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Question

Calculate the amount of (NH4)2SO4 (in gram) which must be added to 1 L solution of 0.2 M NH4OH to yield a solution of pH=9.35 at 25oC.
(Assume volume is not changing during the reaction)
Kb for NH4OH is 1.8×105
Take log 1.8=0.25 and 100.1=1.25

A
5.33 g
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B
5.56 g
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C
8.56 g
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D
10.56 g
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Solution

The correct option is D 10.56 g
Molarity of NH4OH soluiton =moles of NH4OHvolume of solution
Moles of NH4OH=Molarity×Volume=0.2×1=0.2

pOH=14pH at 25oC
pOH=149.35=4.65
pKb=log[Kb]=log[1.8×105]pKb=5log 1.8pKb=50.25=4.75
pOH=pKblog([base][Cation of salt])
log([base][Cation of salt])=4.754.65=0.1
([base][Cation of salt])=10[0.1]=1.25
[base] = moles of NH4OH=0.2
[Cation of salt]=0.21.25=0.16
Moles of (NH+4)=0.16
(NH4)2SO42NH+4+(SO24)
Moles of (NH4)2SO4=moles of NH+42=0.08
Amount of (NH4)2SO4=0.08×132=10.56 g

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