The resistance of 0.1 M weak acid HA in a co | vedclass

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(D) Given: Concentration $C = 0.1 \ M$,Molar conductivity at infinite dilution $\Lambda_{m}^{\circ} = 390 \ S \ cm^2 \ mol^{-1}$,Resistance $R = 2 	i

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(D) Given: Concentration $C = 0.1 \ M$,Molar conductivity at infinite dilution $\Lambda_{m}^{\circ} = 390 \ S \ cm^2 \ mol^{-1}$,Resistance $R = 2 	imes 10^3 \ \Omega$,Cell constant $G^* = 0.78 \ cm^{-1}$.First,calculate the conductivity $K$:$K = \frac{G^*}{R} = \frac{0.78}{2 	imes 10^3} = 3.9 	imes 10^{-4} \ S \ cm^{-1}$.Next,calculate the molar conductivity $\Lambda_{m}$:$\Lambda_{m} = \frac{K 	imes 1000}{C} = \frac{3.9 	imes 10^{-4} 	imes 1000}{0.1} = 3.9 \ S \ cm^2 \ mol^{-1}$.Calculate the degree of dissociation $\alpha$:$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{3.9}{390} = 0.01 = 10^{-2}$.Calculate the concentration of $H^{+}$ ions:$[H^{+}] = C 	imes \alpha = 0.1 	imes 10^{-2} = 10^{-3} \ M$.Finally,calculate the $pH$:$pH = -\log[H^{+}] = -\log(10^{-3}) = 3$.

The resistance of $0.1 \ M$ weak acid $HA$ in a conductivity cell is $2 \times 10^3 \ \Omega$. The cell constant of the cell is $0.78 \ cm^{-1}$ and $\lambda_{m}^{\circ}$ of acid $HA$ is $390 \ S \ cm^2 \ mol^{-1}$. The $pH$ of the solution is

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