Answer
$$[CH_3COOH]_{initial} = 0.148 \space M$$
Work Step by Step
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.
$$\frac{ 1.1 \times 10^{-3} }{ 0.100 } \times 100\% = 1.1 \%$$
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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:
$$K_a = \frac{x^2}{[CH_3COOH]_{initial} - x} $$ $$1.8 \times 10^{-5}= \frac{(0.011[CH_3COOH]_{initial})^2}{[CH_3COOH]_{initial} - 0.011[CH_3COOH]_{initial}}$$
$$1.8 \times 10^{-5} = \frac{1.21 \times 10^{-4}[CH_3COOH]_{initial}^2 }{0.989 [CH_3COOH]_{initial}}$$
$$1.8 \times 10^{-5} = 1.22 \times 10^{-4} [CH_3COOH]_{initial}$$ $$[CH_3COOH]_{initial} = 0.148$$
