Original post by Kvothe the arcane

The major thing that stands out for me is that you have worked out moles for 18.4 cm³but you need to scale up for 250cm³

Okay thank you :smile:

I'm a bit confused though, I thought the 18.4cm^3 had to be used because I am working out how much of it was actually used? :smile:

Reply 5

Kvothe the Arcane

Original post by cjm1998

Okay thank you :smile:

I'm a bit confused though, I thought the 18.4cm^3 had to be used because I am working out how much of it was actually used? :smile:

You haven't used 18.4cm^3 in your calculations.

When you work out the moles of NAOH and use stochiometry to work out the moles of the Succinicacid, you are actually working out the moles of the titre. I.e. how much Succinicacid is used to neutralise the sodium hydroxide.

But... you have to remember that not all of the 2.0g of Succinicacid is used for the neutralization. It's made up to 250cm^3 and a fraction is necessary to neutralise the base. So there will be unused Succinicacid in the flask.

Do you see why we have to "scale up" to work out the formula of Succinicacid?

(edited 8 years ago)

Reply 6

Dr Cobbold

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