How many grams of dry NH4Cl need to be added to 2.00 L of a 0.700 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8×10−5. Express your answer with the appropriate units.
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
POH = 14 - pH
= 14 - 9
= 5
use formula for buffer
pOH = pKb + log ([NH4Cl]/[NH3])
5.0 = 4.7447 + log ([NH4Cl]/[NH3])
log ([NH4Cl]/[NH3]) = 0.2553
[NH4Cl]/[NH3] = 1.8
[NH4Cl]/0.7 = 1.8
[NH4Cl] = 1.26
volume , V = 2 L
use:
number of mol,
n = Molarity * Volume
= 1.26*2
= 2.52 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
= 2.52 mol * 53.49 g/mol
= 1.348*10^2 g
Answer: 1.35*10^2 g
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