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Question 1 Tries remaining: 1 Points out of 1.00 P Flag question What is the percent ionization of a 0.00078 M solution of HF? The Kais 6.4x10-4 Multiple tries are permitted; however, 20% (1/5) point will be deducted for each incorrect response.

Answer 1

HF dissociates as:

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

7.8*10^-4 0 0

7.8*10^-4-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 ((6.4*10^-4)*7.8*10^-4) = 7.065*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.4*10^-4 = x^2/(7.8*10^-4-x)

4.992*10^-7 - 6.4*10^-4 *x = x^2

x^2 + 6.4*10^-4 *x-4.992*10^-7 = 0

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

a = 1

b = 6.4*10^-4

c = -4.992*10^-7

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 = 2.406*10^-6

roots are :

x = 4.556*10^-4 and x = -1.096*10^-3

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

x = 4.556*10^-4

% dissociation = (x*100)/c

= 4.556*10^-4*100/0.0008

= 58.41 %

Answer: 58.4 %