Question

A solution contains 0.45 M HF (ka=6.8x10^-4). Write the dissociation reaction and determine the degree of ionization and he pH of the solution.

Answer 1

HF dissociates as:

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

0.45 0 0

0.45-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 ((6.8*10^-4)*0.45) = 1.749*10^-2

degree of ionisation = x/c = 1.749*10^-2 / 0.45 = 0.0389 (Answer)

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.8*10^-4 = x^2/(0.45-x)

3.06*10^-4 - 6.8*10^-4 *x = x^2

x^2 + 6.8*10^-4 *x-3.06*10^-4 = 0

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

a = 1

b = 6.8*10^-4

c = -3.06*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 = 1.224*10^-3

roots are :

x = 1.716*10^-2 and x = -1.784*10^-2

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

x = 1.716*10^-2

so.[H+] = x = 1.716*10^-2 M

use:

pH = -log [H+]

= -log (1.716*10^-2)

= 1.77 (Answer)