The molar conductivity of 0.007 M acetic acid is 20 S cm 2 mol -1. What is the dissociation constant of acetic acid? Choose the correct option. [ Lambda H +°=350 S cm 2 mol -1, ] [ Lambda CH 3 COO -°=50 S cm 2 mol -1]

Question

The molar conductivity of 0.007 M acetic acid is 20 S cm 2 mol -1. What is the dissociation constant of acetic acid? Choose the correct option. [ Lambda H +°=350 S cm 2 mol -1, ] [ Lambda CH 3 COO -°=50 S cm 2 mol -1] (A) 1.75 × 10-4 mol L -1 (B) 2.50 × 10-4 mol L -1 (C) 1.75 × 10-5 mol L -1 (D) 2.50 × 10-5 mol L -1

Tardigrade Admin · Accepted Answer

Degree of dissociation (α) of CH 3 COOH =( Lambda C / Lambda0)=(20/350+50)=(1/20) K a =( C α2/1-α) 1-α ≈ 1 K a = C α2 K a =0.007 × (1/20) × (1/20) K a =1.75 × 10-5

Q. The molar conductivity of $0.007 M$ acetic acid is $20 S c m^{2} m o l^{- 1}$ . What is the dissociation constant of acetic acid? Choose the correct option. $[Λ_{H^{+}}^{\circ} = 350 S c m^{2} m o l^{- 1},]$
$[Λ_{C H_{3} CO O^{-}}^{\circ} = 50 S c m^{2} m o l^{- 1}]$

$1.75 \times 1 0^{- 4} m o l L^{- 1}$

$2.50 \times 1 0^{- 4} m o l L^{- 1}$

$1.75 \times 1 0^{- 5} m o l L^{- 1}$

$2.50 \times 1 0^{- 5} m o l L^{- 1}$

Degree of dissociation $(α)$ of $C H_{3} COO H$
$= \frac{Λ _{C}}{Λ _{0}} = \frac{20}{350 + 50} = \frac{1}{20}$
$K_{a} = \frac{C α ^{2}}{1 - α}$
$1 - α \approx 1$
$K_{a} = C α^{2}$
$K_{a} = 0.007 \times \frac{1}{20} \times \frac{1}{20}$
$K_{a} = 1.75 \times 1 0^{- 5}$

Q. The molar conductivity of 0.007M acetic acid is 20Scm2mol−1. What is the dissociation constant of acetic acid? Choose the correct option. [ΛH+∘​=350Scm2mol−1,] [ΛCH3​COO−∘​=50Scm2mol−1]

1.75×10−4molL−1

2.50×10−4molL−1

1.75×10−5molL−1

2.50×10−5molL−1