t_Pyro - 14-3-2004 at 02:29
Inorganic qualitative analysis seems to be one topic where every new step leads to initial chaos, then disbelief, then enlightenment. No matter how
well you know the steps and theory of analysis, there's bound to be a test that'll put you to the test and leave you in awe of chemistry
-yet again! 'nuff said! Here are some really interesting results I accidentally came across recently:
Experiment 1. (Actually given to me during the practicals exam)
A green (hygroscopic) salt is given having two or more cations. A water extract is prepared, and used for the following tests: Water extract (w.e.) +
liquor ammonia (dropwise till in excess)= an instantaneous dark blue solution. Flame test=lilac flame. w.e.+ nessler's reagent=green ppt. What
is/are the cation(s) present?
Experiment 2.
A pure white salt "X" is given. The salt, when heated strongly, produces water droplets on the cooler parts of the test tube. Also, a
colourless, odourless gas is produced, that can also be prepared by the action of dilute mineral acid on chalk. A white (crystalline) residue
"Y" was also produced. The residue is dissolved completely in a test tube of warm water. Solid compound "Z", when added to the
solution, produces a white turbidity. When dil. HCl is added by dropping a few drops along the side of the test tube, effervescence takes place, and
the upper part of the solution turns clear. "Z" produces a brick red flame in the flame test. Identify X, Y, and Z.
Experiment 3:
A curdy white ppt is produced when a solution of silver nitrate is added to sodium chloride solution, and this ppt dissolves when a solution of
"A" is added to it in excess. "A" does not fume in the presence of HCl vapours. Identify "A".
somewhat poor test
Polverone - 14-3-2004 at 15:21
Things like experiment 3 presume that the student has a limited amount of knowledge and will appropriately guess what s/he is "supposed" to
know. As far as I can tell, "A" could be most compounds that complex silver well enough to dissolve it (just not aq. ammonia). What is the
student "supposed" to know about compounds that form soluble silver complexes? Is sodium thiosulfate supposed to be your guess, or something
else?
vulture - 14-3-2004 at 15:50
1) One of the cations is K for sure.
2) Calcium's one of them and we're talking about a carbonate.
3) Is not specific enough. It could be thiosulfate, cyanide or maybe even thiocyanide.
unionised - 14-3-2004 at 16:14
Not sure how you ae meant to tell if X is a bicarbonate or a hydrated carbonate.
(and, for what it's worth, CO2 has a definite, rather weak, odour)
t_Pyro - 15-3-2004 at 10:39
The third question is quite a weird one, I must admit, too. What the teacher expected was at least one other substance that could form a soluble
complex with Ag. Well, my guess was NaCN, forming the Ag(CN)<sub>2</sub><sup>-</sup> ion.
unionised:
Close. Very close. However, I wasn't aware of the "odour" of CO<sub>2</sub>. I'll be more alert the next time I
neutralise some acid with baking soda.
Vulture:
1) K is correct. the other cation or cations require(s) more thought, but the answer's right there.
2) Calcium, correct. Carbonate- X, Y, or Z?
3) All correct. Like I said, an odd question, because it has too many answers.
vulture - 15-3-2004 at 13:10
X is the carbonate.
unionised - 15-3-2004 at 15:40
If X is a carbonate then where does the water come from?
(A hydrated salt is not, strictly speaking a pure salt. It is OTOH a pure compound).
Then again, if it is a bicarbonate then you don't need particularly strong heating to liberate CO2.
Nessler's reagent (an alkaline solution of mercuric iodide in KI soln) is used for indicating the presence of ammonia by giving a yellow/ bown
ppt
AFAIK it isn't used for many other things so I guess they think it means there is ammonia in this stuff.
Nickel ammonium sulphate is a common lab reagent (or, at least, it used to be in all the chemistry sets).
I would send those questions back and ask for some better ones
[Edited on 15-3-2004 by unionised]
t_Pyro - 16-3-2004 at 01:14
You're right- X is a bicarbonate, for the reasons that you've just stated. Y=a carbonate. Z=calcium salt. Turbidity=
CaCO<sub>3</sub>. It turns clear because the Calcium Chloride formed on the action of HCl is soluble.
Well, the actual test (Experiment 1) was to identify the mixture given during our practicals exam. I got the lilac flame right away, but there was one
problem. The brand of matchsticks that I use for the flame test seems to have an excess of chlorate, due to which it produces a lilac flame even on
its own. Therefore, a lilac flame is to be treated with suspicion. The platinum loop, on the other hand, does not produce good results during a flame
test.
To make things worse, the borax bead test gave a dark brown bead (hint!), which might have been considered as indistinct. The blue colouration with
liquor ammonia might have been due to Cu, but in that case, the flame test would have revealed all. Surprisingly, the nessler's reagent test
saved me. Yes, it's generally used to test the presence of the ammonium radical, but the positive test is a <i>brown</i> ppt, not
green!
No, you wouldn't send that question back, either, or the examiner would deduct marks for not being thorough enough in the subject.
Go beyond the obvious, and the answer is obvious.
Answers
t_Pyro - 19-3-2004 at 01:30
Expt. 1:
Cations are K<sup>+</sup> and Ni<sup>++</sup>. Nickel readily forms a dark blue complex with ammonia. Also, the colour of the
mixture indicates that nickel or chromium or copper could be present. Nessler's reagent is a mixture of
K<sub>2</sub>HgI<sub>4</sub> and KOH. The OH<sup>-</sup> ions combine with the Ni<sup>++</sup> ions to
form Ni(OH)<sub>2</sub>, which is green in colour.
unionised - 20-3-2004 at 14:59
Just to muddy the waters.
Nesslers reagent contains an excess of iodide. If there were copper present this would oxidise the iodide to iodine and form CuI. If that were to use
up all the iodide in solution then, rather than forming I3- the iodine would precipitate together with this CuI. Presto a brown ppt. (I3- is the
triiodide ion)
OTOH Nesslers reagent + ammonia gives a yellow/ brown colour or ppt depending on how much ammonia there is, that together with ppt of Ni(OH)2 could
give a brown ppt. I still think this is a bit short on information to be a good question.
Has anyone seen any dimethylglyoxime lying about somewhere?
[Edited on 20-3-2004 by unionised]
t_Pyro - 21-3-2004 at 05:35
There was no ammonia present, so there was only a green ppt. Dimethylglyoxime, yes, I know about that test for nickel. The purpose of the question was
to deduce as much as possible, with the given data. This enables one to think beyond the conventional lines.
It's been my observation that not enough attention is paid to the reasons behind the qualitative analysis tests in general, and in the group
separations in particular. I remember asking a teacher whether while adding liquor ammonia and ammonium chloride (Gr. III a), it mattered in which
order I added the reagents. She gave me a funny look, and claimed: "Of course not!" Anybody who's studied the mechanism of the group
separation, though, would know better. Is the situation as bad elsewhere?