Question

For the strong electrolytes NaOH, NaCl and $BaC{l}_2$, the molar ionic conductances at infinite dilution are 248 x ${10}^{-4}$, 126.5 x ${10}^{-4}$ and 280.0 x ${10}^{-4}$ s $m^{2}mo{l}^{-1}$,respectively. Calculate ${\wedge }_m^0$ for $Ba(OH{)}_2$.

Answer 1

${\wedge }_m^o\left(NaOH\right)={\lambda }_{N{a}^{+}}^o+{\lambda }_{O{H}^{-}}^o$

${\wedge }_m^o\left(NaCl\right)={\lambda }_{N{a}^{+}}^o+{\lambda }_{a^{-}}^o$

${\wedge }_m^o\left(BaC{l}_2\right)={\lambda }_{B{a}^{+2}}^o+2{\lambda }_{a^{-}}^o$

$\Rightarrow 2{\wedge }_m^o\left(NaOH\right)-2{\wedge }_m^o(NaCl+{\wedge }_m^o(BaC{l}_2)$

$2{\lambda }_{N{a}^{+}}^o+2{\lambda }_{O{H}^{-}}-(2{\lambda }_{N{a}^{+}}^o+2{\lambda }_{a^{-}}^o)+{\lambda }_{B{a}^{+2}}^o+2{\lambda }_{a^{-}}^o$

${\lambda }_m^{o}Ba(OH{)}_2={\lambda }_{B{a}^{+2}}^o+2{\lambda }_{O{H}^{-}}$

So ${\wedge }_m^o=(2\times 248-2\times 126.5+280)\times {10}^{-4}$

${\wedge }_m^{o}Ba(OH{)}_2=523\times {10}^{-4}$