Question

What is the pH for a 2.0 M solution of HF. Ka= 7.2 x 10^-4

Please show work! Im studying for a test.

Answer 1

HF dissociates as:

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

2 0 0

2-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 ((7.2*10^-4)*2) = 3.795*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)

7.2*10^-4 = x^2/(2-x)

1.44*10^-3 - 7.2*10^-4 *x = x^2

x^2 + 7.2*10^-4 *x-1.44*10^-3 = 0

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

a = 1

b = 7.2*10^-4

c = -1.44*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 = 5.761*10^-3

roots are :

x = 3.759*10^-2 and x = -3.831*10^-2

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

x = 3.759*10^-2

So, [H+] = x = 3.759*10^-2 M

use:

pH = -log [H+]

= -log (3.759*10^-2)

= 1.4249

Answer: 1.42