20 mL of 0.25 M of NH 3 is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. Kb (NH3) = 1.8 × 10 −5. a. 12.50
b. 7.00
c. 4.74
d. 1.12

the pH of the solution after 20 mL HCl was added is 1.12

Option d) 1.12 is the correct answer

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