William Thomas Boone.

A complete course of volumetric analysis for middle and higher forms of schools online

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Excess NaOH found

= (0.0049 X 3 x -3 x 10 x $2\ = 1.252 grm. NaOH.
\ 497

Weight of NaOH used to expel all NH 3

= (2 - 1.252) grm. = 0.748 grm.


Weight of NH 3 expelled = 0.748 x -.? grm.

Percentage of NH 3 in the crystals

0.748 x I7
= 1. x ioo = 31.79 per cent.

This being close to the theoretical value the crystals
may be assumed to be pure.

B. This illustrates the saving of time when solutions


accurately N. and " are used.


31.3 cu. cm. ^ acid were found =

.-. 3 i 3 cu.cm. N: acid = { whole of residual NaOH in the
I0 \fiask = 31.3 cu. cm N. NaOH.

And since 50 cu. cm. N. NaOH were originally taken,
50 cu. cm. 31.3 cu. cm. (i.e. 18.7 cu. cm.) N. NaOH
were required to expel the NH 3 ,

And 18.7 cu. cm. N. NaOH = 0.04 X 18.7 grm.

= 0.748 grm.

Exercise 47. Estimate the percentage weight of
ammonia in a sample of commercial ammonium
sulphate by the "direct" method.

Directions. i. Fit up the apparatus shown in fig. n.

A. Dropping-funnel containing strong solution of NaOH
(5 to 10 grm. of solid dissolved in water). The funnel
does not dip into the liquid in B.

B. Flask (J- to J-l. capacity) contains the weighed-out salt
(i to 2 grm.) dissolved in about 40 cu. cm. water; together
with a piece of granulated zinc to prevent boiling by bump-

c contains small pieces of clean broken glass, ioo cu.
cm. of N. acid are carefully poured through it into flask
D. The moistened glass will now trap any NH.< which
escapes combination with the main bulk of acid in D.


E. The delivery-tube, should be fairly wide and cut off
obliquely at both ends, one of which is shown enlarged at
F. Note that it almost touches the surface of the acid
in D. If this is properly cut and carefully adjusted, the acid
cannot regurgitate into B, and there will be less chance

Fig. ii

of NaOH being carried forward in the steam from B to D.
The tube is conveniently made in two parts joined by a
rubber connection.

The corks used should be of rubber, and all fittings
must be air-tight.

2. Allow the NaOH in A to flow into B, close the stop-
cock, and boil gently until half to two-thirds of the water

(C906) 6


in B has passed over. Rapid boiling tends to drive some
of the alkali over into D and thus vitiates the result.

During this process the delivery-tube passing into D will
probably require some readjustment.

3. When all NH 3 has been driven over into D, with a
jet of distilled water thoroughly wash down all acid from
the pieces of glass in c and from the end of the delivery-

4. The acid in D may be now transferred to a suitable
graduated flask, made up to a known volume, and titrated
with suitable standard alkali.

5. Indicator: methyl orange.

6. The weight of acid which has been neutralized will
be the equivalent of the NH 3 contained in the salt taken.

Notes. (i) The above is known as the "direct method", and
may be used in estimating any ammonium salt.

(ii) Ammonium sulphate is a valuable by-product in the
manufacture of coal-gas. It is useful as a manure on account
of its nitrogen.

Question 47. What weight of nitrogen is there in a
kilogram of the sample of ammonium sulphate you have
just estimated?

Exercise 48. Find the proportion of each salt
present in a mixture of chlorides of ammonium
and sodium.

Directions. i. Determine the weight of NH 3 in about
2 grm. of the mixture by either of the foregoing methods.
If this is x grm., the weight of NH 4 C1 present will be

535* grm.

2. Find the weight of Nad by difference.

Note. This may also be done by estimating the percentage of
chlorine in the mixture. See Questions 76, et seq.

Exercise 49. A solution contains both caustic soda
and free ammonia. Find the weight of each dis-
solved in 1 1.


Directions. i. Find the weight of acid required to neu-
tralize the total alkali in a given volume.

