The specific conductance of 0.0025 M acetic | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The molar conductivity $\wedge_m$ is calculated as: $\wedge_m = \frac{k 	imes 1000}{C}$ Given $k = 5 	imes 10^{-5} \ S \ cm^{-1}$ and $C = 0.002

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(D) The molar conductivity $\wedge_m$ is calculated as: $\wedge_m = \frac{k 	imes 1000}{C}$ Given $k = 5 	imes 10^{-5} \ S \ cm^{-1}$ and $C = 0.0025 \ M$. $\wedge_m = \frac{5 	imes 10^{-5} 	imes 1000}{0.0025} = \frac{0.05}{0.0025} = 20 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation $\alpha$ is given by $\alpha = \frac{\wedge_m}{\wedge_m^\circ} = \frac{20}{400} = 0.05$. The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha}$. Since $\alpha$ is very small,$1 - \alpha \approx 1$,so $K_a \approx C \alpha^2 = 0.0025 	imes (0.05)^2 = 0.0025 	imes 0.0025 = 6.25 	imes 10^{-6} = 62.5 	imes 10^{-7}$. Rounding to the nearest integer,we get $63 	imes 10^{-7}$.

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 specific conductance of $0.0025 \ M$ acetic acid is $5 \times 10^{-5} \ S \ cm^{-1}$ at a certain temperature. The dissociation constant of acetic acid is $...... \times 10^{-7}$. (Nearest integer) Consider limiting molar conductivity of $CH_3COOH$ as $400 \ S \ cm^2 \ mol^{-1}$.

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

$KCl$ $149.9$

$KNO_3$ $145.0$

$HCl$ $426.2$

$NaOAc$ $91.0$

$NaCl$ $126.5$

The molar conductivities $(\lambda_{m}^0)$ at infinite dilution of $KBr$,$HBr$,and $KNH_2$ are $120.5$,$420.6$,and $90.48 \ S \ cm^2 \ mol^{-1}$ respectively. Find the value of $\lambda_{m}^0$ for $NH_3$.

At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

The resistance of $0.05\, M$ solution of oxalic acid is $200\, \Omega$ and the cell constant is $2.0\, cm^{-1}$. The equivalent conductance in $S\, cm^2\, eq^{-1}$ will be:

Which of the following solutions has the maximum conductivity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module