Question

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 solution before any base has been added? 3 points b What would be the pH of the solution after the addition of 20.0 mL of LIOH? 3 points How many mL of the LiOH would be required to reach the halfway point of the titration? 3 points d What is the pH of the solution at the equivalence point? point? 3 points onet e mon er inoo mi tot e What would be the pH of the solution after that addition of 100.0 mL of LiOH? , 3 points

Answer 1

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 pH = -log (5.522105) 7 4.25.

НА 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 ) = 4.5

②. 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 half point of Entration = moles of nich Molarity of Lith 2. 26x103 mel Ord molll. = CORREI. = 22.6 ml.

d. 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 = 4.52215 mol o loo melle = 6:0452 l. Total volume = (0.0452 + 0.0400 e - 0:0852l. Molarity of LiA = 4.52 x 103 mol 0.0852d 0.053 M . LiA Litt HA tot Á theo 0.053 Initial Equilib: 8:053-

Kb = [HA] [CH] [A] = w Ka O'053-X, 1014 02 2.7x168 0.053-2, 3.7x10 = x O'053 92 1.4.4154 [OH = 1.4x154 polt = -log [ot] = -log (1.48104 - 3.9 .pH = 14-3.9 = 101