# calculate the pH and pOH of ca

calculate the pH and pOH of cach of the following soulutions at25 degrees C

a. 0.15 M HNF [Ka(HF) = 3.5 x 10-4]

b. 0.15 m NH3 [Kb (NH3) = 1.8 x10-5]

a)

HF dissociates as:

HF —–> H+ + F-

0.15 0 0

0.15-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 ((3.5*10^-4)*0.15) = 7.246*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

3.5*10^-4 = x^2/(0.15-x)

5.25*10^-5 – 3.5*10^-4 *x = x^2

x^2 + 3.5*10^-4 *x-5.25*10^-5 = 0

a = 1

b = 3.5*10^-4

c = -5.25*10^-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 = 2.101*10^-4

roots are :

x = 7.073*10^-3 and x = -7.423*10^-3

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

x = 7.073*10^-3

So, [H+] = x = 7.073*10^-3 M

use:

pH = -log [H+]

= -log (7.073*10^-3)

= 2.1504

pOH = 14 – pH

= 14 – 2.15

= 11.85

pH = 2.15

pOH = 11.85

b)

NH3 dissociates as:

NH3 +H2O —–> NH4+ + OH-

0.15 0 0

0.15-x x x

Kb = [NH4+][OH-]/[NH3]

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 ((1.8*10^-5)*0.15) = 1.643*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

1.8*10^-5 = x^2/(0.15-x)

2.7*10^-6 – 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-2.7*10^-6 = 0

a = 1

b = 1.8*10^-5

c = -2.7*10^-6

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 = 1.08*10^-5

roots are :

x = 1.634*10^-3 and x = -1.652*10^-3

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

x = 1.634*10^-3

So, [OH-] = x = 1.634*10^-3 M

use:

pOH = -log [OH-]

= -log (1.634*10^-3)

= 2.7867

use:

PH = 14 – pOH

= 14 – 2.7867

= 11.2133

pH = 11.21

pOH = 2.79

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