The molar conductivity of 0.007 M acetic acid is 20 S . What is the dissociation constant of acetic acid? Choose the correct option. [2021]

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Q.
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.

$[\begin{matrix} Λ_{H^{+}}^{\circ} = 350 {S cm}^{2} {mol}^{- 1} \\ Λ_{{CH}_{3} {COO}^{-}}^{\circ} = 50 {S cm}^{2} {mol}^{- 1} \end{matrix}]$ [2021]

1 $2.50 \times 10^{- 5}$ $mol$ $L^{- 1}$

2 $1.75 \times 10^{- 4}$ $mol$ $L^{- 1}$

3 $2.50 \times 10^{- 4}$ $mol$ $L^{- 1}$

4 $1.75 \times 10^{- 5}$ $mol$ $L^{- 1}$

Ans.
(4)

$Λ_{m ({CH}_{3} COOH)}^{\circ} = λ_{H^{+}}^{\circ} + λ_{{CH}_{3} {COO}^{-}}^{\circ}$

$= 350 + 50 = 400$ $S$ ${cm}^{2}$ ${mol}^{- 1}$

Degree of dissociation, $α = \frac{Λ_{m}^{c}}{Λ_{m}^{o}} = \frac{20}{400} = 0.05$

So, dissociation constant, $K_{a} = c α^{2} (for weak electrolytes)$

$= 0.007 {(0.05)}^{2} = 1.75 \times 10^{- 5}$ $mol$ $L^{- 1}$

Solution

1 $2.50 \times 10^{- 5}$ $mol$ $L^{- 1}$

2 $1.75 \times 10^{- 4}$ $mol$ $L^{- 1}$

3 $2.50 \times 10^{- 4}$ $mol$ $L^{- 1}$

4 $1.75 \times 10^{- 5}$ $mol$ $L^{- 1}$

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Solution

Q. The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol-1. What is the dissociation constant of acetic acid Choose the correct option. [ΛH+∘=350 S cm2 mol-1ΛCH3COO-∘=50 S cm2 mol-1] [2021]

1 2.50×10-5 mol L-1

2 1.75×10-4 mol L-1

3 2.50×10-4 mol L-1

4 1.75×10-5 mol L-1

Ans. (4) Λm(CH3COOH)∘ =λH+∘+λCH3COO-∘ =350+50=400 S cm2 mol-1 Degree of dissociation, α=ΛmcΛmo=20400=0.05 So, dissociation constant, Ka=cα2 (for weak electrolytes) =0.007(0.05)2=1.75×10-5 mol L-1

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1 $2.50 \times 10^{- 5}$ $mol$ $L^{- 1}$

2 $1.75 \times 10^{- 4}$ $mol$ $L^{- 1}$

3 $2.50 \times 10^{- 4}$ $mol$ $L^{- 1}$

4 $1.75 \times 10^{- 5}$ $mol$ $L^{- 1}$