Question

Calculate the pH of 510. mL of a 8.16×10^-2-M solution of hydrofluoric acid before and after the addition of 7.65×10^-2 mol of potassium fluoride.

pH befor addition =

pH after addition =

Answer 1

1)

Ka OF HF = 6.6*10^-4

HF dissociates as:

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

8.16*10^-2 0 0

8.16*10^-2-x x x

Ka = [H+][F-]/[HF]

Ka = x*x/(c-x)

6.6*10^-4 = x^2/(8.16*10^-2-x)

5.386*10^-5 - 6.6*10^-4*x = x^2

x^2 + 6.6*10^-4 *x - 5.386*10^-5 = 0

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

a = 1.0

b = 6.6E-4

c = -5.386E-5

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 = 2.159*10^-4

roots are :

x = 7.016*10^-3 and x = -7.676*10^-3

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

x = 7.016*10^-3

so,

[H+] = 7.016*10^-3

use:

pH = -log [H+]

= -log (7.339*10^-3)

= 2.13

Answer: pH before adding 2.13

2)

[HF] = 8.16*10^-2 M

[KF] = number of mol / volume in L

= (7.65*10^-2 mol)/0.510 L

= 0.150 M

Ka = 6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.1805

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.1805+ log {0.15/0.0816}

= 3.44

Answer: pH after adding = 3.44