The Formula of Succinic Acid :: GCSE Chemistry Coursework Investigation

The Formula of succinic Acidsuccinic acid is a diprotic, which means it donates two protons permolecule. Succinic acid can be completely neutralised by sodiumhydroxide. The indicator most suitable for this experiment isphenolphthalein, it is colourless in acids and pinkish in alkalises. Thehalf way stage is about pH 9.3, this is when it ordain either changefrom colourless to a very pale pink or from pink to colourless. Todetermine the relative formula mass of succinic acid I am libe proportionn to doa titration against sodium hydroxide. The equation for the reaction is given below. To contribute got the equationeasier to read, HOOC(CH2)nCOOH will be condensed to H2A because of thetwo hydrogen atoms at either end.H2A+2NaOH Na2A+2H2O(CV) H2A = 1(CV) NaOH 2I am going to use the NaOH as 0.1M because I dont extremity it tooconcentrated, so therefore I am going to use H2A as 0.05M because ofthe ratio 21.In the formula of succinic acid below n is a whole number surrounded by 1and 4. So therefore first I strike to imagine the relative molecularmass of succinic acid.HOOC(CH2)nCOOH H = 1 O = 16 C = 12Mr when n = 1 1+16+16+12 (12+2) 12+16+16+1 = 104Mr when n = 2 1+16+16+12 (12+2) x2 12+16+16+1 = 118Mr when n = 3 1+16+16+12 (12+2) x3 12+16+16+1 = 132Mr when n = 4 1+16+16+12 (12+2) x4 12+16+16+1 = 146From these calculations I can see that I need betwixt 104g and 146g in1 litre to equal 1M. But I want the solution in 250cm3, so therefore Ineed to grant the viewts by 4n = 1 104 = 26g So I need between 26g and 36.5g in 250cm3to make a 1M4 solution.n = 4 146 = 36.5g4I also want to make the solution to 0.05M because of the ratio 21, sotherefore I need to multiply individually 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 fetch with to weigh out the anhydrous succinic acidis from 1.3g to 1.82g,which will make a 0.05M solution in 250cm3.Preparing a standard solution=============================Having work out the weight I can use (1.3g-1.82g), I must weigh outthe solute using an accurate electronic balance that goes to three ten-fold places. I must make sure I clean the balance with a finebrush assuming that it may not have been cleaned after the last timeit was used and set the balance affirm to 0.