Answer
(a) $pH = 3.59$;
(b) Acid to base ratio = $0.45$;
Work Step by Step
(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:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.74 + log(\frac{0.035}{0.05})$
$pH = 3.74 + log(0.7)$
$pH = 3.74 + (-0.15)$
$pH = 3.59$
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(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:
$K_a = \frac{[H_3O^+][HC{O_2}^-]}{[HCO_2H]}$
$1.8 \times 10^{-4} = \frac{8.1 \times 10^{-5}*[HC{O_2}^-]}{[HCO_2H]}$
$\frac{1.8 \times 10^{-4}}{8.1 \times 10^{-5}} = \frac{[HC{O_2}^-]}{[HCO_2H]}$
$\frac{8.1 \times 10^{-5}}{1.8 \times 10^{-4}} = \frac{[HCO_2H]}{[HC{O_2}^-]}$
$0.45 = \frac{[HCO_2H]}{[HC{O_2}^-]}$
