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 9.09-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 Sb3 (aq). The Sb3*(aq) is completely oxidized by 46.9 mL of a 0.110 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is H+(aq) + Bro3 (aq) + Sb3 + (aq) Br-(aq) + Sb5+ (aq)+H2O(1) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number 0

Explanation / Answer

balance:

BrO3- = Br-

Sb+3 = Sb+5

balance O adding H2O

BrO3- = Br- + 3H2O

Sb+3 = Sb+5

balnac eH adding H+

6H+ + BrO3- = Br- + 3H2O

Sb+3 = Sb+5

balance charges

5e- + 6H+ + BrO3- = Br- + 3H2O

Sb+3 = Sb+5 + 2e-

balance e-

10e- + 12H+ + 2BrO3- = 2Br- + 6H2O

5Sb+3 = 5Sb+5 + 10e-

add all

5Sb+3 + 10e- + 12H+ + 2BrO3- = 2Br- + 6H2O + 5Sb+5 + 10e-

cancel common terms

5Sb+3 + 12H+ + 2BrO3- = 2Br- + 6H2O + 5Sb+5

now..

m = 9.09 g of sampl...

mol of KBrO3 used =MV = 0.11*46.9*10^-3 = 0.005159

raito is

2 mol fo BrO3- = 5 mol fo Sb+5

0.005159 mol --> 5/2*0.005159 = 0.0128975 mol of Sb

mass = mol*MW = 0.0128975*121.76 = 1.570 g of Sb

%Sample =mass of Sb/Ttoal mass * 100% = 1.570 / 9.09*100 = 17.271 %

Q2.

Oxidizing agent --> will be reduced READILY, recall that reduction is gain of e-

then..

worst --> Na+, then Fe+2, then I2 finally, the strongest, F2

F2

I2

Fe+2

Na+

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