Back to Directory
NEET CHEMISTRYMedium

The molar conductivity of 0.007 M0.007\text{ M} acetic acid is 20 S cm2 mol−120\text{ S cm}^2\text{ mol}^{-1}. The dissociation constant of acetic acid is: (ΛH+o=350 S cm2 mol−1\Lambda^o_{H^+} = 350\text{ S cm}^2\text{ mol}^{-1} and ΛCH3COO−o=50 S cm2 mol−1\Lambda^o_{CH_3COO^-} = 50\text{ S cm}^2\text{ mol}^{-1})

A

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

B

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

C

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

D

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

Step-by-Step Solution

Given: Concentration, c=0.007 Mc = 0.007\text{ M} Molar conductivity, Λm=20 S cm2 mol−1\Lambda_m = 20\text{ S cm}^2\text{ mol}^{-1} Limiting molar conductivity of H+H^+, ΛH+∘=350 S cm2 mol−1\Lambda^{\circ}_{H^+} = 350\text{ S cm}^2\text{ mol}^{-1} Limiting molar conductivity of CH3COO−CH_3COO^-, ΛCH3COO−∘=50 S cm2 mol−1\Lambda^{\circ}_{CH_3COO^-} = 50\text{ S cm}^2\text{ mol}^{-1}

According to Kohlrausch's law, limiting molar conductivity of acetic acid: Λm∘(CH3COOH)=ΛH+∘+ΛCH3COO−∘\Lambda^{\circ}_m(CH_3COOH) = \Lambda^{\circ}_{H^+} + \Lambda^{\circ}_{CH_3COO^-} Λm∘=350+50=400 S cm2 mol−1\Lambda^{\circ}_m = 350 + 50 = 400\text{ S cm}^2\text{ mol}^{-1}

Degree of dissociation, α=ΛmΛm∘=20400=0.05\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{20}{400} = 0.05

Dissociation constant, Ka=cα21−αK_a = \frac{c\alpha^2}{1 - \alpha} Assuming 1−α≈11 - \alpha \approx 1 (since α\alpha is small): Ka≈cα2=0.007×(0.05)2=0.007×0.0025=1.75×10−5 mol L−1K_a \approx c\alpha^2 = 0.007 \times (0.05)^2 = 0.007 \times 0.0025 = 1.75 \times 10^{-5}\text{ mol L}^{-1}

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: CHEMISTRY Question for NEET | Sushrut