James Freeman Sellers.

An elementary treatise on qualitative chemical analysis online

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Method 2, Hofmann's Separation. Dissolve the
mixed sulphides with HC1 and a crystal of KC1O 3 ,
and evaporate the chlorine and excess of acids. Ar-
range the apparatus for Hofmann's modification of
Marsh's Test and add the dissolved sulphides to the
contents of the generating flask. When the evolution
of the gas has ceased, filter the contents of the test-
tube. The black residue may consist of metallic silver
and SbAg 3 . Test for SbAg 3 by dissolving in tartaric
acid, etc. The filtrate may contain the hypothetical
acid H 3 AsO 3 with excess of AgNO 3 , which can
be tested by producing yellow Ag 3 AsO 3 with dilute
NH 4 OH. Filter the contents of the generating flask,
remove the undissolved zinc, and test the residue for
tin by dissolving . in a small amount of HC1 and then
adding HgCl a .

The reactions involved in the separation of arsenic, antimony,
and tin by Hofmann's method are for :

(a) arsenic, AsH 3 + 3 II 2 O + 6 AgNO 3 = 3 Ag 2 + 6 HNO 3

+ H 3 AsO 3 ;

(6) antimony, SbH g + 3AgNO 3 = SbAg 3 + 3 HNO 3 ;
(c) tin, SnCl 4 + 4H = Sn + 4HC1.

Separation of Sub-group B. Transfer the residue (5),
supposed to consist of the members of the sub-group,
to an evaporating dish, and boil with IINOg 1 (diluted
1 : 2) till the chemical action ceases. Filter. The


residue (a') 1 may be black HgS or a mixture of PIgS
and white Hg(NO 3 ) 2 ' 2 HgS. Dissolve in aqua regia,
boil off the chlorine and excess of acids, and test
with SnCl 2 . The filtrate (a') may contain Pb(NO 3 ) 2 ,
Bi(NO 3 ) 3 , Cu(NO 3 ) 2 , and Cd(NO 3 ) 2 . Concentrate this
filtrate until most of the HNO 3 is driven off, add
dilute H 2 SO 4 , 2 warm gently, and allow to stand for
some time. A white precipitate is PbSO 4 . 3 If lead is
present, add an excess of dilute H 2 SO 4 and evaporate
till all the HNO 3 is expelled. Dilute with water, place
aside to enable the precipitate to settle, and filter off
the insoluble residue (b 1 ). Test it by boiling with
NH 4 C 2 H 3 O 2 and adding K 2 CrO 4 . A yellow precipi-
tate confirms PbCrO 4 . If lead is present, use the
filtrate from PbSO 4 ; if lead is absent, boil off excess
of HNO 3 from filtrate (a 1 ). Add NH 4 OH till alkaline.
A blue fluid confirms the presence of copper and a
white flocculent precipitate, bismuth. Filter. 4 Dissolve
the residue (c 1 ) in a few drops of HC1 and test
for BiOCl 5 by adding an excess of water. If the
filtrate (c r ) is blue, add a dilute solution of KCN care-
fully till the color disappears. Often copper is present
in small quantities, and the blue 6 color is not distinct.
In this event evaporate a small quantity of the filtrate
almost to dry ness, acidify with very dilute HC1, and
add K 4 Fe(CN) 6 . A brown precipitate or coloration
indicates Cu 2 Fe(CN) 6 . If copper is present, add very
little dilute KCN solution. If copper is absent, adding
KCN is not necessary. Pass H 2 S. A yellow 7 precipi-
tate confirms CdS.


CHARACTERISTIC: Insolubility of the hydroxides in alkaline
(NH|OH) solution in the presence of ammonium chloride.

GROUP REAGENT : Ammonium hydroxide with ammonium

Aluminum (salt for study, aluminum sulphate, A1 2 (S0 4 ) 3 ).

1. NH 4 OH precipitates gelatinous aluminum hydrox-
ide, A1(OH) 3 , soluble somewhat in excess of reagent in
the cold, but wholly insoluble if NH 4 C1 is present or if
the solution is boiled.

