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 6.55-g sample of stibnite, antimony, is dissolved in hot, concentrated HCl (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 31.0 mL of a 0.120 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 let us balance the equation.

BrO3- ------> Br- ie +5 ------> -1

therefore we need 6 electrons and to take up 3 oxygens we need 6H+

First Half equation is BrO3- + 6H+ + 6e- ------> Br- +3H2O

Sb3+ -----> Sb5+ ie +3 -----> +5

Second Half equation is Sb3+ -----> Sb5+ +2e- or 3Sb3+ -----> 3Sb5+ +6e

Overall reaction is BrO3- + 6H+ + 3Sb3+- ------> 3Sb5+ + Br- +3H2O

number of moles of BrO3- = concentration X Volume = 0.120 g/L X 0.031 L = 0.00327 mol

From the overall reaction we find that BrO3- and Sb3+ are present in 1:3 ratios

number of moles of Sb3+ = 3 X number of moles of BrO3- =  0.01116 mol

amount of Sb = number of moles of Sb3+ X Molar mass of Sb = 0.01116 mol X 121.76 g mol-1 = 1.3588 g

NB: Molar mass of Sb3+ and Sb will be nearly same as 3 missing electrons do not have large effect on molar mass.

%Sb in the ore = amount of Sb/ total amount of ore X 100 = 1.3588/ 6.55 X 100 = 20.74 %

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