The Ka of a monoprotic weak acid is 8.58 × 10-3. What is the percent ionization of a 0.178 M solution of this acid?
Let the weak acid be HA
HA dissociates as:
HA -----> H+ + A-
0.178 0 0
0.178-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((8.58*10^-3)*0.178) = 3.908*10^-2
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
8.58*10^-3 = x^2/(0.178-x)
1.527*10^-3 - 8.58*10^-3 *x = x^2
x^2 + 8.58*10^-3 *x-1.527*10^-3 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 8.58*10^-3
c = -1.527*10^-3
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 6.183*10^-3
roots are :
x = 3.502*10^-2 and x = -4.36*10^-2
since x can't be negative, the possible value of x is
x = 3.502*10^-2
% dissociation = (x*100)/c
= 3.502*10^-2*100/0.178
= 19.7 %
Answer: 19.7 %
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