Question

For the strong electrolytes $NaOH,NaCl$ and $BaC{l}_2$, the molar ionic conductances at infinite dilution are $248\times {10}^{-4},126.5\times {10}^{-4}$ and $280.0\times {10}^{-4}s.m^2.mo{l}^{-1}$ respectively. Calculate ${{\mathrm{\Lambda }}_m}^{\circ }$ for $Ba{\left(OH\right)}_2$.

Answer 1

Hint:Electrolytes are the substances whose solution in water conducts electric current. Conduction takes place by the movement of ions. Kohlrausch’s law is based on the limiting molar conductivity of an electrolyte.

Complete step by step answer:
Molar conductivity is the conductive power of all the ions produced by one mole of an electrolyte in the given solution. Kohlrausch’s law gives the empirical relationship between molar conductivity and concentration of strong electrolytes. Limiting molar conductivity is due to the migration of cations in one direction and anions in the other under very dilute conditions. The Kohlrausch’s law is expressed in the formula given below:
${{\mathrm{\Lambda }}_m}^{\circ }={\upsilon }_{+}{{\mathrm{\Lambda }}_{+}}^{\circ }+{\upsilon }_{-}{{\mathrm{\Lambda }}_{-}}^{\circ }$, where ${{\mathrm{\Lambda }}_{+}}^{\circ },{{\mathrm{\Lambda }}_{-}}^{\circ }$ are the ionic conductances of the cations and anions at infinite dilution.
It is given that molar ionic conductance of $NaOH$, ${{\mathrm{\Lambda }}_m}^{\circ }\left(NaOH\right)=248\times {10}^{-4}s.m^2.mo{l}^{-1}$
Molar ionic conductance of $NaCl$, ${{\mathrm{\Lambda }}_m}^{\circ }\left(NaCl\right)=126.5\times {10}^{-4}s.m^2.mo{l}^{-1}$
Molar ionic conductance of $BaC{l}_2$, ${{\mathrm{\Lambda }}_m}^{\circ }\left(BaC{l}_2\right)=280.0\times {10}^{-4}s.m^2.mo{l}^{-1}$
Molar ionic conductance of $NaOH$ can be expressed as ${{\mathrm{\Lambda }}_m}^{\circ }\left(NaOH\right)={{\mathrm{\Lambda }}_m}^{\circ }\left(N{a}^{+}\right)+{{\mathrm{\Lambda }}_m}^{\circ }\left(O{H}^{-}\right)$
Molar ionic conductance of $NaCl$ can be expressed as ${{\mathrm{\Lambda }}_m}^{\circ }\left(NaCl\right)={{\mathrm{\Lambda }}_m}^{\circ }\left(N{a}^{+}\right)+{{\mathrm{\Lambda }}_m}^{\circ }\left(C{l}^{-}\right)$
Molar ionic conductance of $BaC{l}_2$ can be expressed as ${{\mathrm{\Lambda }}_m}^{\circ }\left(BaC{l}_2\right)={{\mathrm{\Lambda }}_m}^{\circ }\left(B{a}^{2+}\right)+2{{\mathrm{\Lambda }}_m}^{\circ }\left(C{l}^{-}\right)$
The reaction with respect to the formation of $Ba{\left(OH\right)}_2$ is given below:
$BaC{l}_2+2NaOH\to Ba{\left(OH\right)}_2+2NaCl$
Thus molar ionic conductance of $Ba{\left(OH\right)}_2$ can be expressed as:
${{\mathrm{\Lambda }}_m}^{\circ }\left[Ba{\left(OH\right)}_2\right]={{\mathrm{\Lambda }}_m}^{\circ }\left(BaC{l}_2\right)+2{{\mathrm{\Lambda }}_m}^{\circ }\left(NaOH\right)-2{{\mathrm{\Lambda }}_m}^{\circ }\left(NaCl\right)$
$${{\mathrm{\Lambda }}_m}^{\circ }\left[Ba{\left(OH\right)}_2\right]=\left[{{\mathrm{\Lambda }}_m}^{\circ }\left(B{a}^{+}\right)+2{{\mathrm{\Lambda }}_m}^{\circ }\left(C{l}^{-}\right)\right]+\left[2{{\mathrm{\Lambda }}_m}^{\circ }\left(N{a}^{+}\right)+2{{\mathrm{\Lambda }}_m}^{\circ }\left(O{H}^{-}\right)\right]-\left[2{{\mathrm{\Lambda }}_m}^{\circ }\left(N{a}^{+}\right)+2{{\mathrm{\Lambda }}_m}^{\circ }\left(C{l}^{-}\right)\right]$$
On simplification, we get
$${{\mathrm{\Lambda }}_m}^{\circ }\left[Ba{\left(OH\right)}_2\right]=\left[{{\mathrm{\Lambda }}_m}^{\circ }\left(B{a}^{+}\right)\right]+\left[2{{\mathrm{\Lambda }}_m}^{\circ }\left(O{H}^{-}\right)\right]$$
Substituting the values, we get
$${{\mathrm{\Lambda }}_m}^{\circ }\left[Ba{\left(OH\right)}_2\right]=280.0\times {10}^{-4}s.m^2.mo{l}^{-1}+2\times 248\times {10}^{-4}s.m^2.mo{l}^{-1}-\left(2\times 126.5s.m^2.mo{l}^{-1}\right)$$
Solving,
$${{\mathrm{\Lambda }}_m}^{\circ }\left[Ba{\left(OH\right)}_2\right]=776.0\times {10}^{-4}s.m^2.mo{l}^{-1}-253.0\times 10^{-4}s.m^2.mo{l}^{-1}=523\times 10^{-4}s.m^2.mo{l}^{-1}$$
Molar conductance of $Ba{\left(OH\right)}_2$, $${{\mathrm{\Lambda }}_m}^{\circ }\left[Ba{\left(OH\right)}_2\right]=523\times 10^{-4}s.m^2.mo{l}^{-1}$$

Note: Specific conductance is the ability of an electrolyte solution to conduct electricity. On dilution, the specific conductivity decreases. Equivalent and molar conductance increase with dilution and reaches a maximum value. The conductance of all electrolytes increases with temperature. Kohlrausch’s law can be applied for calculating the degree of dissociation of a weak electrolyte. It can also be used for finding the equivalent conductivity of weak electrolytes.