Answer
$K_a = 9.12 \times 10^{-4}$
Work Step by Step
1. Write the percent ionized equation, and find the $[A^-]$
$\%ionized = \frac{[A^-]}{[HA]_{initial}} \times 100\%$
** In this case, the $[A^-]$ represents the concentration of molecules that have been ionized.
$14\% = \frac{[A^-]}{0.040M} \times 100\%$
$0.14 = \frac{[A^-]}{0.040M}$
$5.6 \times 10^{-3} M = [A^-]$
- Remember: $[H^+] = [A^-] = 5.6 \times 10^{-3}M$
2. Use the ICE table to find the concentrations in equilibrium:
$[HA] = 0.040 - x$
$[H^+] = [A^-] = x$
Since $[A^-]\ and \ [H^+]$ = $5.6 \times 10^{-3}M$:
$[HA] = 0.040 - 5.6 \times 10^{-3} = 0.0344M$
$[H^+] = [A^-] = 5.6 \times 10^{-3}M$
3. Use the Ka equation to find its value:
$K_a = \frac{[A^-][H^+]}{[HA]}$
$K_a = \frac{(5.6 \times 10^{-3})(5.6 \times 10^{-3})}{0.0344}$
$K_a = 9.12 \times 10^{-4}$