Question

What is the pH at the equivalence point of a titration of 0.50 M pyridine with 1.0 M HCI? 17. Data: Kb(pyridine)-1.70x 109 A) 2.54 B) 2.85 C) 4.85 D) 7.84 E) 3.24

Answer 1

At the Equivalence point, all of the pyridine and HCl have reached to form C_5 H_5 N + H^_rightarrow C_5 H_4 NH^+ Molarity of C_5 H_5 NH^+ = 0.50 M Now C_5 H_5 NH^+ hydro in water C_5 H_5 \NH^= + H_2 O rightarrow H_3 O^+ + C_5 H_5 N therefore Ka = -k w/k b = 10^negative 14/1.7 times 10^negative 9 = 5.9 ry 10^negative 6 = x^2/0.50 - x Sine Ka Fo & C_6 H_5 NH^+ is very small compared to 0.50 M, we can ignore the negative x term 5.9 times 10^negative 6 = x^2 . 0.50 = -x = 1.71 times 10^negative 3 therefore x = [H_3 O^+] therefore H^+ = H_3 O^+ = 1.71 times