Question

1. Your supervisor asks you to prepare 250.00 mL of ammonium buffer solution (0.15 M) with pH=9.00, from a concentrated ammonia solution (25%w/w, d=0.91g/cm3) and solid ammonium chloride (98% w/w). (MW of NH3 = 17.03 g/mol and FW of NH4Cl = 53.5 g/mol and pKaNH4+/NH3 =9.24).

a) How many milliliters of the concentrated ammonia solution do you need?

b) How many grams of ammonium chloride do you need?

Answer 1

pH of basic buffer = 14 - (pkb + log(NH4^+/NH3))

pkb of NH3 = 14-9.24 = 4.76

9 = 14 - (4.76+log(x))

x = (NH4^+/NH3) = 1.738

total no of mol of buffer = 250*0.15/1000 = 0.0375 mol

NO of mol of NH3 REQUIRED = x mol

from ratio

NO of mol of NH4^+ REQUIRED = 1.738*x

(x+1.738*x) = 0.0375

x = 0.0137

NO of mol of NH3 REQUIRED = x = 0.0137 mol

Amount of NH3 = 0.0137*17 = 0.2329 g

25% w/w = 25*0.91 = 22.75 %w/v

volume of NH3 solution required = 0.2329*100/22.5 = 1.035 ml

NO of mol of NH4^+ REQUIRED = 1.738*x = 1.738*0.0137 = 0.0238 mol

mass of NH4cl required = = 0.0238*53.5 = 1.2733 g

mass 98%w/w NH4Cl REQUIRED = 1.2733*100/98 = 1.3 g