Question

Calculate the pH for the titration of 100 mL of 0.10 M NH3, Kb = 1.8x10-5, with 0.25 M HBr.

a) Before any acid is added. pH= ?

Answer 1

1)when 0.0 mL of HBr is added

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.1 0 0

0.1-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)*0.1) = 1.342*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/(0.1-x)

1.8*10^-6 - 1.8*10^-5 *x = x^2

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

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

a = 1

b = 1.8*10^-5

c = -1.8*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 = 7.2*10^-6

roots are :

x = 1.333*10^-3 and x = -1.351*10^-3

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

x = 1.333*10^-3

so.[OH-] = x = 1.333*10^-3 M

use:

pOH = -log [OH-]

= -log (1.333*10^-3)

= 2.9

use:

PH = 14 - pOH

= 14 - 2.9

= 11.1

Answer: 11.1