Question

It takes 25 ml of 0.40 M nitric acid solution to completely neutralize 50.0 ml of 0.20 M NH3 solution (i.e. to reach the equivalence point). What is the pH of the equivalence point?(Kb NH3 = 1.8x10^-5)

Answer 1

Moles of nitric acid = 25 x 0.40 = 10 mmol     (mmol stands for millimole)

Moles of ammonia = 50.0 x 0.20 = 10 mmol

NH₃ + HNO₃ NH₄NO₃ + H₂O

1 mole ammonia will react with 1 mole nitric acid to form 1 mole ammonium nitrate.

Therefore, 10 mmol of ammonia will react with 10 mmol of nitric acid to form 10 mmol of ammonium nitrate.

Therefore, moles of ammonium nitrate formed = 10 mmol

Therefore, the moles of ammonium ion (NH₄⁺) = 10 mmol

Total volume of the solution at the equivalence point = (50 + 25) = 75 mL

Concentration of the ammonium ion = 10/75 = 0.133 M

Now, as ammonium nitrate is a salt of strong acid (HNO₃) and weak base (NH₃), its aqueous solution will be acidic as ammonium will produce H₃O⁺ in water.

                                                                 NH₄⁺   + H₂O H₃O⁺ + NH₃

Initial concentration                                0.133                         0            0

Change in concentration                         -X                              X           X

Equilibrium concentration                  (0.133 - X)                       X           X

Given, K_b of NH₃ = 1.8 x 10^-5

Therefore, K_a of ammonium = K_w/K_a = 10^-14/(1.8 x 10^-5) = 5.6 x 10^-10

(K_w = autoprotolysis constant of water = 10^-14)

Now,

K_a = [H₃O⁺][NH₃]/[NH₄⁺]

or, 5.6 x 10^-10 = (X)(X)/(0.133 - X)        

or, 5.6 x 10^-10 = X²/0.133                                  [(0.133 - X) = 0.133 as X<<0.133]

or, X² = 7.4 x 10^-11

or, X = 8.6 x 10^-6

or, [H₃O⁺] = 8.6 x 10^-6

or, -log[H₃O⁺] = -log(8.6 x 10^-6)

or, pH = 5.07

Therefore, pH at the equivalence point = 5.07