What is the molarity of the sodium hydroxide sample? Using the data provided.

Question

Balanced equation: 2 NaOH + H2SO4 ‐‐> Na2SO4 + 2H2O 

Molarity Experiment

A neutralization reaction was performed by reacting sodium hydroxide with sulfuric acid (H₂SO₄) and the following data was taken.

Volume of sodium hydroxide sample	25.0 mL
Volume of sulfuric acid added	28.2 mL
Concentration of sulfuric acid	0.355 M

Answer 1

The molarity of H2SO4  used used is 0.355 M which is equivalent to 0.355 mol/L x 1L/1000 mL = 3.55 x 10^-4 mol/mL. As 28.2 mL H2SO4 was used, the number of moles H2SO4 added was 28.2 mL x 3.55 x 10^-4 mol/mL

= 1.001 x 10^-2 moles H2SO4 added.

The balanced equation shows that 2 moles of NaOH is needed for every mole of H2SO4 used or moles NaOH

= 2.002 x 10^-2 moles NaOH in 25.0 mL.

The molarity of NaOH is 2.002 x 10^-2 moles x 1000 mL/L / 25.0 mL = 0.801 Molar.