1
$\begingroup$

$15~ \textrm{mL}$ of $4 ~\textrm{N}$ hydrochloric acid, $ 25~ \textrm{mL}$ of $2~ \textrm{M}$ $\textrm{H}_3\textrm{PO}_3$ and a certain volume of $2.5~\textrm{M}$ sulphuric acid are mixed together and made upto $2 ~\textrm{L}$.

$30~ \textrm{mL}$ of this acidic mixture exactly neutralises $42.9~\textrm{mL}$ of a sodium carbonate solution which contains $0.1~ \textrm{g}$ of sodium carbonate decahydrate in $10 ~\textrm{mL}$ of solution. The volume of $\textrm{H}_2\textrm{SO}_4$ added is?

I calculated the number of gram equivalents of sodium carbonate and thus calculated the normality of the acidic solution required.

I am having trouble with the statement "is made up to", so I'm assuming that this means that the solution is diluted with water to make a volume of 2 litres.

This seems to be yielding a negative volume. Can you tell me what I'm doing wrong?

$\endgroup$
  • $\begingroup$ The answer will be 8 ml. Deal carefully with molarity & normality ; moles and equivalents. Also observe which are dibasic acid and the sodium carbonate is diacidic base. You will get the result. $\endgroup$ – Soumik Das Jun 8 '18 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.