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 5.51-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 23.2 mL of a 0.135 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3- + Sb3+ = Br- + Sb5+

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

Explanation / Answer

BrO3-: (2x3) + 1= +5 Br- = -1 reducting.

Sb: +3 to +5 oxidating.

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

(Sb3+ ------------> Sb5+ + 2e-)x3 = 3Sb3+ ------------> 3Sb+5 + 6e-

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

molecular mass of Sb: 121.76 g/mol molecular mass of KBrO3: 39.09 + 79.9 + 48 = 167 g/mol

moles of KBrO3 = 0.135 x 0.0232 = 3.132x10-3 moles

1:3 relation ratio. Moles of Sb = 3.132x10-3 moles / 3 = 1.044x10-3  moles of Sb

mass of Sb = 1.044x10-3 x 121.76 g/mol = 0.1271 g of Antimony in the sample

% = (0.1271/5.51)x100 = 2.31%

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