How many grams of dry NH4Cl need to be added to 1.70 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.80? Kb for ammonia is 1.8×10−5.
moles of NH3 = 1.70 x 0.500 = 0.85
pH = 8.80
pOH = 14 - 8.80 = 5.20
Kb = 1.8 x 10^-5
pKb = -log (1.8 x 10^-5)
= 4.74
pOH = pKb + log [salt / base]
5.20 = 4.74 + log [NH4Cl / NH3]
5.20 = 4.74 + log [NH4Cl / 0.85]
moles of NH4Cl = 2.45 mol
mass of NH4Cl = 2.45 x 53.49
mass of NH4Cl = 131 g
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.500...
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