Asked by Jill
assume an unknown sample to be 15% copper, weigh enough dried ore sample to use approximatele 35-45mL of .09M S2O3-. How many grams do I need to measure out to use 40.00mL of the thiosulfate solution?
Answers
Answered by
DrBob222
Note: Your instructor may prefer the use of I3^- instead of I2. If so you may need to adjust the following.
2Cu^2+ + 4I^- ==> CuI2(s) + I2
I2 + S2O3^= ==> S4O6^= + 2I^-
From the above you can see that 1 mol Cu = 1 mol S2O3=.
If you want to use 45 mL, that is 0.045 L and M x L = 0.09 x 0.045 = 0.00405 mols.
g Cu = mols Cu x atomic mass = about 0.00405 x about 64 = about 0.257 g.
0.257/0.15 = about 1.71g. Check my thinking and my math. You need to redo this for 40 mL; I didn't see that until the 45 ml was done. I'm sure you have thought of this but you must NOT go over 50 mL; that means a refill on the buret. That won't ruin the titration but it increases the error.
2Cu^2+ + 4I^- ==> CuI2(s) + I2
I2 + S2O3^= ==> S4O6^= + 2I^-
From the above you can see that 1 mol Cu = 1 mol S2O3=.
If you want to use 45 mL, that is 0.045 L and M x L = 0.09 x 0.045 = 0.00405 mols.
g Cu = mols Cu x atomic mass = about 0.00405 x about 64 = about 0.257 g.
0.257/0.15 = about 1.71g. Check my thinking and my math. You need to redo this for 40 mL; I didn't see that until the 45 ml was done. I'm sure you have thought of this but you must NOT go over 50 mL; that means a refill on the buret. That won't ruin the titration but it increases the error.
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