1) Rank the following titrations in order of increasing pH at the halfway point

Question

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH).

1 2 3 4 5  200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH

1 2 3 4 5  100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH

1 2 3 4 5  200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HI by 0.100 M NaOH

2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH).

1 2 3 4 5  100.0 mL of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) by 0.100 M NaOH

1 2 3 4 5  200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl

1 2 3 4 5  200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH

1 2 3 4 5  100.0 mL of 0.100 M KOH by 0.100 M HCl

Explanation / Answer

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH).

Solution :- larger the kb means higher the pH

So the smallest kb will have lower pH so the order is as follows

  4= 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl

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

3=  100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH
2=   200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl

1 =   100.0 mL of 0.100 M HI by 0.100 M NaOH

2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH).

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

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

1=  200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl

4=  100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH

3 =  100.0 mL of 0.100 M KOH by 0.100 M HCl

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