Question

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.

\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
\ce{\Lambda^{\circ}_{H^+} = 350 S cm^2 mol^{-1}}\\
\ce{\Lambda^{\circ}_{CH_3COO^-} = 50 S cm^2 mol^{-1}}
\end{bmatrix}\]

पर्याय

• 2.50 × 10⁻⁵ mol L⁻¹

• 1.75 × 10⁻⁴ mol L⁻¹

• 2.50 × 10⁻⁴ mol L⁻¹

• 1.75 × 10⁻⁵ mol L⁻¹

Answer 1

1.75 × 10⁻⁵ mol L⁻¹

Explanation:

Given: Λ_m = 20 S cm² mol⁻¹

\[\ce{\Lambda^{\circ}_{m} = \Lambda^{\circ}_{H^+} + \Lambda^{\circ}_{CH_3COO^-}}\]

= 350 + 50

= 400 S cm² mol⁻¹

Concentration (C) = 0.007 mol L⁻¹

Degree of dissociation (α) = \[\ce{\frac{\Lambda_m}{\Lambda^{\circ}_m}}\]

= \[\ce{\frac{20}{400}}\]

= 0.05

K_α ​= Cα²

= 0.007 × (0.05)²

= 0.007 × 0.0025

= 1.75 × 10⁻⁵ mol L⁻¹