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

In the equivalence point, there is no acid,base left so:

(1) 100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH --> forms solution with pH slightly basic

(2) 200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl --> forms solution wit hslightly acidic

(3) 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCl --> slightly acidic pH

(4) 100.0 mL of 0.100 M KOH by 0.100 M HCl --> pH = 7, neutral

(5) 100.0 mL of 0.100 M HNO2 (Ka = 4.0 x 10-4) by 0.100 M NaOH .-->forms slightly basic solution

then

(2) <(3) < (4) < (1) < (5)