Question

A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H 4, 3ClCH2COOH4, and Ka for chloroacetic acid.

Answer 1

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