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 5.41-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 29.2 mL of a 0.105 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

Explanation / Answer

6H+(aq) + BrO3¯(aq) + 3Sb3+(aq) ---> Br¯(aq) + 3Sb5+(aq) + 3H2O(l) moles of bromate in the titration = .155 x .0292 moles = 0.00452 moles these react with 3 x 0.005452 moles of Sb at the eq point = 0.0135 moles of Sb3+ Molar mass of Sb is 121.7g/mol so mass of Sb = .0135 x 121.7 g = 1.642 g % Sb = 1.642 / 5.41 = 0.303 = 30.3%

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