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30 Consider the titration of a 40.0 mL of 0.113 M weak acid HA (Ka=2.7 x 10) with 0.100 M LIOH. a What is the pH of the solut
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Hta & A ag HACA ons Initial Equilibrium O13-2 Kaa [H] [A] [HA] 2.7x108 = = x2 O13-2 x 0.113 92 5:52x10,5 [144] = 5.528105 pHНА Сафн t + A) ha 3:04 Equilib 0.042-2 Kaz 2. Oo42 - 2.1 xia® : 2 о: 64 2. . 3- Axls 5 ри. - los (G4 165 ) = 4.5②. Moles of HA= 0.113 molll x 40 x103 e = 4.58 x103 mol. Half moles of HA= RiR6x10 mol. to reach at Volume of Lion required hd. HA + KTOH LiA & H2O 4.52215 4.528103 4.52.1103 moles of HA = 0.113 mol/l x 40x108 @ - 4.52x103 mol, Volume of LiGH taken =Kb = [HA] [CH] [A] = w Ka O053-X, 1014 02 2.7x168 0.053-2, 3.7x10 = x O053 92 1.4.4154 [OH = 1.4x154 polt = -log [ot] = -lo

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