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- Specific conductance of 0.1 text textM textH textA text is 3.75× 10- 4 text texto texth textm- 1 text textc textm- 1 . If textλ text∞ of textH textA is 250 text texto texth textm- 1 textc textm2 textm texto textl- 1, then dissociation constant textK texta of textH textA is
Q. Specific conductance of 0.1M HA is 3.75×10−4ohm−1cm−1 . If \lambda \infty of HA is 250ohm−1cm2mol−1, then dissociation constant Ka of HA is
Solution:
λm=1000xkm=3.75
α=λmλ∈ftym=3.75250=0.015=1.5×10−2
∴Ka=C(α)2=0.1(0.015)2=2.25×10−5
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