Answer to Question #55920 in Physical Chemistry for palash

Question #55920
At 25 degree celcius λ( 0)H+=3.498*10^-2 and λ(0) 0H- =1.98*10^-2 Sm2mol-1.given specific conductance of water(kappa) is =5.7*10^-6 Sm-1.calculate Kw of water.
Expert's answer
The specific conductance of water is determined by the concentration of [H+] and [OH-]:
k = 1000C λ0(H2O) = 1000C(λ0(H+) + λ0(OH-)), where C – the concentration of of [H+] and [OH-].
Thus, C =k/[1000(λ0(H+) + λ0(OH-))] = 5.7×10-6 Sm-1/ [1000(3.498×10-2 + 1.98×10-2)] =
= 1.04×10-7 mol/L
Finally,
Kw = [H+][OH-] = C2 = 1.0816×10-14

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