Question

Pyridine is a weak organic base, and readily forms a salt with hydrochloric acid.

C5H5N(aq) + HCl(aq) → C5H5NH+(aq) + Cl–(aq)

What is the pH of a 0.075 M solution of pyridinium hydrochloride, [C5H5NH]+Cl–, if Kb for pyridine is 1.5 × 10–9?

pH =_____

Answer 1

ionizes in C5H5NH+--> C5H5N + H+

Let B --> C5H5N  and BH+ = C5H5NH+ for simplicity

the next equilibrium is formed, the conjugate acid and water

BH+(aq) + H2O(l) <-> B(aq) + H3O+(aq)

The equilibrium is best described by Ka, the acid constant

Ka by definition since it is an base:

Ka = [H3O][B]/[BH+]

Ka can be calculated as follows:

Ka = Kw/Kb = (10^-14)/(1.5*10^-9) = 6.67*10^-6

get ICE table:

Initially

[H3O+] = 0

[B] = 0

[BH+] = M

the Change

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = -x

in Equilibrium

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

6.67*10^-6= x*x/(M-x)

solve for x

x^2 + Ka*x - M*Ka= 0

solve for x with quadratic equation

x = H3O+ = 0.0007039

[H3O+]  = 0.0007039

pH = -log([H3O+]) = -log(0.0007039) = 3.15

pH = 3.15