Question

The pH of 0.300M pyridine solution is 10.38 what is kb

Answer 1

               C₅H₅N(aq) + H₂O à C₅H₅NH⁺(aq) +   OH^- (aq)

Initial               0.3-------------------0----------------0-----------------------------0

Equilibrium   (0.3-x)---------------0-----------------(x)--------------------------(x)

Now, Kb= [OH-][ C₅H₅NH⁺]/[ C₅H₅N]

Kb = [x][x]/[0.3-x]

Given that pH of the solution is 10.38

Therefore pH= -log[H⁺]

[H⁺] = -Antilog pH

[H⁺] = -Antilog 10.38 = 4.168 ×10 ^-11

Now we know that Kw = [H⁺][OH^-]

1.00 × 10^-14 = (4.168 ×10 ^-11) [OH^-]

[OH^-] = 0.2399× 10^-3 M

Now from the reaction above we know that there is one to one molar ration between OH^- and C₅H₅NH⁺

Therefore,

Kb = [x][x]/[0.3]

Kb = (0.2399× 10^-3)² /0.3 = 1.91 × 10^-7