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Thread starter 13 years ago
#1
I need a bit if help with this question please can you help me out?

Purity of copper can be found by reacting with with conc. nitric acid, then with excess potassium iodide. Iodine is liberated and is titrated with sodium thiosulphate solution.
Cu + 4HNO3 ---> Cu(NO3)2 + 2NO2 +2H2O
2Cu(2+) + 4I(-) ----> 2CuI +I2
Copper foil weighs 1.74g, made into 250cm3 solution of copper(II)ions. To 25cm3 of this solution excess iodide ions were added, mixture titrated with 0.1mol/dm3 sodium thiosulphate. 26.8cm3 required. % purity of copper foil?

Thanks so much!
0
13 years ago
#2
From the titre value of thiosulphate given, you can work out the moles of thiosulphate that were required.

This then tells you the moles of Iodine that were produced - which tells you the moles of Cu2+ in the 25cm3 aliquot.

Times this by ten to get the moles of Cu2+ in 250cm3 and this gives you the moles of Cu on the copper wire originally.

Multiply this number by the Mr of copper to get the mass of copper in the 1.74g wire.

Percentage purity = ( this mass / 1.74 )*100.

Ask if you need any more help with any individual step.
0
13 years ago
#3
OK

Work out the moles of thiosulphate ions required to react with the liberated iodine,

mol = vol x conc.
= 0.0268 x 0.1
= 2.68 x 10^-3

Use the equation below to work out mole ratio of S2O3^- : I2

I2 + 2S2O3- ----> 2I- + S4O62-

Ratio 2 : 1

therefore, moles I2 Liberated = 1.34 x 10^-2
Then using equation,
2Cu2+ + 4I- ----> 2CuI + I2

you can see that the ratio of I2 : Cu2+ is 1:2

therefore moles of Cu2+ in the 25cm3 portion is 2.68 x 10^-3
(note if you look at equations to start with you can skip the middle part, noticing that the moles of I2 liberated = moles of Cu2+)

in the 250cm3 portion there must be 10x as much copper,

therefore moles Cu2+ = 2.68 x 10^-2

In equation,
Cu + 4HNO3 -----> Cu(No3)2 + 2NO2 + 2H2O

the moles of Cu2+ ions [in Cu(NO3)2] is in the ratio of 1:1 with the copper,

therefore moles of copper in the foil = 2.68 x 10^-2

using mol = mass/RFM

0.0268 = mass/64

therefore mass of Cu = 1.7152

Work out percentage
mass obtained/original mass x 100
= 1.7152/1.74 x 100 = 98.6%

Can someone please check this, its late and mistakes are likely
0
13 years ago
#4
(Original post by Hawk)
Can someone please check this, its late and mistakes are likely
Yes - looks right.
0
Thread starter 13 years ago
#5
Wow, that's really helpful- thnks so much, Hawk.
(hmm...can't seem to be able to give a same person a reputation again...)
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