Answer
(a) $pH = 4.64$
(b) $pH = 9.68$
Work Step by Step
(a)
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 6.3 \times 10^{- 5})$
$pKa = 4.20$
2. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.033}{0.012}$
- 2.8: It is.
3. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.033}{6.3 \times 10^{-5}} = 524$
- $ \frac{0.012}{6.3 \times 10^{-5}} = 290$
4. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.20 + log(\frac{0.033}{0.012})$
$pH = 4.20 + log(2.8)$
$pH = 4.20 + 0.44$
$pH = 4.64$
(b)
1. Since $NH{_4}^+$ is the conjugate acid of $NH_3$ , 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:
- $\frac{[Base]}{[Acid]} = \frac{0.409}{0.153}$
- 2.67: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.409}{5.6 \times 10^{-10}} = 7.29\times 10^{8}$
- $ \frac{0.153}{5.6 \times 10^{-10}} = 2.73\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.409}{0.153})$
$pH = 9.25 + log(2.67)$
$pH = 9.25 + 0.426$
$pH = 9.68$
