Question

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.210 M pyridine, C5H5N(aq) with 0.210 M HBr(aq).

after addition of 19.0 mL of HBr

after addition of 35.0 mL of HBr

Answer 1

initially

millimoles of pyridine = 25.0 x 0.210 = 5.25

pKb of pyridine = 8.75 (standard value)

1) millimoles of HBr added = 19.0 x 0.210 = 3.99

after HBr added

millimoles of pyridine = 5.25 - 3.99 = 1.26

millimoles of salt formed = 3.99

total volume = 19.0 + 25.0 = 44.0 mL

[pyridine] = 1.26 / 44 = 0.029 M

[salt] = 3.99 / 44 = 0.091 M

pOH = pKb + log [salt] / [pyridine]

pOH = 8.75 + log [0.091] / [0.029]

pOH = 9.25

pH = 14 - 9.25

pH = 4.75

2) millimoles of HBr added = 35.0 x 0.210 = 7.35

after Hbr added

HBr left = 7.35 - 5.25 = 2.10

total volume = 25.0 + 35.0 = 60.0 mL

[HBr] = 2.10 / 60.0 = 0.035 M

as HBr strong acid

[HBr] = [H⁺] = 0.035 M

pH = - log [H⁺]

pH = - log [0.035]

pH = 1.46