Question

For the strong electrolytes NaOH, NaCl and $BaC{l}_2,$ the molar ionic conductances at infinite dilution are $248.1\times {10}^{-4},126.5\times {10}^{-4}and280.0\times {10}^{-4}S{m}^{2}mo{l}^{-1},$ respectively. Calculate ${\Lambda }_m^{\circ }\text{for}Ba(OH{)}_2$:

• A. $523.2\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• B. $550.4\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• C. $600.5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• D. $500.0\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is A $523.2\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

$BaC{l}_2+2NaOH⟶Ba(OH{)}_2+2NaCl$

Thus, according to Kohlrausch's law of independent migration of ions,

${\Lambda }_{mBa(OH{)}_2}^{\circ }={\Lambda }_{mBaC{l}_2}^{\circ }+2{\Lambda }_{mNaOH}^{\circ }-2{\Lambda }_{mNaCl}^{\circ }$

$=(280+2\times 248.1-2\times 126.5)\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
$=523.2\times {10}^{-4}S{m}^{2}mo{l}^{-1}$