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Question

At 25C, λ0(H+)=3.498×102 and , λ0(OH)=1.98×102Sm2mol1. Given for water κ=5.7×106Sm1, Calculate Kw.

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Solution

λH2O at 25°=K×1000C
C= Concentration of H2O at 25°C=55.55M
λH2O=5.7×106×100055.55=1.02×104Sm2mol1
λH2O=λH++λOH=(3.498+1.98)×102=5.478×102Sm2mol1
α=λλ°=1.02×1045.478×102=1.86×103
Kw=[H+][OH][H2O]=Cα×CαC(1α)=Cα21α
α<Cl
Kw=Cα2=55.55×(1.86×103)2=1.92×104

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