Answer
$K_a = 1.4 \times 10^{-4} $
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CH_3CH(OH)COOH ]& [ CH_3CH(OH)COO^- ]& [ H_3O^+ ]\\
Initial& 0.0284 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.0284 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH_3CH(OH)COO^- ][ H^+ ]}{[ CH_3CH(OH)COOH ]}$$
$$K_a = \frac{(x)(x)}{[ CH_3CH(OH)COOH ]_{initial} - x}$$
3. Use the percent ionization to find x:
$$Percent \space ionization = \frac{x}{[ CH_3CH(OH)COOH ]_{initial}} \times 100\% $$
$$x = \frac{ 6.7 \% \times 0.0284 }{100\%}$$
$x = 1.9 \times 10^{-3} $
4. Substitute the value of x and calculate the $K_a$:
$$K_a = \frac{( 1.9 \times 10^{-3} )^2}{ 0.0284 - 1.9 \times 10^{-3} }$$
$K_a = 1.4 \times 10^{-4} $
