William Thomas Boone.

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

. (page 9 of 11)
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Exercise 108. Make 250 eu. em. each of ^ NaOH
and ^ KMn0 4 . Standardize each with oxalic acid,
pure crystals of which are provided.

Exercise 109. A given powder consists of a mix-
ture of oxalic acid (H,C. 2 4 , 2H 2 0) and potassium
oxalate (K 2 C,0 4 , 2H 2 0). Find the proportion of
each present.

Directions. i. Estimate free oxalic acid with standard
alkali. See Ex. 54.

2. Estimate total oxalate with standard KMnO 4 . See
Exs. 61 and 62.

3. Weight of potassium salt will be - ? that of the
combined acid.

Exercise 110. A solution of hydrochloric acid has
been partially neutralized by potash. Find the
weight of each originally used per litre.

Directions. i. Total chlorine, and hence total HC1 used,
obtained by titration with standard AgNO 3 .

2. Estimate free HC1 with standard alkali.

3. Calculate the weight of KOH equivalent to the HC1

Exercise 111. A solution of caustic potash has
been partly neutralized with hydrochloric acid.
Find the weight of KOH per litre originally present.

Directions. i. Estimate the free alkali with standard

2. Estimate chlorine, acidifying (or neutralizing) each
portion titrated with HNO 3 .

Total KOH will be the sum of that found free and that
equivalent to the chlorine.

Exercise 112. Solutions of potassium chloride and
potassium bicarbonate have been mixed. Find the
weight of each in 1 1.


Exercise 113. A solution contains both hydro-
chloric and sulphuric acids. Find the weight of
each present per litre.

Directions. i. Find total acid-forming hydrogen with
standard alkali.

2. Find that hydrogen combined with chlorine with
standard AgNO 3 .

Note. Consider whether there is any advantage in using the
portions titrated in (i) for titration with AgNO 3 in (2).

Exercise 114. Find the percentage of available
chlorine in some fresh bleaching- powder by its
oxidizing action on a ferrous salt.

Directions. i. To about 10 grm. of Mohr's salt (ferrous
ammonium sulphate crystals) add about 25 cu. cm. dilute
H 2 SO 4 ; dissolve, and make up to 250 cu. cm. with boiled-
out water.

2. Prepare solution of bleaching powder. (See Ex. 81.)
Dissolve up i to 2 grm. of the powder to 250 cu. cm. of
solution, and titrate as soon as possible.

3. To 25 cu. cm. of the acid ferrous solution add 25
cu. cm. of the bleaching solution and mix well; then find
excess of ferrous iron.


Ca(OCl,) + 4 FeSO 4 + 2 H 2 SO 4

= 2 Fe 2 (S0 4 ) 3 + CaCl 2 + 2 H 2 O

shows 56 grm. of ferrous iron are oxidized by 8 grm. of
oixygecn = 3-5.5 grin, available chlorine.

Exercise 115. Estimate the percentage of barium
in some pure crystals of baHum chloride. How
does it accord with the formula BaCL, 2H,0?

Method I. By standard Na.,CO r See Exs. 39 and 40.
The solution should be neutral.
Method II. By standard K 2 Cr.,O 7 .
Directions. The solution of Bad., must be made alka-


line with AmOH, free from carbonate, and heated to 70 C.
Titrate with standard K 2 Cr 2 O 7 . The end-point is reached
when no further cloudiness is produced, and the clear
liquid shows a trace of yellow colour due to slight excess
of K 2 Cr 2 O 7 . Equation

2 BaCl 2 + K 2 Cr 2 O 7 + 2 NH 3 + H 2 O
416.6 294.2

= 2 BaCrO 4 + 2 KC1 + 2NH 4 C1

shows relationship between the reacting masses of barium
salt and dichromate.

N.B. It should be noticed that i 1. of K 2 Cr 2 O r , as used in
the above or similar estimations, must contain 7.355 grm. K 2 Cr 2 O 7 .

Exercise 116. Estimate the percentage of lead in
some "white lead".

Directions. Dissolve i to 2 grm. in acetic acid, and
precipitate the lead by a known volume of standard K. 2 Cr 2 O 7
taken in good excess. Boil, filter off precipitate, and at once
wash well with hot water. Determine excess of K 2 Cr 2 O 7
in filtrate and washings by adding excess of KI, then

titrating with - - Na 2 S 2 O 3 . See Ex. 79.

294.2 grm. of K 2 Cr 2 O 7 = 414.2 grm. Pb; and equations
given in Ex. 78 and 79 show relationship between iodine
set free and K 2 Cr 2 O 7 , &c.

