Question

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

100.0 mL of 0.100 M KOH by 0.100 M HCl

200.0 mL of 0.100 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) by 0.100 M NaOH

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

100.0 mL of 0.100 M HI by 0.100 M NaOH

100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) 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).

100.0 mL of 0.100 M KOH by 0.100 M HCl

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

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

100.0 mL of 0.100 M C₂H₅NH₂ (K_b = 5.6 x 10^-4) by 0.100 M HCl

200.0 mL of 0.100 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) by 0.100 M NaOH

Answer 1