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Question

(A) According to Kohlrausch's law of independent migration of ions:$\Lambda_m^\infty(Ba(OH)_2) = \lambda^\infty(Ba^{2+}) + 2\lambda^\infty(OH^-) = 523

Vedclass · Accepted Answer

$251.4$

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(A) According to Kohlrausch's law of independent migration of ions:$\Lambda_m^\infty(Ba(OH)_2) = \lambda^\infty(Ba^{2+}) + 2\lambda^\infty(OH^-) = 523.28 \ \dots (i)$$\Lambda_m^\infty(BaCl_2) = \lambda^\infty(Ba^{2+}) + 2\lambda^\infty(Cl^-) = 280.0 \ \dots (ii)$$\Lambda_m^\infty(NH_4Cl) = \lambda^\infty(NH_4^+) + \lambda^\infty(Cl^-) = 129.8 \ \dots (iii)$We need to find $\Lambda_m^\infty(NH_4OH) = \lambda^\infty(NH_4^+) + \lambda^\infty(OH^-)$.Using the equations: $\Lambda_m^\infty(NH_4OH) = \Lambda_m^\infty(NH_4Cl) + \frac{1}{2}\Lambda_m^\infty(Ba(OH)_2) - \frac{1}{2}\Lambda_m^\infty(BaCl_2)$$\Lambda_m^\infty(NH_4OH) = 129.8 + \frac{523.28}{2} - \frac{280.0}{2}$$\Lambda_m^\infty(NH_4OH) = 129.8 + 261.64 - 140.0 = 251.44 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

Calculate the molar conductivity of $NH_4OH$ at infinite dilution,given that the molar conductivities of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ are $523.28$,$280.0$,and $129.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

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