What is the pH of a 1.2 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 1.2 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

First, we need to write out the expression for the equilibrium constant (KS1U1B1bS1U1B0) using the equation for the dissociation of pyridine:

KS1U1B1bS1U1B0 = [HS1U1B15S1U1B0NHS1U1P1+][OHS1U1P1-S1S1P0]/[HS1U1B15S1U1B0N][HS1U1B12S1U1B0O]

The pH of the solution can be calculated using the Henderson-Hasselbalch equation, which relates the pH to the concentration of the acid and its conjugate base:

pH = pKa + log([A-]/[HA])

In this case, pyridine (C5H5N) is acting as a base and its conjugate acid is HS1U1B15S1U1B0NHS1U1P1+. The pKa of pyridine is 5.25. We can assume that at equilibrium, the concentration of pyridine (base) is equal to the concentration of its conjugate acid.

Let x be the concentration of [HS1U1B15S1U1B0NHS1U1P1+]. Then,
[HS1U1B15S1U1B0N] = [C5H5N] - x
[OHS1U1P1-S1S1P0] = x
[HS1U1B12S1U1B0O] is assumed to be large enough (infinite) compared to [C5H5N] and x, so it can be approximated as [HS1U1B15S1U1B0N].

Substituting these into the equilibrium constant expression, we get:

1.9 × 10S1U1P1-9S1S1P0 = x^2/([C5H5N] - x)

Solving for x using the quadratic formula, we get:

x = [HS1U1B15S1U1B0NHS1U1P1+] = 5.38 × 10S1U1P1-S1S1P0 M

Using the Henderson-Hasselbalch equation, we get:

pH = 5.25 + log(5.38 × 10S1U1P1-S1S1P0/1.2) = 9.68

Therefore, the answer is C) 9.68.

What Is the PH of a 1