The volume of 0.1 N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH− in aqueous solution is:

  1. 200 mL
  2. 400 mL
  3. 600 mL
  4. 800 mL

Since it is a dibasic acid, it has two ionisable hydrogen atoms and hence its n-factor is 2. I simply used $$V\cdot 0.1\cdot 2=0.04$$ and got volume as 200 mL. But however according to the answer key it is given as 400 mL. Now the answer key is not of the highest quality and often has errors, so it can be challenged (but you need to pay a decent sum of money for the chance that they may look into it). Am I right to challenge this question or am I missing something?

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1 Answer 1


Unfortunately you are missing something. The concentration is already given as an equivalence concentration, thus taking the dibasic nature of the acid into account. Therefore you have been given $c(\ce{H+})=0.1~\mathrm{mol\,L^{-1}}$, not $c(\ce{H2A})=0.1~\mathrm{mol\,L^{-1}}$. In the latter case your calculation would be correct.

The use of normality is ambiguous and it is not recommended to be used any more.


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