Answer to Question #171829 in Organic Chemistry for imy

Question

If a 1.25 g sample zinc oxide, 95% ZnO, were treated with 50mL of 1.1230N sulfuric acid

in the usual way, what volume of 0.9765N sodium hydroxide would be required in the

back titration?

Answer 1

ZnO + H₂SO₄ = ZnSO₄ + H₂O

n(H2SO4) = n(ZnO) = 0.95*(1.25 g)/(81.38 g/mol) = 0.0146 mol

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

Molarity of H₂SO₄ if half its normality, so

n(NaOH) = 2*n(H₂SO₄)_left = 2*((0.05 L)*(1.1230/2 M) - 0.0146 mol) = 0.02695 mol

Molarity of NaOH if the same as its normality, so

V(NaOH) = (0.02695 mol)/(0.9765 M) = 0.0276 L = 27.6 mL