Exercise 117. Find the weight of NaOCl dissolved
in 1 1. of the hypochlorite solution given.

Directions. Use standard As 4 O 6 . See Ex. 86.

Exercise 118. The organic acid provided reduces
potassium permanganate. Find its equivalent both
as acid and reducing agent.

Directions. See " Equivalent of Organic Acids", p. 69,
also Ex. 53, &c.

(C906) 10


Exercise 119. A solution contains both sulphuric
and oxalic acids. Find the weight of each present
in 11.

Directions. i. Find the total alkali necessary to neu-
tralize both acids.

2. Find weight of oxalic with standard KMnO 4 , and
calculate that of alkali used to neutralize this acid.

3. Difference between results of (i) and (2) will be the
equivalent of the sulphuric acid present.

Exercise 120. Some sulphuric acid had half its
equivalent of potassium hydroxide added to it, to-
gether with a small quantity of oxalic acid crystals
(H 2 C 2 4 , 2 H 2 0). The solution was made up to 2 1.
Find the weight of each reagent originally used.

Solid potassium permanganate and ^ KOH are

Exercise 121. Find the equivalent of mercury in
mercuric compounds by making use of the reaction
between mercuric chloride and sodium hydroxide.

Directions. Dissolve 2 to 3 grm. of pure HgCl 2 . Pre-
cipitate HgO with N. NaOH used in good excess. Find

excess of NaOH with - HC1. The gramme-equivalent of

mercury is that weight corresponding to 40 grm. NaOH.

HgCl 2 + 2 NaOH = HgO + 2 NaCl + H 2 O.

Exercise 122. See if the result obtained in Ex.
121 is confirmed by titrating a solution of mercuric
chloride with standard AgN0 3 .

Exercise 123. Find the proportion in which mer-
curic chloride and potassium iodide react to form
the compound Hgl,, 2KI.

Directions. i. Prepare solutions from the purest salts

obtainable: KI ; N ^ HgCl 2 .
10 20


2. HgCl 2 solution in burette. First trace of permanent
red colour (HgI 2 ) shows end-point; as any excess of
HgCl 9 after all the KI has entered into the compound at
once decomposes the latter, according- to the equation

HgCl 2 + HgI 2 , 2 KI = 2 HgI 2 + 2 KC1.
Note. Personne uses this method to estimate mercury.

Exercise 124. A given solution contains ferrous
oxalate and oxalic acid. Find the weight of each
present per litre. You are provided with solid

potassium permanganate and ^ alkali.

Directions. Use the alkali to estimate the free oxalic

acid; and - KMnO, to estimate total oxalate.

Calculate the weight of KMnO 4 used by the free oxalic
acid. The remainder has reacted with ferrous oxalate
according to the equation

6 KMnO 4 + 10 Fe(COO) 2 + 24 H 2 SO 4

948 1438
= 5 Fe 2 (S0 4 ), + 3 K 2 S0 4 + 6 MnSO 4 + 24 H,O.

Exercise 125. Estimate the percentage of sodium
nitrite in the sample provided by making use of its
reducing action towards potassium permanganate.

Directions. i. Weigh out about a gramme of the nitrite
from a weighing-bottle, and dissolve up to 100 cu. cm.

2. Prepare some KMnO 4 .

3. Mix together 10 cu. cm. nitrite solution with 200
cu. cm. of the KMnO 4 solution, and add about 5 cu. cm.
dilute H SO 4 . Warm to 40, and allow a few minutes
for the reaction to become complete. Note. The reaction
takes place very slowly in cold solution.

4. Find excess of KMnCX with - oxalic acid. The



proportion between the nitrite and the KMnO 4 reduced
by it is shown in the equation

2 KMnO 4 + 5 NaNO, + 3 H 2 SO 4

= K 2 SO 4 + 2 MnSO 4 + 5 NaNO 3 + 3 H 2 O.

What objections are there to titrating- the warmed nitrite
directly with KMnO 4 ?

Exercise 126. Estimate the percentage of pure
tin in a commercial sample.

Directions (Lowenthal's Method). i. Dissolve about
0.5 grm. of the tin in 5 cu. cm. concentrated solution of
ferric chloride containing- a little free HC1. The tin is
present as chloride, and has reduced its equivalent of ferric
to ferrous iron. Use a 25o-cu.-cm. flask, and prevent
oxidation by adding a pinch of NaHCO 3 to expel air.
Warm gently till the solution is complete; then cool, and
add cold boiled-out water up to the mark.

2. Estimate the ferrous iron with - KMnO 4 . Equation


SnCl 2 +Fe 2 Cl 6 = 2FeCl 2 + SnCl 4 shows that 2 x 55.8grm.
Fe" = 119 grm. Sn. (See also Ex. 129, and Precautions
in using Permanganate, p. 75.)

