Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.230A pyridine, C5H5N(aq) with 0.230M HBr(aq):

a.) before addition of any HBr

b.) after addition of 12.5 mL of HBr

c.) after addition of 20.0 mL of HBr

d.) after addition of 25.0 mL of HBr

e.) after addition of 36.0 mL of HBr

Answer 1

a) For only pyridine:

Kb = 1.7 x 10^-9

Pyridine aqueous equilibrium:

C₅H₅N + H₂O <-> C₅H₅NH⁺ + OH^-

Kb = 1.7 x 10^-9 = [C₅H₅NH⁺] [OH^-] / [C5H5N]

Using concentrations in equilibrium:

1.7 x 10^-9 = x² / [0.23 - x]

Concentration of OH^-= x = 0.0000197729

pOH = -log [OH^-] = - log (0.0000197729) = 4.7

pH = 14 - 4.7 = 9.296

b) After adding 12.5 mL of 0.23M of HBr:

C₅H₅N + HBr <-> C₅H₅NH⁺ + Br^-

Moles of pyridine = 0.00575

Moles of HBr = 0.002875

Remainder of pyridine = 0.00575 - 0.002875 = 0.002875

We use Henderson-Hasselbach eq.

pOH = pKb + log ([BH⁺]/[B]) = -log(1.7 x 10^-9) + log (0.002875/0.002875) = 8.76955

pH = 14 - 8.76955 = 5.23

c) Moles of pyridine = 0.00575

Moles of HBr = 0.0046

Remainder of pyridine = 0.00575 - 0.0046 = 0.00115

We use Henderson-Hasselbach eq.

pOH = pKb + log ([BH⁺]/[B]) = -log(1.7 x 10^-9) + log (0.0046/0.00115) = 9.3716

pH = 14 - 9.3716 = 4.6284

d) We now have a solution of pyridine ion on water:

C₅H₅NH⁺ + H₂O -> C₅H₅N + H₃O⁺

Ka for pyridine = (1x10^-14 / 1.7x10^-9) = 5.8823 x 10^-6

5.8823 x 10^-6 = x² / (0.115 - x)

x = [H₃O⁺] = 0.000819639

pH = -log(0.000819639) = 3.0864

e) We are left with pure HBr:

0.036 L * 0.23 = 0.00828

Remaining moles of HBr: 0.00828 - 0.00575 = 0.00253 moles of HBr

Concentration = 0.00253 / 0.061 = 0.04147

pH = -log(0.04147) = 1.3822