Determine the volume in milliliters of a 0.260 M NaOH solution needed to neutralize a 295 mL solution of 0.173 M HCl and 0.346 M H2SO4.

Answer 1

This problem is a little different from most because you are neutralizing TWO acids (HCl and H2SO4) instead of just one acid. The concepts are the same, but we have to look at the total number of hydrogens (H+) that are reacting.

HCl + NaOH ==> NaCl + H2O .. ONE hydrogen reacts from each HCl

H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H2O .. TWO hydrogens react from each H₂SO₄

So, first we will determine TOTAL moles of hydrogens reacting:

For HCl: 295 ml x 1 L / 1000 ml x 0.173 mol / L = 0.05104 mols HCl x 1 H/mol = 0.05104 mols H⁺

For H₂SO₄: 295 ml x 1L/1000 ml x 0.346 mol/L = 0.1021 mol H₂SO₄ x 2H/mol = 0.2041 mol H⁺

Total moles H⁺ = 0.05104 + 0.2041 = 0.2551 mols H⁺

It takes ONE mole of NaOH to react with each mole of H⁺: OH^- + H⁺ ==> H2O

moles NaOH needed = 0.2551 mols H+ x 1 mol OH- / mol H+ = 0.2551 mols NaOH needed

Volume NaOH = 0.2551 mols NaOH x 1 L / 0.260 mols = 0.981 L = 981 mls