What mass of ammonium chloride should be added to 2.45 L of a 0.145 M NH3 in order to obtain a buffer with a pH of 9.45?
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
POH = 14 - pH
= 14 - 9.45
= 4.55
use formula for buffer
pOH = pKb + log ([NH4Cl]/[NH3])
4.550000000000001 = 4.7447 + log ([NH4Cl]/[NH3])
log ([NH4Cl]/[NH3]) = -0.1947
[NH4Cl]/[NH3] = 0.6387
[NH4Cl]/0.145 = 0.6387
[NH4Cl] = 0.0926
volume , V = 2.45 L
use:
number of mol,
n = Molarity * Volume
= 9.261*10^-2*2.45
= 0.2269 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
= 0.2269 mol * 53.49 g/mol
= 12.14 g
Answer: 12.1 g
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