Chapter 5, Problem 5.56P

(a)

Interpretation Introduction

Interpretation:

A verification whether the molar conductivity follows the Kohlrausch law has to be done.  The value of the limiting molar conductivity has to be calculated. 

Concept Introduction:

Kohlrausch law:

The molar conductivity of an electrolyte at infinite dilution is equal to the sum of the individual conductances of the anions and cations. 

${\Lambda }_m^{\circ }=\lambda \left(Cation\right)+\lambda \left(Anaion\right)$

Answer to Problem 5.56P

The limiting molar conductivity is found to be $126{\Omega }^{-1}mo{l}^{-1}c{m}^2$.

Explanation of Solution

According to Kohlrausch law, Molar conductivity of a strong electrolyte weakly depends on concentration.  On dilution, there is a regular increase in the molar conductivity, due to the decrease in solute-solute interaction. 

Molar conductivity can be mathematically represented as

${\Lambda }_m=\frac{Cellcons\tan t}{R\times c}$

Where,

$\begin{array}{l}R=\mathrm{Re}sis\tan ce\\ \\ c=Concentration\\ \\ Cellcons\tan t=\frac{l}{A}\\ \\ l=Lengthofthecell\\ \\ A=Areaofcross-\sec tion\end{array}$

Substituting the first set of values and cell constant in the above equation and solving for ${\Lambda }_m$

$\begin{array}{l}{\Lambda }_m=\frac{0.2063\times {10}^3}{3314\times 0.00050}{\Omega }^{-1}mo{l}^{-1}c{m}^2\\ \\ =124.5021{\Omega }^{-1}mo{l}^{-1}c{m}^2.\end{array}$

Similarly, the rest of the calculation can be done as shown above.

$\begin{array}{ccc}R/\Omega & c/\left(mold{m}^{-3}\right)& {\Lambda }_m/\left({\Omega }^{-1}mo{l}^{-1}c{m}^2\right)\\ 3314& 0.00050& 124.5021\\ 1669& 0.0010& 123.6069\\ 342.1& 0.0050& 120.6080\\ 172.5& 0.010& 119.5942\\ 89.08& 0.020& 115.7947\\ 37.14& 0.050& 111.0931\end{array}$

It is clear from the table that molar conductivity of a strong electrolyte weakly depends on concentration as all the values came nearly same.  Also it shows that on dilution, there is a regular increase in the molar conductivity. 

The graph between ${\Lambda }_m$ and $\sqrt{c}$ will make it more clear. The equation of Kohlrausch law is

${\Lambda }_m={\Lambda }_m^{\circ }-\kappa \sqrt{c}$

Where,

$\begin{array}{l}{\Lambda }_m^{\circ }=\lim itingmolarconductivity\\ \\ \kappa =acons\tan t\end{array}$

$\begin{array}{cc}\sqrt{c}/\left(mold{m}^{-3}\right)& {\Lambda }_m/\left({\Omega }^{-1}mo{l}^{-1}c{m}^2\right)\\ 0.0223& 124.5021\\ 0.0316& 123.6069\\ 0.0707& 120.6068\\ 0.1& 119.5942\\ 0.1414& 115.7947\\ 0.2236& 111.0931\end{array}$

Figure.1

Upon extrapolation the graph, where it will touch on the Y-axis that will be the limiting molar conductivity.  Hence, the limiting molar conductivity is around $126{\Omega }^{-1}mo{l}^{-1}c{m}^2$.

(b)

Interpretation Introduction

Interpretation:

The value of coefficient $\kappa$ has to be determined. 

Concept Introduction:

Kohlrausch law:

The molar conductivity of an electrolyte at infinite dilution is equal to the sum of the individual conductances of the anions and cations. 

${\Lambda }_m^{\circ }=\lambda \left(Cation\right)+\lambda \left(Anaion\right)$

Answer to Problem 5.56P

The value of coefficient $\kappa$ has been determined to be $69.5323\times 10^3{\Omega }^{-1}mo{l}^{-2}c{m}^5$.

