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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