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 5.67-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 23.8 mL of a 0.135 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

Solution :-

Lets first write the balanced reaction equation.

BrO3-(aq) + Sb3+(aq) -----> Br- (aq) + Sb5+(aq)

Balancing reaction by half reaction method

BrO3-(aq)   --- > Br- (aq)                             reduction

Sb3+(aq) -----> Sb5+(aq)                                 oxidation

Lets balance the oxygens by adding the H2O to the deficient side

BrO3-(aq)   --- > Br- (aq) +3H2O(l)

Sb3+(aq) ----> Sb5+(aq)

Now lets balance hydrogens by adding H+ to the deficient side

6H+(aq) + BrO3-(aq)   --- > Br- (aq) +3H2O(l)

Sb3+(aq) -----> Sb5+(aq)

Now lets balance the electronic charges by adding electrons to the deficient side

6e- + 6H+(aq) + BrO3-(aq)   --- > Br- (aq) +3H2O(l)

Sb3+(aq) -----> Sb5+(aq)+ 2e-

Now multiply the equations by such numbers so that the electrons in the both equation get balanced.

So multiply equation 1 by 1 and multiply equation 2 by 3 and then add both equations to get the final equation.

Then we get

6e- + 6H+(aq) + BrO3-(aq)   --- > Br- (aq) +3H2O(l)

3Sb3+(aq) -----> 3Sb5+(aq)+ 6e-

___________________________________________

6H+(aq) + BrO3-(aq) + 3Sb3+(aq) ----> Br- (aq) +3H2O(l) + 3Sb5+(aq)

So the balanced reaction equation is as follows

6H+(aq) + BrO3-(aq) + 3Sb3+(aq) -----> Br- (aq) +3H2O(l) + 3Sb5+(aq)

Now lets calculate the moles of the KBrO3 using its molarity and volume

Moles = molarity * volume in liter

Moles of KBrO3 = 0.135 mol per L * 0.0238 L                              (23.8 ml = 0.0238 L)

                            = 0.003213 mol KBrO3

Now using the mole ratio of the KBrO3 and Sb3+ lets calculate the moles of the Sb3+

Mole ratio is 1 :3 for KBrO3 to Sb3+

Therefore moles of Sb3+ are calculated as

(0.003213 mol KBrO3 * 3 mol Sb3+ )/ 1 mol KBrO3 = 0.009639 mol Sb3+

Now lets convert moles of Sb3+ to its mass

Mass = moles * molar mass

Mass of Sb3+ = 0.009639 mol * 121.76 g per mol

                       = 1.1736 g Sb3+

So the sample of the mass of antimony present in the stibnite sample = 1.1736 g

Now lets calculate the percentage of the Antimony

% antimony = (mass of antimony / mass of stibnite) *100%

                      = (1.1736 g / 5.67 g)*100%

                      = 20.70 % Antimony (Sb3+)

So the mass of antimony = 1.1736 g

Percent of antimony = 20.70 %

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