Chemistry: The Central Science (13th Edition)

The ratio of $HC{O_3}^-$ to $H_2CO_3$ when the pH is 7.1 is : 5.4.
1. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 7.1}$ $[H_3O^+] = 7.943 \times 10^{- 8}M$ 2. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][HC{O_3}^-]}{[H_2CO_3]}$ $4.3 \times 10^{-7} = \frac{7.943 \times 10^{-8}*[HC{O_3}^-]}{[H_2CO_3]}$ $\frac{4.3 \times 10^{-7}}{7.943 \times 10^{-8}} = \frac{[HC{O_3}^-]}{[H_2CO_3]}$ $5.413 = \frac{[HC{O_3}^-]}{[H_2CO_3]}$