Question

Molar conductivity at infinite dilution for NH4Cl, NaOH and NaCl solution at 298K are respectively 129.8, 218.4 and 108.9 Scm2 mol-1 and ^m for 10-2 M solution of NH4OH is 9.33 Scm2 mol-1. Calculate the degree of dissociation of NH4OH

Answer 1

${\Lambda}_{m}^{0}(NH_4OH)={\lambda}(NH_4^+)+{\lambda}(OH^-)\\ ={\lambda}(NH_4^+)+{\lambda}(Cl^-)+{\lambda}(Na^+)+{\lambda}(OH^-)-{\lambda}(Na^+)-{\lambda}(Cl^-)\\ ={\Lambda}_{m}^{0}(NH_4Cl)+{\Lambda}_{m}^{0}(NaOH)-{\Lambda}_{m}^{0}(NaCl)\\ =129.8+218.4-108.9\\ =239.3 Scm^2mol^{-1}\\ \alpha =\dfrac{{\Lambda}_{m}(NH_4OH)}{{\Lambda}_{m}^{0}(NH_4OH)}\\ =\dfrac{9.33}{239.3}=0.039$

Answer 2

Answer: The degree of dissociation of $NH_4OH$ is 0.0389.

Explanation: To calculate the molar conductivity at infinite dilution of $NH_4OH$, we use Kohlrausch's Law.

$\Lambda ^o_{NH_4OH}=\Lambda ^o_{NH_4Cl}+\Lambda ^o_{NaOH}-\Lambda ^o_{NaCl}$

$\Lambda ^o_{NH_4Cl}=129.8Scm^2mol^{-1}\\\Lambda ^o_{NaOH}=218.4Scm^2mol^{-1}\\\Lambda ^o_{NaCl}=108.9Scm^2mol^{-1}$

Putting the values in above equation, we get

$\Lambda ^o_{NH_4OH}=129.8+218.4-108.9\\\Lambda^o_{NH_4OH}=239.9Scm^2mol^{-1}$

We are given molar conductivity of $NH_4OH$,

$\Lambda^m_{NH_4OH}=9.33Scm^2mol^{-1}$

Now, to calculate the degree of association, we use the formula:

$\alpha =\frac{\Lambda ^m_{NH_4OH}}{\Lambda ^o_{NH_4OH}}$

Putting the values in above equation, we get

$\alpha =\frac{9.33}{239.9}\\\alpha =0.0389$