Answer
$[H^+] = 0.011M$
$[ClCH_2COO^-] = 0.011M$
$[ClCH_2COOH]= 0.089M$
$Ka = 1.36 \times 10^{-3}$
Work Step by Step
1. Drawing the ICE table, we get:
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
2. Find "x", using the percent ionization:
11.0% = 0.11
$ 0.11 = \frac{x}{0.100M}$
$0.011M = x$
3. Calculate the acid concentration (equilibrium):
$[HA] = 0.100M - 0.011M$
$[HA] = 0.089M$
4. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][A^-]}{[HA]}$
$Ka = \frac{(0.011)*(0.011)}{0.089}$
$Ka = 1.36 \times 10^{-3}$
