The molar conductivity of a solution of a wea | vedclass

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(C) For a weak acid,the degree of ionization $\alpha$ is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.Given $\Lambda_m(HX) = \frac{1}{10} \Lambda

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$3$

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(C) For a weak acid,the degree of ionization $\alpha$ is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.Given $\Lambda_m(HX) = \frac{1}{10} \Lambda_m(HY)$,and assuming $\Lambda_m^0(HX) \approx \Lambda_m^0(HY)$ (since $\lambda_{X^{-}}^0 \approx \lambda_{Y^{-}}^0$),we have $\alpha_{HX} = \frac{1}{10} \alpha_{HY}$.The dissociation constant $K_a$ for a weak acid is $K_a = C \alpha^2$.For $HX$: $K_a(HX) = C_1 \alpha_1^2 = 0.01 	imes (\frac{1}{10} \alpha_{HY})^2 = 0.01 	imes \frac{1}{100} \alpha_{HY}^2 = 10^{-4} \alpha_{HY}^2$.For $HY$: $K_a(HY) = C_2 \alpha_2^2 = 0.10 	imes \alpha_{HY}^2 = 10^{-1} \alpha_{HY}^2$.Taking the ratio: $\frac{K_a(HX)}{K_a(HY)} = \frac{10^{-4} \alpha_{HY}^2}{10^{-1} \alpha_{HY}^2} = 10^{-3}$.Therefore,$pK_a(HX) - pK_a(HY) = -\log(K_a(HX)) + \log(K_a(HY)) = -\log(\frac{K_a(HX)}{K_a(HY)}) = -\log(10^{-3}) = 3$.

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+}\|Cu$	$(II)$ $0.059$
$(C)$ $Zn\|Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

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