First slide

Applications of Kohlraush law

Question

At T(K) the molar ionic conductivities of NH₄⁺ and OH^– at infinite dilution are 72 and 198 Scm² mol^–1 respectively. The molar conductivity of 0.01M NH₄OH solution at the same temperature is found to be 9 Scm² mol^–1. The percentage dissociation of NH₄OH at this concentration is

Easy

Solution

$\alpha = \frac{{{\lambda _C}}}{{{\lambda _o}}}$

$\alpha = \frac{9}{{\lambda _{N{H_4}^ + }^\infty + \lambda _{O{H^ - }}^\infty }} = \frac{9}{{270}} = 0.033 \Rightarrow \% \alpha = 0.033 \times 100 = 3.3\%$

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