Answer
(b) $$K_a = 2.1 \times 10^{-6}$$
Work Step by Step
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ H^+][ A^- ]}{[ HA ]}$$
2. At equilibrium, these are the concentrations of each compound:
$[ HA ] = 0.10 - x$
$[ H^+ ] = x$
$[ A^- ] = x$
Since $[H^+] = 4.6 \times 10^{-4}$, $x = 4.6 \times 10^{-4}$
$[HA] \approx 0.10 M$
$[H^+] = [A^-] = 4.6 \times 10^{-4} M$
$$K_a = \frac{(4.6 \times 10^{-4})(4.6 \times 10^{-4})}{(0.10)} = 2.1 \times 10^{-6}$$
