## Chemistry 9th Edition

1. Find the $[H_3O^+]$ value: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.0}$ $[H_3O^+] = 1.0 \times 10^{- 3}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3CO_2^-(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $CH_3CO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [CH_3CO_2^-] = 0 + x = x$ -$[CH_3CO_2H] = [CH_3CO_2H]_{initial} - x$ For approximation, we are going to consider $[CH_3CO_2H]_{initial} = [CH_3CO_2H]$ 3. Now, use the Ka and x values and equation to find the initial concentration value. $Ka = \frac{[H_3O^+][CH_3CO_2^-]}{ [Initial CH_3CO_2H] - x}$ $1.8\times 10^{- 5}= \frac{[x^2]}{ [Initial CH_3CO_2H] - x}$ $1.8\times 10^{- 5}= \frac{( 1.0\times 10^{- 3})^2}{[Initial CH_3CO_2H] - 1.0 \times 10^{- 3}}$ $[Initial CH_3CO_2H] - 1.0\times 10^{- 3} = \frac{ 1.0 \times 10^{- 6}}{ 1.8\times 10^{- 5}}$ $[Initial CH_3CO_2H] = \frac{ 1.0 \times 10^{- 6}}{ 1.8\times 10^{- 5}} + 1.0\times 10^{- 3}$ $[Initial CH_3CO_2H] = 0.057M$