2. Repeat (i) after boiling off the ammonia, when the
weight of NaOH can be found.

3. The weight of NH 3 can be found from the difference
in the weights of acid used in (i) and (2).


In comparison with mineral acids these are "weak",
and it will be well to examine the behaviour of some
of their salts containing strong and weak bases with
different indicators.

Exercise 50. Find the effect of dilute solutions
of acetates, &c., of such bases as sodium, ammonium,
lead, &c., upon litmus, methyl orange, and phenol-

Exercise 51. Assuming the acidity of the vinegar
provided to be wholly due to acetic acid, find its
percentage weight in the sample.

Directions. i. Use - NaOH, free from Na.,CO 3 .

2. Equation

CH 3 .COOH + NaOH = CH 3 .COONa + H 2 O.

3. Indicator: phenolphthalein. Why?

The colour change can be better followed by dropping
the vinegar into the alkali. Should the vinegar be so dark
coloured as to obscure the end-point, a strip of paper dipped
in the indicator may be used externally.

4. Do not assume that i cu. cm. of the vinegar weighs
i grm. Find the weight of a definite volume.

5. Percentage required = weight of acid x joo

weight of vinegar

The equivalent of an organic acid is the weight
required to neutralize one equivalent of alkali.


In carrying out this determination the most reliable
indicator is phenolphthalein. But since caustic soda
or potash is apt to contain carbonate, and the pink
colour due to this indicator disappears when only half
the carbonate has been decomposed, standard baryta
is preferred, as its clear solution can contain no car-
bonate, and any formed by reaction with carbon dioxide
can be immediately detected.

Exercise 52. Prepare some decinormal barium

Directions. i. Introduce 4 to 5 grm. of the finely pow-
dered crystallized hydroxide into about 200 cu. cm. of
distilled water contained in a flask. Cork, and allow to
stand for a couple of days, shaking up occasionally. The

saturated solution produced will be about '-.


2. Decant the clear liquid into a clean flask fitted with
rubber cork.


3. Find its exact strength with - - HC1. Indicator:


Equation Ba(OH) 2 + 2 HC1 = BaCl 2 + 2 H 2 O.

171.4 72.9

4. Dilute to ~ strength with freshly boiled and cooled


distilled water.

5. Test your result gravimetrically.

Measure out 25 cu. cm. into a weighed dish, and add
dilute H 2 SO 4 until slightly acid. Evaporate to dryness on
a water bath, then ignite at dull red heat for five to ten
minutes over a Bunsen flame. Cool in exsiccator and
weigh. The weight of Ba(OH) 2 in the solution taken was

17M t h at O f the sulphate formed.
2 33-4

* The stock bottle for baryta water should be fitted with


rubber stopper and "calcium chloride tube" containing dry soda-

Exercise 53. Make some one pep cent (approx.)
solution of tartarie acid, and determine the equi-
valent of the acid.

Directions. i. Dissolve up about 2 grm. of crystals,
accurately weighed, to make 200 cu. cm. of solution.


2. Titrate with -Ba(OH) 2 . Indicator: phenolphthalein.

3. Since barium has diad valency, the equivalent of the
acid will be that weight in grammes which neutralizes
half the gramme formula weight (85.7) of Ba(OH) 2 . (See
notes, p. 152.)

Basicity and Molecular Weight of Organic Acids.
When a definite weight of some particular alkali is
neutralized by acid, a salt is produced which may
often be easily crystallized out and identified. Some
acids invariably yield the same salt no matter in what
proportions acid and alkali are mixed, and the reagent
in excess remains mixed with the salt. There are
others which can be induced to form more than one
distinct salt when the acid is in excess of that neces-
sary to form the normal salt. In every such case,
however, it is found that the quantity of acid required
to form "acid salts" bears a simple ratio to that
necessary for the production of the normal salt, pro-
vided the same weight of alkali is used. Acids may
thus be classified as monobasic, dibasic, &c., accord-
ing to the number of distinct salts each can form with
some particular alkali. In the case of a dibasic acid,
the quantity found necessary to form the acid salt is
just double that required to form the normal salt.

