Q. Specific conductance of 0.1HA is 3.75×104ohm1cm1 . If \lambda \infty  of HA is 250ohm1cm2mol1, then dissociation constant Ka of HA is

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Solution:

λm=1000xkm=3.75
α=λmλftym=3.75250=0.015=1.5×102
Ka=C(α)2=0.1(0.015)2=2.25×105