## Chemistry: The Central Science (13th Edition)

The pH of this solution is equal to $4.14$;
1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 4})$ $pKa = 3.745$ 2. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.25}{0.1}$ - 2.5: It is. 3. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.25}{1.8 \times 10^{-4}} = 1389$ - $\frac{0.1}{1.8 \times 10^{-4}} = 555.6$ 4. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 3.745 + log(\frac{0.25}{0.1})$ $pH = 3.745 + 0.3979$ $pH = 4.143$