The conductivity of a 0.00241 M acetic acid solution is $7.896\times10^{-5}$ S cm$^{-1}$. Calculate its molar conductivity in this solution. If $\Lambda^{\circ}_m$ for acetic acid is 390.5 S cm$^2$ mol$^{-1}$, what would be its dissociation constant?

CBSE Class 12 · 2008 · Redox Reactions and ElectrochemistryLong Answer

The conductivity of a 0.00241 M acetic acid solution is

7.896 \times 1 0^{- 5}

S cm

^{- 1}

. Calculate its molar conductivity in this solution. If

Λ_{m}^{\circ}

for acetic acid is 390.5 S cm

^{2}

mol

^{- 1}

, what would be its dissociation constant?

Answer: Λ_m ≈ 32.76 S cm² mol⁻¹; Ka ≈ 1.85 × 10⁻⁵

Molar conductivity:

Λ_{m} = \frac{κ \times 1000}{C} = \frac{7.896 \times 1 0 ^{- 5} \times 1000}{0.00241}

Λ_{m} = 32.76 S c m^{2} mo l^{- 1}

Degree of dissociation:

α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{32.76}{390.5} = 0.0839

Dissociation constant:

K_{a} = \frac{C α ^{2}}{1 - α} = \frac{0.00241 \times ( 0.0839 ) ^{2}}{1 - 0.0839}

K_{a} = 1.85 \times 1 0^{- 5}

CBSE Class 12 Board · 2008 · Subjective