Question

1. Strong Acid versus Strong Base Derive a titration curve for the titration of 50.0 mL of 0.150 M HCl with 0.100 M NaOH. Calculate the pH for the following volumes of NaOH (0 mL, 25 mL, 50 mL, 70 mL, 75 mL, 80 mL, 90 mL, 100 mL). pH Volume of NaOH, in milliters 0 25 50 70 75 80 90 100 L(g) (h) pH at the equivalence point Specify your choice of indicator

Answer 1

0 Strong Acid V. Strong Base Hiration : 50 mL 0.150M HU = 775150810-3 LX 0.15M = 7.5610-3 moles . and 0.1M Naolt has taken. As Hal & Naolt are strong acid & bone respectively 80 neutralization reaction takes place. So, the reactionis Hul(99) + Nach(99) → Nadl(09) +H2014 or simply, H(aq) + OH- (aq) → H₂O (99). * So in the reaction medium 7.5x 103 moles of 14t in present with the addition of otno. of moles of oH increase & neutralization reaction. will going on, thus, pH will increase gradually. so a when volume of Naolt =0mL moles of H+ = 50x103 LX0.15M = 7.5x10-3 moles 7.5x10 3 moles = 0.15M 50x10-3L pH = -log[H+ = 0.82 b) 25 m Lò NaOH (0 | M) moles of H+ = 7.5x103, moles of OH = 25x10 3 LXO.IM 14103 = 2.5x10 3 Adler .. Remaining H+ = (7.5-2.5 x 10 3 = = 5x10-3 moles Total volm.of non. (50+25) ML = 15X10 3L [H+2 = 5:100% = 0.067 pH = -log[H +7 = 1.17

c) 50mL O IM Naolt Remaining At = [(7.5x10-3) - (50x10 3 +0.1) 7 = 2.5810-3 moles. ruth - 2.5810-3 moles = 0.025 (500 +50) x 10^3 pH = 1.602 7omL of 0.1m NaOH Remaining H+ = [ (7.5*103)-(70x 10 3 x 0.1)] = 0.5810-3 moles [H+ = 4.1678103 .PH = 2.38 e) 75 mL o 1M Naot: (715x10-3) - (75x105x0.1) moles o moles (50+75) x 103 L [H+2 -75X10-3) - (15X10-3x o. pH As [tty is o, thm, neutralization reaction has finished so, ir soln. Only H₂O is and saltis present. Thus, pit at this condition ist pH=7 80mLO 1M Naolt ar in the solm. there is no Ht, only oHe conc. will decide pH of the rom. pon Excess oH = (80X 10°3x0:1) 6.5X10-3) = 0,5 X 10-3 moles. = 3.868103 (80+50) X10 3 = & pol = 2.414 pH = 11.585 g) 90 mL 0 M Naolt : [04 2 = 0.5X1003 10-3 qox10 "x0:1)-7. 5X10) 0-0108 0.0107 [OH-] = 190 +50) X10-3 o pH = 14-pott = 14-1.918 2 12.03 (12.03)

moles ) 100 mL O. IM Naot: [OH-] = (100x10 -3 0.1)-(7.5810) (100+50) x 103 = 0.0167 moles • po pH = 14-polt = 12.22 o Titration Curre : 25 su volim 75 of Naol 100 -/ gügg Nolm of Naoh 0 ML 25 mL 50 mL 70 mL 75 mL 80 mL 90 mL 100mL pll 0.82 1.17 1.602 2.38 7 11.585 12.03 12.22 8) h) o ® pH at equivalence point =7 Any kind of indicator we can use. e.g. phenolpthalim