Question

How many grams of dry NH4Cl need to be added to 2.20 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.62? Kb for ammonia is 1.8×10−5.

Answer 1

Kb = 1.8*10^-5

pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745

POH = 14 - pH
= 14 - 8.62
= 5.38

use formula for buffer
pOH = pKb + log ([NH4Cl]/[NH3])
5.380000000000001 = 4.7447 + log ([NH4Cl]/[NH3])
log ([NH4Cl]/[NH3]) = 0.6353
[NH4Cl]/[NH3] = 4.3179
[NH4Cl]/0.5 = 4.3179
[NH4Cl] = 2.1589
volume , V = 2.2 L


use:
number of mol,
n = Molarity * Volume
= 2.159*2.2
= 4.75 mol

Molar mass of NH4Cl,
MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)
= 1*14.01 + 4*1.008 + 1*35.45
= 53.492 g/mol

use:
mass of NH4Cl,
m = number of mol * molar mass
= 4.75 mol * 53.49 g/mol
= 254 g
Answer: 254 g