The molar conductivity of 0.007 M acetic acid is 20 S cm^2 mol^-1 . What is the dissociation constant of acetic acid? [Given _m^(H^+) = 350 S cm^2 mol - Chemistry MCQ NEET 2021 2026 | ExamTest.live

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The molar conductivity of 0.007 M acetic acid is 20 S cm^{2} mol^{- 1} . What is the dissociation constant of acetic acid? [Given Λ_{m}^{\circ} (H^{+}) = 350 S cm^{2} mol^{- 1} and Λ_{m}^{\circ} (CH_{3} COO^{-}) = 50 S cm^{2} mol^{- 1}]

Solution & Step-by-step Explanation

Step 1: Find limiting molar conductivity (Λ_{m}^{\circ}): Λ_{m}^{\circ} (CH_{3} COOH) = 350 + 50 = 400 S cm^{2} mol^{- 1} Step 2: Find degree of dissociation (α): α = \frac{Λ _{m}^{c}}{Λ _{m}^{\circ}} = \frac{20}{400} = 0.05 Step 3: Calculate dissociation constant (K_{a}): K_{a} = C α^{2} = 0.007 \times (0.05)^{2} K_{a} = 0.007 \times 0.0025 = 1.75 \times 1 0^{- 5} mol/L

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The molar conductivity of 0.007 M acetic acid is 20 S cm^{2} mol^{- 1} . What is the dissociation constant of acetic acid? [Given Λ_{m}^{\circ} (H^{+}) = 350 S cm^{2} mol^{- 1} and Λ_{m}^{\circ} (CH_{3} COO^{-}) = 50 S cm^{2} mol^{- 1}]

1.75 \times 1 0^{- 5} mol L^{- 1}

2.50 \times 1 0^{- 5} mol L^{- 1}

1.75 \times 1 0^{- 4} mol L^{- 1}

2.50 \times 1 0^{- 4} mol L^{- 1}

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