Answer
The pH of this solution is equal to $9.68$.
Work Step by Step
1. Since $(CH_3)_3NH^+$ is the conjugate acid of $(CH_3)_3N$ , we can calculate its kb by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 6.4\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 6.4\times 10^{- 5}}$
$K_a = 1.563\times 10^{- 10}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.563 \times 10^{- 10})$
$pKa = 9.806$
3. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.075}{0.1}$
- 0.75: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.075}{1.562 \times 10^{-10}} = 4.8\times 10^{8}$
- $ \frac{0.1}{1.562 \times 10^{-10}} = 6.4\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.806 + log(\frac{0.075}{0.1})$
$pH = 9.806 + -0.1249$
$pH = 9.681$
