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coolnufc
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I have been trying to find the concentration of copper in a copper sulphate solution and i was wondering if someone could check my calculations and maybe set me straight on a few things. Rep will be given of course.
Titration
I put 25cm^3 of copper sulphate solution into a conical flask, and added 15cm^3 of 10% potassium iodide solution to the conical flask. I then titrated this solution against sodium thiosulphate.
The average titre was 24cm^3
I have the equations for the addition of potassium iodide solution:
2Cu^2+ + 4I^- --> 2CuI + I2
And then the titration:
I2 + 2S2O3^2- --> S4O6^2- + 2I^-
Because I2 has the same moles in both equations, i worked out the concentration of S2O3, and then thought that would be the same as Cu2+ as they are in the same moles. But i think this is really wrong. So help would be appreciated. Anyway, this is what i did.
m=cxv
0.1 x (24/1000) = 0.0024moles.
c = m/v
conc. CuSO4 = 0.0024 / (25/1000) = 0.096moldm^-3
Gravimetric
I used zinc to displace the copper to create a solid copper precipitate.
Copper mass = 0.21g
Moles = 0.21 / 63.5 = 3.31 x 10^-3 moles
3.31 x 10^-3 moles in 25cm^3 of CuSO4
So in 1dm^3 --> (3.31 x 10^-3) x 40 = 0.1324 mol dm^-3
Titration
I put 25cm^3 of copper sulphate solution into a conical flask, and added 15cm^3 of 10% potassium iodide solution to the conical flask. I then titrated this solution against sodium thiosulphate.
The average titre was 24cm^3
I have the equations for the addition of potassium iodide solution:
2Cu^2+ + 4I^- --> 2CuI + I2
And then the titration:
I2 + 2S2O3^2- --> S4O6^2- + 2I^-
Because I2 has the same moles in both equations, i worked out the concentration of S2O3, and then thought that would be the same as Cu2+ as they are in the same moles. But i think this is really wrong. So help would be appreciated. Anyway, this is what i did.
m=cxv
0.1 x (24/1000) = 0.0024moles.
c = m/v
conc. CuSO4 = 0.0024 / (25/1000) = 0.096moldm^-3
Gravimetric
I used zinc to displace the copper to create a solid copper precipitate.
Copper mass = 0.21g
Moles = 0.21 / 63.5 = 3.31 x 10^-3 moles
3.31 x 10^-3 moles in 25cm^3 of CuSO4
So in 1dm^3 --> (3.31 x 10^-3) x 40 = 0.1324 mol dm^-3