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 8.79-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCI(aq) and passed over a reducing agent so that all the antimony is in the form Sb^3+(ag). The Sb^3+(aq) is completely oxidized by 38.4 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is H^+ (aq) + BrO^- 3(aq) + Sb^3+ (aq) right arrow Br^-(aq) + Sb^5+(aq) + H_2O(l) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

6H+ + BrO3^- + 3Sb^+3 ................> Br^- + 3Sb^+5 + 3H2O

According to the balanced equation

1 mole of BrO3- is required 3mole of Sb^+3

no of moles of BrO3- = Molarity *vol. of the sol. in lt = 0.13*0.0384 = 4.99*10^-3 moles

so 4.99*10^-3 moles of BrO3- can oxidises 3*4.99*10^-3 moles od Sb^+3

so amount of Sb^+3 = no.of moles *(mol.wt of SbCl3)

                              = 3*4.99*10^-3*228 = 3.413 gm.

% of Sb^+3 in sample = (3.413/8.79)*100 = 38.8 % or 39%

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