The Formula of Succinic Acid

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Recommended: Empirical formula for succinic acid

The Formula of Succinic Acid

Succinic acid is a diprotic, which means it donates two protons per

molecule. Succinic acid can be completely neutralised by sodium

hydroxide. The indicator most suitable for this experiment is

phenolphthalein, it is colourless in acids and pink in alkalises. The

half way stage is about pH 9.3, this is when it will either change

from colourless to a very pale pink or from pink to colourless. To

determine the relative formula mass of succinic acid I am going to do

a titration against sodium hydroxide.

The equation for the reaction is given below. To make the equation

easier to read, HOOC(CH2)nCOOH will be condensed to H2A because of the

two hydrogen atoms at either end.

H2A+2NaOH à Na2A+2H2O

(CV) H2A = 1

(CV) NaOH 2

I am going to use the NaOH as 0.1M because I don’t want it too

concentrated, so therefore I am going to use H2A as 0.05M because of

the ratio 2:1.

In the formula of succinic acid below n is a whole number between 1

and 4. So therefore first I need to calculate the relative molecular

mass of succinic acid.

HOOC(CH2)nCOOH H = 1 O = 16 C = 12

Mr when n = 1 1+16+16+12 (12+2) 12+16+16+1 = 104

Mr when n = 2 1+16+16+12 [(12+2) x2] 12+16+16+1 = 118

Mr when n = 3 1+16+16+12 [(12+2) x3] 12+16+16+1 = 132

Mr when n = 4 1+16+16+12 [(12+2) x4] 12+16+16+1 = 146

From these calculations I can see that I need between 104g and 146g in

1 litre to equal 1M. But I want the solution in 250cm3, so therefore I

need to divide the weights by 4:

n = 1 104 = 26g So I need between 26g and 36.5g in 250cm3

to make a 1M

4 solution.

n = 4 146 = 36.5g

4

I also want to make the solution to 0.05M because of the ratio 2:1, so

therefore I need to multiply each weight by 0.05.

n = 1 26 x 0.05 = 1.3g

n = 4 36.5 x 0.05 = 1.8205g

So the range I can work with to weigh out the anhydrous succinic acid

is from 1.3g to 1.82g,which will make a 0.05M solution in 250cm3.

Preparing a standard solution

=============================

Having calculated the weight I can use (1.3g-1.82g), I must weigh out

the solute using an accurate electronic balance that goes to three

decimal places. I must make sure I clean the balance with a fine

brush assuming that it may not have been cleaned after the last time

it was used and set the balance back to 0.

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