The number of salts which an organic acid is cap-
able of forming with an alkali thus furnishes a clue
to its molecular weight. Since monobasic acids can


form one salt only with an alkali, it is assumed that
their molecules contain only one hydrogen atom re-
placeable by metal ; and consequently that weight of
the acid (in grammes) which can be neutralized by
40 grm. of NaOH is assumed to be its molecular
weight. The molecular weight of a monobasic acid
is thus equal to its equivalent. In the case of a dibasic
acid, since the acid salts derived from it must contain
at least one atom of sodium (or other monad radical),
and that proportionately twice as much sodium hy-
droxide is required to produce the normal salt, it
follows that the acid contains at least two replaceable
hydrogen atoms, and its gramme molecular weight
may be assumed to be that neutralized by 80 grm. of
NaOH. In other words, it is assumed to be twice
the equivalent.

Speaking generally then, the molecular weight of
an organic acid = its equivalent x basicity. (For
other methods of determining basicity see works on
Physical Chemistry.)

Question 48. The formula given to tartaric acid may
be written C 4 H 6 O 6 . From its equivalent (found in Ex. 53)
what would you conclude its basicity to be?

Exercise 54. Oxalic acid whose formula may be
represented as H>C.Q 4 crystallizes with two molecules
of water. Find its basicity.

Directions. i. 3 grm. of crystals dissolved up to 250

cu. cm. will work with ' alkali.


2. Indicator: phenolphthalein.

molecular weight

3. Basicity = - -. p-


(See notes, p. 152.)
Note on the term "Equivalent". It is interesting to remember



that the idea of equivalence was first suggested by the discovery
in the latter part of the eighteenth century, that different weights
of various bases were necessary to neutralize one and the same
weight of acid. These different weights were said to be "equi-
valent" in neutralizing power. Further research showed a similar
relationship between elements. Wollaston used the term for the
proportional weights in which elements combine, and fixed that
of hydrogen as unity, because its proportional weight in a com-
pound was found to be invariably less than that of any other

Question 49. The formula for acetic acid is H 4 C 2 O 2 ,
and it is found that 2. grm. of sodium hydroxide neutralize
3 grm. of the acid. (a) Find what weight of the acid
should be contained in a litre of normal strength, (b) What
inference do you draw as to the basicity of the acid?

Question 50. On titrating an aqueous solution of a
dibasic organic acid containing i grm. in 250 cu. cm. with


- baryta water, it was found that 25 cu. cm. of the acid

solution required 22.2 cu. cm. of Ba(OH) 2 to neutralize

it. What is the formula weight of the acid?

Question 51. Two solutions A and B of the same acid
were prepared, A containing 8 grm. and B 14 grm. B was
partly neutralized by the addition of 8 grm. of alkali, and
each solution was made up to i 1. On titrating the acid
solutions with unstandardized caustic soda it was found
that 20 cu. cm. of the latter neutralized 10 cu. cm. of A
and 40 cu. cm. of B. Find the ratio between the weights
of acid and alkali necessary to neutralize each other.

Question 52. In another case it was found that B was
alkaline, and that 5 cu. cm. of A were necessary to neu-
tralize 25 cu. cm. of B. What is the ratio in this case?

Question 53. Some solutions of caustic potash and
baryta water were titrated against acid, when it was found
that 50 cu. cm. of baryta water and 40 cu. cm. of the potash
were necessary to neutralize equal volumes of the acid.
Compare the number of atoms of metal present in equal
volumes of the solutions.



Methods depending on Direct
Oxidation and Reduction

The chief compounds used in volumetric estima-
tions involving direct oxidation are potassium per-
manganate and potassium dichromate, both of which
are readily reduced to other substances. In each case
the change is sharp and easily detected.

