Lambda _m^circ for NaCl,HCl and NaAc are 126. | vedclass

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(A) According to Kohlrausch's law of independent migration of ions:$\Lambda _{m(HAc)}^\circ = \lambda _{H^{+}}^\circ + \lambda _{Ac^{-}}^\circ$We can

Vedclass · Accepted Answer

$390.5 \ S \ cm^2 \ mol^{-1}$

Vedclass · Answer

(A) According to Kohlrausch's law of independent migration of ions:$\Lambda _{m(HAc)}^\circ = \lambda _{H^{+}}^\circ + \lambda _{Ac^{-}}^\circ$We can express this in terms of the given values:$\Lambda _{m(HAc)}^\circ = (\lambda _{H^{+}}^\circ + \lambda _{Cl^{-}}^\circ) + (\lambda _{Na^{+}}^\circ + \lambda _{Ac^{-}}^\circ) - (\lambda _{Na^{+}}^\circ + \lambda _{Cl^{-}}^\circ)$$\Lambda _{m(HAc)}^\circ = \Lambda _{m(HCl)}^\circ + \Lambda _{m(NaAc)}^\circ - \Lambda _{m(NaCl)}^\circ$Substituting the given values:$\Lambda _{m(HAc)}^\circ = 425.9 + 91.0 - 126.4$$\Lambda _{m(HAc)}^\circ = 390.5 \ S \ cm^2 \ mol^{-1}$

$\Lambda _m^\circ$ for $NaCl$,$HCl$ and $NaAc$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda _m^\circ$ for $HAc$.

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