If a 0.250 g sample of zinc oxide (95%) was treated with 50 ml of 1.123 N sulfuric acid,

Question

what volume of 0.977 N sodium hydroxide would be required in the residual titration?

Answer 1

Reaction 1: ZnO + H₂SO₄ ==> ZnSO₄ + H₂O

moles ZnO present = 0.250 g ZnO x 1 mol ZnO / 81.38 g = 0.003072 mols x 0.95 = 0.002918 mols ZnO

moles H₂SO₄ used = 50 ml x 1 L / 1000 ml x 0.5615 mol / L = 0.0281 mols H₂SO₄

(note: 1.123 N H₂SO₄ x 1/2 = 0.5615 M H₂SO₄ = 0.5615 mol/L)

According to the balanced equation, 0.002918 mols ZnO would require 0.002198 mols H₂SO₄ (1:1).

But 0.0281 mols H₂SO₄ were added. This leaves an excess H₂SO₄ of 0.0281 - 0.002198 = 0.0259 mols

Reaction 2: H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O

moles H₂SO₄ = 0.0259 mols

moles NaOH needed = 0.0259 mol H₂SO₄ x 2 mol NaOH/mol H₂SO₄ = 0.0518 mols NaOH needed

volume NaOH needed = 0.0518 mols x 1 L/0.977 mols = 0.0530 L = 53 mls