Pyridine, C5H5N, is a bad-smelling liquid that is a weak base in water. Its pKb is 8.77. What is the pH of a 0.20 M aqueous solution of this compound

pH = 9.26

Explanation:

Let's make the reaction of this weak base:

C₅H₅N  +  H₂O  ⇄    C₅H₅NH⁺  +  OH⁻     Kb

We do not know the Kb, but we can discover it from pKb

Kb = 10^⁻pKb  →  10⁻⁸'⁷⁷ = 1.70×10⁻⁹

So, let's make the expression for Kb

Kb = [OH⁻] . [C₅H₅NH⁺] / [C₅H₅N]

In the equilibrum, we would have  0.20 moles from the begining - x (react)

So we would proudce x moles of OH⁻ and x moles of C₅H₅NH⁺.

In conclussion:

1.70×10⁻⁹ = x . x / (0.20 - x)

To avoid the quadratic formula we can miss the x from the substraction

1.70×10⁻⁹ = x² / 0.20

1.70×10⁻⁹ . 0.20 = x²

We avoid the x from the (0.20-x), because the base is so concentrated, and Kb is very small.

x = √(1.70×10⁻⁹ . 0.20) → 1.84×10⁻⁵

That's the value of [OH⁻], so we can calculate pOH

- log [OH⁻] = pOH → - log 1.84×10⁻⁵ = 4.73

Then pH = 14 - pOH

14 - 4.73 = 9.26

c. 75.7

i think that is the right