Answer
$[BrCH_2COO^-] and [H_3O^+] = 0.0132M$
$[BrCH_2COOH] = 0.0868M$
$Ka = 2.007\times 10^{- 3}$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [BrCH_2COO^-] = x$
-$[BrCH_2COOH] = [BrCH_2COOH]_{initial} - x$
2. Write the percent ionization equation, and find 'x':
%ionization = $\frac{x}{[Initial Acid]} \times 100$
13.2= $\frac{x}{ 0.1} \times 100$
0.132= $\frac{x}{ 0.1}$
$ 0.0132= x$
Therefore: $[BrCH_2COO^-] and [H_3O^+] = 0.0132M$
And, $[BrCH_2COOH] = 0.0868M$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][BrCH_2COO^-]}{ [BrCH_2COOH]}$
$Ka = \frac{x^2}{[InitialBrCH_2COOH] - x}$
$Ka = \frac{( 0.0132)^2}{ 0.1- 0.0132}$
$Ka = \frac{ 1.742\times 10^{- 4}}{ 0.0868}$
$Ka = 2.007\times 10^{- 3}$
