The quantity of antimony in a sample can be determined by an oxidation-reduction

Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 8.33-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 33.8 mL of a 0.140 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3-(aq) + Sb3+(aq) --> Br-(aq) + Sb5+(aq)

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

Explanation / Answer

First we balance the equation,

BrO3-(aq) + Sb3+(aq) --> Br-(aq) + Sb5+(aq)

Half reactions are,

3Sb3+ ------> 3Sb5+ + 6e-

BrO3- + 6H+ + 6e- ------> Br- + 3H2O

Overall reaction,

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

According to the stoichiometry of the reaction 3 moles of Sb reacts with 1 mole of BrO3-

Given, The Sb3+ (aq) is completely oxidized by 33.8 mL of a 0.140 M aqueous solution of KBrO3(aq)

Moles = Molarity x Volume (L)

=> Moles of BrO3- = 0.14 x 0.0338 = 0.004732 moles

=> Moles of Sb3+ required = 3 x 0.004732 = 0.0142 moles

Molar Mass of Sb = 121.76 g / mol

=> Mass of Sb3+ required = 0.0142 x 121.76 = 1.7285 g = Amount of antimony in the sample in grams

% Antimony in ore = 1.7285 x 100 / 8.33 = 20.75 %

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