Question

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 (K_a = 1.3 x 10^-5) 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

100.0 mL of 0.100 M KOH 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

1 2 3 4 5  100.0 mL of 0.100 M NH₃ (K_b = 1.8 x1 0^-5) by 0.100 M HCl

Answer 1

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

at equivalence point

pH = 7+1/2(pka+logC)

c = concentration of salt = base = 0.1*100/200 = 0.05 M

pka of acid = -log ka = -log(1.3*10^-5) = 4.88

pH = 7+1/2(4.88+log0.05) = 8.79 (basic)

2. 100.0 mL of 0.100 M C2H5NH2 (Kb = 5.6 x 10-4) by 0.100 M HCl

acidc solution pkb = -log(5.6*10^(-4)) = 3.25

pH = 7 - 1/2(pkb+logC)

if C is same , by increasing pka ,pH decreases.

3.100.0 mL of 0.100 M KOH by 0.100 M HCl

neutral

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

basic pka = -log(7.2*10^(-4)) = 3.14

pH = 7 + 1/2(pka+logC)

if C is same , by increasing pka ,pH increases.

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

acidc   pkb = -log(1.8*10^(-5))= 4.74

so that

order of pH

1>4>3>2>5