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Copper watch

1. I have a few questions i need help with please:

1. Show how copper(II) sulphate is acidic with use of an equation.
2. Copper ore, CuCO3.xCu(OH)2, was analysed as follows:
a. dissolved completely in acid.
b. 76 cm^3 of CO2 was produced at r.t.p
c. Remaining solution made up to 50cm^3
d. to 10cm^3 of this solution, excess KI was added.
e. Iodine liberated reacted with 25.4 cm^3 of 0.05M sodium thiosulphate solution.

Calculate x.

3. Why are copper(III) compounds rare?
4. What properties might copper(III) compounds have and why?

For question 2, i worked out the moles of Cu(2+) and moles of CO2, but what do i do then?

Thank you
2. (Original post by Shacas)
I have a few questions i need help with please:

1. Show how copper(II) sulphate is acidic with use of an equation.
What sort of reactions to confirm a substance is an acid? - the hint is given in the next question really. it starts with a carb - ....... - ate - figure it out

2. Copper ore, CuCO3 x Cu(OH)2, was analysed as follows:
a. dissolved completely in acid.
b. 76 cm^3 of CO2 was produced at r.t.p
c. Remaining solution made up to 50cm^3
d. to 10cm^3 of this solution, excess KI was added.
e. Iodine liberated reacted with 25.4 cm^3 of 0.05M sodium thiosulphate solution.

Calculate x.

3. Why are copper(III) compounds rare?
4. What properties might copper(III) compounds have and why?

For question 2, i worked out the moles of Cu(2+) and moles of CO2, but what do i do then?

Thank you
First thing for second part of the question is to write the balanced equation for the reaction. Do that first, then, you'd see that one of them must be limiting and the other is in excess(in most cases). You have to use values for the limiting reactant as they are the one that is used up completely in the reaction.

Then, the rest should be quite easy.
3. (Original post by shengoc)
First thing for second part of the question is to write the balanced equation for the reaction. Do that first, then, you'd see that one of them must be limiting and the other is in excess(in most cases). You have to use values for the limiting reactant as they are the one that is used up completely in the reaction.

Then, the rest should be quite easy.

Thanks for that. I understand number 1. For number 2, how can i write a balanced equation if i don't know the value of x?
4. (Original post by Shacas)
I have a few questions i need help with please:

1. Show how copper(II) sulphate is acidic with use of an equation.
2. Copper ore, CuCO3.xCu(OH)2, was analysed as follows:
a. dissolved completely in acid.In how much acid, what acid is used?
b. 76 cm^3 of CO2 was produced at r.t.p
c. Remaining solution(what solution are you talking about) made up to 50cm^3
d. to 10cm^3 of this solution, excess KI was added.
e. Iodine liberated reacted with 25.4 cm^3 of 0.05M sodium thiosulphate solution.

Calculate x.

3. Why are copper(III) compounds rare?
4. What properties might copper(III) compounds have and why?

For question 2, i worked out the moles of Cu(2+) and moles of CO2, but what do i do then?

Thank you
Can you post the whole question, not cut up pieces of it. Your questions seem incomplete. Like what acid is used, could be mono or di - protonated acid.
5. (Original post by shengoc)
Can you post the whole question, not cut up pieces of it. Your questions seem incomplete. Like what acid is used, could be mono or di - protonated acid.
That is all i am given. The solution i am guessing is after the copper ore has been dissolved in acid and gives off CO2.
6. (Original post by Shacas)
I have a few questions i need help with please:

1. Show how copper(II) sulphate is acidic with use of an equation.
Here you need to show the hydrolysis/dissociation of the hexaaquacopper (II) ion

2. Copper ore, CuCO3.xCu(OH)2, was analysed as follows:
a. dissolved completely in acid.
It doesn't matter what type of acid as it's in excess. What matters is the volume of CO2 produced which tells you (via the equation) the moles of CuCO3 in the mixture.

All of the Cu(OH)2 will also react making solube copper (II) ions for further analysis. (below)

b. 76 cm^3 of CO2 was produced at r.t.p
c. Remaining solution made up to 50cm^3
d. to 10cm^3 of this solution, excess KI was added.
e. Iodine liberated reacted with 25.4 cm^3 of 0.05M sodium thiosulphate solution.

Calculate x.
This tells you the moles of Cu2+ ions by the stoichiometry of the redox reactions.

All of the Cu2+ reacts with excess KI:

2Cu2+ + 2KI --> 2Cu+ + I2

hence 1 mole of iodine is produced per 2 moles of copper (II).

The iodine is then titrated with thiosulphate:

I2 + 2S2O32- --> S4O62- + 2I-

1 mole iodine per 2 moles thiosulphate, therefore overall moles of thiosulphate = moles of copper (II)

For question 2, i worked out the moles of Cu(2+) and moles of CO2, but what do i do then?

Thank you
If you have total moles of Cu2+ and moles of CuCO3 then as:
CuCO3.xCu(OH)2

simple subtraction will give you moles of Cu(OH)2

you then have mole ratio of CuCO3 to Cu(OH)2 and hence can find 'x'
7. (Original post by charco)
Here you need to show the hydrolysis/dissociation of the hexaaquacopper (II) ion

It doesn't matter what type of acid as it's in excess. What matters is the volume of CO2 produced which tells you (via the equation) the moles of CuCO3 in the mixture.