2. NaOH acts like NH 4 OH, except that A1(OH) 3 is
completely dissolved in excess of reagent, forming
sodium aluminate, Na 3 AlO 3 . This in turn is recon-
verted into insoluble A1(OH) 3 if the solution is boiled
with NH 4 C1.

3. BaC0 3 , suspended in water, precipitates aluminum
completely in the cold as A1(OH) 3 mixed with a basic
salt, probably A1(OH)CO 3 .

4. (NH 4 ) 2 S precipitates A1(OH) 3 with evolution of H 2 S.

5. HNajjPC^ precipitates white aluminum phosphate,
A1PO 4 -H 2 O, soluble in alkalies in the absence of
NH 4 C1 and in HC1 and HNO 3 ; insoluble in HC 2 H 3 O 2 .

6. Spectrum (see Special Method for Aluminum,
p. 62).




The reactions between many salts and alkalies in the presence of
ammonium compounds demand the following further explanation.
Soluble alkalies react with salts of many metals to produce
hydroxides of various solubilities. Some of the more insoluble
of these behave as weak acids in the presence of strong bases and
combine with them to form new classes of soluble salts.
The following is a type of this class of reactions :

ZnCl 2 + 2 NaOH = Zn(OH) 2 + 2 NaCl ;
Zn(OH) 2 + 2NaOH = Na 2 ZnO 2 (sodium zincate) + 2H 2 O.

Ammonia behaves much like the other soluble bases, provided
certain precautions are observed. The following reactions can
occur :

ZnCl 2 + 2NH 4 OH = Zn(OH) 2 + 2NH 4 C1;
Zn(OH) 2 + 2NH 4 OH = (NH 4 ) 2 ZnO 2 (ammonium zincate) + 2 H 2 O.

The last reaction is interrupted in two ways :

First: NH 3 splits off easily, thus allowing Zn(OH) 2 to be
reclaimed. Second: the by-product, NH 4 C1, strongly influences
the reaction and redissolves the precipitated Zn(HO) 2 .

In order to obtain (NH 4 ) 2 ZnO 2 it is necessary to add NH 4 OH
to a cold solution of ZnCl 2 , to avoid decomposition of (NH 4 ) 2 ZnO 2
into NH 3 and Zn(OH) 2 , and to filter off the solution of NH 4 C1.
When an excess of NH 4 C1 or other ammonium salt is added, all
of these conditions of solubility are modified. This applies to
the reactions with the hydroxides of the alkali metals as well
as ammonia.

In the case of the hydroxides of the bivalent metals, Fe(OH) 2 ,
Zn(OH) 2 , Mn(OH) 2 , Co(OH) 2 , Ni(OH) 2 , Ba(OH) 2 , Sr(OH) 2 ,
Ca(OH) 2 , and Mg(OH) 2 , the tendency of the alkalies to redissolve
them is greatly augmented by NH 4 C1. Hence these hydroxides
are not precipitated in the presence of ammonium salts.

In the case of the hydroxides of the trivalent metals, A1(OH) 3 ,
O(OH) 3 , and Fe(OH) 3 , the tendency of the alkalies to redissolve
them is counteracted by NH 4 C1.


Hence these hydroxides are completely precipitated in the
presence of ammonium salts.

Two theories are now sanctioned by good authorities for the
interpretation of the influence of ammonium salts in the reactions
just mentioned.

1. Double Salts Theory

(a) Salts of Bivalent Metals. The following pairs of reac-
tions will explain this theory with reference to the salts of biva-
lent metals :

ZnCl 2 + 2NaOH (or NH 4 OH) = Zn(OH) 2 + 2NaCl,
Zn(OII) 2 + 4NH 4 C1= ZnCr 2 -2]SrH 4 Cl (a soluble double salt)

+ 2NII 4 OH;

MgCl 2 + 2 NaOH = Mg(OH) 2 + 2NaCl,
Mg(OH), + 3NH 4 C1 - MgCl 2 - NH 4 C1 + 2NH 4 OH.

