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Question

Chemistry Question on Electrochemistry

Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning

A

Λm2CKaΛm+KaΛm2=0\Lambda_m^2 C - K_a \Lambda_m + K_a \Lambda_m^{\circ 2} = 0

B

ΛmΛm+AC12=0\Lambda_m - \Lambda_m^{\circ} + AC \frac{1}{2} = 0

C

ΛmΛmAC12=0\Lambda_m - \Lambda_m^{\circ} - AC \frac{1}{2} = 0

D

Λm2C+KaΛm2KaΛm=0\Lambda_m^2 C + K_a \Lambda_m^{\circ 2} - K_a \Lambda_m^{\circ} = 0

Answer

Λm2CKaΛm+KaΛm2=0\Lambda_m^2 C - K_a \Lambda_m + K_a \Lambda_m^{\circ 2} = 0

Explanation

Solution

The relationship between molar conductivity Λm\Lambda_m, molar conductivity at infinite dilution Λm\Lambda_m^\circ, and concentration CC for a weak electrolyte can be derived from the dissociation equilibrium. The correct equation involves the dissociation constant KaK_a and accounts for the variation of Λm\Lambda_m with concentration. For weak electrolytes, the molar conductivity Λm\Lambda_m is related to the degree of dissociation α\alpha as:
α=ΛmΛm.\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}.
The dissociation constant KaK_a is expressed as:
Ka=Cα21α.K_a = \frac{C\alpha^2}{1 - \alpha}.
Substituting α=ΛmΛm\alpha =\frac{\Lambda_m}{\Lambda_m^\circ} into the equation:
Ka=C(ΛmΛm)21ΛmΛm.K_a = \frac{C \left(\frac{\Lambda_m}{\Lambda_m^\circ}\right)^2}{1 - \frac{\Lambda_m}{\Lambda_m^\circ}}.
Simplifying and rearranging, the equation becomes:
Λm2CKaΛm2+KaΛmΛm=0.\Lambda_m^2 C - K_a \Lambda_m^{\circ 2} + K_a \Lambda_m \Lambda_m^\circ = 0.
This is the equation that correctly represents the relationship between molar conductivity, concentration, and dissociation constant for a weak electrolyte.
Final Answer: (1)