Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions.
Ka(HF)=6.3×10−4

A. 0.31 molL−1 HF

B. 0.31 molL−1 NaF

C. a mixture that is 0.31 molL−1 in HF and 0.31 molL−1 in NaF

Answer 1

A)

HF dissociates as:

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

0.31 0 0

0.31-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)*0.31) = 1.397*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/(0.31-x)

1.953*10^-4 - 6.3*10^-4 *x = x^2

x^2 + 6.3*10^-4 *x-1.953*10^-4 = 0

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

a = 1

b = 6.3*10^-4

c = -1.953*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 = 7.816*10^-4

roots are :

x = 1.366*10^-2 and x = -1.429*10^-2

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

x = 1.366*10^-2

So, [H+] = x = 1.366*10^-2 M

use:

pH = -log [H+]

= -log (1.366*10^-2)

= 1.8644

Answer: 1.86

B)

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/6.3*10^-4

Kb = 1.587*10^-11

F- dissociates as

F- + H2O -----> HF + OH-

0.31 0 0

0.31-x x x

Kb = [HF][OH-]/[F-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.587*10^-11)*0.31) = 2.218*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.218*10^-6 M

use:

pOH = -log [OH-]

= -log (2.218*10^-6)

= 5.654

use:

PH = 14 - pOH

= 14 - 5.654

= 8.346

Answer: 8.35

C)

Ka = 6.3*10^-4

pKa = - log (Ka)

= - log(6.3*10^-4)

= 3.201

use:

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

= 3.201+ log {0.31/0.31}

= 3.201

Answer: 3.20