Analytical Chem istry - DePauw University

452 Analytical Chemistry 2.00. 1005 MNaOH× 0. 04268 LNaOH = 4.289× 10 −3mol NaOHwhich gives the analyte’s equivalent weight asganalyte 0.2521 gEW = == 58.78 g/equivalent− 3equiv. analyte 4.289×10 equiv.The possible formula weights for the weak acid aren = 1: FW = EW = 5878 . g/equivalentn = 2:FW = 2× EW = 117.6 g/equivalentn = 3: FW = 3× EW = 176.3 g/equivalentIf the analyte is a monoprotic weak acid, then its formula weight is 58.78g/mol, eliminating ascorbic acid as a possibility. If it is a diprotic weak acid,then the analyte’s formula weight is either 58.78 g/mol or 117.6 g/mol, dependingon whether the titration is to the first or second equivalence point.Succinic acid, with a formula weight of 118.1 g/mole is a possibility, butmalonic acid is not. If the analyte is a triprotic weak acid, then its formulaweight is 58.78 g/mol, 117.6 g/mol, or 176.3 g/mol. None of these valuesis close to the formula weight for citric acid, eliminating it as a possibility.Only succinic acid provides a possible match.Practice Exercise 9.10Figure 9.21 shows the potentiometric titration curve for the titration of a0.500-g sample an unknown weak acid. The titrant is 0.1032 M NaOH.What is the weak acid’s equivalent weight?Click here to review your answer to this exercise.1412108pH6Figure 9.21 Titration curve for Practice Exercise 9.10.4200 20 40 60 80 100 120Volume of NaOH (mL)