Question

What volume of 2.50 M NaOH is required to neutralize 85.0 mL of 1.35 M H2SO4

Answer 1

Sol.

Reaction :

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

So , at equivalence point , Millimoles of NaOH = 2 × Millimoles of H2SO4 = 2 × Conc. of H2SO4 × Volume of H2SO4 = 2 × 1.35 × 85.0 = 229.5 mmol

Now , Conc. of NaOH = 2.50 M

So , Volume of NaOH = Millimoles of NaOH / Conc. of NaOH

= 229.5 / 2.50

= 91.8 mL

Therefore , Volume of NaOH required is  91.8 mL