Co3+ react with NH3 to form a complex cation with formula [Co(NH3)n]3+(aq).
In an experiment to determine the value of n in the formula, excess ammonia was added to 25.00 cm3 of 0.133 M Co(NO3)3 (aq). The solution was diluted to 50 cm3 with deionized water. The diluted solution was then shaken with trichloromethane until equilibrium was attained. 25 cm3 of the organic layer are required 7.5 cm3 of 0.047 M H2SO4 (aq) for complete neutralization, while 25 cm3 of the aqueous layer required 32.15 cm3 of 1.07 M HCl (aq) for complete neutralization.
Determine the value of n in the formula.
Given partition coefficient of ammonia between water and trichloromethane is 24.8.
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partition coefficient+comlex
2009-04-24 7:14 am
回答 (2)
2009-04-24 7:34 pm
✔ 最佳答案
NH3 in the organic layer after extraction: NH3(CHCl3)
NH3 in the aqueous layer after extraction: [Co(NH3)n]3+(aq) and NH3(aq)
Assuming that all the Co3+ ions are converted to [Co(NH3)n]3+(aq) complex ions.
Consider the titration of the organic layer:
NH3 + 2H2SO4 → (NH4)2SO4
Mole ratio NH3 : H2SO4 = 2 : 1
No. of moles of H2SO4 used = 0.047 x (7.5/1000) = 0.0003525 mol
No. of moles of NH3 in the organic layer = 0.0003525 x 2 = 0.000705 mol
[NH3(CHCl3)] = 0.000705/(25/1000) = 0.0282 mol dm-3
Consider the extraction:
[NH3(aq))]/[NH3(CHCl3)] = 24.8
[NH3(aq)] = 0.0282 x 24.8 = 0.6994 mol dm-3
Concentration of free NH3(aq) in the aqueous layer = 0.6994 mol dm-3
Consider the titration of the aqueous layer:
NH3 + HCl → NH4Cl
Mole ratio NH3 : HCl = 1 : 1
No. of moles of HCl used = 1.07 x (32.15/1000) = 0.0344 mol
No. of moles of NH3 being titrated = 0.0344 mol
Consider the 50 cm3 of the aqueous layer after extraction:
No. of moles of free NH3(aq) = 0.6994 x (50/1000) = 0.03497 mol
Total no. of moles of NH3 = 0.0344 x (50/25) = 0.0688 mol
No. of moles of NH3 in the complex = 0.0688 - 0.03497 = 0.03383 mol
No. of moles of Co3+ = 0.133 x (25/1000) = 0.003325 mol
Mole ratio Co3+ : NH3
= 0.003325 : 0.03383
≈ 1 : 10
Hence, n = 10
=
2009-04-24 6:39 pm
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