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 !

NH₃  + H₂O → NH₄⁺ + OH^-

 

K_b  = [NH₄⁺ ] [OH^- ]    =  1.77 x 10^-5

            [NH₃]

Before neutralization,

[NH₄⁺ ]  = [OH^-] = x

Thus

K_b  =   x²    = 1.77x 10^-5

         0.10

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

 

 

Now K_w = [H⁺][OH^-]

So [H⁺]  = K_w/ [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