1. Tardigrade
  2. Question
  3. Chemistry
  4. The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is text0 text.005 textS textm text2 textmo textl text-1 and the limiting molar conductivity of HA is text0 text.005 textS textm text2 textmo textl text-1 at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka) of HA at this temperature is

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Q. The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is $ \text{0}\text{.005}\,\text{S}\,{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}} $ and the limiting molar conductivity of HA is $ \text{0}\text{.005}\,\text{S}\,{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}} $ at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant $ ({{K}_{a}}) $ of HA at this temperature is

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Solution:

$ {{\lambda }_{{{m}^{o}}}}=0.05S{{m}^{2}}mo{{l}^{-1}} $ C= 0.009 m $ {{\lambda }_{m}}=0.05S\,{{m}^{2}}mo{{l}^{-1}} $ $ \alpha =\frac{{{\lambda }_{m}}}{{{\lambda }_{{{m}^{o}}}}}=0.1 $ For weak acid, $ {{K}_{a}}=\frac{{{\alpha }^{2}}c}{1-\alpha }=\frac{{{(0.1)}^{2}}\times 0.009}{(1-0.1)}={{10}^{-4}} $