Exercise 127. Assuming- that no iron is present,
estimate the percentage of metallic zinc in zinc

Directions. i. Weigh out accurately about 3 grm. of
iodine into a small strong bottle fitted with a glass stopper,
and dissolve in water with the aid of KI.

2. Add about 0.5 grm. of the zinc dust, ground very
finely, and accurately weighed.

3. Tie down the stopper securely, and place the bottle
in a water-bath, kept near boiling-point, for one hour.
Occasionally shake up the contents of the bottle. By
this time the metallic zinc will have formed ZnI 2 . The
oxide of zinc will remain unacted upon.


4. Allow the bottle to cool; transfer its contents with
the rinsings to a 2OO-cu.-cm. graduated flask, and add
water to the mark. Find the excess of iodine with
standard Na.,S 2 O 3 .

The weight of metallic zinc in the dust taken was

5*5 that of the iodine used up.

Exercise 128. Repeat the estimation carried out
in Ex. 127, using- a gasometrie method.

Directions. Find the volume of the hydrogen evolved
when 0.5 to 0.75 grm. of the zinc dust is treated with sul-
phuric acid. (See apparatus used for determining equiva-
lents of metals.) Reduce volume of hydrogen to S.T.P.

Remember that 22.4 1. of hydrogen (at S.T.P.) are
evolved by action of 65.4 grm. of zinc.

Exercise 129. Estimate the proportion of metal-
lic iron in some "reduced iron" (ferrum redactum).

Method (A). Directions. i. Dissolve about 2.5 grm. of
HgCl 2 in 50 cu. cm. of water contained in a glass-stoppered
bottle. Add about 0.5 grm. of the iron powder accurately
weighed out; then digest for one hour (at least), as given
111*13) of exercise 127. Iron oxide is unacted upon.

2. Make a solution of MnSO 4 : 10 grm. of crystals dis-
solved up to 50 cu. cm. roughly.

3. Cool the contents of the bottle, and filter quickly into
loo-cu.-cm. graduated flask. Wash out bottle with boiled-
out water, allowing the rinsings to pass into the flask.
Make up to the mark with boiled-out water; and proceed
to titrate the ferrous solution at once.

4. To each 10 cu. cm. add 10 cu. cm. (about) of dilute
H 2 SO 4 , and the same quantity of MnSO 4 solution, and

titrate with KMnO 4 .

Note. The KMnO 4 should be dropped in slowly. The end-
point colour is not of the same tint as when no MnSO 4 is used.
Of what use is this salt?


Method (B). Directions. i. Make a hot (short of boil-
ing) solution of (roughly) 5 grm. of CuSO 4 , 5 H 2 O in 75
cu. cm. water contained in a small flask preferably stop-
pered. Add 0.5 to i.o grm. of the iron powder accurately
weighed. Stand for about 15 minutes with occasional
shaking. Keep out air as much as possible.

2. Quickly filter off the precipitated copper, &c. (iron
oxide is acted upon) into loo-cu.-cm. measuring-flask.
Rinse out with boiled-out water, and pass into the flask.
Make up to the mark, when cold, with dilute H SO 4 .

3. Titrate the ferrous solution with * K 2 Cr 2 O 7 .

Exercise 130. Repeat the estimation of metallic
iron in "reduced iron" by a gasometrie method.

Directions. See Ex. 128.


Notes on Electrolytic Dissociation
and Indicators

When an electric current passes through certain
solutions, especially when water is the solvent, the
substance dissolved gradually separates into two
portions with unlike chemical properties. One portion
gathers round the positive terminal and the other
round the negative. Solutions of compounds which
behave in this way are called " electrolytes"; and
since the disrupted components appear to move
through the solution (to one or other terminal or
"electrode") they were called by Faraday "ions"
(efyu = to go). Metals, ammonium, and the acid-


forming hydrogen of acids make their appearance at
the " cathode" and are referred to as "kations",
while those constituents found at the " anode" are
known as "anions".