(&) Salts of Trivalent Metals. The application of equations
analogous to those of the preceding paragraph would lead us to
expect the following reactions to occur with hydroxides of triva-
lent metals :

(NII 4 ) 8 A1O 8 + 3 HC1 + NH 4 C1,

A1C1 8 NH 4 C1 + 3 NH 4 OH.

But the hypothetical, soluble bodies, (NTI 4 ) 8 A1O 8 and
A1C1 3 -NH 4 C1, are not known to exist, and their non-existence
is taken to explain the failure of the hydroxides of trivalent
metals to dissolve in the solutions of ammonium salts.

2. Ionic Theory

Another explanation of the part played by ammonium salts is
based upon the simple ionic principle that the addition of an ion
in common with one in the solute decreases the dissociation of
the latter.

(a) Salts of Bivalent Metals. When NaOH or NH 4 OH is
added in excess to the solution of a bivalent metal, a precipitate


is formed which readily dissolves on the addition of NH 4 C1. The
explanation is that the addition of the common ion, NH 4 , sup-
presses the negative ion, OH, thus driving the dissociated ions
into undissociated and inactive molecules of NH 4 OH. As the
hydroxides of the bivalent metals in question are usually moder-
ately well dissociated, NH 4 C1 would not only suppress the free
OH ions of any excess of NH 4 OH, but also those of the hydrox-
ides themselves. Of course it must be understood that if NaOH
is used, this reaction first occurs :

NaOH + NH 4 C1 = NaCl + NH 4 OH.

For application of the principle the important case of magne-
sium salts with NH 4 OH and NH 4 C1 is considered :
Substituting in the equation a b = c-k, 1 the equation

NH 4 x OH = NH 4 OH x jfc, or N ^^ H = fc, is obtained.

NH 4 , OH, NH 4 OH, and k are, respectively, the positive and nega-
tive ions, the undissociated molecules, and the ionization constant
for NH 4 OH. Now when NH 4 C1 is introduced the number of
NH 4 ions is greatly increased, and the result is to suppress the
OH ions. By letting x = number of NH 4 ions added, and y
those of the undissociated molecules of NH 4 OH (resulting from
the addition of NH 4 C1), the equation becomes :

(NH 4 + x -y} (OH - y) . .
NH 4 OH + y

This decreases the number of OH ions and leaves only (OH y)
for unit volume. It is owing to this disappearance of OH ions
that Mg(OH) 2 is not precipitated, too little molecular Mg(OH) 2
being formed to oversaturate the solution.

(6) Salts of Trivalent Metals. When NaOH, NH 4 C1, and
A1C1 3 are brought together the following reactions probably
occur :

A1C1 3 + 3 NaOH = A1(OH) 8 + 3 NaCl ;
NaOH + NH 4 C1 = NaCl + NII 4 OH.


Now A1(OH) 3 , unlike Mg(OH) 2 and hydroxides of some biva-
lent metals, is a very weak base and, consequently, very poorly
dissociated. Hence," when an excess of NH 4 C1 is added, it sup-
presses only the OH ions of NH 4 OH not those of A1(OH) 3 .
Thus NH 4 C1 not only does not affect the precipitated hydroxide
but also destroys the power of NH 4 OH to dissolve it.

Chromium (salt for study, chromium sulphate, Cr 2 (S0 4 ) 3 ).

1. NH^OH 1 gives reactions similar to 1, under alumi-

2. NaOH gives reactions similar to 2, under aluminum.

3. BaC0 3 gives reactions similar to 3, under alumi-
num, except that the precipitation requires more time
for its completion.

4. (NH 4 ) 2 S gives reactions similar to 4, under alumi-

5. Fusion with fusion mixture on a platinum foil, or
with sodium dioxide, Na 2 O 2 , on thick silver foil, gives
a soluble yellow mass, containing sodium chromate,
Na 2 CrO 4 .

