Chapter 16, Problem 16B.5P

(a)

Interpretation Introduction

Interpretation:

The Kohlrausch’s law has to be verified for the molar conductivity.  The limiting molar conductivity ${\Lambda }_{\text{m}}^{\text{o}}$  and coefficient $k$ have to be calculated.

Concept introduction:

Conductivity is a property of a material that depends upon the charge carriers in a sample.  Molar conductivity is the conductivity per unit molar concentration.  According to Kohlrausch’s law, the molar conducivities of the dilute solutions containing strong electrolytes depend upon the square root of the concentration.  The conductivity of a solution when the concentration tends to zero is known as the limiting molar conductivity.  Limiting molar conductivity is the sum of contributions from individual ions.

Answer to Problem 16B.5P

The graph between molar conductivity and $c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ gives a straight line which proves that that molar conductivity follows the Kohlrausch’s law.

The value of limiting molar conductivity $\left({\Lambda }_{\text{m}}^{\text{o}}\right)$ is $\underset{\_}{0.125Sd{m}^{3}c{m}^{-1}mo{l}^{-1}}$.

The value of Kohlrausch’s constant $\left(k\right)$ is $\underset{\_}{0.066\frac{Sd{m}^{3}c{m}^{-1}mo{l}^{-1}}{{\left(mold{m}^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$.

Explanation of Solution

The equation for the Kohlrausch’s law is shown below.

    ${\Lambda }_{\text{m}}={\Lambda }_{\text{m}}^{\text{o}}-k{c}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$        (1)

Where,

• ${\Lambda }_{\text{m}}$ is the molar conductivity.
• ${\Lambda }_{\text{m}}^{\text{o}}$ is the limiting molar conductivity.
• $k$ is the Kohlrausch’s constant.
• $c$ is the molar concentration.

The molar conductivity is calculated by the formula shown below.

    ${\Lambda }_{\text{m}}=\frac{\kappa }{c}$        (2)

Where,

• ${\Lambda }_{\text{m}}$ is the molar conductivity.
• $\kappa$ is the conductivity.
• $c$ is the molar concentration.

The conductivity of a solution is defined as shown below.

  $\kappa =\frac{C}{R}$        (3)

Where,

• $C$ is cell constant.
• $R$ is the resistance.

Substitute $\kappa =\frac{C}{R}$ in equation (2).

    ${\Lambda }_{\text{m}}=\frac{C}{Rc}$        (4)

The given data is shown below.

$c/\left(\text{mol}{\text{dm}}^{-3}\right)$	$R/\Omega$
$0.00050$	$3314$
$0.0010$	$1669$
$0.0050$	$342.1$
$0.010$	$174.1$
$0.020$	$89.08$
$0.050$	$37.14$

The value of cell constant is given as $0.206\text{3}{\text{cm}}^{-1}$.

The molar conductivity at each concentration is calculated using equation (4) as shown below.

$c/\left(\text{mol}{\text{dm}}^{-3}\right)$	$R/\Omega$	${\Lambda }_{\text{m}}=\frac{C}{Rc}\left(\text{S}{\text{dm}}^{\text{3}}{\text{cm}}^{-1}{\text{mol}}^{-1}\right)$
$0.00050$	$3314$	$0.1245$
$0.0010$	$1669$	$0.1236$
$0.0050$	$342.1$	$0.1206$
$0.010$	$174.1$	$0.1184$
$0.020$	$89.08$	$0.1158$
$0.050$	$37.14$	$0.1111$

The value of $c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is calculated as shown below.

$c/\left(\text{mol}{\text{dm}}^{-3}\right)$	$R/\Omega$	${\Lambda }_{\text{m}}=\frac{C}{Rc}\left(\text{S}{\text{dm}}^{\text{3}}{\text{cm}}^{-1}{\text{mol}}^{-1}\right)$	$c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
$0.00050$	$3314$	$0.1245$	$0.0224$
$0.0010$	$1669$	$0.1236$	$0.0316$
$0.0050$	$342.1$	$0.1206$	$0.0707$
$0.010$	$174.1$	$0.1184$	$0.1000$
$0.020$	$89.08$	$0.1158$	$0.1414$
$0.050$	$37.14$	$0.1111$	$0.2236$

Now plot a graph between ${\Lambda }_{\text{m}}$ and $c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ as shown below.

Figure 1

The equation obtained from the graph is $\text{y}=-0.066\text{x}+0.125$.  In the graph, $\text{y-axis}$ represents ${\Lambda }_{\text{m}}$ and $\text{x-axis}$ represents $c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.  Therefore, the equation can be written as ${\Lambda }_{\text{m}}=0.125-0.066c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Compare the equation ${\Lambda }_{\text{m}}=0.125-0.066c^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ with equation (1) to obtain the value of ${\Lambda }_{\text{m}}^{\text{o}}$ and $k$.

Therefore, the value of ${\Lambda }_{\text{m}}^{\text{o}}$ is $\underset{\_}{0.125Sd{m}^{3}c{m}^{-1}mo{l}^{-1}}$ and the value of Kohlrausch’s constant $\left(k\right)$ is $\underset{\_}{0.066\frac{Sd{m}^{3}c{m}^{-1}mo{l}^{-1}}{{\left(mold{m}^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

(b)

Interpretation Introduction

Interpretation:

The value of molar conductivity, conductivity and resistance of the $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution has to be calculated.

