Lambda_m^0(NH_4OH) is equal to: | vedclass

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(B) According to Kohlrausch's law of independent migration of ions:$\Lambda_m^0(NH_4OH) = \lambda^0(NH_4^+) + \lambda^0(OH^-)$To obtain this, we use s

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$\Lambda_m^0(NH_4Cl) + \Lambda_m^0(NaOH) - \Lambda_m^0(NaCl)$

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(B) According to Kohlrausch's law of independent migration of ions:$\Lambda_m^0(NH_4OH) = \lambda^0(NH_4^+) + \lambda^0(OH^-)$To obtain this, we use strong electrolytes:$\Lambda_m^0(NH_4Cl) = \lambda^0(NH_4^+) + \lambda^0(Cl^-)$$\Lambda_m^0(NaOH) = \lambda^0(Na^+) + \lambda^0(OH^-)$$\Lambda_m^0(NaCl) = \lambda^0(Na^+) + \lambda^0(Cl^-)$Adding the first two and subtracting the third:$\Lambda_m^0(NH_4Cl) + \Lambda_m^0(NaOH) - \Lambda_m^0(NaCl) = (\lambda^0(NH_4^+) + \lambda^0(Cl^-)) + (\lambda^0(Na^+) + \lambda^0(OH^-)) - (\lambda^0(Na^+) + \lambda^0(Cl^-))$$= \lambda^0(NH_4^+) + \lambda^0(OH^-) = \Lambda_m^0(NH_4OH)$.

$\Lambda_m^0(NH_4OH)$ is equal to:

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