The conductivity of a 0.05 , M solution of a | vedclass

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Question

(A) First,calculate the molar conductivity $(\lambda_{m})$ using the formula: $\lambda_{m} = \frac{1000 	imes \kappa}{M}$$\lambda_{m} = \frac{1000

Vedclass · Accepted Answer

$8 	imes 10^{-5}$

Vedclass · Answer

(A) First,calculate the molar conductivity $(\lambda_{m})$ using the formula: $\lambda_{m} = \frac{1000 	imes \kappa}{M}$$\lambda_{m} = \frac{1000 	imes 10^{-3}}{0.05} = 20 \, S \, cm^{2} \, mol^{-1}$Next,calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\infty}} = \frac{20}{500} = 0.04$Finally,calculate the dissociation constant $(K_{a})$ using the formula: $K_{a} = C \alpha^{2}$$K_{a} = 0.05 	imes (0.04)^{2} = 0.05 	imes 0.0016 = 8 	imes 10^{-5}$

The conductivity of a $0.05 \, M$ solution of a weak monobasic acid is $10^{-3} \, S \, cm^{-1}$. If $\lambda_{m}^{\infty}$ for the weak acid is $500 \, S \, cm^{2} \, mol^{-1}$,calculate the $K_{a}$ of the weak monobasic acid.

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