Question

1) BOH is a weak Base. Kb = 2 x 10^-5. The pH of a 0.1M solution is?

2) A 0.1M solution of BCl is prepared. Kb for BOH is 6 x 10-6 . What is the solution’s pH?

Answer 1

1)

BOH dissociates as:

BOH +H2O -----> B+ + OH-

0.1 0 0

0.1-x x x

Kb = [B+][OH-]/[BOH]

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 ((2*10^-5)*0.1) = 1.414*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)

2*10^-5 = x^2/(0.1-x)

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

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

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

a = 1

b = 2*10^-5

c = -2*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 = 8*10^-6

roots are :

x = 1.404*10^-3 and x = -1.424*10^-3

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

x = 1.404*10^-3

so.[OH-] = x = 1.404*10^-3 M

use:

pOH = -log [OH-]

= -log (1.404*10^-3)

= 2.8526

use:

PH = 14 - pOH

= 14 - 2.85

= 11.15

Answer: 11.15

2)

use:

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

Ka = (1.0*10^-14)/6*10^-6

Ka = 1.667*10^-9

B+ + H2O -----> BOH + H+

0.1 0 0

0.1-x x x

Ka = [H+][BOH]/[B+]

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 ((1.667*10^-9)*0.1) = 1.291*10^-5

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

so, x = 1.291*10^-5 M

so.[H+] = x = 1.291*10^-5 M

use:

pH = -log [H+]

= -log (1.291*10^-5)

= 4.89

Answer: 4.89