Conductivity of $0.00241 \, M$ acetic acid is $7.896 \times 10^{-5} \, S \, cm^{-1}$. Calculate its molar conductivity. If $\Lambda_m^o$ for acetic acid is $390.5 \, S \, cm^2 \, mol^{-1}$,what is its dissociation constant?
(N/A) Given,$\kappa = 7.896 \times 10^{-5} \, S \, cm^{-1}$,$c = 0.00241 \, mol \, L^{-1}$.
Molar conductivity,$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} = 32.76 \, S \, cm^2 \, mol^{-1}$.
Degree of dissociation,$\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{32.76}{390.5} = 0.084$.
Dissociation constant,$K_a = \frac{c \alpha^2}{1 - \alpha} = \frac{0.00241 \times (0.084)^2}{1 - 0.084} = \frac{0.00241 \times 0.007056}{0.916} = 1.86 \times 10^{-5} \, mol \, L^{-1}$.
Molar conductivity,$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} = 32.76 \, S \, cm^2 \, mol^{-1}$.
Degree of dissociation,$\alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{32.76}{390.5} = 0.084$.
Dissociation constant,$K_a = \frac{c \alpha^2}{1 - \alpha} = \frac{0.00241 \times (0.084)^2}{1 - 0.084} = \frac{0.00241 \times 0.007056}{0.916} = 1.86 \times 10^{-5} \, mol \, L^{-1}$.
Explore More
Similar Questions
The $pK_{b}$ of dimethylamine and $pK_{a}$ of acetic acid are $3.27$ and $4.77$ respectively at $T(K)$. The correct option for the $pH$ of dimethylammonium acetate solution is :
MediumNEET 2021
View SolutionWhat is the $[OH^{-}]$ in the final solution prepared by mixing $20.0 \ mL$ of $0.050 \ M \ HCl$ with $30.0 \ mL$ of $0.10 \ M \ Ba(OH)_2$? $...... \ M$
MediumAIPMT 2009
View SolutionHydrogen ion concentration of an aqueous solution is $1 \times 10^{-4} \ M$. The solution is diluted with an equal volume of water. The hydroxyl ion concentration of the resultant solution in terms of $mol \ dm^{-3}$ is
Medium
View SolutionThe $pH$ of $0.1 \ M$ solution of the following salts increases in the order:
DifficultIIT 1999
View Solution$1 \times 10^{-5} \ S \ M$ $AgNO_3$ is added to $1 \ L$ of a saturated solution of $AgBr$. The conductivity of this solution at $298 \ K$ is $......... \times 10^{-8} \ S \ m^{-1}$.
Given: $K_{sp}(AgBr) = 4.9 \times 10^{-13}$ at $298 \ K$,$\lambda^0_{Ag^{+}} = 6 \times 10^{-3} \ S \ m^2 \ mol^{-1}$,$\lambda^0_{Br^{-}} = 8 \times 10^{-3} \ S \ m^2 \ mol^{-1}$,$\lambda^0_{NO_3^-} = 7 \times 10^{-3} \ S \ m^2 \ mol^{-1}$.
Given: $K_{sp}(AgBr) = 4.9 \times 10^{-13}$ at $298 \ K$,$\lambda^0_{Ag^{+}} = 6 \times 10^{-3} \ S \ m^2 \ mol^{-1}$,$\lambda^0_{Br^{-}} = 8 \times 10^{-3} \ S \ m^2 \ mol^{-1}$,$\lambda^0_{NO_3^-} = 7 \times 10^{-3} \ S \ m^2 \ mol^{-1}$.
Vedclass 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