Answer
$K_a = 6.31 \times 10^{-5}$
Work Step by Step
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} $
$[HA] \approx 0.010M$
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(6.31 \times 10^{-7})}{0.01}$
$K_a = 6.31 \times 10^{-5}$