Calculate the pH of a o.10 M ammonia solution . Calculate the pH after 50. mL of this solution is treated with 25 mL of 0.10 M HCl . The dissociation constant of ammonia , Kb = 1.77 x 1-^-5 
 
( this is a textbook example question .I am not able to understand the method used in textbook . So please help me by giving some easy method to solve this . ) 
 
Thank you !

Asked by Saravanan | 24th Oct, 2014, 09:48: AM

Expert Answer:

NH3  + H2O → NH4+ + OH-

 

Kb  = [NH4+ ] [OH- ]    =  1.77 x 10-5

            [NH3]

Before neutralization,

[NH4+ ]  = [OH-] = x

Thus

K=   x2    = 1.77x 10-5

         0.10

Hence x = 1.33 x 10-5 = [OH-]

 

 

Now Kw = [H+][OH-]

So [H+]  = Kw/ [OH-] = 10-14 / 1.33x 10-12 = 7.51 x 10-12

        

p H = - log(7.51 x 10-12) = 11.12

On addition of 25 ml of 0.1 M HCl solution to 50 ml of 0.1 M ammonia solution , 2.5 mol of ammonia are neutralized. The resulting 75 ml solution contains the remaining unneutralised 2.5 mol of NH3 molecules and 2.5 mol of NH4+.

 

                                         NH3    + HCl  → NH4+   + Cl-

                                        2.5         2.5          0             0

 

At equilibrium                     0            0            2.5        2.5

The resulting 75 ml of solution contains 2.5 mol of NH4+ ions and 2.5 mol of uneutralised NH3 molecules. This NH3 exists in the following equilibrium.

NH4OH     ⇌      NH4+     + OH-

0.033 M- y      y                  y

Where y = [OH-] = [NH4+]

The final 75ml solution after neutralization already contains 2.5 mol NH4+ ions, thus total concentration of NH4+ ions is given as:

[NH4+]  = 0.033 + y

As y is small, [NH4OH] = 0.033 M and [NH4+] = 0.033M

Kb =  [NH4+]  [OH- = y (0.033)  = 1.77x 10-5 M

             [NH4OH]          (0.033)

 Thus , y = 1.77 x 10-5 = [OH-]

[H+] = 10-14             = 0.56 x 10-9

         1.77 x 10-5

 

Hence p H = 9.24

Answered by Arvind Diwale | 25th Oct, 2014, 05:08: PM