Question

At$18^\circ {\text{C}}$, the mobilities of ${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ and ${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ ions are $6.6 \times {10^{ - 4}}$ and $5.7 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{vol}}{{\text{t}}^{{\text{ - 1}}}}{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$ at infinite dilution. Calculate the equivalent conductance of ammonium chlorate solution.

Answer 1

Hint: Using the ionic mobilities calculate the ionic conductance of both the ions. Use the Kohlrausch law and Calculate the equivalent conductance of ammonium chlorate solution.

Formula Used :
${\lambda _{^ + }} = F \times {U_{^ + }}$
${\lambda _ - } = F \times {U_ - }$
$\Lambda 0 = {\lambda _{^ + }} + {\lambda _{^ - }}$

Complete step by step answer:
Using the given mobilities of ${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ and ${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ ions calculate the ionic conductance of ${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ and ${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ ions as follows:
${\lambda _{^ + }} = F \times {U_{^ + }}$
Here,
${\lambda _{^ + }}$= ionic conductance of ${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ions
$F$ = Charge = 96500 C
${U_{^ + }}$ = mobility of ${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ ion =$6.6 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{vol}}{{\text{t}}^{{\text{ - 1}}}}{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$
So, ${\lambda _{^ + }} = 96500{\text{ C}} \times 6.6 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{vol}}{{\text{t}}^{{\text{ - 1}}}}{\text{se}}{{\text{c}}^{{\text{ - 1}}}} = 63.69{\text{c}}{{\text{m}}^{\text{2}}}{\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$
${\lambda _ - } = F \times {U_ - }$
Here,
${\lambda _ - }$ = ionic conductance of${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ ions
${U_ - }$ = mobility of ${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ ion =$5.7 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{vol}}{{\text{t}}^{{\text{ - 1}}}}{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$
So, ${\lambda _ - } = 96500{\text{ C}} \times 5.7 \times {10^{ - 4}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{vol}}{{\text{t}}^{{\text{ - 1}}}}{\text{se}}{{\text{c}}^{{\text{ - 1}}}} = 55{\text{c}}{{\text{m}}^{\text{2}}}{\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$
Kohlrausch law states that the equivalent conductance ($\Lambda 0$) of any electrolyte at infinite dilution is the sum of the ionic conductance of the cation and anion given by the electrolytes at infinite dilution.
$\Lambda 0 = {\lambda _{^ + }} + {\lambda _{^ - }}$
Now we have ionic conductance of the cation (${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$) and anion (${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$) so calculate the equivalent conductance ammonium chlorate solution as follows:
$\Lambda 0 = 63.69{\text{c}}{{\text{m}}^{\text{2}}}{\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}} + 55{\text{c}}{{\text{m}}^{\text{2}}}{\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$
$\Lambda 0 = 118.69{\text{c}}{{\text{m}}^{\text{2}}}{\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$

**Thus, the equivalent conductance of ammonium chlorate solution is $118.69{\text{c}}{{\text{m}}^{\text{2}}}{\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$.

Note: **
Ionic conductance is directly proportional to the ionic mobility. The greater the mobility of ions greater is the ionic conductance. The equivalent conductance of an electrolyte also depends on the number of each type of ions present in the solution. Ammonium chlorate produces one ${\text{N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ and one ${\text{Cl}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ ion in the solution.