6. Na^ heated with a solution of a chromium salt
gives yellow Na 2 CrO 4 .

7. Borax bead with both oxidizing and reducing flames
gives a yellow-green coloration of sodium chromium
metaborate, Na 6 Cr 2 (BO 2 ) 12 .

Reactions 5 and 6 illustrate the conversion of chromium as a
base-producing element to chromium as an acid-producing element.

There are two classes of chromium compounds derived from
the two oxides Cr 2 O 3 and CrO 3 .

Chromic oxide, Cr 2 O 3 , is basic and forms salts with acids:
Cr 2 O 3 + 6 HC1 = 2 CrCl 3 + 3 H 2 O. By oxidation Cr 2 O 3 is changed
to chromium trioxide, CrO 3 , which is an anhydride and forms


salts with bases: CrO 3 + 2NaOH = Na 2 CrO 4 + H 2 O. The oxi-
dation of Cr 2 O 3 in solution may be accomplished by Na 2 O 2 or
by hydrogen dioxide, H 2 O 2 . CrO 3 is a strong oxidizing agent
and is easily reduced to Cr 2 O 3 by various reagents, namely,
H 2 S, S0 2 , HC1, and many organic compounds. If H 2 S is passed
through an acidified (HC1) solution of potassium dichromate,
K 2 Cr 2 O 7 , there will be a change of color from red to green :
K 2 Cr 2 O 7 + 3H 2 S + 8HC1 = 2CrCl 3 + 2KC1 + 3S + 7H 2 O.

In this case chromium is changed from the acid to the basic


Borax (sodium tetraborate, Na 2 B 4 O 7 - 10H 2 O), like Na 2 CO 3 , is
both an inactive flux, as in normal sodium borate, Na 3 BO 3 , and
an active chemical agent, as in boric acid, H 3 BO 3 . (See Fusion,
p. 43.) When borax is heated, it loses its water of crystallization
and fuses to a clear bead on the platinum wire. If a metallic
oxide or salt is fused with the clear bead, a double borate is
formed, which is often colored. The reaction can be easily
understood by a review of the principal hydroxyl acids of boron.
Water unites with boric oxide, B 2 O 3 , increasing in an arithmet-
ical progression to form a systematic chain of polyboric acids,
of which the following are important in this connection :

2 B 2 O 3 + H 2 O = H 2 B 4 O 7 , dihydroxyl tetraboric acid ;

2B 2 O 3 + 2H 2 O = 4HBO 2 , metaboric acid;

2B 2 O 3 + 3H 2 O = H 6 B 4 O 9 , hexahydroxyl tetraboric acid;

2B 2 O 3 + 4H 2 O = 2H 4 B 2 O 5 , diboric acid;

2B 2 O 3 + 5H 2 O = H 10 B 4 O n , dekahydroxyl tetraboric acid;

2B 2 O 3 + 6H 2 O = 4H 3 BO 3 , (normal) orthoboric acid.

From these equations it can be seen that any succeeding acid
in the list can be formed from the next preceding by the addition
of one molecule of water ; and the reverse is also true, that any
higher acid can be formed from the next lower.

The corresponding salts of these acids are formed in a
similar manner, provided the factor to be added is a metallic


oxide instead of water. The most important salts are given for
illustration :

Na 2 B 4 O 7 , sodium tetraborate (borax) ;

Na 2 B 4 O 7 + Na 2 O = 4NaBO 2 , metaborate;
Na 2 B 4 O 7 + 3 Na 2 O = 2 Na 4 B 2 O 5 , diborate ;
Na 2 B 4 O 7 + 5Na 2 O = 4Na 8 BO 8 , " orthoborate.

This further explains the statement above, that borax is both
an inactive flux, like Na 3 BO 3 , and an active chemical agent, like

o o

It is equivalent to four molecules of boric acid with one mole-
cule of water replaced by sodium oxide, and the other five mole-
cules of water displaced :

4H 3 BO 3 + Na 2 O = Na 2 B 4 O 7 + 6H 2 O.