Potassium Permanganate (KMnO 4 ). The reaction
between this compound and the reducing agent hy-
drogen sulphide may here be recalled. The odour of
the one and the purple colour of the other disappeared
together, and it was evident that new substances were
formed. The addition of dilute sulphuric acid pre-
vented the formation of the dark precipitate of man-
ganese hydroxide, and allowed the colour change
to be readily followed. A standard solution of per-
manganate can be used in the estimation of several
substances which tend to pass into a more highly
oxidized condition; but its commonest use is for
estimating ferrous iron.

Exercise 55. Find the effect of dropping- some
dilute permanganate solution into ferrous and ferric
solutions separately with and without the previous
addition of dilute sulphuric acid; also into dilute
and concentrated sulphuric acid alone.

Directions. i. Use FeSO 4 and FeCl 3 .
2. Add the KMnO 4 till the solutions become permanently

There appears to be no reaction between perman-


ganate and ferric solutions, whether acid is present
or not, as the purple colour is retained from the first
drop. This is not so with ferrous solutions, as the
colour is instantly destroyed and does not appear
until a certain quantity of the permanganate has been
added, when the appearance resembles that in the
ferric solution. We may therefore assume that at
this point there is no ferrous iron remaining. Dilute
sulphuric acid has no apparent action on perman-
ganate, but it prevents precipitation of oxides of
manganese and iron in the above reaction, and allows
the colour change to be clearly seen. Ferrous sul-
phate and permanganate may be supposed initially
to break up thus

loFeO, SO 3 + K 2 O, Mn 2 O 7

= 5 Fe 2 O 8 + 2 MnO + K 2 O + ioSO 3 .

Ten equivalents only of SO 3 are supplied by the
ferrous sulphate, whereas eighteen are required to
convert the whole of the oxides to soluble sulphates.
Eight equivalents must therefore be supplied by added
sulphuric acid. A great excess of this acid must be
avoided, as it is capable of reducing the permanganate
to green man ganate. Strong sulphuric acid decom-
poses solid permanganate on heating sometimes with
violence, when oxygen is liberated:

4KMnO 4 -f 8H 2 SO 4

= 4 KHSO 4 + 4 MnSO 4 + 6 H,O + 5 O 2 .

The reaction between ferrous sulphate and potassium
permanganate in the presence of dilute sulphuric acid
may now be written

2 KMnO 4 -f 10 FeSO 4 + 8 H 2 SO 4

^Ti 558.4 (Fe")

= 5Fe,(S0 4 ) 3 + K. 2 S0 4 + 2 MnSO 4 + 8H 2 O;


316.1 grm. of the permanganate being necessary to
supply the oxygen required by 558.4 grm. of ferrous
iron for its conversion to the ferric condition. It will
be evident, therefore, that when the strength of a
solution of either salt is known, that of the other can
be easily determined. It need hardly be pointed out
that in the estimation of the total iron in a solution
by this method, care must be taken to have it wholly
in the ferrous condition. Now it might appear that
any standard solution of permanganate can be easily
prepared by dissolving the required weight of solid
up to the proper volume. Unfortunately, however,
the salt is not very soluble (it requires fifteen times
its weight of water for complete solution); and, in
addition, the solid being apt to suffer slight decom-
position, the formula KMnO 4 cannot be relied upon
as representing its exact composition. It is usual
to prepare a solution somewhat above decinormal
strength, determine its exact value by titration against
some specially prepared ferrous solution, then dilute
it to the required standard. The ferrous- solution may
be made by dissolving in sulphuric acid a known
weight of the purest iron wire obtainable, taking
proper precautions to prevent oxidation by air. The
excess of acid present does not matter. Or, a known
weight of pure ferrous salt may be dissolved up to
a given volume with similar precautions. Ferrous
ammonium sulphate is usually considered the best
for this purpose, as in the presence of the ammonium
salt the ferrous sulphate does not oxidize so readily.
(See also p. 28.)

The chief points to be remembered in using per-
manganate are

i. The salt, whether solid or in solution, is apt to
suffer some decomposition on keeping. Dr. Friend


says that, unless pure, a solution is apt to alter in
value in a few days.

2. The solid does not dissolve very readily; and
undissolved particles may easily escape detection in
the highly coloured liquid.

