The Ka of a monoprotic weak acid is 8.01 × 10-3. What is the percent ionization of a 0.156 M solution of this acid?
HA dissociates as:
HA -----> H+ + A-
0.156 0 0
0.156-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 ((8.01*10^-3)*0.156) = 3.535*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)
8.01*10^-3 = x^2/(0.156-x)
1.25*10^-3 - 8.01*10^-3 *x = x^2
x^2 + 8.01*10^-3 *x-1.25*10^-3 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 8.01*10^-3
c = -1.25*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 = 5.062*10^-3
roots are :
x = 3.157*10^-2 and x = -3.958*10^-2
since x can't be negative, the possible value of x is
x = 3.157*10^-2
% dissociation = (x*100)/c
= 3.157*10^-2*100/0.156
= 20.2 %
Answer: 20.2 %
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