## Chemistry (4th Edition)

The $\frac{[Acid]}{[Base]}$ ratio in this carbonic acid buffer is equal to $0.024$
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: $K_a = \frac{[H_3O^+][HC{O_3}^-]}{[H_2CO_3]}$ $4.2 \times 10^{-7} = \frac{1 \times 10^{-8}*[HC{O_3}^-]}{[H_2CO_3]}$ $\frac{4.2 \times 10^{-7}}{1 \times 10^{-8}} = \frac{[HC{O_3}^-]}{[H_2CO_3]}$ $\frac{1 \times 10^{-8}}{4.2 \times 10^{-7}} = \frac{[H_2CO_3]}{[HC{O_3}^-]}$ $0.0238 = \frac{[H_2CO_3]}{[HC{O_3}^-]}$