3. It is decomposed by rubber, paper, and other
organic matter, as well as by ordinary reducing
agents. Filtering a standard solution through paper,
or using it in an ordinary Mohr's burette, therefore
reduces its strength.

4. It reacts with hydrochloric acid, and liberates
chlorine, unless the acid is present in very small
quantity, or unless manganous or ammonium sul-
phate is present as well. About 0.5 grm. added to
25 cu. cm. of ferrous solution is found to be sufficient
to prevent this reaction.

K 9 O, Mn 2 O 7 + i6HCl

= 2 MnCl 2 + 2 KC1 + 5 C1 2 + 8 H 2 O.

Exercise 56. Make solutions of ferrous sulphate
and ferrous ammonium sulphate, and test for ferric

Direction. Use potassium ferrocyanide (K 4 FeC 6 N 6 ) and
ammonium sulphocyanide (NH 4 SCN).

Exercise 57. Prepare about 20 grm. of "Mohr's
salt" (ferrous ammonium sulphate crystals) in a
finely crystalline condition.

Directions, i. Dissolve formular proportions of the
purest crystals of ferrous sulphate and ammonium sulphate
separately in the smallest quantity of moderately hot dis-
tilled water recently boiled, and add a few drops of dilute
H 2 SO 4 . Formula weight of FeSO 4 , 7 H 2 O = 278; that
of (NH 4 ) 2 SO 4 = 132.2. gV f these weights in grm.
should yield 18 to 20 grm. of the required crystals.


2. Mix the solutions and cool whilst constantly agitating,
so as to produce small crystals.

3. After standing about an hour, filter and dry the
crystals thoroughly by pressure between filter paper, then
expose them to air until they do not clog together.

4. Preserve in a stoppered bottle.

5. Formula: Fe(NH 4 ) 2 (SO 4 ),, 6 H 2 O.

Question 54. What weight of iron is contained in a
formula weight of Mohr's salt? What proportion of the
salt consists of iron? What weight of the crystals should
be dissolved up to 250 cu. cm. to make a solution equiva-
lent to KMnO 4 ?

316.06 grm. KMnO 4 lose 80 grm. of oxygen in oxidizing
558.4 grm. of ferrous iron. .*. i cu. cm. ' solution =

0.00316 grm. KMnO 4 , and the iron ammonium solution
must contain 0.00558 grm. Fe" which is contained in

0.004102 grm. of the crystals. 250 cu. cm. of - -' ferrous

ammonium sulphate will therefore require 10.2 grm. of

Exercise 58. Make a litre of ^ permanganate

solution, standardizing* it with your own ferrous
ammonium sulphate.

Directions. i. Grind the KMnO 4 to a fine powder to
ensure quick solution, and dissolve 3.5 grm. up to a litre.

N.B. Be careful that the solution is complete before
titrating. To ascertain this, allow the liquid to stand a
few minutes, and decant it into another vessel. Do not
filter through paper.

2. Use recently boiled distilled water to dissolve the
FeAm.,SO 4 , 6 H 2 O. Take care that it is quite cold before
making up to the mark.

3. KMnO 4 in burette with glass tap.


4. Add a few cubic centimetres of dilute H 2 SO 4 to each
portion -of ferrous solution titrated. A brown precipitate
shows insufficiency of acid.

5. Drop in the KMnO 4 till a permanent faint-pink colour
is obtained.

6. Read to the surface of the KMnO 4 .


7. If you have made your ferrous solution exactly -,


25 cu. cm. should require less than 25 cu. cm. of the
KMnO 4 solution.

8. To make the necessary dilution

Suppose 25 cu. cm. - ' Fe" solution = 22.5 cu. cm.

KMnO 4 . Evidently 22.5 cu. cm. contain as much KMnO 4

as 25 cu. cm. - KMnO 4 . So 22.5 cu. cm. must be made

up to 25 cu. cm., or 900 cu. cm. to i 1.

Exercise 59. With dilute sulphuric acid and iron
wire whose percentage purity is known, make a
solution of ferrous salt, and with it check the
result obtained in the last exercise.

