Question

Question For the polyprotic acid hitidines"), the hydrogena in bold can be fout. When He istrated with NOH and it takes 45.0 mL to remove all protons positie. In the trution curve, how many more required to reach the second p? Another way of asking this is when presented with attration curve of His where on the should look (which volume of NaOH, or wants to determine the second piks? H + K CH & ĐH H CH "HN A. 45 mL 8.37.5 mL 6.30 ml D. 15 ml E. 22.5 ml

Answer 1

are three bydrogen that is NaHis it2 + Naolt -> Na, ts H et H2o. it Naolt Naz Hist & H2O Answer 122-55mL There lost. So, the ili) NatHISH geachons happens in steps His H3 + Naolt - Na HigH2 + H₂O + + st IS The point where ist reaction ends equivalence point. Second equivalence point occurs when and reaction completes ard equivalence point occurs when reaction is completed. The moles or volume of NaOH required in each reaction same. So, volume required fox one eauivalence point - 1 / 3 x X45 x is x total 3 - 15 ml Each reaction starts with an acid HIS H₃ + in second in second HIS H₂ in HISH) and equivalence point ist ID 3rd

all acid is converted ist Hista HISH 200 reaches cohen. to ets base. (in in third His ) At the midpoint of the reaction, the solution contains equal amount of acid and its conjugate base At this point, the pH of solution will be pka of the and with which reaction started. ie midpoint of forst reaction, gives the pka of His Hat . This is oka, midpoint of second seaction. gives pka of His Hg. This is plaz Similarry 3rd reaction & midpoint gives kaze cohen midpoint occurs exactly balf of bose is added for complete reac- tion. So, first midpoot is at 15 - 15 = 7.5 mL

god 15 + 9.5 = 22.5ml SH midpoint at 3rd midpoint at at 30+7.5 = 37.5 equivalence point at 15 equivalence point at 15+15=30ml equivalence point at 30115=456L mu 2nd 3rd