Advertisements
Advertisements
Question
How would you determine the molar conductivity at infinite dilution of NH4OH with the help of Kohlrausch’s law?
Advertisements
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
\[\ce{\Lambda{^{\circ}_{NH_4OH}} = \frac{\Lambda{^{\circ}_{Ba(OH)_2}} + 2\Lambda{^{\circ}_{NH_4Cl}} - \Lambda{^{\circ}_{BaCl_2}}}{2}}\]
⇒ \[\ce{\Lambda{^{\circ}_{NH_4OH}}}\]
= \[\ce{\Lambda{^{\circ}_{NH^+_4}} + \Lambda^{\circ}_{OH^-}}\]
