The Ka of a monoprtic weak acid is 8.75 × 10^-3. What is the percent ionization of a 0.191 M solution of this acid?
HA dissociates as:
HA -----> H+ + A-
0.191 0 0
0.191-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 ((8.75*10^-3)*0.191) = 4.088*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)
8.75*10^-3 = x^2/(0.191-x)
1.671*10^-3 - 8.75*10^-3 *x = x^2
x^2 + 8.75*10^-3 *x-1.671*10^-3 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 8.75*10^-3
c = -1.671*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 = 6.762*10^-3
roots are :
x = 3.674*10^-2 and x = -4.549*10^-2
since x can't be negative, the possible value of x is
x = 3.674*10^-2
% dissociation = (x*100)/c
= 3.674*10^-2*100/0.191
= 19.2 %
Answer: 19.2 %
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