Answer:
[CICH2COOH] = 0.089 M
[CICH2COO-] = 0.011 M
[H+] = 0.011 M
Ka = 1.36 x 10^-3
Explanation:
The reaction
CICH2COOH ⇄ H+ (aq) + CICH2COO- (aq)
The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.
let us calculate Ka
First , find change in concentration
since , 11% ionized
change in concentration = 0.10 X 11% = 0.011 M
Initial Concentration of CICH2COOH = 0.10 M
change in concentration of CICH2COOH = - 0.011 M
Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m
Initial Concentration of CICH2COO- = 0 M
change in concentration of CICH2COO- = + 0.011 M
Equilibrium Concentration of CICH2COO- = 0.011 M
Initial Concentration of H+ = 0 M
change in concentration of H+ = + 0.011 M
Equilibrium Concentration of H+ = 0.011 M
Therefore,
[CICH2COOH] = 0.089 M
[CICH2COO-] = 0.011 M
[H+] = 0.011 M
Ka = [H+][CICH2COO-] /[CICH2COOH]
Ka = (0.011 * 0.011) / (0.089)
Ka = 1.36 x 10^-3
