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

the old alchemists made significant contributions to science even though there starting premises were off a bit. their conclusions after failure would steer them into better directions for investigation. science is no more than close observation and the initial mistakes result in rethinking and retrying experiments.