All of the Cu(OH)2 will also react making solube copper (II) ions for further analysis. (below)

b. 76 cm^3 of CO2 was produced at r.t.p

This tells you the moles of Cu2+ ions by the stoichiometry of the redox reactions.

All of the Cu2+ reacts with excess KI:

2Cu2+ + 2KI --> 2Cu+ + I2

hence 1 mole of iodine is produced per 2 moles of copper (II).

The iodine is then titrated with thiosulphate:

I2 + 2S2O32- --> S4O62- + 2I-

1 mole iodine per 2 moles thiosulphate, therefore overall moles of thiosulphate = moles of copper (II)

If you have total moles of Cu2+ and moles of CuCO3 then as:
CuCO3.xCu(OH)2

simple subtraction will give you moles of Cu(OH)2

you then have mole ratio of CuCO3 to Cu(OH)2 and hence can find 'x'

Thank you for that.
You say that from moles of CO2, i can work out moles of CuCO3 via the equation, but what is the equation?
8. (Original post by Shacas)
Thank you for that.
You say that from moles of CO2, i can work out moles of CuCO3 via the equation, but what is the equation?
CuCO3 + 2H+ --> Cu2+ + CO2 + H2O
9. (Original post by charco)
CuCO3 + 2H+ --> Cu2+ + CO2 + H2O
So moles of Cu(2+) = 1.27x10^-3
moles of CO2 = 3.17x10^-3 that implies moles of CuCO3 = 3.17x10^-3

Is that correct so far?

But then what do i subtract?
10. (Original post by charco)
CuCO3 + 2H+ --> Cu2+ + CO2 + H2O
Ok, i got moles of CuCO3 = 3.17x10^-3
does that mean moles of Cu(OH)2 = 3.18x10^-3.

So n is roughly = 1?

Thank you
11. (Original post by Shacas)
So moles of Cu(2+) = 1.27x10^-3
moles of CO2 = 3.17x10^-3 that implies moles of CuCO3 = 3.17x10^-3

Is that correct so far?

But then what do i subtract?
The total moles of Cu2+ come from the moles of CuCO3 + the moles of Cu(OH)2

So if you subtract the moles of CuCO3 from the total moles of Cu2+ you get the moles of Cu(OH)2
12. (Original post by charco)
The total moles of Cu2+ come from the moles of CuCO3 + the moles of Cu(OH)2

So if you subtract the moles of CuCO3 from the total moles of Cu2+ you get the moles of Cu(OH)2
Thank you, is what i posted above correct?

Also, have you got any idea on the other two questions? as i can't seem to find anything about them on the internet.
13. (Original post by Shacas)
Thank you, is what i posted above correct?

Also, have you got any idea on the other two questions? as i can't seem to find anything about them on the internet.

Copper ore, CuCO3.xCu(OH)2, was analysed as follows:
a. dissolved completely in acid.
b. 76 cm^3 of CO2 was produced at r.t.p
c. Remaining solution made up to 50cm^3
d. to 10cm^3 of this solution, excess KI was added.
e. Iodine liberated reacted with 25.4 cm^3 of 0.05M sodium thiosulphate solution.
Moles of CO2 = 76/24000 = 0.00317
Moles of thiosulphate = 0.05 x 0.0254 = 0.00127
but this came from only 10cm3 of solution therefore whole solution contains 0.00127 x 5 = 0.00635

Hence moles of Cu2+ = 0.00635
moles of CuCO3 = 0.00317
Therefore moles of Cu(OH)2 = 0.00635 - 0.00317 = 0.00318

So yes the ratio of CuCO3 to Cu(OH)2 is 1:1 and 'x' = 1

3. Why are copper(III) compounds rare?
4. What properties might copper(III) compounds have and why?
3. The third ionisation energy for copper is too high to be compensated for by the extra lattice energy of Cu(III) salts

4. Cu(III) would have a configuration of [Ar] 4s0 3d8

The compounds would be acidic by hydrolysis (high polarising power of high charge density Cu(III))
Paramagnetic (unpaired d electrons)
Coloured (partially occupied 'd' orbitals)
They would be oxidising agents (easily accept electrons to form Cu2+)
They would form complex ions (as other copper states, typical transition metal property)
Possible catalytic activity
14. (Original post by charco)
Moles of CO2 = 76/24000 = 0.00317
Moles of thiosulphate = 0.05 x 0.0254 = 0.00127
but this came from only 10cm3 of solution therefore whole solution contains 0.00127 x 5 = 0.00635

Hence moles of Cu2+ = 0.00635
moles of CuCO3 = 0.00317
Therefore moles of Cu(OH)2 = 0.00635 - 0.00317 = 0.00318

So yes the ratio of CuCO3 to Cu(OH)2 is 1:1 and 'x' = 1

3. The third ionisation energy for copper is too high to be compensated for by the extra lattice energy of Cu(III) salts

4. Cu(III) would have a configuration of [Ar] 4s0 3d8

The compounds would be acidic by hydrolysis (high polarising power of high charge density Cu(III))
Paramagnetic (unpaired d electrons)
Coloured (partially occupied 'd' orbitals)
They would be oxidising agents (easily accept electrons to form Cu2+)
They would form complex ions (as other copper states, typical transition metal property)
Possible catalytic activity
Thank you so much for your help.
15. (Original post by Shacas)
Thank you so much for your help.
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