Thus it is in part a salt and in part an anhydride.
If an oxide of a heavy metal is substituted for Na 2 O, double
borates are formed:

Na 2 B 4 O 7 4- CoO = Na 2 Co(BO 2 ) 4 , sodium cobaltous metaborate;
Na 2 B 4 O 7 + 3 CoO = Na 2 Co 3 (B 2 O 5 ) 2 , " cobaltous diborate ;
Na 2 B 4 O 7 + 5 CoO = Na 2 Co 5 (BO 3 ) 4 , " cobaltous orthoborate.

Or, in the case of the triad element chromium :

3Na 2 B 4 O 7 + Cr 2 O 3 = Na c Cr 2 (BO 2 ) 12 , sodium chromic metaborate,

The metaborate, then, is the first product formed by adding a
small quantity of the oxide to an excess of borax, while the ortho-
borate is the last formed by adding a larger amount of the oxide.
What the actual composition of a given bead, made without
weighing its components, may be can only be determined by
a quantitative analysis. Probably every bead contains more or
less of each of a large number of double borates. However, as
the colors can be seen best by using small quantities of the
oxides with a large excess of borax, the metaborates predomi-
nate, and as such the beads are usually represented.


The same laws which apply to the formation of polyborates
apply to the formation of polyphosphates and silicates. Hence
the vast number of natural and artificial silicates can be traced
to their corresponding acids.

Ferrous Iron, Fe" (salt for study, ammonium ferrous
sulphate, (NH 4 ) 2 Fe"(S0 4 ) 2 ).

1. NH 4 OH and NaOH precipitate ferrous hydroxide,
Fe(OH) 2 , which oxidizes quickly to brown ferric
hydroxide, Fe(OII) 3 . NH 4 C1 partly prevents the pre-
cipitation by NH 4 OH, and partly that by NaOH.

2. (NH 4 ) 2 S precipitates black ferrous sulphide, FeS.

3. K 4 Fe(CN) 6 precipitates white potassium ferrous
ferrocyanide, K 2 Fe"Fe(CN) 6 , which rapidly oxidizes to
Prussian blue.

4. K 3 Fe(CN) 6 precipitates Turnbull's blue, ferrous
ferricyanide, Fe 3 "(Fe(CN) 6 ) 2 .

5. Borax bead with the oxidizing flame gives a yellow
coloration, Na 6 Fe 2 '"(BO 2 ) 12 ; with the reducing flame,
a green coloration, Na 2 Fe"(BO 2 ) 4 .

Ferric Iron, Fe'" (salt for study, ferric chloride, FeCl 3 ).

1. NH 4 OH and NaOH precipitate brown ferric hydrox-
ide, Fe(OH) 3 , insoluble in excess of reagents. NH 4 C1
does not prevent the precipitation either by NH 4 OH or
by NaOH.

2. BaC0 3 precipitates a brown basic salt, Fe 2 O(CO 3 ) 2 .

3. H 2 S reduces ferric to ferrous salts and precipi-
tates free sulphur : -

2 FeCl 3 + H 2 S - 2FeCl 2 + 2HC1 + S.


The precipitate of sulphur formed in this reaction is sometimes
mistaken for the sulphides of certain members of Group II.
This confusion will not occur if it is recalled that the members of
Group II are all colored, whereas a precipitate of finely divided
sulphur is white.

4. (NH 4 ) 2 S reduces ferric to ferrous salts and precipi-
tates ferrous sulphide :

2FeCl 8 + 3 (NH 4 ) 2 S = 2FeS + 6NH 4 C1 + S.

