Question

Determine the volume of a 0.125 M NaOH solution needed to react with 35.0 mL of a 0.623 M sulfuric acid solution. SHOW WORK.

H₂SO₄(aq) + 2NaOH(aq) ---> Na₂SO₄(aq) + 2H₂O(l)

Answer 1

neutrilization reaction between NaOH and H₂SO₄ is

H₂SO_4(aq) + 2NaOH_(aq)   Na₂SO_4(aq) + 2H₂O_(l)

no. of mole = molarity volume of solution in liter

no. of mole of H₂SO₄ = 0.623 0.035 = 0.021805 mole

According to reaction 1 mole of H₂SO₄ react with 2 mole of NaOH therefore to react with 0.021805 mole of H₂SO₄

required mole of NaOH = 0.021805 2 = 0.04361 mole

volume of solution in liter = no. of mole / molarity

volume of solution of NaOH = 0.04361 / 0.125 = 0.34888 liter = 348.88 ml

348.88 ml NaOH required.