Calculate the amount of (NH4)2SO4 (in gram) which must be added to 1L solution of 0.2MNH4OH to yield a solution of pH=9.35 at 25oC.
(Assume volume is not changing during the reaction) Kb for NH4OH is 1.8×10−5
Take log1.8=0.25and100.1=1.25
A
5.33g
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B
5.56g
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C
8.56g
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D
10.56g
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Solution
The correct option is D10.56g Molarity of NH4OH soluiton =moles of NH4OHvolume of solution Moles ofNH4OH=Molarity×Volume=0.2×1=0.2
pOH=14−pH at 25oC pOH=14−9.35−=4.65 pKb=−log[Kb]=−log[1.8×10−5]pKb=5−log1.8pKb=5−0.25=4.75 pOH=pKb−log([base][Cation of salt]) log([base][Cation of salt])=4.75−4.65=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)2SO4⇌2NH+4+(SO2−4)
Moles of (NH4)2SO4=molesofNH+42=0.08
Amount of (NH4)2SO4=0.08×132=10.56g