## Chemistry 12th Edition

$K_a = 6.31 \times 10^{-5}$
1. Calculate $[H^+]$ $[H^+] = 10^{-pH} = 10^{-6.20} = 6.31 \times 10^{-7}M$ - Remember: $[H^+] = [A^-] = x$ ** x = concentration of molecules that ionized. 2. Use the Ka formula to calculate the Ka value: ** $[HA] = [HA]_{initial} - x$ $[HA] = 0.010 - 6.31 \times 10^{-7}$ $[HA] \approx 0.010M$ $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(6.31 \times 10^{-7})}{0.01}$ $K_a = 6.31 \times 10^{-5}$