A 8.3 mL sample of 0.180 M sulfuric acid (H2SO4) is titrated with 0.250 M NaOH Calculate the volume, in milliliters, of NaOH solution that will be used.

Answer 1

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

moles H2SO4 present = 8.3 ml x 1 L / 1000 ml x 0.180 mol / L = 0.001494 mols H2SO4

moles NaOH needed = 0.001494 mol H2SO4 x 2 mol NaOH / mol H2SO4 = 0.002988 mol NaOH

Volume of 0.250 M NaOH:

x L (0.250 mol/L) = 0.002988 mols

x = 0.011952 L

x = 11.95 mls = 12.0 mls (3 sig. figs.)