Question

Given the limiting molar conductivities of NaCl, HCl and NaAc are 120.4, 425.9 and 91

Scm2mol-1

. What should be that value for HAc​

Answer 1

Answer:

Limiting molar conductivity for HAc = 396.5 S cm² mol⁻¹

Explanation:

Given:

$\sf \Lambda^0\:for\:NaCl=120.4\:S\:cm^2\:mol^{-1}$

$\sf \Lambda^0\:for\:HCl=425.9\:S\:cm^2\:mol^{-1}$

$\sf \Lambda^0\:for\:NaAc=91\:S\:cm^2\:mol^{-1}$

To Find:

$\sf \Lambda^0\:for\:HAc$

Solution:

According to Kohlrausch's law we know that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of cations and anions of the electrolyte.

Therefore from given,

$\sf \Lambda^0\:for\:NaCl=\lambda_{Na^+}^0+\lambda_{Cl^-}^0=120.4\:S\:cm^2\:mol^{-1}---(1)$

$\sf \Lambda^0\:for\:HCl=\lambda_{H^+}^0+\lambda_{Cl^-}^0=425.9\:S\:cm^2\:mol^{-1}---(2)$

$\sf \Lambda^0\:for\:NaAc=\lambda_{Na^+}^0+\lambda_{Ac^-}^0=91\:S\:cm^2\:mol^{-1}---(3)$

Now we need to find the limiting molar conductivity of HAc, that is,

$\sf \Lambda^0\:for\:HAc=\lambda_{H+}^0+\lambda_{Ac^-}^0$

Now add equations 2 and 3,

$\sf \lambda_{H^+}^0+\lambda_{Cl^-}^0+\lambda_{Na^+}^0+\lambda_{Ac^-}^0=425.9\:S\:cm^2\:mol^{-1}+91\:S\:cm^2\:mol^{-1}$

$\sf \lambda_{H^+}^0+\lambda_{Cl^-}^0+\lambda_{Na^+}^0+\lambda_{Ac^-}^0=516.9\:S\:cm^2\:mol^{-1}---(4)$

Now subtract equation 1 from equation 4,

$\sf \lambda_{H^+}^0+\lambda_{Cl^-}^0+\lambda_{Na^+}^0+\lambda_{Ac^-}^0-\lambda_{Na^+}^0-\lambda_{Cl^-}^0=516.9\:S\:cm^2\:mol^{-1}-120.4\:Sc\:cm^2\:mol^{-1}$

$\sf \lambda_{H^+}^0+\lambda_{Ac^-}^0=396.5\:S\:cm^2\:mol^{-1}$

$\implies \sf \Lambda^0\:for\:HAc$

Therefore the limiting molar conductivity of HAc is 396.5 S cm² mol⁻¹