Question

1.) Calculate the percent ionization of a 0.419 M solution of hydrofluoric acid.

% ionization = ????

2.) in the laboratory a student measures the percent ionization of a 0.463 M solution of acetylsalicylic acid (aspirin), HC9H7O4, to be 2.46%.

Calculate the value of Ka from this experimental data.

Ka = ????

Answer 1

1)
Ka of HF = 6.6*10^-4

HF dissociates as:

HF          ----->     H+   + F-
0.419                 0         0
0.419-x               x         x


Ka = [H+][F-]/[HF]
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 ((6.6*10^-4)*0.419) = 1.663*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)
6.6*10^-4 = x^2/(0.419-x)
2.765*10^-4 - 6.6*10^-4 *x = x^2
x^2 + 6.6*10^-4 *x-2.765*10^-4 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.6*10^-4
c = -2.765*10^-4

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 = 1.107*10^-3

roots are :
x = 1.63*10^-2 and x = -1.696*10^-2

since x can't be negative, the possible value of x is
x = 1.63*10^-2

% dissociation = (x*100)/c
= 1.63*10^-2*100/0.419
= 3.891 %
Answer: 3.89 %

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