Given the following molar conductivities at 2 | vedclass

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(B) First,calculate the limiting molar conductivity of crotonic acid $(HC)$ using Kohlrausch's law: $\Lambda_{m}^{\infty}(HC) = \Lambda_{m}^{\infty}(H

Vedclass · Accepted Answer

$1.11 	imes 10^{-5}$

Vedclass · Answer

(B) First,calculate the limiting molar conductivity of crotonic acid $(HC)$ using Kohlrausch's law: $\Lambda_{m}^{\infty}(HC) = \Lambda_{m}^{\infty}(HCl) + \Lambda_{m}^{\infty}(NaC) - \Lambda_{m}^{\infty}(NaCl)$ $= (426 + 83 - 126)\,\Omega^{-1}\,cm^{2}\,mol^{-1} = 383\,\Omega^{-1}\,cm^{2}\,mol^{-1}$. Next,calculate the molar conductivity of the $0.001\,M$ solution: $\Lambda_{m} = \frac{\kappa 	imes 1000}{C} = \frac{3.83 	imes 10^{-5} 	imes 1000}{0.001} = 38.3\,\Omega^{-1}\,cm^{2}\,mol^{-1}$. Calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\infty}} = \frac{38.3}{383} = 0.1$. Finally,calculate the ionization constant $(K_{a})$: $K_{a} = \frac{C\alpha^{2}}{1 - \alpha} = \frac{0.001 	imes (0.1)^{2}}{1 - 0.1} = \frac{10^{-3} 	imes 10^{-2}}{0.9} = \frac{10^{-5}}{0.9} \approx 1.11 	imes 10^{-5}$.

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