Answer
$K_a = 1.5 \times 10^{-5} $
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CH_3CH_2CH_2COOH ]& [ CH_3CH_2CH_2COO^- ]& [ H_3O^+ ]\\
Initial& 0.15 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.15 -x& x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ CH_3CH_2CH_2COO^- ][ H^+ ]}{[ CH_3CH_2CH_2COOH ]}$$
$$K_a = \frac{(x)(x)}{[ CH_3CH_2CH_2COOH ]_{initial} - x}$$
3. Using the $H_3O^+$ concentration:
$[H_3O^+] = x = 1.51 \times 10^{-3} $
4. Substitute the value of x and calculate the $K_a$:
$$K_a = \frac{( 1.51 \times 10^{-3} )^2}{ 0.15 - 1.51 \times 10^{-3} }$$
$K_a = 1.5 \times 10^{-5} $
