The molar conductivity of $0.025 \ mol \ L^{-1}$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.
(N/A) Given:
$C = 0.025 \ mol \ L^{-1}$
$\Lambda_m = 46.1 \ S \ cm^2 \ mol^{-1}$
$\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$
$\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$
Step $1$: Calculate molar conductivity at infinite dilution $\Lambda_m^o(HCOOH)$:
$\Lambda_m^o(HCOOH) = \lambda^o(H^{+}) + \lambda^o(HCOO^{-}) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$
Step $2$: Calculate degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{46.1}{404.2} \approx 0.114$
Step $3$: Calculate dissociation constant $(K_a)$:
$K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.025 \times (0.114)^2}{1 - 0.114} = \frac{0.025 \times 0.012996}{0.886} \approx 3.67 \times 10^{-4} \ mol \ L^{-1}$
$C = 0.025 \ mol \ L^{-1}$
$\Lambda_m = 46.1 \ S \ cm^2 \ mol^{-1}$
$\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$
$\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$
Step $1$: Calculate molar conductivity at infinite dilution $\Lambda_m^o(HCOOH)$:
$\Lambda_m^o(HCOOH) = \lambda^o(H^{+}) + \lambda^o(HCOO^{-}) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$
Step $2$: Calculate degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{46.1}{404.2} \approx 0.114$
Step $3$: Calculate dissociation constant $(K_a)$:
$K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.025 \times (0.114)^2}{1 - 0.114} = \frac{0.025 \times 0.012996}{0.886} \approx 3.67 \times 10^{-4} \ mol \ L^{-1}$
Explore More
Similar Questions
Consider the following cell reaction:
$Cd_{(s)} + Hg_{2}SO_{4(s)} + \frac{9}{5}H_{2}O_{(l)} \rightleftharpoons CdSO_{4} \cdot \frac{9}{5}H_{2}O_{(s)} + 2Hg_{(l)}$
The value of $E_{\text{cell}}^{0}$ is $4.315 \ V$ at $25^{\circ}C$. If $\Delta H^{\circ} = -825.2 \ kJ \ mol^{-1}$,the standard entropy change $\Delta S^{\circ}$ in $J \ K^{-1} \ mol^{-1}$ is ........ . (Nearest integer) [Given: Faraday constant $= 96487 \ C \ mol^{-1}$]
$Cd_{(s)} + Hg_{2}SO_{4(s)} + \frac{9}{5}H_{2}O_{(l)} \rightleftharpoons CdSO_{4} \cdot \frac{9}{5}H_{2}O_{(s)} + 2Hg_{(l)}$
The value of $E_{\text{cell}}^{0}$ is $4.315 \ V$ at $25^{\circ}C$. If $\Delta H^{\circ} = -825.2 \ kJ \ mol^{-1}$,the standard entropy change $\Delta S^{\circ}$ in $J \ K^{-1} \ mol^{-1}$ is ........ . (Nearest integer) [Given: Faraday constant $= 96487 \ C \ mol^{-1}$]
Given the values of $ᴧ^{0}$ for $KCl, KNO_3, HCl, NaOAc,$ and $NaCl$ are $149.9, 146.0, 426.2, 91.0,$ and $226.5 \, S \, cm^2 \, eq^{-1}$ respectively,calculate the value of $ᴧ^{0}$ for $HOAc$.
Medium
View SolutionGiven that $Ni^{2+}/Ni = -0.25 \ V$; $Cu^{2+}/Cu = 0.34 \ V$; $Ag^{+}/Ag = 0.80 \ V$; $Zn^{2+}/Zn = -0.76 \ V$. Which of the following reactions under standard conditions will not take place in the specified direction?
Medium
View SolutionThe values of the limiting molar conductivity $(\lambda^0)$ for $NaCl$,$HCl$ and $NaOAc$ are $126.4$,$425.9$ and $91.0 \, S \, cm^2 \, mol^{-1}$,respectively. For $HOAc$,$\Lambda^0$ in $S \, cm^2 \, mol^{-1}$ is
MediumKVPY 2011
View SolutionDuring electrolysis of aqueous $NaOH$,$4 \ g$ of $O_2$ gas is liberated at $NTP$ at the anode. The volume of $H_2$ gas liberated at the cathode in $litres$ is $..............$
MediumAIPMT 1998
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo