1. Strong Acid versus Strong Base Derive a titration curve for the titration of 50.0 mL...

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