Answer
(a)
$$[H_3O^+] = 6.0 \times 10^{-3}$$
$$[OH^-] = 1.7 \times 10^{-12} \space M$$
$$pH = 2.22 $$
$$pOH = 11.78$$
(b)
$K_a = 1.9 \times 10^{-4} $
Work Step by Step
(a)
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HA ]& [ A^- ]& [ H_3O^+ ]\\
Initial& 0.20 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.20 -x&x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ A^- ][ H^+ ]}{[ HA ]}$$
$$K_a = \frac{(x)(x)}{[ HA ]_{initial} - x}$$
3. Use the percent ionization to find x:
$$Percent \space ionization = \frac{x}{[ HA ]_{initial}} \times 100\% $$
$$x = \frac{ 3.0 \% \times 0.20 }{100\%}$$
$x = 6.0 \times 10^{-3} $
$$[H_3O^+] = x = 6.0 \times 10^{-3}$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 6.0 \times 10^{-3} } = 1.7 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 6.0 \times 10^{-3} ) = 2.22 $$
$$pOH = 14 - 2.22 = 11.78$$
(b)
4. Substitute the value of x and calculate the $K_a$:
$$K_a = \frac{( 6.0 \times 10^{-3} )^2}{ 0.20 - 6.0 \times 10^{-3} }$$
$K_a = 1.9 \times 10^{-4} $