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Q.

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 of HX at concentration 'C'.
${\land^{\circ}}_{m}$ = Limiting molar conductivity of HX.
c = Molar concentration of HX.
$k_{a}$ = Dissociation constant of Hx]

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a

$1 / (K_{a} {\land^{\circ}}_{m})$

b

$K_{a} {\land^{\circ}}_{m} / 2$

c

$2 K_{a} {\land^{\circ}}_{m}$

d

$K_{a} {\land^{\circ}}_{m}$

answer is A.

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Detailed Solution

For weak acid, Question Image

Question Image
Divide by $' \land_{m}'$

Plot has slope y-intercept = $\frac{1}{\land_{0}}$ = P Then,

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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 of HX at concentration 'C'.∧∘m = Limiting molar conductivity of HX.c = Molar concentration of HX.ka = Dissociation constant of Hx]