What is the pH of a 0.30 M pyridine solution that has K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH^-(aq) .

Question

What is the pH of a 0.30 M pyridine solution that has K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq).

Quizplus · Accepted Answer

The dissociation constant of the pyridine can be written as:
KS1U1B1bS1U1B0 = [CS1U1B15S1U1B0HS1U1B15S1U1B0NHS1U1P1+S1S1P0][OHS1U1P1-S1S1P0]/[HS1U1B15S1U1B0N]

At equilibrium, the concentration of pyridine remaining unreacted is (0.30 - [HS1U1B15S1U1B0N]), while the concentration of [HS1U1B15S1U1B0NHS1U1P1+S1S1P0] and [OHS1U1P1-S1S1P0] are equal and can be represented as x.

So, the dissociation constant can be written as:
1.9 × 10S1U1P1-9S1S1P0 = xS1U1P2/(0.30 - x)

Using the quadratic equation to solve for x, we get:
x = 1.37 x 10S1U1P1-S1S1P0

Therefore, the concentration of [HS1U1B15S1U1B0NHS1U1P1+S1S1P0] and [OHS1U1P1-S1S1P0] is 1.37 x 10S1U1P1-S1S1P0 M, and the concentration of pyridine remaining unreacted is (0.30 - 1.37 x 10S1U1P1-S1S1P0) M.

To calculate pH, we need to find the concentration of [H+] ion in solution. Since pyridine is a weak base, we can assume that the contribution of [OH-] to the solution is negligible. Hence,

[HS1U1B15S1U1B0N][H+] /[CS1U1B15S1U1B0HS1U1B15S1U1B0NHS1U1P1+S1S1P0]= K_a

[H+]= S1U1P0K_a/[HS1U1B15S1U1B0N]

[H+]= S1U1P0K_a/(0.30 - 1.37 x 10S1U1P1-S1S1P0)

Substituting the values, we get:
[H+] = 9.38 x 10S1U1P1-S1S1P0 M

Taking the negative logarithm of [H+],

pH = -log[H+]
pH = -log(9.38 x 10S1U1P1-S1S1P0)
pH = 9.38

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