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 5.97-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl and passed over a reducing agent so that all the antimony is in the form Sb3+. The Sb3+ is completely oxidized by 30.8 mL of a 0.110 M aqueous solution of KBrO3. The potential of the solution is being measured using a platinum wire and a S.H.E standard elcetrode.

The unbalanced equation for the reaction is: BrO3- + Sb3+ -----> Br- + Sb5+

a) Calculate the potential of the solution after 15 mL of KBrO3 is added

b) Calculate the potential after 45.8 mL of KBrO3 is added

c) Calculate the amount of antimony in the sample and its percentrage in the ore.

Explanation / Answer

Two important points to note:

1) Before the equivalence point, the potential of the system, Esys is calculated by the Sb3+/Sb5+ system.

2) After the equivalence point, Esys is calculated by BrO3-/Br- system.

The two half reactions are:

Sb5+ + 2e- <===> Sb3+ , E0Sb = +0.818 V

BrO3- + 6 H+ + 6e- <===> Br- + 3 H2O, E0BrO3 = +0.61 V

The balanced equation is

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

The reactions are fully reversible and for this to happen,

Esys = ESb = EBrO3

(c) We shall calculate the answer to part (c) first.

Molar ratio of BrO3- and Sb5+ is 1: 3. The amount of BrO3- added is (0.0308 L)(0.110 mole/L) = 0.003388 mole

The amount of Sb5+ that have been neutralized is (3 mole Sb5+)/(1 mole BrO3-)*0.003388 mole = 0.010164 mole.

Molar mass of antimony is 121.76 gm/mole.

Therefore, amount of Sb present in the ore is (0.010164 mole)*121.76 gm/mole = 1.2371.24 gm (ans)

The percentage of Sb in the ore is (1.24 gm/ 5.97 gm)*100 = 20.72920.73 (ans)

Now we answer parts (a) and (b).

(a) 15 mL of BrO3- is added; moles of bromate added = (0.015 L)(0.110 mole/L) = 0.00165 mole.

Let total volume of the solution be V1 L.

[Sb3+] = 0.00165 mole/V1 L = 0.00165/V1 M

[Sb5+] = (0.010164 – 0.00165) mole/V1 L = 0.008514/V1 M

Esys = ESb = E0Sb – 0.0591/2*log10[Sb3+]/[Sb5+] = 0.818 V – 0.02955V*log100.00165/0.008514= {0.818 – 0.02955*(- 0.7126)}V = (0.818 + 0.0210)V = 0.839 V (ans)

(b) In this case, we have an excess of bromate; the molar concentration of bromide will be equal to the concentration of Sb5+ at the end point (since bromide is formed by the neutralization reaction).

Let the total volume of the solution be V2 L.

[Br-] = 0.010164 mole/ V2 L = 0.010164/V2 M

[BrO3- ] = (0.0458 – 0.0308 L)(0.110 mole/L)/V2 L = 0.00165/V2 L

Esys = EBrO3 = E0BrO3 – 0.0591/1*log10[Br-]/[BrO3-] = 0.61 V – 0.0591 V*log100.010164/0.00165 = (0.61 – 0.0591*0.789)V = (0.61 – 0.0467)V = 0.5633 V (ans).

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