AL chem titration [challenge!]

2009-07-18 6:59 pm

To determine the copper cotent of a brass screw, a screw of mass 3.75g was dissolved in concentrated nitric(V) acid to give copper(II) ions. This solution was then made up to 250 cm^3 with distilled water.

25 cm^3 of this solution was pipetted into a conical flask. After neutralisation by excess sodium carbonate a faint blue precipitate was obtained. This was then dissolved by a small amountof ethanoic acid. An excess of potassium iodide was then added to this now neutral solution of copper(II) ions.

This solution was then titrated against a 0.2 M solution of sodium thiosulphate using an iodine indicator. 21.10 cm^3 was requried to reach the end point. What was the percentage of copper in the brass screw.

CC Lee answer: 71,5%

回答 (1)

老爺子

2009-07-18 8:10 pm

✔ 最佳答案

Consider the titration of the iodine solution against 0.2 M solution of sodium thiosulphate using starch indicator (NOT iodine indicator):

I2 + 2Na2S2O3 → 2NaI + 2Na2S4O6
Mole ratio I2 : Na2S2O3 = 1 : 2

No. of moles of Na2S2O3 = 0.2 x (21.10/1000) = 0.00422 mol
No. of moles of I2 in the solution = 0.00422 x (1/2) = 0.00211 mol

Consider the formation of I2 from Cu2+ and excess KI:
2Cu2+ + 4I- → 2CuI + I2
Mole ratio Cu2+ : I2 = 2 : 1

No. of moles of I2 formed = 0.00211 mol
No. of moles of Cu2+ used = 0.00211 x 2 = 0.00422 mol

Consider the preparation of 250 cm3 of Cu2+ solution:
3Cu + 8H+ + 2NO3- → 3Cu2+ + 4H2O + 2NO
Mole ratio Cu : Cu2+ = 1 : 1

No. of moles of Cu2+ in 25 cm3 solution = 0.00422 mol
No. of moles of Cu2+ in 250 cm3 solution = 0.00422 x (250/25) = 0.0422 mol
No. of moles of Cu used = 0.0422 mol
Molar mass of Cu = 63.5 g/mol
Mass of Cu used = 0.0422 x 63.5 = 2.68 g

Mass percentage of Cu in the screw = (2.68/3.75) x 100% = 71.5%

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