What methods can be used to determine iron(II) in a solution?
610466
Spectrophotometric methods, or those based on the absorption of light by an analytical species, are some of the simplest and most common methods used in quantitative analysis. The amount of light which is absorbed by the sample (A) is correlated with the concentration (c) of the analytical species present using Beer's Law (A = e bc; where e is the molar absorptivity for the analytical species and b is the pathlength of the sample cell). \n\nIn this experiment, the analytical species, iron, will be studied using a colorimetric method in which a reagent is added to transform iron into a highly absorbing complex. Iron is often found in nature in reasonably high concentrations and may enter the hydrosphere through the weathering of iron salts and minerals. Both iron (II) and iron (III) ions are commonly found dissolved in water either in a colloidal form or as inorganic or organic complexes. There are also multiple anthropogenic sources of iron in the environment such as canneries, tanneries, textile mills, and metal cleaning operations. \n\nA relatively simple but sensitive colorimetric procedure for the determination of iron involves first reducing all iron present in the sample to iron (II) ion using hydroxylamine hydrochloride and then complexing with 1,10-phenanthroline in a solution buffered at low pH:\n\nPart 1: Preparation of Iron Standard Solutions\n\nWeigh 351 mg of high quality ferrous ammonium sulfate hexahydrate (FeSO4(NH4)2SO4 Ÿ 6H2O) and quantitatively transfer to a 500 mL volumetric flask. Add 50 mL of distilled water followed by 1 mL of concentrated sulfuric acid. Dilute to the mark with distilled water and mix thoroughly. Using this solution, prepare five standard solutions of iron (II) ion (about 10 - 15 mL should be sufficient) ranging in concentration from 0.5 to 5 ppm.\n\nPart 2: Preparing a Beer's Law Plot\n\nTransfer a 5 mL aliquot of the lowest concentration iron standard solution (probably 0.5 ppm) to a 125 mL Erlenmeyer flask and test the pH with test paper. If the pH is greater than 4.5, add enough 0.2 M sulfuric acid to bring the pH to about 3.5, counting the number of drops that you have added. \n\nAgain, counting the drops that you add, add sodium citrate (259 g/L) buffer to bring the pH to about 4.5. Pipet 1 mL of 10% hydroxylamine hydrochloride and 3 mL of 0.3% 1,10-phenanthroline into the sample and mix the solution well. Allow about 5 min for development of color.\n\nUse the same number of drops of sulfuric acid and sodium citrate for your remaining standard solutions. For each solution, add 1 mL of 10% hydroxylamine hydrochloride and 3 mL of 0.3% 1,10 phenanthroline and mix the solutions well. Allow about 5 min for development of color for each solution.\n\nScan the absorbance of each of the solutions from 300 to 600 nm using the Shimadzu ultraviolet-visible spectrometer, noting the absorbance value of each solution at 510 nm. For each sample, use water as your reference blank.\n\nPart 3: Analysis of samples\n\nWater samples should be measured for each member of your group in addition to determining the iron concentration in an unknown water sample provided by the Instructor. Treat the samples in the same way as the standards: using a 5 mL sample, add sulfuric acid initially, if necessary, to adjust the pH, followed by citrate buffer, reducing agent (1 mL of 10% hydroxylamine hydrochloride), and the indicator (3 mL of 0.3% 1,10 phenanthroline). Mix the solutions well and allow about 5 minutes for development of the complex. \n\nScan the absorbance of each of your sample solutions from 300 to 600 nm using the Shimadzu ultraviolet-visible spectrometer, noting the absorbance value of each solution at 510 nm. Using your Beer's Law plots prepared in Part 2 and the measured absorbances of your samples, determine the iron concentrations in each of your water samples.
799105
Assume you have the salt AxBy and the molar solubility is s.\n\nThen\n\nAxBy <=> xA^(+y) +yB(^-x)\ns mole/lt\ngive... .. .. ..xs .. .. .. ..ys\n\nKsp=([A(+y)]^x) *[B(-x)]^y =(xs)^x * (ys)^y=>\nKsp= (x^x)*(y^y)*s^(x+y)\n\ne.g. for Fe3(PO4)2 tou have x=3, y=2, so\nKsp=(3^3)*(2^2)*s^(3+2)= 36s^5
770385
FeCl3+3NaOH>>Fe(OH)3+3NaCl
770691
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
256406
4Fe3+ + 9O2 ->  6Fe2O3
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
491963
it's either copper or iron.  i don't remem ber exactly which one.
831216
Cu (s) + Pb(NO3)2 (aq) ---------> No Reaction. \nCopper will not replace lead.\n\nreaction VI in the attached link.
393459
Take solutions of each metal ion and pieces of each metal.  Put metal pieces in each solution that does not contain that metal ion and observe whether any reaction is taking place.  Tabulate all observations and you should be able to see quickly which are the most and least reactive metals.
824286
That depends on how much iron is added.\nCopper will continue to replace iron as long as any iron is left.\nIf xs iron was present, no copper would be left in solution...you would wind up with copper solid, suspended in iron chloride  soln...with any xs iron solid still unreacted.\nBut if you started with less than one mol of iron for each mol of copper, then yes, some copper would remain in soln.
537646
The thing is that iron III hydroxide isn't soluble in water.  It just sits there and doesn't make many hydroxide ions, which are necessary for a strongly basic solution.  Barium hydroxide, however, is soluble and breaks up into barium ions and hydroxide ions, giving a strongly basic solution.
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
378767
You would complete dissolve it in strong acids and then separate the metals by qualitative analysis techniques.  Not simple and very expensive.
883128
Use a magnet. Iron is attracted to it, lead is not.
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O