Explanation of Solution

The slope of the graph between ${\Lambda }_m$ and $\sqrt{c}$ will give the value of the coefficient $\kappa$.

Figure.2

Slope of the graph:

$Slope=\frac{119.5942-111}{0.1-0.2236}=-\mathrm{69.5323.}$

After considering carefully, the unit of $\kappa$ will be ${\text{Ω}}^{\text{-1}}{\text{mol}}^{\text{-2}}{\text{cm}}^5$. So ${10}^3$ will be multiplied with the answer to get the exact value of $\kappa$.

Therefore, the value of coefficient $\kappa$ has been determined to be $69.5323\times 10^3{\Omega }^{-1}mo{l}^{-2}c{m}^5$.

(c)

Interpretation Introduction

Interpretation:

The molar conductivity, conductivity and the resistance of $NaI\left(aq\right)$ has to be determined.

Concept Introduction:

Kohlrausch law:

The molar conductivity of an electrolyte at infinite dilution is equal to the sum of the individual conductances of the anions and cations. 

${\Lambda }_m^{\circ }=\lambda \left(Cation\right)+\lambda \left(Anaion\right)$

Answer to Problem 5.56P

The molar conductivity, conductivity and the resistance of $NaI\left(aq\right)$ has been determined to be $11.9947mS{m}^{2}mo{l}^{-1}$, $0.119947mS{m}^{-1}$ and $17.1992\Omega$ respectively. 

Explanation of Solution

(I)

Given Data:

$\begin{array}{l}\lambda \left(N{a}^{+}\right)=5.01mS{m}^{2}mo{l}^{-1}\\ \\ \lambda \left(I^{-}\right)=7.68mS{m}^{2}mo{l}^{-1}\\ \\ \left[NaI\right]=0.010mold{m}^{-3}\end{array}$

The limiting molar conductivity of $NaI\left(aq\right)$ can be calculated as

$\begin{array}{l}{\Lambda }_m^{\circ }\left(NaI\right)=\lambda \left(N{a}^{+}\right)+\lambda \left(I^{-}\right)\\ \\ =\left(5.01+7.68\right)mS{m}^{2}mo{l}^{-1}\\ \\ =12.69mS{m}^{2}mo{l}^{-1}.\end{array}$

Now, molar conductivity

$\begin{array}{l}{\Lambda }_m={\Lambda }_m^{\circ }-\kappa \sqrt{c}\\ \\ =12.69mS{m}^{2}mo{l}^{-1}-\left(69.5323\times {10}^3\right)\left({\text{Ω}}^{\text{-1}}{\text{mol}}^{\text{-2}}{\text{cm}}^5\right)\left(\sqrt{0.010}\right)\left(mold{m}^{-3}\right)\\ \\ =\left(12.69-0.6953\right)mS{m}^{2}mo{l}^{-1}\\ \\ =11.9947mS{m}^{2}mo{l}^{-1}.\end{array}$

Therefore, the molar conductivity of $NaI\left(aq\right)$ has been found out to be $11.9947mS{m}^{2}mo{l}^{-1}$.

(II)

Calculation of conductivity:

Molar conductivity can be mathematically represented as

${\Lambda }_m=\frac{{\rm K}}{c}$

Where,

${\rm K}=Conductivity$

Hence, the conductivity can be calculated as

$\begin{array}{l}{\rm K}={\Lambda }_m\times c\\ \\ =\left(11.9947\times 0.010\right)mS{m}^{-1}\\ \\ =0.119947mS{m}^{-1}.\end{array}$

Therefore, the conductivity of $NaI\left(aq\right)$ has been found out to be $0.119947mS{m}^{-1}$.

(III)

The conductivity can be further simplified as

${\rm K}=\frac{1}{R}\times Cellcons\tan t$

Hence, the resistance can be calculated as

$\begin{array}{l}R=\frac{Cellcons\tan t}{K}\\ \\ =\frac{0.2063c{m}^{-1}}{0.119947mS{m}^{-1}}\\ \\ =17.1992\Omega .\end{array}$

Therefore, the resistance of $NaI\left(aq\right)$ has been found out to be $17.1992\Omega$.