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 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
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 HCl 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

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 C₂H₅NH₂ (K_b = 5.6 x 10^-4) 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 NH₃ (K_b = 1.8 x1 0^-5) 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 HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

Answer 1

1)

at half way point . pH = pKa

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

pOH = pKb = -log Kb = 3.25

pH = 10.5
b)

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

pH = 4.74 = pKa
c) 100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

pH = 3.14 = pKa
d) 100.0 mL of 0.100 M HCl by 0.100 M NaOH

pH = 1.48
e) 100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl

pH = 9.26

order :

100.0 mL of 0.100 M C₂H₅NH₂ (K_b = 5.6 x 10^-4) by 0.100 M HCl   ---------------------> 5
200.0 mL of 0.100 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) by 0.100 M NaOH -------------------> 3
100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH     ------------------------> 2
100.0 mL of 0.100 M HCl by 0.100 M NaOH -----------------------------------------------> 1
100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl ---------------------------> 4

2)

here weak acid + strong base pH > 7

        weak base + strong acid pH < 7

        strong acid + strong base pH = 7

100.0 mL of 0.100 M C₂H₅NH₂ (K_b = 5.6 x 10^-4) by 0.100 M HCl ---------------------> 3
100.0 mL of 0.100 M KOH by 0.100 M HCl ------------------------------------------------> 2
100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl ---------------------------> 1
200.0 mL of 0.100 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) by 0.100 M NaOH------------------> 5
100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH -------------------------> 4