Question

How would you make a 750.0 mL of a buffer at pH 9.00 if you had 0.48 M ammonia and a bottle of solid ammonium chloride? Kb = 1.8 x 10 -5 (show all reactions and work, with units)

Answer 1

NH₄⁺ ---> NH₃ + H⁺

Ka = [NH₃] [H⁺] / [NH₄⁺]

pH = 9 ⇒ [H⁺] = 10^-9

Ka = Kw/Kb = 10^-14 / 1.8 x 10^-5 = 5/9 x 10^-9 M

[NH₃]/[NH₄⁺] = 5/9 x 10^-9 / 10^-9 = 5/9

Assuming negligible volume change in solution due to addition of solid NH₄Cl

[NH₃] = 0.48 M

[NH₄⁺] = 0.48 * 9/5 = 0.86 M

Moles of NH₄Cl required = 0.86 * 750 /1000 = 0.65 moles

Mass of NH₄Cl required = 0.65 * 53.5 = 34.7 grams

By adding 34.7 grams of solid ammonium chloride to 750 mL of 0.48 M ammonia solution, required buffer of pH = 9 can be made.