The molar conductivity of 0.007 M acetic aci | vedclass

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Question

(C) First,calculate the limiting molar conductivity of acetic acid:$\Lambda_{m}^{\circ}(CH_{3}COOH) = \Lambda_{H^{+}}^{\circ} + \Lambda_{CH_{3}COO^{-}

Vedclass · Accepted Answer

$1.75$

Vedclass · Answer

(C) First,calculate the limiting molar conductivity of acetic acid:$\Lambda_{m}^{\circ}(CH_{3}COOH) = \Lambda_{H^{+}}^{\circ} + \Lambda_{CH_{3}COO^{-}}^{\circ} = 350 + 50 = 400 \ S \ cm^{2} \ mol^{-1}$Next,calculate the degree of dissociation $(\alpha)$:$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{20}{400} = 0.05$Finally,calculate the dissociation constant $(K_{a})$ using the formula $K_{a} = C \alpha^{2} / (1 - \alpha)$. Since $\alpha$ is very small,we use $K_{a} \approx C \alpha^{2}$:$K_{a} = 0.007 	imes (0.05)^{2}$$K_{a} = 7 	imes 10^{-3} 	imes 25 	imes 10^{-4}$$K_{a} = 175 	imes 10^{-7} = 1.75 	imes 10^{-5} \ mol \ L^{-1}$Thus,the value is $1.75$.

Electrolyte	$\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$
$KCl$	$149.9$
$KNO_3$	$145.0$
$HCl$	$426.2$
$NaOAc$	$91.0$
$NaCl$	$126.5$

The molar conductivity of $0.007 \ M$ acetic acid is $20 \ S \ cm^{2} \ mol^{-1}$. What is the dissociation constant of acetic acid? (In $\times 10^{-5} \ mol \ L^{-1}$)
$[\Lambda_{H^{+}}^{\circ}=350 \ S \ cm^{2} \ mol^{-1}, \Lambda_{CH_{3}COO^{-}}^{\circ}=50 \ S \ cm^{2} \ mol^{-1}]$

Similar Questions

Calculate $\Lambda _{HOAc}^{\infty }$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_2O$ at $25\ ^oC$.

Electrolyte $\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$

$KCl$ $149.9$

$KNO_3$ $145.0$

$HCl$ $426.2$

$NaOAc$ $91.0$

$NaCl$ $126.5$

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The molar conductivity of $0.007 \ M$ acetic acid is $20 \ S \ cm^{2} \ mol^{-1}$. What is the dissociation constant of acetic acid? (In $\times 10^{-5} \ mol \ L^{-1}$)$[\Lambda_{H^{+}}^{\circ}=350 \ S \ cm^{2} \ mol^{-1}, \Lambda_{CH_{3}COO^{-}}^{\circ}=50 \ S \ cm^{2} \ mol^{-1}]$