Question

From the following molar conductivities at infinite dilution:
$\lambda _m^\circ$ for A{l_2}{\left( {S{O_4}} \right)_3}$$$$ = 858\,S\,\,c{m^2}\,mo{l^{ - 1}}
$\lambda _m^\circ$ for $N{H_4}OH = 238.3\,S\,\,c{m^2}\,mo{l^{ - 1}}$
$\lambda _m^\circ$ for ${\left( {N{H_4}} \right)_2} = 238.4\,S\,\,c{m^2}\,mo{l^{ - 1}}$
Calculate $\lambda _m^\circ$ for $Al{\left( {OH} \right)_3}$
(A) $715.2\,S\,\,c{m^2}\,mo{l^{ - 1}}$
(B) $1575.6\,S\,\,c{m^2}\,mo{l^{ - 1}}$
(C) $786.3\,S\,\,c{m^2}\,mo{l^{ - 1}}$
(D) $157.56\,S\,\,c{m^2}\,mo{l^{ - 1}}$

Answer 1

We will use the concept explained by Kohlrausch’s law to calculate the molar conductivity of $Al{\left( {OH} \right)_3}$. This law explains that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the cations and anions.

Complete step by step answer:
Kohlrausch’s law defines that the dissociation of an electrolyte is complete in the infinite dilution and thus the ions constituting the electrolyte contributes.
Thus, the molar conductivity at infinite dilution of electrolyte is the algebraic sum of the molar conductivity of its ions.
$\lambda _m^\circ = \lambda _ + ^\circ + \lambda _ - ^\circ$
Where, $\lambda _m^\circ$ is molar conductivity of the electrolyte
$\lambda _ + ^\circ$ is molar conductivity of the cation
$\lambda _ - ^\circ$ is molar conductivity of anion
Now in the given question we are given with
$\lambda _m^\circ$ for A{l_2}{\left( {S{O_4}} \right)_3}$$$$ = 858\,S\,\,c{m^2}\,mo{l^{ - 1}}
$\lambda _m^\circ$ for $N{H_4}OH = 238.3\,S\,\,c{m^2}\,mo{l^{ - 1}}$
$\lambda _m^\circ$ for ${\left( {N{H_4}} \right)_2} = 238.4\,S\,\,c{m^2}\,mo{l^{ - 1}}$
The respective electrolyte dissociates as:
(A) $A{l_2}{\left( {S{O_4}} \right)_3} \rightleftharpoons 2A{l^{3 + }} + 3SO_4^{2 - }$
(B) $N{H_4}OH \rightleftharpoons NH\,_4^ + + O{H^ - }$
(C) ${\left( {N{H_4}} \right)_2}S{O_4} \rightleftharpoons 2NH\,_4^ + + SO_4^{2 - }$
(D) $Al\left( {O{H_3}} \right) \rightleftharpoons A{l^{3 + }} + 3O{H^ - }$
Multiplying (B) with $6$ and (C) with $3$and then adding (A) and (B) we will get,
B. $6N{H_4}OH \rightleftharpoons 6NH\,_4^ + + 6O{H^ - }$
C. $3{\left( {N{H_4}} \right)_2}S{O_4} \rightleftharpoons 6NH\,_4^ + + 3SO_4^{2 - }$
(A) $+$ (B):
$A{l_2}{\left( {S{O_4}} \right)_3} + 6N{H_4}OH \rightleftharpoons 2A{l^{3 + }} + 3SO_4^{2 - } + 6NH\,_4^ + + 6O{H^ - }$
Now subtracting (C) from [(A) $+$ (B)]
$3N{H_4}S{O_4} \rightleftharpoons 6NH\,_4^ + + 3SO_4^{2 - }$
$\left[ {A{l_2}{{\left( {S{O_4}} \right)}_3} + 6N{H_4}OH} \right] - 3{\left( {N{H_4}} \right)_2}S{O_4} \rightleftharpoons \left[ {2A{l^{3 + }} + 3SO_4^{2 - } + 6NH\,_4^ + + 6O{H^ - }} \right] - \left[ {6NH\,_4^ + + 3SO_4^{2 - }} \right]$
On simplifying
$2Al{\left( {OH} \right)_3} \rightleftharpoons 2A{l^{3 + }} + 6O{H^ - }$
Similar calculations will be made with molar conductivities:
$2\left[ {\lambda _m^\circ \left( {Al{{\left( {OH} \right)}_3}} \right)} \right] = \lambda _m^\circ A{l_2}{\left( {S{O_4}} \right)_3} + 6\left( {\lambda _m^\circ N{H_4}OH} \right) - 3\left( {\lambda _m^\circ {{\left( {N{H_4}} \right)}_2}S{O_4}} \right)$
$= 858\,S\,c{m^2}mo{l^{ - 1}} + 6 \times 238.3\,S\,c{m^2}mo{l^{ - 1}} - 3 \times 238.48\,S\,c{m^2}mo{l^{ - 1}}$
$= 1572.6\,S\,c{m^2}mo{l^{ - 1}}$
$\lambda _m^\circ \left( {Al{{\left( {OH} \right)}_3}} \right) = \dfrac{{1572.6}}{2}\,S\,c{m^2}mo{l^{ - 1}} = 786.3\,S\,c{m^2}mo{l^{ - 1}}$
Hence, option C is correct.

Additional information: Molar conductivity is the conductance property of a solution. It can be defined as the conducting power of all the ions that are formed by dissolving a mole of electrolyte in a solution. It determines the efficiency of a given electrolyte in conducting electricity in a solution. The molar conductivity increases with decrease in concentration or dilution both for weak and strong electrolytes.

Note: The units of molecular conductivity is $S\,c{m^2}mo{l^{ - 1}}$. The molar conductivity increases with increase in the dilution. Moreover, molar conductivity is not a constant value. In case of specific conductivity, the conductivity increases as the concentration of the electrolyte increases.