Question

The absolute velocity (ionic mobility) of ${\text{A}}{{\text{g}}^ + }$ is $5.7 \times {10^{ - 4}}{\text{cm }}{{\text{s}}^ - }$ and of ${\text{NO}}_3^ -$ $6.9 \times {10^{ - 4}}{\text{cm }}{{\text{s}}^ - }$ assuming complete dissociation, calculate the equivalent conductivity of 0.01M ${\text{AgN}}{{\text{O}}_3}$ at infinite dilution.
A. 121.5 ${\text{Sc}}{{\text{m}}^2}{\text{e}}{{\text{q}}^{ - 1}}$
B. 1.2 ${\text{Sc}}{{\text{m}}^2}{\text{e}}{{\text{q}}^{ - 1}}$
C. 0.12 ${\text{Sc}}{{\text{m}}^2}{\text{e}}{{\text{q}}^{ - 1}}$
D. None of the above

Answer 1

Since the question is stating infinite dilution and complete dissociation, we can say that ${\text{AgN}}{{\text{O}}_3}$ is a strong electrolyte and the molar conductivity at infinite dilution is called limiting molar conductivity. Recall the Kohlrausch law of independent migration of ions. And calculate limiting molar conductivity.

Complete Step by step answer: Let’s first understand the Kohlarsh law of independent migration of ions, the law states that limiting molar conductivity of an electrolyte can be represented by the sum of the individual contribution of the anion and cation of the electrolyte. i.e., $\wedge _{\text{m}}^ \circ$ = ${{\lambda }}_ + ^ \circ {{ + \lambda }}_ - ^ \circ$ (where, ${{\lambda }}_ + ^ \circ$ and ${{\lambda }}_ - ^ \circ$ are the limiting molar of cation and anion respectively)
We know that the ionic conductance and ionic mobility are directly proportional to each other
$\Rightarrow$ ${{\lambda }}_ + ^ \circ$ = F${{\text{U}}_ + }$ and,
$\Rightarrow$ ${{\lambda }}_ - ^ \circ$ = F${{\text{U}}_ - }$
Since we are given the velocities of silver cation and nitrate anion for ${\text{AgN}}{{\text{O}}_3}$ electrolyte. We can now calculate the limiting molar conductivity.
$\Rightarrow$ $\wedge _{\text{m}}^ \circ$ = F (${{\text{U}}_{{\text{A}}{{\text{g}}^ + }}}$ + ${{\text{U}}_{{\text{NO}}_3^ - }}$ )
$\Rightarrow$ $\wedge _{\text{m}}^ \circ$ = 96485 ($5.7 \times {10^{ - 4}}$+ $6.9 \times {10^{ - 4}}$ )
$\Rightarrow$ $\wedge _{\text{m}}^ \circ$ = 96485 (12.6 $\times$ ${10^{ - 4}}$), [ here we have used the addition formula ${\text{a}} \times {\text{1}}{{\text{0}}^{\text{n}}}{\text{ b}} \times {\text{1}}{{\text{0}}^{\text{n}}} = {\text{ (a + b)}} \times {\text{1}}{{\text{0}}^{\text{n}}}$]
$\Rightarrow$ $\wedge _{\text{m}}^ \circ$ = 1215711 $\times$ ${10^{ - 4}}$
$\Rightarrow$ $\wedge _{\text{m}}^ \circ$ = 121.6 (we have only taken to the 1st decimal point, by rounding off the value from 121.57)
Now, we will calculate the value of molar conductance using the following formula,
$\Rightarrow$ ${ \wedge _{\text{m}}} = \wedge _{\text{m}}^ \circ {\text{ }} - {\text{A}}\sqrt {\text{c}}$ (the value of A for ${\text{AgN}}{{\text{O}}_3}$electrolyte is 1 since the charges on the cation and anion produce after dissociation is 1-1 )
$\Rightarrow$ ${ \wedge _{\text{m}}} = 121.6{\text{ }} - 1\sqrt {0.01}$
$\Rightarrow$ ${ \wedge _{\text{m}}} = 121.6{\text{ }} - 0.1$ ( the value of $\sqrt {0.01}$ = 0.1 )
$\Rightarrow$ ${ \wedge _{\text{m}}} = 121.5$
Now, let’s calculate the value of equivalent conductivity;
$\Rightarrow$ ${ \wedge _{{\text{eq}}}}$ = $\dfrac{{{ \wedge _{\text{m}}}}}{{\text{n}}}$
$\Rightarrow$ ${ \wedge _{{\text{eq}}}}$ = $\dfrac{{{ \wedge _{\text{m}}}}}{{\text{n}}}$ (the value of, $\Rightarrow$ n (valence) = molecular mass/ equivalent mass = 170/170 = 1)
$\Rightarrow$ ${ \wedge _{{\text{eq}}}}$ = $\dfrac{{121.5}}{1}$
$\Rightarrow$ ${ \wedge _{{\text{eq}}}}$ = 121.5 ${\text{Sc}}{{\text{m}}^2}{\text{e}}{{\text{q}}^{ - 1}}$

Hence, the correct answer is option (A)i.e., 121.5 ${\text{Sc}}{{\text{m}}^2}{\text{e}}{{\text{q}}^{ - 1}}$

Note: Remember we have given the value ionic mobility (velocity), and not the limiting molar conductivity, otherwise if we just add up the values limiting molar conductivity of corresponding anions and cation we can get the limiting molar conductivity ${\text{AgN}}{{\text{O}}_3}$. But in this question, we had to use the concept of ionic mobility and ionic conductance to calculate limiting molar conductivity.