The equivalent conductances of two strong electrolytes at infinite dilution in H_2O (where ions move freely through a solution) at 25oC are given below: \wedge _CH_3COONa^o = 91.0 S cm2/equiv \wedge _HCl^o = 426.2 S cm2/equiv What additional information/quantity one needs to calculate \wedge ^o of an aqueous solution of acetic acid?

\wedge ^o of chloroacetic acid (C/ CH_2COOH )

The limiting equivalent conductance of H^ + ( \wedge _H^ + ^o)

Electrochemistry PYQ — JEE Main 2007 · Shift 0 · Q28 with Solution

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Correct answer: B

We need the limiting equivalent conductance of acetic acid, $\Lambda^\circ_{\mathrm{CH_3COOH}}$ .
Use Kohlrausch’s law of independent migration of ions: $\Lambda^\circ = \lambda^\circ_+ + \lambda^\circ_-$

Therefore, $\Lambda^\circ_{\mathrm{CH_3COOH}} = \lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{CH_3COO^-}}$
Given: $\Lambda^\circ_{\mathrm{CH_3COONa}} = \lambda^\circ_{\mathrm{CH_3COO^-}} + \lambda^\circ_{\mathrm{Na^+}} = 91.0$ $\Lambda^\circ_{\mathrm{HCl}} = \lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{Cl^-}} = 426.2$
If we add these two: $\Lambda^\circ_{\mathrm{CH_3COONa}} + \Lambda^\circ_{\mathrm{HCl}} = \lambda^\circ_{\mathrm{CH_3COO^-}} + \lambda^\circ_{\mathrm{Na^+}} + \lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{Cl^-}}$

To obtain $\lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{CH_3COO^-}}$ , we must subtract $\lambda^\circ_{\mathrm{Na^+}} + \lambda^\circ_{\mathrm{Cl^-}} = \Lambda^\circ_{\mathrm{NaCl}}$
Hence, $\Lambda^\circ_{\mathrm{CH_3COOH}} = \Lambda^\circ_{\mathrm{CH_3COONa}} + \Lambda^\circ_{\mathrm{HCl}} - \Lambda^\circ_{\mathrm{NaCl}}$
Therefore, the additional quantity required is: $\Lambda^\circ_{\mathrm{NaCl}}$
Check options:

A: Not needed.
B: Needed directly from Kohlrausch’s law.
C: Not sufficient with given data.
D: If $\lambda^\circ_{\mathrm{H^+}}$ alone were given, we would still need $\lambda^\circ_{\mathrm{CH_3COO^-}}$ , which is not directly available without $\lambda^\circ_{\mathrm{Na^+}}$ or another suitable salt.

So the correct option is B.

View written solutionFree

Correct answer: B

We need the limiting equivalent conductance of acetic acid, $\Lambda^\circ_{\mathrm{CH_3COOH}}$ .
Use Kohlrausch’s law of independent migration of ions: $\Lambda^\circ = \lambda^\circ_+ + \lambda^\circ_-$

Therefore, $\Lambda^\circ_{\mathrm{CH_3COOH}} = \lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{CH_3COO^-}}$
Given: $\Lambda^\circ_{\mathrm{CH_3COONa}} = \lambda^\circ_{\mathrm{CH_3COO^-}} + \lambda^\circ_{\mathrm{Na^+}} = 91.0$ $\Lambda^\circ_{\mathrm{HCl}} = \lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{Cl^-}} = 426.2$
If we add these two: $\Lambda^\circ_{\mathrm{CH_3COONa}} + \Lambda^\circ_{\mathrm{HCl}} = \lambda^\circ_{\mathrm{CH_3COO^-}} + \lambda^\circ_{\mathrm{Na^+}} + \lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{Cl^-}}$

To obtain $\lambda^\circ_{\mathrm{H^+}} + \lambda^\circ_{\mathrm{CH_3COO^-}}$ , we must subtract $\lambda^\circ_{\mathrm{Na^+}} + \lambda^\circ_{\mathrm{Cl^-}} = \Lambda^\circ_{\mathrm{NaCl}}$
Hence, $\Lambda^\circ_{\mathrm{CH_3COOH}} = \Lambda^\circ_{\mathrm{CH_3COONa}} + \Lambda^\circ_{\mathrm{HCl}} - \Lambda^\circ_{\mathrm{NaCl}}$
Therefore, the additional quantity required is: $\Lambda^\circ_{\mathrm{NaCl}}$
Check options:

A: Not needed.
B: Needed directly from Kohlrausch’s law.
C: Not sufficient with given data.
D: If $\lambda^\circ_{\mathrm{H^+}}$ alone were given, we would still need $\lambda^\circ_{\mathrm{CH_3COO^-}}$ , which is not directly available without $\lambda^\circ_{\mathrm{Na^+}}$ or another suitable salt.

So the correct option is B.