Directions. i. Make 100 or 200 cu. cm. This solution
easily oxidizes, and is made as required.

2. Accurately weigh sufficient wire cut into pieces.
[Piano \vire or fine binding wire may be assumed to contain

99.6 per cent Fe. For 100 cu. cm. - - ferrous solution,

100 x 0.00558 grm. will be required; i.e. about 0.5 grin,
will do very well.]

3. Calculate the weight of pure iron in the quantity of
wire taken.

4. To prevent the formation of ferric compound during
solution (by contact with air). Fit the flask with cork and
exit tube, &c., as shown either in A or B (fig. 12). One-third
fill it with dilute H 2 SO 4 , and drop in i or 2 grm. of pure
NaHCO.,, and then the iron wire. When the evolution of


CO., slackens replace cork, &c. The expelled gas will
drive all air out of the flask. Solution will be quickened
by gently heating.

In A the expelled hydrogen is bubbled through freshly
boiled water contained in the beaker. When solution is
complete, some of this water may be allowed to pass back-
wards into the flask. B is a Bunsen valve, i.e. a piece of


Fig. i a

rubber tubing with a clean slit cut lengthwise about half
an inch long. It is slipped over the projecting tube, while
the free end is stopped with the glass rod. This arrange-
ment allows hydrogen and steam to escape, while it prevents
air from entering, since the edges of the slit close up.

5. When the iron has dissolved cool quickly, and make
up to the mark with distilled water recently boiled and
cooled. The carbon particles may be neglected.

Note. If carefully prepared, this solution should give a more
reliable result than the ferrous ammonium solution.


Exercise 60. Prepare some pure crystals of
oxalic acid from cane sugar.

Directions. i. Do this in a fume closet.

2. In a porcelain dish heat 70 cu. cm. of strong HNO 3
to about 100 C. ; then having removed it from the flame,
add 15 grm. of cane sugar.

3. After the reaction has ceased, evaporate to a quarter
its bulk, and set aside till cold.

4. Separate the crystals, and dry them on filter paper.

5. Redissolve them in the smallest possible quantity of
hot water, and allow the crystals to form again. This
may be repeated.

6. Dry the crystals thoroughly, and keep in a stoppered

Question 55. Oxalic acid is dibasic, and its crystals
are represented by the formula H 2 C 2 O 4 , 2 H 2 O. What


weiefht should be taken to make i 1. of - acid?


Exercise 61. Test the purity of your crystals by
titration against standard alkali.

Directions. i. Weigh out sufficient crystals to make

-50 cu. cm. of acid presumably '. This will be sufficient

for this and the next exercise.


2. Use - NaOH and phenolphthalein.

3. 80 grm. NaOH = 126 grm. of crystals
H 2 C 2 O 4 , 2 H 2 O + 2 NaOH = Na. 2 C 2 O 4 + 4 H 2 O.

^6 80

Note. Decinormal oxalic acid is not stable, and should be
prepared as required. Stock solution should be of N. strength.

Exercise 62. Find the weight of oxalic acid
necessary to reduce 3*16 grm. of potassium per-



Directions. i. ' KMnO, in burette.

2. Use the oxalic solution prepared in the last exercise.
To each portion add a little dilute H 2 SO 4 or HC1. Half
the titrations might be made with each acid, noting- any
difference in the results.

3. Befoce titrating, heat each portion to 60, as the re-
action takes place very slowly in the cold.

4. Equation expressing the reaction :

2 KMn0 4 + 5 (H 2 C 2 4 , 2 H 2) + 3 H 2 SO 4
316 630

= K 2 SO 4 + 2 MnSO 4 -f ioCO 2 + 18 H 2 O.

N.B. Accurate work in the last exercise will show that

the reaction is complete when equal volumes of ' KMnO 4

1 2 3 5 7 8 9 10 11

Online LibraryWilliam Thomas BooneA complete course of volumetric analysis for middle and higher forms of schools → online text (page 5 of 11)