Question

2) A sample of an unknown organic acid weighing 0.500 grams is dissolved in water to make 100 ml of solution. It is then titrated with 0.100 N base. Calculate the equivalent weight of the acid from the following data, which list the absorbance of the solution at volumes of base added: 0 ml = 0.72; 4.00 ml = 0.63; 7.91 ml = 0.55; 12.10 ml = 0.46; 16.15 ml = 0.40; 20.02 ml = 0.40; 24.15 ml 0.39.

Answer 1

(let's consider volume is proportional to mols)

total amount absorbed=3.55

total volume of bse spent=84.33ml

number of acid mols spent= 0.1 x 84.33 x 10^-3

                                      = 8.433x 10^-3 mol

(3.5/84.33)x 5=0.20g

percentage of acid in original sample= (0.2/5) x 100% =4%