Question

Calculate the molar conductivity of NH₄OH at infinite dilution. λ°m of following are given.
1) Ba(OH)₂ = 523.3
2) BaCl₂ = 280
3) NH₄Cl = 130

Answer 1

${\Lambda}_{m}^{\circ}(NH_4OH)\\ =\lambda (NH_4^+)+\lambda (OH^-)\\ =\lambda (NH_4^+)+\lambda (Cl^-)\\ +\dfrac{1}{2}(\lambda (Ba^{2+})+2\lambda (OH^-))\\ -\dfrac{1}{2}(\lambda (Ba^{2+})+2\lambda (Cl^-))\\ ={\Lambda}_{m}^{\circ}(NH_4Cl)+\dfrac{1}{2}{\Lambda}_{m}^{\circ}(Ba(OH)_2)-\dfrac{1}{2}{\Lambda}_{m}^{\circ}(BaCl_2)\\ =130+\dfrac{1}{2}\times 523.3-\dfrac{1}{2}\times 280\\ =251.65 Scm^2mol^{-1}$