Answer
The pH of this buffer is equal to $8.89$
Work Step by Step
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}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.556\times 10^{- 10}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.556 \times 10^{- 10})$
$pKa = 9.255$
3. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.15}{0.35}$
- 0.4286: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.15}{5.556 \times 10^{-10}} = 2.7\times 10^{8}$
- $ \frac{0.35}{5.556 \times 10^{-10}} = 6.3\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.255 + log(\frac{0.15}{0.35})$
$pH = 9.255 + log(0.4286)$
$pH = 9.255 + ( -0.3679)$
$pH = 8.887$
