Answer:
0.00091%
Explanation:
The fraction of 4-chlorobutanoic acid that would dissociatr in aqueos solution is a function of Ionization percentage. It is obtained by
Ka = [dissociated acid] / [original acid] x 100%
The equation of the reaction is
C₂HClCOOH + H20 (aq) = C₂HClCOO⁻ + H₃O⁺ pka =4.69
But pKa = - log Ka
4.69 = - Log Ka
10⁻⁴⁶⁹ = Ka
Taking the antilog of the equation we get the ionization constant
Ka = 0.0000204 M
=2.04 x 10⁻⁵M
At the beginning of the reaction we have the following concentrations
1.8M C₂HClCOOH : 0M (zero molar) C₂HClCOO⁻ : 0M (zero molar) H₃O⁺
At equilibrium, we have,
(1.8M -x) xM C₂HClCOO⁻ and xM H₃0⁺
Therefore,
Ka = [C2HClCOO-] [H30+] / [C2HClCOOH],
Inputing the value of Ka
Ka .[C2HClCOOH] = [CHCLCOO-] [H3O+]
0.0000204 (1.8 - X) = (x).(x)
x² = (0.00003672 - 0.0000204 X)
= (3.672 x10-5 - 2.04 x10⁻⁵X)
x² +2.04x10⁻⁵ x = 3.672 x 10⁻⁵
x² + 2.04 x 10⁻⁵x - 3.672 x 10⁻⁵ = 0
x = 0.00001632
= 1.632 x 10⁻⁵
Inputing back into equation 1
1.8 - x = [H3O+]
1.8 - 0.00001632 = 1.7999837
It therefore implies that only 0.00001632M of 4-chloroutanoic acid dissciated at equilibrium, we can now calculate the percentage dissociation by
Percentage dissociation = 0.00001632 / 1.8M x 100%
= (1.632 x 10⁻⁵/1.8 ) x 100%
= 0.00090667%
= 0.00091%