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Question: A 10g mixture of \(C{u_2}S\) and CuS was treated with 200mL of 0.75M \(Mn{O_4}^ - \) in acid solutio...

A 10g mixture of Cu2SC{u_2}S and CuS was treated with 200mL of 0.75M MnO4−Mn{O_4}^ - in acid solution producing SO2,Mn2+,Cu2+S{O_2},M{n^{2 + }},C{u^{2 + }}. The SO2S{O_2} was boiled off and the excess of MnO4−Mn{O_4}^ - was titrated with 175mL of 1M Fe2+F{e^{2 + }} solution. Calculate the percentage of CuS in the original mixture (To the nearest integer).

Explanation

Solution

First find the amount of MnO4−Mn{O_4}^ - solution used against 1M Fe2+F{e^{2 + }} solution. Then find the number of electrons involved in oxidation and reduction reactions. Use mole equivalents to calculate the composition of the given mixture.

Complete answer:
We will first find the amount of 0.75M MnO4−Mn{O_4}^ - solution that was titrated with a given amount of 1MFe2+F{e^{2 + }} solution. Then, we will find the used mole equivalents of MnO4−Mn{O_4}^ - for the reaction with the given mixture of compounds. From it, we will find the % of CuS in the mixture.
- We are given that 200mL of 0.75M MnO4−Mn{O_4}^ - solution was added to the mixture. We know that in this compound, manganese undergo reduction as below.
Mn7+→Mn2++5e−M{n^{7 + }} \to M{n^{2 + }} + 5{e^ - }
Thus, we can see that 5 electrons are produced in the reaction.
So, we can say that number of mole equivalents of MnO4−Mn{O_4}^ - present in the solution = Volume ×\times Concentration ×\times Number of electrons
Mole equivalents of MnO4−Mn{O_4}^ - = 200×0.75×5=750200 \times 0.75 \times 5 = 750
Now, it is given that the remaining MnO4−Mn{O_4}^ - is reacted with Fe2+F{e^{2 + }} solution. We know that Fe2+F{e^{2 + }} undergoes oxidation in the following way.
Fe2+→Fe3++e−F{e^{2 + }} \to F{e^{3 + }} + {e^ - }
We can say that number of mole equivalents of Fe2+F{e^{2 + }} = Volume ×\times Concentration ×\times Number of electrons
Thus, mole equivalents of Fe2+F{e^{2 + }} = 175×1×1=175175 \times 1 \times 1 = 175
So, we can say that number of mole equivalents of MnO4−Mn{O_4}^ - used in the reaction with mixture = total mole equivalents of MnO4−Mn{O_4}^ - - mole equivalents of Fe2+F{e^{2 + }}
Thus, mole equivalents of MnO4−Mn{O_4}^ - used in the reaction mixture = 750 – 175 = 575 mole equivalents
Now, we will find the number of electrons required to oxidize both Cu2SC{u_2}S and CuS.
- For Cu2SC{u_2}S,
2Cu+→2Cu2++2e−2C{u^ + } \to 2C{u^{2 + }} + 2{e^ - }
S2−→S4++6e−{S^{2 - }} \to {S^{4 + }} + 6{e^ - }
For CuS,
S2−→S4++6e−{S^{2 - }} \to {S^{4 + }} + 6{e^ - }
Now, assume that the masses of Cu2SC{u_2}S and CuS is x and y respectively.
So, we can say that
x+y=10gm ....(1)x + y = 10gm{\text{ }}....{\text{(1)}}
To find the mole equivalent of CuS and Cu2SC{u_2}S, we will use the following formula.
Mole equivalents = Weight ×1000×Number of electrons involvedMolecular weight{\text{Mole equivalents = }}\dfrac{{{\text{Weight }} \times {\text{1000}} \times {\text{Number of electrons involved}}}}{{{\text{Molecular weight}}}}
Now,for Cu2SC{u_2}S, we can write the above formula as
Mole equivalents = x×1000×8159.2{\text{Mole equivalents = }}\dfrac{{x \times 1000 \times 8}}{{159.2}}
For CuS,
Mole equivalents = y×1000×695.6{\text{Mole equivalents = }}\dfrac{{y \times 1000 \times 6}}{{95.6}}
Now, we can write that
Mole equivalents of MnO4Mn{O_4} used = Mole equivalents of Cu2SC{u_2}S + Mole equivalents of CuS
575 = x×1000×8159.2\dfrac{{x \times 1000 \times 8}}{{159.2}} +y×1000×695.6\dfrac{{y \times 1000 \times 6}}{{95.6}} ……(2)
Now, as we solve the equation (1) and (2), we obtain that
x = 4.206gm and y = 5.794gm
Now, we can write that % of CuS in the mixture = Mass of CuSMass of mixture×100=5.79410×100=57.94%\dfrac{{{\text{Mass of CuS}}}}{{{\text{Mass of mixture}}}} \times 100 = \dfrac{{5.794}}{{10}} \times 100 = 57.94\%

Thus, we can conclude that % of CuS in the mixture is 58%.

Note:
Do not forget that as the number of electrons involved in particular oxidation and reduction reactions are different, we need to multiply the concentration of the compound by the number of electrons involved in order to find the mole equivalents.