Answer
$pH = 9.25$
Work Step by Step
1. Since $N{H_4}^+$ is the conjugate acid of $N{H_3}$ , we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.6\times 10^{- 10}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa = 9.25$
3. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.2}{0.2}$
- 1: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.2}{5.6 \times 10^{-10}} = 3.6\times 10^{8}$
- $ \frac{0.2}{5.6 \times 10^{-10}} = 3.6\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.2}{0.2})$
$pH = 9.25 + log(1)$
$pH = 9.25 + 0$
$pH = 9.25$
