Question

How many gram of ammonia should be added to 500mL of 1M ammonium chloride to give a pH 9?(Ka ammonium=5.6x10-10)

Answer 1

NH3 = 17 g/mole

pH = 9–––> [H+] = 10^-9 M ––––> [OH–] = 10^-5 M

Ka*Kb = Kwater

(5.6×10^-10)*Kb = 10^-14

Kb = 1.8×10^-5 ( Kb of NH3)

[OH–] = nNH3*Kb / nNH4Cl

10^-5 = nNH3*1.8×10^-5 / 0.5

nNH3 = 0.28 mole

0.28*17 = 4.76 g of NH3