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  3. Chemistry
  4. A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is:

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Q. A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is:

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Solution:

For weak electrolytes, according to Ostwalds dilution law $ \alpha =\sqrt{K.V} $ here, $ \alpha =0.01%=0.0001=1\times {{10}^{-4}} $ $ V=\frac{1}{C}=\frac{1}{1.0}=1litre $ $ \therefore $ $ {{K}_{a}}=\frac{{{\alpha }^{2}}}{V}=\frac{{{(1\times {{10}^{-4}})}^{2}}}{1} $ $ =1\times {{10}^{-8}} $