What is the pH at the equivalence point of the titration of 50.0 mL of 0.300 M NH3(aq) with 0.150 M HCl(aq)?

pH = 5.19, at the equivalence point

Explanation:

This is titration of a weak base and a strong acid. pH in the equivalence point will be acid. To determine this pH, we know that in in the equivalence point:

mmoles base = mmoles of base

50 mL  . 0.3M = 0.150 M . Volume of acid

Volume of acid = (50 . 0.3)/ 0.15 → 100 mL

Total volume of equivalence point is 50 mL + 150 mL = 200 mL

Neutralization reaction is:

NH₃ + HCl → NH₄Cl

NH₃ +  H₃O⁺ ⇄  NH₄⁺  +  H₂O

15          15

0           0            15

Ammonia's Kb = 1.8×10⁻⁵

In the equivalence point, we have only ammonium, beacuse all the amonia reacted to the protons. Let's make the equilibrium

      NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺      Ka

I:      15            -             -            -

Eq:  15-x                        x           x

Ka = 1×10⁻¹⁴ / Kb → 5.55×10⁻¹⁰

[NH₄⁺] = 15 mmoles / 200 mL = 0.075 M

In the equilibrium, we can avoid the 15 - x, because Ka is soo small.

Expression for Ka is [ NH₃ ] .  [H₃O⁺] /  [NH₄⁺] = 5.55×10⁻¹⁰

[H₃O⁺] = √(5.55×10⁻¹⁰ . 0.075)

[H₃O⁺] =  6.45×10⁻⁶

- log [H₃O⁺] =  pH → 5.19

hope this !

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