Question

Calculate the [H ] and pH of a 0.000501 M hydrofluoric acid solution. Keep in mind that the Ka of hydrofluoric acid is 6.80 × 10-5.

Answer 1

1)
HF dissociates as:

HF          ----->     H+   + F-
5.01*10^-4                 0         0
5.01*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 ((6.8*10^-5)*5.01*10^-4) = 1.846*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)
6.8*10^-5 = x^2/(5.01*10^-4-x)
3.407*10^-8 - 6.8*10^-5 *x = x^2
x^2 + 6.8*10^-5 *x-3.407*10^-8 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.8*10^-5
c = -3.407*10^-8

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 = 1.409*10^-7

roots are :
x = 1.537*10^-4 and x = -2.217*10^-4

since x can't be negative, the possible value of x is
x = 1.537*10^-4

So, [H+] = x = 1.537*10^-4 M
Answer: 1.54*10^-4 M

2)
use:
pH = -log [H+]
= -log (1.537*10^-4)
= 3.8134
Answer: 3.813