Question

2)Rank the following titrations in order of increasing pH at the equivalence point of the titration...

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

Homework Answers

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)

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