Question

Calculate the pH at the stoichiometric point when 50 mL of 0.090 M pyridine is titrated with 0.32 M HCl.

Answer 1

moles of C5H5N = molarity x volume

= 0.090 mol/L x 0.0500 L

= 0.00450 mol

moles of HCl required to reach the equivalence point

= 0.00450 mol

Volume of HCl = 0.00450 mol / 0.32 M = 0.0140625 L

total volume = 0.0500 + 0.0140625 =0.0640625 L

Moles of C5H5NH+ formed = 0.00450 mol

[C5H5NH+] = 0.00450 mol / 0.0640625 L = 0.0702 M

C5H5NH+ <=> C5H5N + H+

Ka = Kw/Kb = 1.0 x 10^-14 / 1.7 x 10^-9

=5.9 x 10^-6

Equilibrium constant expression of the reaction

Ka = [C5H5N] [H+] /[C5H5NH+]

5.9 x 10^-6 = x^2 / 0.0702-x

x = [H+]= 0.000644 M

pH = - log [H+] = - log (0.000644)

pH = 3.19