Answer:
the pH of the solution after 20 mL HCl was added is 1.12
Option d) 1.12 is the correct answer
Step-by-step explanation:
Firstly; number of mmols of NH₃ before addition of HCl = M x V = 0.25 x 20 = 5
Also number of mmols pf HCl added = M x V = 0.40 x 20 = 8
Number of mmols of NH₄Cl formed
= number of mmols of NH3 before addition of HCl = 5
Number of mmoles of HCl remained = 8 – 5 = 3
Therefore Total volume of solution = 20 + 20 = 40 mL
Molarity of HCl = 3 / 40 = 0.075 M
Molarity of NH₄Cl = 5 / 40 = 0.125 M
Therefore
Ka x Kb = 10⁻¹⁴
Ka x 1.8 x 10⁻⁵ = 10⁻¹⁴
Ka (NH₄Cl) = 5.56 x 10⁻¹⁰
[H+] from HCl = 0.075 M
[H+] from NH₄Cl = (Ka x C)^1/2
= (5.56 x 10⁻¹⁰ x 0.125)^1/2 = 8.34 x 10⁻⁶
Hence, [H+] from NH₄Cl is negligible.
pH = -log 0.075
pH = 1.12
Therefore the pH of the solution after 20 mL HCl was added is 1.12
Option d) 1.12 is the correct answer