Question

The molar conductivities at infinite dilution of $KCl,KN{{O}_{3}}$ and $AgN{{O}_{3}}$ at 298K are $0.01499m{{\hom }^{2}}/mo{{l}^{-1}},0.01250m{{\hom }^{2}}/mo{{l}^{-1}}\And 0.01334m{{\hom }^{2}}/mo{{l}^{-1}}$ respectively. What is the molar conductivity of AgCl infinite dilution at this temperature?

Answer 1

The ratio of measure conductivity (specific conductance) of an electrolyte solution to its molar concentration is known as molar conductivity. The molar conductivity of weak electrolytes depends on the concentration of the solution. The more dilute a solution, the greater its molar conductivity, due to increased ionic dissociation.

Complete step by step solution:
The molar conductivity of each ionic species is proportional to its electrical mobility per unit electric field.
According to Kohlraush’s law states that “ at infinite dilution, when dissociation is complete, each ion makes a definite contribution towards equivalent conductance of electrolyte conductance of the electrolyte irrespective of the nature of the ion with which it is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of the contribution of its anions and cations”.
${{\lambda }_{eq}}^{\infty }={{\lambda }_{+}}^{\infty }+{{\lambda }_{-}}^{\infty }$
Where ${{\lambda }_{eq}}^{\infty }$ = molar conductivity of given salt at infinite dilution,
${{\lambda }_{+}}^{\infty }$ = molar conductivity of cation,
${{\lambda }_{-}}^{\infty }$ = molar conductivity of anion
Apply Kohlraush’s law to given salts $KCl,KN{{O}_{3}}$ and $AgN{{O}_{3}}$respectively as follows,

& {{\lambda }^{\infty }}_{KCl}={{\lambda }^{\infty }}({{K}^{+}})+{{\lambda }^{\infty }}(C{{l}^{-}}) \\\ & {{\lambda }^{\infty }}_{KN{{O}_{3}}}={{\lambda }^{\infty }}({{K}^{+}})+{{\lambda }^{\infty }}(N{{O}_{3}}^{-}) \\\ & {{\lambda }^{\infty }}_{AgN{{O}_{3}}}={{\lambda }^{\infty }}(A{{g}^{+}})+{{\lambda }^{\infty }}(N{{O}_{3}}^{-}) \\\ \end{aligned}$$ From the above equations, let us calculate the molar conductivity of AgCl by substituting the given values using the above equations, $$\begin{aligned} & {{\lambda }^{\infty }}_{AgCl}={{\lambda }^{\infty }}_{AgN{{O}_{3}}}-{{\lambda }^{\infty }}_{KN{{O}_{3}}}+{{\lambda }^{\infty }}_{KCl} \\\ & {{\lambda }^{\infty }}_{AgCl}=0.01334-0.01250+0.001499=0.002339m{{\hom }^{2}}/mol \\\ \end{aligned}$$ Hence, the molar conductivity of AgCl at infinite dilutions at 298K is $$0.002339m{{\hom }^{2}}/mol$$ **Note:** $O{{H}^{-}}$ ions will have the highest molar conductivity amongst all anions. Greater in size high conductivity at infinite dilution but cations ${{H}^{+}}$ has the highest conductivity amongst all cations. As the number of ions of complex compounds when dissolved in water increases, molar conductivity also increases.