about molarity question I want details explanation thx..

Question

In an experiment ,1.00M sodium hydroxide solution was added to 25.0cm^3 of 1.00M sulphuric acid until the acid was completely neutralized.What is the concentration of sodium sulphate in the resulting solution?

Answer 1

2NaOH + H2SO4 → Na2SO4 + 2H2O

Mole ratio NaOH : H2SO4 = 2 : 1
No. of moles of H2SO4 used = MV = 1 x (25 x 10-3) = 0.025 mol
No. of moles of NaOH used = 2 x 0.025 = 0.05 mol
Volume of NaOH used = (No. of moles) / M = 0.05 / 1 = 50 cm3

Mole ratio H2SO4 : Na2SO4 = 1 : 1
No. of moles of H2SO4 used = 0.025 mol
No. of moles of Na2SO4 formed = 0.025 mol
Volume of final solution = 50 + 25 = 75 cm3
Concentration of Na2SO4 = (No. of moles) / Volume = 0.025 / (75 x 10-3) = 0.333 mol dm-3