Answer
The $\frac{[Acid]}{[Base]}$ ratio in this carbonic acid buffer is equal to $0.024$
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 8}$
$[H_3O^+] = 1 \times 10^{- 8}$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][HC{O_3}^-]}{[H_2CO_3]}$
$4.2 \times 10^{-7} = \frac{1 \times 10^{-8}*[HC{O_3}^-]}{[H_2CO_3]}$
$\frac{4.2 \times 10^{-7}}{1 \times 10^{-8}} = \frac{[HC{O_3}^-]}{[H_2CO_3]}$
$\frac{1 \times 10^{-8}}{4.2 \times 10^{-7}} = \frac{[H_2CO_3]}{[HC{O_3}^-]}$
$0.0238 = \frac{[H_2CO_3]}{[HC{O_3}^-]}$
