The conductivity of $0.001028 \, mol \, L^{-1}$ acetic acid is $4.95 \times 10^{-5} \, S \, cm^{-1}$. Calculate its dissociation constant if $\Lambda_m^\circ$ for acetic acid is $390.5 \, S \, cm^2 \, mol^{-1}$.
(N/A) First,calculate the molar conductivity $(\Lambda_m)$ using the formula $\Lambda_m = \frac{\kappa \times 1000}{c}$.
$\Lambda_m = \frac{4.95 \times 10^{-5} \, S \, cm^{-1} \times 1000 \, cm^3 \, L^{-1}}{0.001028 \, mol \, L^{-1}} = 48.15 \, S \, cm^2 \, mol^{-1}$.
Next,calculate the degree of dissociation $(\alpha)$ using $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
$\alpha = \frac{48.15}{390.5} = 0.1233$.
Finally,calculate the dissociation constant $(K_a)$ using $K_a = \frac{c \alpha^2}{1 - \alpha}$.
$K_a = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233} = 1.78 \times 10^{-5} \, mol \, L^{-1}$.
$\Lambda_m = \frac{4.95 \times 10^{-5} \, S \, cm^{-1} \times 1000 \, cm^3 \, L^{-1}}{0.001028 \, mol \, L^{-1}} = 48.15 \, S \, cm^2 \, mol^{-1}$.
Next,calculate the degree of dissociation $(\alpha)$ using $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
$\alpha = \frac{48.15}{390.5} = 0.1233$.
Finally,calculate the dissociation constant $(K_a)$ using $K_a = \frac{c \alpha^2}{1 - \alpha}$.
$K_a = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233} = 1.78 \times 10^{-5} \, mol \, L^{-1}$.
Explore More
Similar Questions
For a spontaneous reaction,determine the values of $\Delta G^o$,equilibrium constant $K$,and $E^o_{cell}$ respectively.
Medium
View SolutionWhich of the following is not correct?
Given below are two statements:
Statement $I$: For $KI$,molar conductivity increases steeply with dilution.
Statement $II$: For carbonic acid,molar conductivity increases slowly with dilution.
In the light of the above statements,choose the correct answer from the options given below:
Statement $I$: For $KI$,molar conductivity increases steeply with dilution.
Statement $II$: For carbonic acid,molar conductivity increases slowly with dilution.
In the light of the above statements,choose the correct answer from the options given below:
Match List-$I$ with List-$II$.
List-$I$ List-$II$ $A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ $I$. Primary battery $B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ $II$. Discharging of secondary battery $C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ $III$. Fuel cell $D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ $IV$. Charging of secondary battery
Choose the correct answer from the options given below.
| List-$I$ | List-$II$ |
| $A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ | $I$. Primary battery |
| $B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ | $II$. Discharging of secondary battery |
| $C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ | $III$. Fuel cell |
| $D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ | $IV$. Charging of secondary battery |
Choose the correct answer from the options given below.
At $298 \ K$,the equilibrium constant is $2 \times 10^{15}$ for the reaction:
$Cu_{(s)} + 2 Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$
The equilibrium constant for the reaction $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$ is $x \times 10^{-8}$. The value of $x$ is (Nearest Integer).
$Cu_{(s)} + 2 Ag^{+}_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2 Ag_{(s)}$
The equilibrium constant for the reaction $\frac{1}{2} Cu^{2+}_{(aq)} + Ag_{(s)} \rightleftharpoons \frac{1}{2} Cu_{(s)} + Ag^{+}_{(aq)}$ is $x \times 10^{-8}$. The value of $x$ is (Nearest Integer).
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