Question

Determine the pH of a 0.200 M solution of hydrofluoric acid (HF) (Ka = 6.3 x 10^-4)

a. 4.20

b. 3.90

c. 2.02

d. 0.69

*please show process*

Answer 1

HF dissociates as:

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

0.2 0 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 ((6.3*10^-4)*0.2) = 1.122*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/(0.2-x)

1.26*10^-4 - 6.3*10^-4 *x = x^2

x^2 + 6.3*10^-4 *x-1.26*10^-4 = 0

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

a = 1

b = 6.3*10^-4

c = -1.26*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 = 5.044*10^-4

roots are :

x = 1.091*10^-2 and x = -1.154*10^-2

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

x = 1.091*10^-2

So, [H+] = x = 1.091*10^-2 M

use:

pH = -log [H+]

= -log (1.091*10^-2)

= 1.96

Answer: d