Concept introduction:

As mentioned in concept introduction in part (a).

Answer to Problem 16B.5P

The value of molar conductivity of $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is $\underset{\_}{12.03mS{m}^{2}mo{l}^{-1}}$.

The value of conductivity of $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is $\underset{\_}{120.3mS{m}^{-1}}$.

The value of resistance of the $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is $\underset{\_}{171.49\Omega }$.

Explanation of Solution

The value of molar conductivity of the $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is calculated by the formula shown below.

    ${\Lambda }_{\text{m}}={\Lambda }_{\text{m}}^{\text{o}}-k{c}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$        (1)

Where,

• ${\Lambda }_{\text{m}}$ is the molar conductivity.
• ${\Lambda }_{\text{m}}^{\text{o}}$ is the limiting molar conductivity.
• $k$ is the Kohlrausch’s constant.
• $c$ is the molar concentration.

The molar conductivity of $\text{NaI}$ solution is calculated as shown below.

    ${\Lambda }_{\text{m}}^{\text{o}}=\text{λ}\left({\text{Na}}^{+}\right)+\text{λ}\left({\text{I}}^{-}\right)$        (2)

Where,

• $\text{λ}\left({\text{Na}}^{+}\right)$ is the molar conductivity of sodium ion.
• $\text{λ}\left({\text{I}}^{-}\right)$ is the molar conductivity of iodide ion.

The value of $\text{λ}\left({\text{Na}}^{+}\right)$ and $\text{λ}\left({\text{I}}^{-}\right)$ is $5.0\text{1}\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}$ and $\text{7}\text{.68}\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}$ respectively.

Substitute the value of $\text{λ}\left({\text{Na}}^{+}\right)$ and $\text{λ}\left({\text{I}}^{-}\right)$ in equation (2).

    $\begin{array}{c}{\Lambda }_{\text{m}}^{\text{o}}=\text{λ}\left({\text{Na}}^{+}\right)+\text{λ}\left({\text{I}}^{-}\right)\\ =5.0\text{1}\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}+7.68\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}\\ =12.69\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}\end{array}$

The value of Kohlrausch’s constant $\left(k\right)$ and $c$ is $0.066\frac{\text{S}{\text{dm}}^{\text{3}}{\text{cm}}^{-1}{\text{mol}}^{-1}}{{\left(\text{mol}{\text{dm}}^{-3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ and $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}$ respectively.

Substitute the value of Kohlrausch’s constant $\left(k\right)$ and $c$ in equation (1).

    $\begin{array}{c}{\Lambda }_{\text{m}}={\Lambda }_{\text{m}}^{\text{o}}-k{c}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =12.69\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}-0.066\frac{\text{S}{\text{dm}}^{\text{3}}{\text{cm}}^{-1}{\text{mol}}^{-1}}{{\left(\text{mol}{\text{dm}}^{-3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\times {\left(\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =12.69\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}-0.0066\text{S}\left(\frac{{10}^3\text{mS}}{1\text{S}}\right){\text{dm}}^{\text{3}}\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1{\text{dm}}^{\text{3}}}\right){\text{cm}}^{-1}\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right){\text{mol}}^{-1}\\ =12.69\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}-0.66\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}\\ =\underset{\_}{12.03mS{m}^{2}mo{l}^{-1}}\end{array}$

Therefore, the value of molar conductivity of $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is $\underset{\_}{12.03mS{m}^{2}mo{l}^{-1}}$.

The conductivity is calculated by the formula shown below.

    $\kappa ={\Lambda }_{\text{m}}c$        (3)

Where,

• ${\Lambda }_{\text{m}}$ is the molar conductivity.
• $\kappa$ is the conductivity.
• $c$ is the molar concentration.

The value of $c$ is $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}$.

Substitute the value of $c$ and ${\Lambda }_{\text{m}}$ in equation (3).

    $\begin{array}{c}\kappa ={\Lambda }_{\text{m}}c\\ =12.03\text{mS}{\text{m}}^{\text{2}}{\text{mol}}^{-1}\times \text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\left(\frac{1{\text{dm}}^{\text{3}}}{{10}^{-3}{\text{m}}^{\text{3}}}\right)\\ =\underset{\_}{120.3mS{m}^{-1}}\end{array}$

Therefore, the value of conductivity of $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is $\underset{\_}{120.3mS{m}^{-1}}$.

The resistance of a solution is calculated as shown below.

  $R=\frac{C}{\kappa }$        (4)

Where,

• $C$ is cell constant.
• $R$ is the resistance.
• $\kappa$ is the conductivity.

The value of cell constant $\left(C\right)$ is given as $0.206\text{3}{\text{cm}}^{-1}$.

Substitute the value of $C$ and $\kappa$ in equation (4)

    $\begin{array}{c}R=\frac{C}{\kappa }\\ =\frac{0.206\text{3}{\text{cm}}^{-1}\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)}{120.3\text{mS}\left(\frac{{10}^{-3}\text{S}}{1\text{mS}}\right){\text{m}}^{-1}}\\ =\frac{20630}{120.3}{\text{S}}^{-1}\left(\frac{1\Omega }{1{\text{S}}^{-1}}\right)\\ =\underset{\_}{171.49\Omega }\end{array}$

Therefore, the value of resistance of the $\text{0}\text{.010}\text{mol}{\text{dm}}^{-3}\text{NaI}$ solution is $\underset{\_}{171.49\Omega }$.