Map General Chemistr sity Science Books presented by Sapling Lea Donald McQuarri

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Map General Chemistr sity Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 5.85-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 31.6 mL of a 0.105 M aqueous solution of KBro3(aq). The unbalanced equation for the reaction is Broa (aq) sbs (aq) Br (aq) sbs+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore Number Number

Explanation / Answer

Let us balance the given redox transformation,

BrO3- (aq.) + Sb3+ (aq.) --------> Br (aq.) + Sb5+ (aq.) ……………. (Unbalanced)

OHR: Sb3+ (aq.) ---------> Sb5+ (aq.)

RHR: BrO3- (aq.) ------------> Br (aq.)

Steps for balancing,

i) Balancing of atoms other than O and H,

OHR: no need

RHR: no need.

ii) Balancing of O and H,

OHR: no need

RHR: BrO3- (aq.) + 6H+ (aq.) ------------> Br (aq.) + 3H2O

iii) Balancing of charge,

OHR: Sb3+ (aq.) ---------> Sb5+ (aq.) + 2e-

RHR: BrO3- (aq.) + 6H+ (aq.) + 6 e- ------------> Br (aq.) + 3H2O

iv) Balancing of electron number,

OHR: needed to be multiplied by 3

OHR: 3Sb3+ (aq.) ---------> 3Sb5+ (aq.) + 6e-

v) Adding OHR and RHR,

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

6 electrons cancelled mutually then,

3Sb3+ (aq.) + BrO3- (aq.) + 6H+ (aq.) ---------> 3Sb5+ (aq.) + Br (aq.) + 3H2O …………(Balanced redox transformation)

From stoichiometry of above balanced redox titration it’s clear that,

1 equivalent of Sb3+ = 3 equivalent of BrO3- (aq.) i.e. KBrO3

At equivalence point,

[Sb3+] = 3 [BrO3] …………. (1)

In redox titration at equilibrium,

For BrO3- i.e. KBrO3; V1 = 31.6 mL and M = 0.105 M

Hence, milimoles of KBrO3 = V1 x M1= 31.6 x 0.105 = 3.318 milimoles.

Hence using eq. (1),

[Sb3+] = 3 x 3.318 milimoles

[Sb3+] = 9.954 milimoles.

Let the volume of solution containing Sb3+ be 1000 mL and molarity be M

Then,

Milimoles of Sb3+ = 1000 x M = 9.954 milimoles

Hence, M = 9.954/1000 = 9.954 x 10-3 M.

Atomic mass of Sb = 121.76 g

Hence,

Mass of Sb in stibnite sample = Molarity x Atomic mass of Sb

Mass of Sb in stibnite sample = 9.954 x 10-3 x 121.76

Mass of Sb in stibnite sample = 1.212 g.

Amount of Sb in stibnite sample = 1.212 g.

Now,

Amount of stibnite sample = 5.85 g. and amount of Sb = 1.212g

Then,

% Sb amount = [(Amount of Sb) / (Amount of Stibnite)] x 100

% Sb amount = (1.212/5.85)x100

% Sb amount = 20.72 %

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