Question

Equivalent conductance of NaCl, HCl, and $C{{H}_{3}}COONa$ at infinite dilution are 126.45, 426.16 and 91 $oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$ respectively. The equivalent conductance of $C{{H}_{3}}COOH$ at infinite dilution would be:
(A)101.38 $oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$
(B)253.62 $oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$
(C)389.71 $oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$
(D)678.0 $oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$

Answer 1

Hint: To understand the concept behind this question we should know equivalent conductance and its effect on dilution. Understand the question well and being aware of Kohlrausch's law may help us in solving this question.

Complete step by step answer:
Equivalent conductance defined as the conducting power of all the ions produced by one gram equivalent of a given solution. It is denoted by ($\wedge $).
\[\wedge =KV\]
Where, K = Specific conductance
V = Volume in cc. containing in eq. of the electrolyte.
Now, let's see equivalent conductance at infinite dilution: Equivalent conductance increases with increase in dilution but after a certain limit it becomes constant that is beyond this point equivalent conductance does not increase. The maximum value of equivalent conductance is known as equivalent conductance at infinite dilution. It is denoted by (${{\wedge }_{\infty }}$).
To solve it we should Kohlrausch's law: At infinite dilution an ionic species that cations or anions contributes fixed value at given temperature towards equivalent conductance at the electrolyte irrespective of the other ionic species in combination with it. This contribution is known as equivalent ionic conductance at infinite dilution. The equivalent ionic conductance at infinite dilution for cation is denoted by $\lambda _{C}^{\circ }$and for anion it denoted by\[\lambda _{a}^{\circ }\].
According to kohlrausch's law, the equivalent conductance of an electrolyte at infinite dilution is equal to the sum of equivalent ionic conductance of cations and anions.
\[\wedge _{\infty }^{\circ }=\lambda _{c}^{\circ }+\lambda _{a}^{\circ }\]
Now, let's solve the problem,
Given, $\wedge {{_{\infty }^{\circ }}_{Nacl}}$= 126.45$oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$
$\wedge {{_{\infty }^{\circ }}_{HCl}}$= 426.16$oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$
$\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COONa}}$ = 91$oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}$
From kohlrausch's law ,
\[\wedge _{\infty }^{\circ }=\lambda _{c}^{\circ }+\lambda _{a}^{\circ }\]
So, applying this
\[\wedge {{_{\infty }^{\circ }}_{NaCl}}=\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }\] -equation 1
\[\wedge {{_{\infty }^{\circ }}_{HCl}}=\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }\] -equation 2
\[\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COONa}}=\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }\] -equation 3
We know that,
\[\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COOH}}=\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }\] -equation 4
Firstly, add equation 3 and equation 2 then subtract equation 1
\[\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COONa}}+\wedge {{_{\infty }^{\circ }}_{HCl}}-\wedge {{_{\infty }^{\circ }}_{NaCl}}=\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }+\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }-(\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ })\]On solving this,
\[\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COONa}}+\wedge {{_{\infty }^{\circ }}_{HCl}}-\wedge {{_{\infty }^{\circ }}_{NaCl}}=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }+\lambda _{{{H}^{+}}}^{\circ }\]
From equation 4, we know that \[\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COOH}}=\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }\]
\[\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COONa}}+\wedge {{_{\infty }^{\circ }}_{HCl}}-\wedge {{_{\infty }^{\circ }}_{NaCl}}=\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COOH}}\] - equation 5
Now substituting the value of $\wedge {{_{\infty }^{\circ }}_{Nacl}}$, $\wedge {{_{\infty }^{\circ }}_{HCl}}$and $\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COONa}}$ in equation 5
\[91+426.16-(126.45)=\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COOH}}\]
\[390.71=\wedge {{_{\infty }^{\circ }}_{C{{H}_{3}}COOH}}\]

Thus option C is the correct answer.

Note: Be thorough with the concept of equivalent conductance. The kohlrausch's law also can be defined in terms of molar conductance, It will be almost similar to it. so, don't confuse it with both definitions of kohlrausch's law. This law is applied to find degree of dissociation, molar conductance and equivalent conductance.