Question

How many grams of dry NH4Cl need to be added to 1.50 L of a 0.100 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.82? Kb for ammonia is 1.8×10−5. Express your answer with the appropriate units.

Answer 1

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

POH = 14 - pH

= 14 - 8.82

= 5.18

use formula for buffer

pOH = pKb + log ([NH4Cl]/[NH3])

5.18 = 4.7447 + log ([NH4Cl]/[NH3])

log ([NH4Cl]/[NH3]) = 0.4353

[NH4Cl]/[NH3] = 2.7244

[NH4Cl]/0.1 = 2.7244

[NH4Cl] = 0.2724

volume , V = 1.5 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.2724*1.5

= 0.4087 mol

Molar mass of NH4Cl = 1*MM(N) + 4*MM(H) + 1*MM(Cl)

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

we have below equation to be used:

mass of NH4Cl,

m = number of mol * molar mass

= 0.4087 mol * 53.492 g/mol

= 21.9 g

Answer: 21.9 g