## Chemistry 12th Edition

$K_a = 9.12 \times 10^{-4}$
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: $K_a = \frac{[A^-][H^+]}{[HA]}$ $K_a = \frac{(5.6 \times 10^{-3})(5.6 \times 10^{-3})}{0.0344}$ $K_a = 9.12 \times 10^{-4}$