Molarity of NH3 solution = moles of NH3/ volume of solution
moles of NH3=0.2×0.5=0.1
pOH=14−pH
pOH=14−9.44=4.56
pKb=−log[Kb]=−log[1.78×10−5]=4.74
pOH=pKb−log[salt][base]
log[salt][base]=4.74−4.56=0.18
[salt][base]=antilog[0.18]=1.51
[base]= moles of NH3=0.1
[salt]=1.510.1=0.07
Moles of (NH4)+=0.07
(NH4)2SO4⇌2NH4++(SO4)−2
moles of (NH4)2SO4=20.07=0.035
amount of (NH4)2SO4=0.035×132.14=4.6249g
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