Question

Based on figure 2, the titration curve of 25.0 mL of phosphoric acid with 0.1082 M NaOH, determine the following:

1) The molarity of the acid

2) the Ka1 and pKa1 of the acid

3) the Ka2 and pKa2 of the acid

Figure 2. Typical Polyprotic Acid Titration, pH vs. ml NaOH 12.00 Hd ml NaOH

Answer 1

Solution olume a phosphorie acid Concentation of Naot = 25ml 0.1082 M 1. The molarity of the acid Volume of the ibant = 28 milhas L Concennetion of Nach & 0.1082M multiply Molarity X Volume 1 x 0.1082 Lecc This giver amount of titant chemical added to reach first equivalence point = 0.0025 802, mole no. of miles of analyle = no. of moles of tikant foreach first equivalence point 0.00228 02 molee Volume of analyte = 25ml = 25 L -0.025L Molanty = nosof moles 0.002902 0.025 Ilotume int = - Molarity of the acid. = 0.89 769 in 0.095 216 M .

2 acid the Kal and Pkal of the PH= pka ot log [base] [acid [bar] & [acid] pH , pka - as much bar as and i pka = 2.20 ka a 10-2 = 0,0063 pH = Pka 2 = 7-2 Kaz= 10 = 6.30 810-8