Plotting 1/∧m against c∧m for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is [∧m= molar conductivity∧m0= limiting molar conductivityc = molar concentration Ka= dissociation constant of HX]

∧m=Ka=cα21−αKa=c∧m∧m021−∧m∧m0Ka−Ka.∧m∧m0=c∧m∧m021∧m=c∧mKa.∧m02+1∧m0Ka∧m−Ka∧m0=c∧m∧m02+1∧m0S=1Ka.∧m02P=1∧m0PS=1∧m0×Ka.∧m02=Ka.∧m0

Plotting 1/∧m against c∧m for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is [∧m= molar conductivity∧m0= limiting molar conductivityc = molar concentration Ka= dissociation constant of HX]

∧m=Ka=cα21−αKa=c∧m∧m021−∧m∧m0Ka−Ka.∧m∧m0=c∧m∧m021∧m=c∧mKa.∧m02+1∧m0Ka∧m−Ka∧m0=c∧m∧m02+1∧m0S=1Ka.∧m02P=1∧m0PS=1∧m0×Ka.∧m02=Ka.∧m0

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Plotting $1 / \land_{m}$ against $c \land_{m}$ for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is
[ $\land_{m}$ = molar conductivity
$\land_{m}^{0}$ = limiting molar conductivity
c = molar concentration
$K_{a}$ = dissociation constant of HX]

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$K_{a} \land_{m}^{0}$

$K_{a} \land_{m}^{0} / 2$

$2 K_{a} \land_{m}^{0}$

$1 / (K_{a} \land_{m}^{0})$

answer is A.

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Detailed Solution

$\land_{m} = K_{a} = \frac{{cα}^{2}}{1 - α}$
$K_{a} = \frac{c {(\frac{\land_{m}}{\land_{m}^{0}})}^{2}}{1 - \frac{\land_{m}}{\land_{m}^{0}}}$
$K_{a} - K_{a} . \frac{\land_{m}}{\land_{m}^{0}} = c {(\frac{\land_{m}}{\land_{m}^{0}})}^{2}$

$\frac{1}{\land_{m}} = \frac{c \land_{m}}{K_{a} . {(\land_{m}^{0})}^{2}} + \frac{1}{\land_{m}^{0}}$ $\frac{K_{a}}{\land_{m}} - \frac{K_{a}}{\land_{m}^{0}} = \frac{c \land_{m}}{{(\land_{m}^{0})}^{2}} + \frac{1}{\land_{m}^{0}}$
$S = \frac{1}{K_{a} . {(\land_{m}^{0})}^{2}}$
$P = \frac{1}{\land_{m}^{0}}$
$\frac{P}{S} = \frac{1}{\land_{m}^{0}} \times K_{a} . {(\land_{m}^{0})}^{2}$
$= K_{a} . \land_{m}^{0}$