Answer
The pH of this solution is equal to $5.28$.
Work Step by Step
1. Since $C_5H_5NH^+$ is the conjugate acid of $(C_5H_5N$ , we can calculate its kb by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.7\times 10^{- 9} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.7\times 10^{- 9}}$
$K_a = 5.882\times 10^{- 6}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.882 \times 10^{- 6})$
$pKa = 5.23$
3. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.51}{0.45}$
- 1.133: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.51}{5.881 \times 10^{-6}} = 8.67\times 10^{4}$
- $ \frac{0.45}{5.881 \times 10^{-6}} = 7.65\times 10^{4}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 5.23 + log(\frac{0.51}{0.45})$
$pH = 5.23 + 0.05436$
$pH = 5.284$
