Answer:
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
