# Stoichiometry yielding negative volume

$$15~ \textrm{mL}$$ 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?

• 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. – Soumik Das Jun 8 '18 at 13:34