Answer
(a) pH = 11.13
(b) pH = 8.97
Work Step by Step
Equations needed:
$pH + pOH = 14$
$[OH^-] = 10^{-pOH}$
$Kb = \frac{[OH^-][B^+]}{[BOH]}$**
**All these concentrations are the values in the equilibrium.
$[OH^-] = [B^+] = x$
$[BOH] = [BOH]_{initial} - x$
Kb (Ammonia) = $1.8 \times 10^{-5}$
Kb (Pyridine) = $1.7 \times 10^{-9}$
(a)
Since the concentration has a large value, compared to the ka, we can assume:
$[BOH] \approx [BOH]_{initial}$
Therefore:
$1.8 \times 10^{-5} = \frac{x * x}{0.10}$
$1.8 \times 10^{-6} = x^2$
$x = 1.34 \times 10^{-3}M = [OH^-]$~
- Now, find the pH:
$pOH = -log[OH^-]$
$pOH = -log( 1.34 \times 10^{- 3})$
$pOH = 2.87$
$pH + pOH = 14$
$pH + 2.87 = 14$
$pH = 11.13$
(b)
Since the concentration has a large value, compared to the ka, we can assume:
$[BOH] \approx [BOH]_{initial}$
Therefore:
$1.7 \times 10^{-9} = \frac{x * x}{0.05}$
$8.5 \times 10^{-11} = x^2$
$x = 9.22 \times 10^{-6} = [OH^-]$
- Now, find the pH:
$pOH = -log[OH^-]$
$pOH = -log( 9.22 \times 10^{- 6})$
$pOH = 5.03$
$pH + pOH = 14$
$pH + 5.03 = 14$
$pH = 8.97$
