Equivalence conductance at infinite dilution of NH4Cl, NaOH and NaCl are 129.8, 217.4 and 108.9 Ω−1

Question

Equivalence conductance at infinite dilution of NH4Cl, NaOH and NaCl are 129.8, 217.4 and 108.9 Ω−1 cm2 mol−1, respectively. If the equivalent conductance of 0.01 N solution of NH4OH is 9.532 Ω−1 cm2 mol−1, then the degree of dissociation of NH4OH at this temperature is

Accepted Answer

$\Lambda_{eq\left( NH_{4}OH \right)}^{o} = \Lambda_{eq\left( NH_{4}Cl \right)}^{o} + \Lambda_{eq(NaOH)}^{o} - \Lambda_{eq(NaCl)}^{o}$

= 129.8 + 217.4 – 108.9 = 238.3 ohm^–1 cm² mol^–1

Now, $\alpha = \frac{\lambda_{eq}}{\lambda_{eq}^{o}} = \frac{9.532}{238.3} = 0.04 = 4\%$

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