5. K 4 Fe(CN) 6 precipitates Prussian blue, ferric ferro-
cyanide, Fe 4 '"(Fe(CN) 6 ) 3 .

6. K 3 Fe(CN) 6 gives a brown coloration.

7. KCNS gives a deep-red coloration, ferric sulpho-
cyanate, Fe(CNS) 3 . 1

8. Borax bead gives the same results as with ferrous


The separation of the members of this group is based
upon the facts that Cr(OH) 3 is oxidized to soluble
Na 2 CrO 4 by means of Na 2 O 2 or by fusion with the
mixture of Na 2 CO 3 and KNO 3 , and that A1(OH) 8 is
soluble in an excess of NaOH. Two methods of mak-
ing the separation are given, of which the first and
simpler is to be employed in the absence of phosphoric,
boric, silicic, and hydrofluoric acids; whereas the sec-
ond and more complicated method is to be followed in
the presence of these bodies. 2

Boil off all traces of H 2 S, testing for its removal by
holding above the liquid a strip of paper moistened
with AgNO 3 or Pb(NO 3 ) 2 . It must be driven off


completely; since, if it were allowed to remain, the
members of Group IV would be precipitated out of due
course upon the addition of NH 4 OH, the precipitant
for the members of Group III. Next test for iron by
adding K 3 Fe(CN) 6 to a small portion of the solution;
and in case it is present add a few drops of HNO 3 and
boil until the reaction for ferrous compounds disappears.
The oxidation of iron to the ferric state at this point
is necessary, since, if left in the ferrous state, it
would not be precipitated by NH 4 OH in presence
of NH 4 C1.

The solution should now be tested for oxalic acid
or other organic matter by evaporating a small portion
of the solution to dryness, and heating the residue in
a closed tube connected with a small rubber delivery
tube, through which any gas that may be evolved can
be conducted into lime water. A charred residue in
the closed tube and a white precipitate in the lime
water indicate the presence of organic compounds. 1
If such are found, evaporate the whole solution to
dryness and heat the residue with the addition of a
few drops of sulphuric acid, until the organic matter
is thoroughly decomposed. Cover the residue with

1 In testing for organic matter blackening is not conclusive. Many inor-
ganic salts, among them those of iron, nickel, cobalt, and manganese,
blacken when heated. Furthermore, a failure to blacken is not an evidence
of the absence of organic compounds, since some of them which contain a
large per cent of oxygen oxalates in particular do not char, but give
off all their carbon as oxides. The lime-water test, too, is not absolute,
though more reliable than that by charring. Compounds evolving oxides
of sulphur also whiten lime water. But both tests are good signs ; and as
gentle ignition is also the means of eliminating silicic acid, it is best to evap-
orate and ignite the solution even when the presence of organic acids is
doubtful. ^


concentrated HC1, evaporated almost to dryness, add
water and a few drops of concentrated HNO 3 , and boil.
Filter the solution from the separated carbon and silica.
Phosphoric acid 1 must next be tested for by warm-
ing a small portion of the nitrate with an excess of
HNO 3 solution of (NH 4 ) 2 MoO 4 . Should it be present,
barium must also be tested for at this stage by making
a small portion alkaline with NH 4 OH, acidifying with
HC 2 H 3 O 2 , and testing for barium with K 2 CrO 4 . Next
add NH 4 C1, 2 boil, and add NH 4 OH till its odor persists.
Filter quickly while hot. Reserve nitrate (a) for subse-
quent groups. Redissolve residue (a) in least quantity
of HC1, nearly neutralize with Na 2 CO 3 , transfer to a
stoppered flask, and add when cold a large excess of
suspended BaCO 3 . 3 Shake from time to time, and filter
after 15 minutes. If phosphoric acid is present, combine
nitrates (a) and (ft) 4 ; if absent, test nitrate (b) for man-
ganese by evaporating to dryness and fusing with Na 2 CO 3
and KNO 3 . Residue (b) may consist of basic salts of
Group III, if phosphoric acid is absent ; or, if present, it
may also contain phosphates of metals of Groups III,
IV, V, and of magnesium.

