Question

What is the percent ionization for a 0.300M HF solution? Ka for HF is 6.3×10-4.

Answer 1

HF dissociates as:

HF -----> H+ + F-

0.300 0 0

0.300-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 ((6.3*10^-4)*0.3) = 1.375*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)

6.3*10^-4 = x^2/(0.3-x)

1.89*10^-4 - 6.3*10^-4 *x = x^2

x^2 + 6.3*10^-4 *x-1.89*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.3*10^-4

c = -1.89*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 = 7.564*10^-4

roots are :

x = 1.344*10^-2 and x = -1.407*10^-2

since x can't be negative, the possible value of x is

x = 1.344*10^-2

% dissociation = (x*100)/c

= 0.0134*100/0.3

= 4.48 %

Answer: 4.48 %