Solveeit Logo
Login

Question

Chemistry Question on Conductance

The molar conductivity of 0.007 M0.007 \,M acetic acid is 20 S cm2mol−120\, S\, cm ^{2} mol ^{-1}. What is the dissociation constant of acetic acid? Choose the correct option. [ΛH+∘=350 S cm2mol−1,ΛCH3COO−∘=50 S cm2mol−1]\left[\Lambda_{ H ^{+}}^{\circ}=350 \,S \,cm ^{2} mol ^{-1}, \Lambda_{ CH _{3} COO ^{-}}^{\circ}=50\, S\, cm ^{2} mol ^{-1}\right]

A

1.75×10−4 mol L−11.75 \times 10^{-4}\, mol\, L ^{-1}

B

2.50×10−4 mol L−12.50 \times 10^{-4}\, mol\, L ^{-1}

C

1.75×10−5 mol L−11.75 \times 10^{-5}\, mol\, L ^{-1}

D

2.50×10−5 mol L−12.50 \times 10^{-5} \, mol\, L ^{-1}

Answer

1.75×10−5 mol L−11.75 \times 10^{-5}\, mol\, L ^{-1}

Explanation

Solution

Degree of dissociation (α)(\alpha) of CH3COOHCH _{3} COOH
=ΛCΛ0=20350+50=120=\frac{\Lambda_{ C }}{\Lambda_{0}}=\frac{20}{350+50}=\frac{1}{20}
Ka=Cα21−αK _{ a }=\frac{ C \alpha^{2}}{1-\alpha}
1−α≈11-\alpha \approx 1
Ka=Cα2K _{ a }= C \alpha^{2}
Ka=0.007×120×120K _{ a }=0.007 \times \frac{1}{20} \times \frac{1}{20}
Ka=1.75×10−5K _{ a }=1.75 \times 10^{-5}