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Chemistry Question on Equilibrium

The equivalent conductance of M32\frac{ M }{32} solution of a weak monobasic acid is 8.0 mho cm28.0\, mho\, cm ^{2} and at infinite dilution is 400 mho cm2400\, mho\, cm ^{2}, The dissociation constant of this acid is -

A

1.25×10−51.25 \times 10^{-5}

B

1.25×10−61.25 \times 10^{-6}

C

6.25×10−46.25 \times 10^{-4}

D

1.25×10−41.25 \times 10^{-4}

Answer

1.25×10−51.25 \times 10^{-5}

Explanation

Solution

α=ACA∞=8400\alpha= \frac{A_{C}}{A_{\infty}}=\frac{8}{400} Ka=Cα2 K_{a} =C \alpha^{2} =132×8400×8400=\frac{1}{32} \times \frac{8}{400} \times \frac{8}{400} =1.25×10−5=1.25 \times 10^{-5}