Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions. (Ka(HF)=6.8×10^−4)

a) 0.15 M HF

b) 0.15 M NaF

c) a mixture that is 0.15 M in HF and 0.15 M in NaF

Answer 1

a)

Lets write the dissociation equation of HF

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

0.15 0 0

0.15-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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^-4)*0.15) = 1.01*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)

6.8*10^-4 = x^2/(0.15-x)

1.02*10^-4 - 6.8*10^-4 *x = x^2

x^2 + 6.8*10^-4 *x-1.02*10^-4 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 6.8*10^-4

c = -1.02*10^-4

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 4.085*10^-4

putting value of d, solution can be written as:

x = {-6.8*10^-4 + √(4.085*10^-4)}/2

x = {-6.8*10^-4 - √(4.085*10^-4)}/2

solutions are :

x = 9.765*10^-3 and x = -1.045*10^-2

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

x = 9.765*10^-3

So, [H+] = x = 9.765*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (9.765*10^-3)

= 2.01

Answer: 2.01

b)

we have below equation to be used:

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)/6.8*10^-4

Kb = 1.471*10^-11

F- dissociates as

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

0.15 0 0

0.15-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.471*10^-11)*0.15) = 1.485*10^-6

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

so, x = 1.485*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.485*10^-6)

= 5.83

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.83

= 8.18

Answer: 8.18

c)

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 3.167+ log {0.15/0.15}

= 3.167

Answer: 3.17