When 25.0 ml of sulfuric acid solution was completely neutralized in a titration with 0.05 m naoh solution, it took 18.3 ml of the naoh(aq) to complete the job. the reaction is:
naoh(aq) + h2so4(aq) → na2so4(aq) + h2o(l)
what was the molarity of the sulfuric acid solution?

Balanced equation will be as follows

2NaOH + H2SO4=Na2SO4 + 2H2O
from MaVa/MbVb=Na/Nb    where Ma is molarity of acid=?
                                                   Va is volume of acid=25.0 mL
                                                   Mb is molarity of base= 0.05 M
                                                   Vb is volume of base=18.3 mL
                                                    Nb number of moles of base= 2
                                                     Na is number of moles of acid =1

Ma*25 mL / 18.3mL*0.05M = 1/2
Ma=18.3*0.05M / 50 = 0.0183M

The molarity of sulphuric acid = 0.0183M

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