Question

The mobilities for ${\mathbf{Na}}^{+}$and ${\mathbf{Cl}}^{-}$in aqueous solution are given in Table 22.7. Calculate the conductivity of a 0.10-M solution of NaCl in water at 25°C.

Short answer

$\lambda {°}_{solution}=126.4\Omega c{m}^2.mo{l}^{-1}$

Step-by-step solution

Conductivity of a solution

A solution can conduct electricity when it contains free ions in its solution.

For example, $\mathbf{aq}\mathbf{.}\mathbf{NaCl}$solution.

The conductivity of NaCl solution

When mobilities of ions are given then the conductivity of that ion in solution can be calculated by using the formula:

$\lambda {°}_{solution}=\lambda {°}_{+}+\lambda {°}_{-}$

$\begin{array}{l}\lambda {°}_{+}=F{\upsilon }_{+}\\ \lambda {°}_{-}=F{\upsilon }_{-}\end{array}$

Where, $\lambda {°}_{+}$is the conductivity of cation, $\lambda {°}_{-}$is the conductivity of anion.

F is 96485 Coulombs/mol.

${\upsilon }_{+}$is the mobility of cation and ${\upsilon }_{-}$is the mobility of anion.

Calculation

• From table 22.7, the mobilities of $N{a}^{+}$and $C{l}^{-}$ are $5.19\times {10}^{-4}c{m}^2.V^{-1}.s^{-1}$ and $7.91\times {10}^{-4}c{m}^2.V^{-1}.s^{-1}$

Molar conductivity of $N{a}^{+}$ :

$\begin{array}{c}\lambda {°}_{N{a}^{+}}=F{\upsilon }_{N{a}^{+}}\\ =96485\times 5.19\times {10}^{-4}\\ =50.1\Omega .c{m}^2.mo{l}^{-1}\end{array}$

role="math" localid="1663702768211" $\begin{array}{c}\lambda {°}_{C{l}^{-}}=F{\upsilon }_{C{l}^{-}}\\ =96485\times 7.91\times {10}^{-4}\\ =76.3\Omega .c{m}^2.mo{l}^{-1}\end{array}$

$\begin{array}{c}\lambda {°}_{solution}=\lambda {°}_{+}+\lambda {°}_{-}\\ =(50.1+76.3)\Omega c{m}^2.mo{l}^{-1}\\ =126.4\Omega c{m}^2.mo{l}^{-1}\end{array}$