Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on June 25, 2012, 10:17:42 AM
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From the wikipedia page:
4th analytical group of cations
The fourth group of cations include Zn2+, Ni2+, Co2+, and Mn2+. This group is determined by the addition of ammonium chloride, ammonium hydroxide, and hydrogen sulfide gas to the solution of the salt. A colored precipitate indicates Mn2+...
Is is really neccessary to add hydrogen sulfide gas or I can add it as a liquid, because I don't know how to add it as a gas, or is there any better way to determine Mn2+ ions?
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Ther best method use oxidation to HMnO4 with NaBiO3 (the best reagent), Pb3O4 or PbO2 in nitric acid. Moreover this can be done using the whole mixture of cations without separation any of them.
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From what I know at least since seventies instead of gaseous H2S everyone uses thioacetamide.
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I am preparing for the next level (more ions) of qualitative inorganic analisys. I need to make a chart, but I got stuck on the IVth group: Mn2+ and Zn2+. Unfortunately, I won't be allowed to use thioacetamide and as I found out, neither H2S is allowed.
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The fourth group of cations include Zn2+, Ni2+, Co2+, and Mn2+. This group is determined by the addition of ammonium chloride, ammonium hydroxide, and hydrogen sulfide gas to the solution of the salt. A colored precipitate indicates Mn2+...[/sub]
NiS and CoS are black. How can you see white or pale pink precipitate?
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From the fourth group I only need to know the determination processes for Mn2+ and Zn2+ for the test. I need to find another way for proving those catinos because I won't be able to use H2S(g), so I'm stuck now.
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I told you about Mn2+ test.
How this cations react with an excess of NaOH?
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Thanks, good one, white precipitate soluble in acids not in excess of NaOH.
Now, I need only for the Zn2+ cation something good.
Edit: Found similar for Zn2+ only it will dissolve in excess of NaOH. Ca2+ and Mg2+ make insoluble hidroxydes witn NaOH, too, but they won't dissolve in excess of NaOH, right?
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http://faculty.coloradomtn.edu/jeschofnig/class/class_jeschof/ch2-lb8.htm
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I found a very nice way for Mn2+ with HNO3 and PbO2. Tomorrow I should finish the whole list.
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http://faculty.coloradomtn.edu/jeschofnig/class/class_jeschof/ch2-lb8.htm
Is that reaction characteristic only for Zn2+ and not for the alkali and earth-alkali metals, because I'm planning to test their presence right after Zn?
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Supernatant after treatment with an excess of NaOH should be used.
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I will have plenty of the unkown salt so I can make a new solution. Ca, Ba, Mg, Na and K don't react with K4[Fe(CN)6]?
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https://www.msu.edu/~grahamm9/General%20Solubility%20Rules%20for%20Ionic%20Cmpds.%20in%20H20.htm
There are better methods for qualitative analysis of Ca, Ba, Mg cations than using ferrocyanides.
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I didn't mean to determine them with ferrocyanides (I am planning to do it on fire), but I don't understand how to be sure if there is Zn2+ present in the solution or one of the earth alkali metals if they all make white precipitates with ferrocyanides. A mistake could happen.
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You should know, at least basis of qualititative analytical chemistry.
http://fch.upol.cz/skripta/fcc_and_zvem_english/anal/anal_teorie.htm
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Thanks for the link, I know the basics, but found out something new on the link. The best way to determine Zn2+ would be to add sulfide water, because on the link that you gave me it is said that ZnS is the only white sulfide. Only anions left.
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There are different procedures for single cations, there are systematic procedure for mixture.
For some mixtures you can use shortcuts, but not for all. In the systematic procedure for mixture the zinc cation may be together with Al3+ and chromate only - then both sulfide and ferrocyanide works.
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I will be given a salt made from one cation and one anion only, no mixtures.
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On your like isn't written that Al gives a precipitate with NH3. In my book it is said that Al gives a white precipitate with NH4Cl and NH3 buffer solution and Iron gives a redish-brown precipitate. What is correct?
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Even NH4Cl itself forms Al(OH)3 from Al salts but in colloidal form almost invisible (solution of Al salts are often additionaly acidified - then nothing happens).
Pure ammonia or ammonia buffer (NH3/NH4Cl) forms visible colourless precipitates (often gelatinous).
Iron gives a redish-brown precipitate
correct
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Ok, thanks.
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I finished my list, but I have some doubts. Can someone check it if I write it here?