# Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677a: 13

The ${[CH_3CO_2H]}$ to ${[CH_3C{O_2}^-]}$ ratio is equal to $0.56$;

#### Work Step by Step

1. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 5}$ $[H_3O^+] = 1 \times 10^{- 5}$ 2. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][CH_3C{O_2}^-]}{[CH_3CO_2H]}$ $1.8 \times 10^{-5} = \frac{1 \times 10^{-5}*[CH_3C{O_2}^-]}{[CH_3CO_2H]}$ $\frac{1.8 \times 10^{-5}}{1 \times 10^{-5}} = \frac{[CH_3C{O_2}^-]}{[CH_3CO_2H]}$ $\frac{1 \times 10^{-5}}{1.8 \times 10^{-5}} = \frac{[CH_3CO_2H]}{[CH_3C{O_2}^-]}$ $0.56 = \frac{[CH_3CO_2H]}{[CH_3C{O_2}^-]}$

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