Calculate $\Lambda_{\mathrm{HOAc}}^{\infty}$ Using appropriate molar conductances of the electrolytes listed above at infinite dilution in $\mathrm{H}_{2}\mathrm{O}$ at $25^{\circ}\mathrm{C}$ | Electrolyte | KCl | $\mathrm{KNO}_{3}$ | HCl | NaOAc | NaCl | |-------------|-----|----------------------|-----|-------|------| | $\Lambda^{\infty}$ (S cm² mol⁻¹) | 149.9 | 145.0 | 426.2 | 91.0 | 126.5 |

JEE Main · 2005 · 08 May 2005 · Offline · Redox Reactions and Electrochemistry

Calculate

Λ_{HOAc}^{\infty}

Using appropriate molar conductances of the electrolytes listed above at infinite dilution in

H_{2} O

2 5^{\circ} C

Electrolyte	KCl	$KNO_{3}$	HCl	NaOAc	NaCl
$Λ^{\infty}$ (S cm² mol⁻¹)	149.9	145.0	426.2	91.0	126.5

(A)517.2
(B)552.7
(C)390.7✓ Correct
(D)217.5

Answer: (C)

Solution

Kohlrausch's law of independent ionic migration lets us assemble a weak acid's limiting conductance from strong-electrolyte data. Acetic acid can be built as

Λ_{HOAc}^{\infty} = Λ_{HCl}^{\infty} + Λ_{NaOAc}^{\infty} - Λ_{NaCl}^{\infty}

, because combining the

H X^{+}

and

OAc X^{-}

contributions while cancelling the

Na X^{+}

and

Cl X^{-}

leaves just the ions of

HOAc

. Putting in the numbers,

426.2 + 91.0 - 126.5 = 390.7

S cm

^{2}

mol

^{- 1}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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JEE Main · 2005 · 08 May 2005 · Offline · Redox Reactions and Electrochemistry

Solution

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