What is the pH of a 0.30 M pyridine solution that has a Kb = 1.9 × 10-9 ? The equation for the dissociation of pyridine isGroup of answer choices9.388.7210.384.62

To solve this problem, we need to use the formula for the base ionization constant (Kb) and the definition of pH.

Step 1: Write the ionization equation for pyridine (C5H5N): C5H5N + H2O ⇌ C5H5NH+ + OH-

Step 2: Set up the Kb expression: Kb = [C5H5NH+][OH-] / [C5H5N]

Step 3: Set up an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations.

Initial: [C5H5N] = 0.30 M [C5H5NH+] = 0 [OH-] = 0

Change: [C5H5N] = -x [C5H5NH+] = +x [OH-] = +x

Equilibrium: [C5H5N] = 0.30 - x [C5H5NH+] = x [OH-] = x

Step 4: Substitute the equilibrium concentrations into the Kb expression and solve for x: 1.9 × 10^-9 = x^2 / (0.30 - x)

Assuming x is much smaller than 0.30, the equation simplifies to: 1.9 × 10^-9 = x^2 / 0.30

Solving for x gives the concentration of OH-, which is approximately 7.4 × 10^-6 M.

Step 5: Use the definition of pOH to find the pOH of the solution: pOH = -log[OH-] = -log(7.4 × 10^-6) = 5.13

Step 6: Use the relationship between pH and pOH to find the pH of the solution: pH = 14 - pOH = 14 - 5.13 = 8.87

So, the pH of the 0.30 M pyridine solution is approximately 8.87.