How many mL of 1.00 M sulfuric acid are required to neutralize 25.0 mL of 4.00 M NaOH?

Question

How many mL of 1.00 M sulfuric acid are required to neutralize 25.0 mL of 4.00 M NaOH?

(start by writing the balanced equation for this acid-base reaction)

Answer 1

Write the correctly balanced equation:

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O ... balanced equation

moles of NaOH present = 25.0 ml x 1 L/1000 ml x 4.00 mol/L = 0.1 moles NaOH

moles H2SO4 needed = 0.1 moles NaOH x 1 mol H2SO4/2 mol NaOH = 0.05 moles H2SO4 needed

volume H2SO4 needed: (x L)(1.00 mol/L) = 0.05 moles

x = 0.05 L = 50 ml of H2SO4 needed