Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 7.51-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3 (aq). The Sb3 (aq) is completely oxidized by 37.1 mL of a 0.115 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3-(aq)+Sb+3(aq)-->Br-(aq)+Sb+5(aq) (unbalanced)

Calculate the amount of antimony in the sample and its percentage in the ore.

Answer 1

BrO3- + 3 Sb3+ + 6 H+ -----> Br- + 3 Sb5+ + 3 H2O

(0.0371 L) x (0.115 mol/L BrO3-) x (3 mol Sb / 1 mol BrO3-) x (121.760 g Sb/mol) = 1.55 g Sb

(1.55 g Sb) / (7.51 g) = 0.207 = 20.7% Sb