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How would you determine the molar conductivity at infinite dilution of NH4OH with the help of Kohlrausch’s law?

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Question

How would you determine the molar conductivity at infinite dilution of NH4OH with the help of Kohlrausch’s law?

Very Long Answer
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Solution

To calculate `Lambda_m^circ`(NH4OH), the method uses known values of molar conductivities at infinite dilution of related strong electrolytes and applies Kohlrausch’s Law of Independent Ionic Migration, which states:

`Lambda_m^circ ("Electrolyte") = lambda_+^circ + lambda_-^circ`

At infinite dilution, we know that,

\[\ce{\Lambda{^{\circ}_{Ba(OH)_2}} = \Lambda{^{\circ}_{Ba^{2+}}} + 2\Lambda^{\circ}_{OH^-}}\]

\[\ce{\Lambda{^{\circ}_{Ba(Cl)_2}} = \Lambda{^{\circ}_{Ba^{2+}}} + 2\Lambda^{\circ}_{Cl^-}}\]

\[\ce{\Lambda{^{\circ}_{NH_4Cl}} = \Lambda{^{\circ}_{NH{^{+}_{4}}}} + \Lambda^{\circ}_{Cl^-}}\]

After using the above equations we will reach

\[\ce{\Lambda{^{\circ}_{NH_4OH}} = \frac{\Lambda{^{\circ}_{Ba(OH)_2}} + 2\Lambda{^{\circ}_{NH_4Cl}} - \Lambda{^{\circ}_{BaCl_2}}}{2}}\]

⇒ \[\ce{\Lambda{^{\circ}_{NH_4OH}}}\]

= \[\ce{\Lambda{^{\circ}_{NH^+_4}} + \Lambda^{\circ}_{OH^-}}\]

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