What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq). A)4.62 B)8.72 C)9.38 D)10.38

Question

Quizplus · Accepted Answer

First, let's write out the equilibrium expression for the dissociation of pyridine:

Kw = [NH+][OH-] = [NH][HPO4-]/[NH2PO4]

The value given, KS1U1B1bS1U1B0 = 1.9 × 10S1U1P1-9S1S1P0, is actually the equilibrium constant for the reaction written in the reverse direction (the association of pyridinium and hydroxide ions to form pyridine and water). To use this value, we need to take the inverse to get the equilibrium constant for the dissociation of pyridine:

Kdiss = 1/KS1U1B1bS1U1B0 = 5.26 × 10S1U1P1S1S1P0

Now, let's set up the ICE table for the dissociation of pyridine:

CS1U1B15S1U1B0HS1U1B15S1U1B0N(aq)+ HS1U1B12S1U1B0O(l)⇌ CS1U1B15S1U1B0HS1U1B15S1U1B0NHS1U1P1+S1S1P0(aq)+ OHS1U1P1-S1S1P0(aq)
I: 0.30 M      0 M              0 M
C: -x          +x              +x
E: 0.30-x      x               x

Next, we can plug the equilibrium concentrations into the expression for Kdiss and solve for x:

Kdiss = [NH+][OH-]/[NH]
5.26 × 10S1U1P1S1S1P0 = xS1S1P0S2/(0.30-x)
x = 1.80 × 10S1U1P1S1S1P0

Finally, we can calculate the pH:

pH = -log[H+]
[H+] = x = 1.80 × 10S1U1P1S1S1P0 M
pH = 9.38

Therefore, the answer is choice C.

What Is the PH of a 0