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 9.33-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 50.4 mL of a 0.105 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

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

use oxidation numbers to balance this equation.

Consider first the bromate(V) ion BrO3. It is converted to Br

BrO3Br

+51

Bromine has changed from +5 to -1. This will require the addition of 6 electrons:

BrO3+6e6Br

To take account of 3 oxygens we will need 6H+ ions to form water:

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

This is our first half - equation.

Now for the second:

Sb3+Sb5+

+3+5

Antimony has gone from +3 to +5. This means it needs to give out 2 electrons:

Sb3+Sb5++2e

The problem here is that the electrons don't balance.

Multiply by 3

3Sb3+3Sb5++6e

Now we can add both sides of each equation together

BrO3+6H+ +6e+3Sb3+Br+3H2O+3Sb5++6e

You can see here that the 6 electrons have now cancelled from both sides to give us:

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

1 mole of BrO3- reacts with 3 mole of Sb3+

50.4 mL of 0.105 M of BrO3- is completely oxidize Sb3+

no. of moles of BrO3- = 0.105*50.4 mL/1000 = 0.0053 mol

which is eqivalent to 3 moles of Sb3+ = 3 * 0.0053 mol =0.0159 moles

thus, no. of moles of Sb3+ present in ore = 0.0159 g * 121 g/mol = weight of Sb3+ =1.9239 g

weight of Sb3+ present in ore = 1.9239 g

weight of ore = 9.33 g

% of Sb = weight of Sb3+ * 100 / weight of ore = 1.9239 * 100 / 9.33 = 20.6%

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