## Chemistry and Chemical Reactivity (9th Edition)

(a) $pH = 3.59$; (b) Acid to base ratio = $0.45$;
(a) 1. Calculate the pKa value: $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 4})$ $pKa = 3.74$ 2. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.035}{0.05}$ - 0.7: It is. 3. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.035}{1.8 \times 10^{-4}} = 294$ - $\frac{0.05}{1.8 \times 10^{-4}} = 388$ 4. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 3.74 + log(\frac{0.035}{0.05})$ $pH = 3.74 + log(0.7)$ $pH = 3.74 + (-0.15)$ $pH = 3.59$ ----- (b) 3.59 + 0.50 = 4.09 1. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.09}$ $[H_3O^+] = 8.1 \times 10^{- 5}$ 2. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][HC{O_2}^-]}{[HCO_2H]}$ $1.8 \times 10^{-4} = \frac{8.1 \times 10^{-5}*[HC{O_2}^-]}{[HCO_2H]}$ $\frac{1.8 \times 10^{-4}}{8.1 \times 10^{-5}} = \frac{[HC{O_2}^-]}{[HCO_2H]}$ $\frac{8.1 \times 10^{-5}}{1.8 \times 10^{-4}} = \frac{[HCO_2H]}{[HC{O_2}^-]}$ $0.45 = \frac{[HCO_2H]}{[HC{O_2}^-]}$