In Absence of Phosphates

Method 1. Thoroughly wash the residue (b) and trans-
fer to a test-tube. Add a small quantity of water and
some bits of Na 2 O 2 , and boil till effervescence ceases.
Filter and wash. The residue (<?) may be brown Fe(OH) 3 ,
whose identity can be confirmed by dissolving in dilute
HC1 and testing with K 4 Fe(CN) 6 . The filtrate (c) may
contain yellow Na 2 CrO 4 and Na 3 AlO 3 .


The following equations explain the formation of the
soluble salts :

3 Na 2 O 2 + 2 Cr(OH) 8 = 2 Na 2 CrO 4 + 2 H 2 O + 2 NaOH ;
6 NaOH + 2 A1(OH) 8 = 2 Na 3 AlO 3 + 6 H 2 O.

Divide the filtrate (c) into two parts. Acidify the
one with HC 2 H 3 O 2 and test for chromium by adding
Pb(C 2 H 8 O 2 ) 2 . Acidify the other part with dilute HC1,
and while boiling test for aluminum by adding an
excess of NH 4 OH, or to the second part of the
filtrate add some NH 4 C1 and boil. After cooling, a
gelatinous precipitate confirms presence of aluminum.

Method 2. Dry residue (b) and fuse in a platinum
crucible or foil with an excess of fusion mixture. The
cooled mass is triturated in a mortar, preferably a
glass one, is digested with water for fifteen minutes,
and then is filtered. The residue (c) is tested for iron,
and the filtrate (c) for chromium and aluminum, as in
Method 1. In case of doubt test the solution for
aluminum with the spectroscope.

In Presence of Phosphates

Dissolve residue (b) in a little HC1, nearly neutralize
by cautious addition of Na 2 CO 3 , add NaC 2 H 3 O 2 and
HC 2 H 3 O 2 , and boil and filter. The filtrate (c) may con-
tain phosphates of the metals of Groups IV and V,
and of magnesium. The residue (c) may consist of the
phosphates of aluminum, chromium, and iron. To the
filtrate (c) add dilute FeCl 3 , drop by drop, until a
red coloration appears. At this point, precipitation of


FePO 4 is completed. The red color indicates the
formation of Fe(C 2 H 3 O 2 ) 3 , an excess of which would
redissolve the FePO 4 . Hence the solution should now
be boiled in order to change any excess of Fe(C 2 H 3 O 2 ) 3
to an insoluble basic acetate, FeO(C 2 H 3 O 2 ). It is neces-
sary to filter the mixture with the pump while hot, as
FeO(C 2 H 3 O 2 ) redissolves to Fe(C 2 H 3 O 2 ) 3 on cooling.
The filtrate (d) may contain the chlorides of Groups IV
and V, and of magnesium. This filtrate should be
combined with filtrate (a) and afterwards tested for sub-
sequent groups. The residue (c), possibly consisting of
FePO 4 and some FeO(C 2 H 3 O 2 ), should be rejected.

The separation of the members of Group III, which
may be present as phosphates in residue (5), can now
be made without difficulty by either of the procedures,
Methods 1 and 2.

The following further explanations may be given concerning
the method of analysis in presence of phosphates :

It will be seen, on reference to the Table of Solubilities given
on p. 34, that the phosphates of Mg, Ba, Sr, Ca, Co, Ni, Mn, and
Zn are all more or less insoluble in water or alkaline solutions.
If present in an acid solution, they therefore will be thrown out
upon neutralization ; and if this solution be under examination
for the members of Group III, they will separate simultaneously
with the hydroxides of this group. Hence the necessity of follow-
ing the modified method of analysis in their presence. They play
so important a part in the separation of the members of Group III
that the student is advised to perform in advance the exercises
with phosphoric acid, given on p. 135, in order that he may
have some practical knowledge of their reactions.

In removing phosphoric acid after almost neutralizing the HC1
solution with Na 2 CO 3 , NaC 2 H 3 O 2 is added to destroy the solvent
effect of HC1, which dissolves all of the phosphates. HC 2 H 3 O 2 ,


which is set free from NaC 2 H 3 O 2 by metathesis with HC1, dissolves

1 2 3 4 5 7 9 10 11 12

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