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 8.51-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+ (aq). The Sb^3+(aq) is completely oxidized by 34.5 mL of a 0.140 M aqueous solution of KBrO_3(aq). The unbalanced equation for the reaction is BrO_3^- (aq) + Sb^3+ (aq) rightarrow Br^-(aq) + Sb^5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

First we have to balance the equation. We do that by balancing each half-reaction separately, then combining them.

BrO3- ==> Br- . . .Add 3 H2O to the right side to balance O.
BrO3- ==> Br- + 3H2O . . .add 6H+ (acidic solution) to the left side to balance H.
BrO3- + 6H+ ==> Br- + 3H2O . . .add 6e- to the left side to balance the charge.
BrO3- + 6H+ + 6e- ==> Br- + 3H2O

Sb3+ ==> Sb5+ . . .Add 2e- to the right side to balance the charge.
Sb3+ ==> Sb5+ + 2e-

Multiply the Sb equation by 3 to give 6e- on the right side. Then add the new equation to the BrO3- equation; the 6e- will cancel.

.3Sb3+ ==> 3Sb5+ + 6e-
+BrO3- + 6H+ + 6e- ==> Br- + 3H2O
===================================
3Sb3+ + BrO3- + 6H+ ==> 3Sb5+ + Br- + 3H2O


moles BrO3- added = M BrO3- x L BrO3- = (0.140)(0.0345) = 0.00483 moles BrO3-

The balanced equation tells us that 1 mole of BrO3- reacts with 3 moles of Sb3+.

0.00483 moles BrO3- x (3 mole Sb3+ / 1 mole BrO3-) = 0.01449 moles Sb3+

From the periodic table, the molar mass of Sb (or Sb3+; the 3 missing electrons have very little effect on the mass ) = 121.8 g.

0.01449 moles Sb x (121.8 g Sb / 1 mole Sb) = 1.76488 g Sb

%Sb = (g Sb / g ore) x 100 = (1.76488 / 8.51) x 100 = 20.73 %Sb

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