Question

2)Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH 200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCl 100.0 mL of 0.100 M KOH by 0.100 M HCl 100.0 mL of 0.100 M HNO2 (Ka = 4.0 x 10-4) by 0.100 M NaOH

Answer 1

First to all the pH at equivalence point of strong acid and base, is in general terms, 7, so the pH for KOH and HCl would be 7.

For HC3H5O2:

100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH
makes 0.100 moles NaC3H5O2 / 0.200 litres = 0.05 Molar C3H5O2-, which does a hydrolysis:

C3H5O2- in water --> HC3H5O2 & OH-

Kh = Kw/Ka =1e-14 / 1.3e-5 = 7.7e-10

7.7e-10 = [HC3H5O2] [OH-] / [ C3H5O2-]

7.7e-10 = [x] [x] / [ 0.0500]

x =[OH] = 6.2e-6

pOH =5.21.... pH =8.79

For (C₂H₅)₂NH:

200.0 mL of 0.100 M (C₂H₅)₂NH (K_b = 1.3 x 10^-3) by 0.100 M HCl
makes 0.200 moles (C₂H₅)₂NH2+ / 0.400 litres = 0.05 Molar (C₂H₅)₂NH2+, which does a hydrolysis:

(C₂H₅)₂NH2+, in water --> C₂H₅)₂NH & H3O+

Kh = Kw/Kb =1e-14 / 1.3e-3 = 7.69e-12

7.7e-12 = [(C₂H₅)₂NH, ] [H3O+] / [ (C₂H₅)₂NH2+]

7.7e-12 = [x] [x] / [ 0.0500]

x =[H⁺] = 6.2e-7

pH =6.2

For HNO2

100.0 mL of 0.100 M HNO₂ (K_a = 4.0 x 10^-4) by 0.100 M NaOH

produces 0.05 Molar NO2-, which hydrolyzes:

NO2- in water --> HNO2 & OH-

Kh = Kw/Ka = 1e-14 / 4.0 x 10^-4= 2.5 e-11
2.5e-11 = [HNO2] [OH] / [NO2-]

2.5 e-11 = [x] [x] / [0.05]

x = [OH] = 1.1 e-6

pOH = 5.9..... pH = 8.1

For NH₃

100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl

Kh = Kw / Kb = 1.8e-5

1.8e-5 = [x] [x] / 0.05

x= H+ = 9.5 e-3

pH = 2.02

order of increasing pH at the equivalence point

1------100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl

2-----200.0 mL of 0.100 M (C₂H₅)₂NH (K_b = 1.3 x 10^-3) by 0.100 M HCl

3-----100.0 mL of 0.100 M KOH by 0.100 M HCl
4-----100.0 mL of 0.100 M HNO₂ (K_a = 4.0 x 10^-4) by 0.100 M NaOH

5----100.0 mL of 0.100 M HC₃H₅O₂ (K_a = 1.3 x 10^-5) by 0.100 M NaOH

Hope this helps