## General Chemistry: Principles and Modern Applications (10th Edition)

$$[CH_3COOH]_{initial} = 0.148 \space M$$
1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ CH_3CH_2CO_2H ]& [ CH_3CH_2CO{_2}^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH_3CH_2CO{_2}^- ][ H_3O^+ ]}{[ CH_3CH_2CO_2H ]}$$ $$K_a = \frac{(x)(x)}{[ CH_3CH_2CO_2H ]_{initial} - x}$$ 3. Assuming $0.100 \gt\gt x:$ $$K_a = \frac{x^2}{[ CH_3CH_2CO_2H ]_{initial}}$$ $$x = \sqrt{K_a \times [ CH_3CH_2CO_2H ]_{initial}} = \sqrt{ 1.3 \times 10^{-5} \times 0.100 }$$ $x = 1.1 \times 10^{-3}$ 4. Calculate the percent ionization. $$\frac{ 1.1 \times 10^{-3} }{ 0.100 } \times 100\% = 1.1 \%$$ ---- Now, we are analyzing the ionization of acetic acid. 5. Find the molarity of the acetic acid solution. $$1.1 \% = \frac{x}{[CH_3COOH]_{initial}} \times 100\%$$ $$x = 0.011[CH_3COOH]_{initial}$$ 6. Write the Ka expression for the equilibrium of acetic acid: $$K_a = \frac{x^2}{[CH_3COOH]_{initial} - x}$$ $$1.8 \times 10^{-5}= \frac{(0.011[CH_3COOH]_{initial})^2}{[CH_3COOH]_{initial} - 0.011[CH_3COOH]_{initial}}$$ $$1.8 \times 10^{-5} = \frac{1.21 \times 10^{-4}[CH_3COOH]_{initial}^2 }{0.989 [CH_3COOH]_{initial}}$$ $$1.8 \times 10^{-5} = 1.22 \times 10^{-4} [CH_3COOH]_{initial}$$ $$[CH_3COOH]_{initial} = 0.148$$