Question

a student was conducting a titration

48.92 mL of a 0.1089 M solution of acetic acid to a 200 ml beaker. The student then added 3 ant then added 35.8 mL of deionized water. After adding the water to the solution, the mixture was stirred to ensure proper mixing of solution Calculate the theoretical phat the following points of calibration curve. a) Initial pl, before addition of any NaOH? Be sure to consider the change in volume due to the addition of deionized water. Initial pH= b) After addition of 46.871 ml of 0.1089 M NaOH (before equivalence point) c) At the equivalence point d) After the addition of 72.4899 mL of 0.1089 M NaOH (after equivalence point)

La=1.8•10^-5

Answer 1

Ass er to he roden: I H20 48. 42 ml 6 0.10 89 M 4eOH 358 ml Total volume= 84.72 ml Conc o Aco= 48.92x0-1089 84.72 0. 063M 4 or addation Alaot Moles of 0.063x 84.72 x1o 5. 34X153 AeoH CiHts COOH + HO CH Lo0 Hg0 + 0. 063 1 Msles E. (0.069-x) Ka= 1.8 x105 Ka-log (0) Ka= -5 4.74 (o.063-) .0.063-h 3 1.8x1D O.063 1.134 x106 ON 1064x 163 OV 1 064 x10 2.97 Ph

)Addilion o 46. 871 m 0-108 9 (M) NaoH: Moles s Naon NaoH Cttg Cag Na+ H20 5. 34x163 46.871X01089X1D 51OxiD 5.10x163 S10X153 S.1OX10 5.10 X16 3 0.24 X10 5.10x103 Salt-1e CoyNa7510x1 Actd = Tog logit= 04XID Tsalt7 pH 4.74+log /10i 3 0-24X153 PH-4.74+ 132 o.06 ) At ayivaln bat CH2 COOH NaoH Cb CoONa + ,0 5.34x10 5.34X163 I - 34XID 5.34X1D 5.34x10 3 5.3 4x10 84-72 X103 LCHy COONA 0.063 M M = COOHt OH CHa COONA +Hz0 0.063 E (0.063-x) Ka. Ko114 8 x10 K1014 55%XI6 10 Ko

oy 5.57 x1010 O.663-he 0.063 (x-690.8 5.56x100 0.063 -0 3 50X10 5.91x10 K1OH-5. 91m6 log (5.91x165) 5.2.3 PHt pOH = 14 PH 14-5.23 g.77 72-4899 ml e of MoTes NaoH= Nasy: 0-1089m 72-4 99X 0-10 t9x 16 3 7.89 X103 CH COOH + NaoH CH COONA + H, 0 7.89x10 5.34x1D - 5.34x 534X10 -5 34x10 5.34X16 3 E 2.5SxIb 3 Lo 2 55x10 0.030M 84-72 X16 3

poH= ... 153 H14-POH 12.48