JEE Advanced levelSingle correct optionDifficulty 3 of 3ie045
For 0.05M of a weak monobasic acid, the molar conductivity is 7.8Scm2mol−1 and the limiting molar conductivity is 390Scm2mol−1. Taking α=Λm/Λm∘ and Ka=cα2, Ka is
A2.0×10−5
B1.0×10−3
C4.0×10−5
D1.0×10−5
Answer: A
α=7.8/390=0.02. Ka=cα2=0.05×(0.02)2=2.0×10−5. Using cα=10−3 gives [HX+], not Ka.