This behaviour may be explained by supposing
that the ions resulting from each molecule are op-
positely electrically charged. When the oppositely
electrified terminals of a battery are introduced into
the solution, the positively charged ions (kations) are
consequently attracted to one terminal the kathode
and repelled from the other the anode ; whilst those
negatively charged (anions) are attracted to the anode
and repelled from the kathode. Contact with the
electrodes neutralizes the attracted ions; they are
"ions" no longer. The conductivity of a solution
thus appears to depend upon the number of ions it
contains ; and this number will vary with the quantity
of substance dissolved, and with the readiness with
which its molecules electrolytically dissociate (or

Normal salt solutions, and those of certain acids
and bases, conduct the electric current most readily;
from which it must be assumed that these contain the
greatest proportion of dissociated molecules. Such
acids and bases are called " strong", and the common
mineral acids and caustic soda and potash serve as
examples. There is evidence that most of the ordinary
bench reagents contain considerably more than half
their molecules in the ionized state. But whether the
electrolyte be a good or poor conductor, i.e. whether
it may be designated " strong" or "weak", the pro-
portion of its dissociated molecules increases with
dilution until all assume this condition. Even water
itself, from the fact that it conducts the electric cur-
rent, though very feebly, must be looked upon as


containing at least some of its molecules in a dis-
sociated state.

From this point of view chemical reactions are ex-
plained as due to reactions between the ions, which
when removed from solution are replaced by others
derived from previously undissociated molecules, until
one of the reagents is exhausted. It is usual to dis-
tingush kations by dots and anions by dashes the
number of dots and dashes corresponding with the
valency. The ions provided by sodium hydroxide are
represented as Na* and OH'; and those of hydrogen
sulphate as aH* and SO/'. The neutralization of
caustic soda by hydrochloric acid, since both are
" strong", may be represented thus

H' + Cl' + Na' + OH' = Na' + Cl' + H 2 O.

One of the products being a normal salt and remain-
ing in solution is represented in the dissociated state;
but the other product water whose tendency to dis-
sociate is so small, is represented as a molecule.
Weak bases and weak acids are also similarly repre-
sented, although their tendency to dissociate is much
greater than water.

The ion common to all dissociated acids is hydrion
(H*), and to it their acid properties are ascribed.
Similarly, hydroxidion (OH) 7 is common to all dis-
sociated alkalis, and is responsible for their alka-
linity. Neutralization is theoretically attained when
acid and alkali have been mixed so as to produce
equivalent quantities of hydrions and hydroxidions;
for the small proportion of ions which do not unite to
form water molecules is probably no greater than the
water itself contained. We should consequently ex-
pect an aqueous solution of a normal salt also to be
neutral, since none of its ions consist either of hydrion


or hydroxidion. This is only true of those normal
salts resulting from reactions between strong acids
and strong bases ; such as, sodium chloride, potas-
sium sulphate, &c. Where a weak acid or weak
base takes part in their formation, solutions showing
distinct acidity or alkalinity are obtained. Suppose
the acid is weak. Solution dissociates more or less
of the salt, and produces momentarily, we may sup-
pose, an equivalent number of kations and anions;
but the tendency of the acid to ionize being small, its
anions largely unite with hydrions derived from the
water, and become molecules of acid. The base, how-
ever, being strong, comparatively few molecules form;
and, as a result, a sufficient excess of hydroxidions
(also derived from water) remains to affect the in-
dicator. Potassium cyanide serves as an example of
this behaviour. The molecule ionizes into K* and
CN' when the anion combines with H* (derived
from the water) to form molecules of HCN. The
OH' also derived from the water is responsible for
the alkaline reaction. Borax and, to a lesser extent,
sodium acetate behave similarly.

Normal sodium phosphate, although derived from
a strong base and moderately strong acid, appears to
behave exceptionally as its solution is strongly alka-
line. An explanation is afforded by regarding the
salt as derived from the strong base caustic soda and
a weak acid, HNa 2 PO 4 .


According to the above hypothesis, indicators ol
acids and alkalis must possess a weakly acid or basic
character themselves, and show slight tendency to


dissociate, although their salts may do so compara-
tively freely. Their marked colour-changes are due
to differences in colour between their undissociated
molecules and ions derived from them. The addition
of fairly strong acid, or comparative abundance of
hydrogen ions, furnishes a far greater opportunity
than existed previously for what anions of the indi-
cator are present to meet hydrions and thereby form
acid molecules. Whatever proportion of molecules
of the indicator was previously dissociated, it is now
much smaller, and any colour exhibited will be
due to these molecules. On the other hand, the
introduction of a strong base, or abundance of
hydroxidions, lessens the proportion of hydrions
through the formation of water molecules (whose
tendency to dissociate is extremely slight); and
whatever colour is now seen is due to anions of the

Although the composition of many indicators is
unknown, that of one of the commonest is well estab-
lished, and may be used to illustrate the application
of the hypothesis. Phenolphthalein possesses the
properties of a very weak acid, and its solution in
water or in the presence of acid is colourless. Its
molecules are therefore assumed to be colourless.
With alkalis it forms salts whose solutions are red.
This colour must be due to anions of phenolphthalein,
for other salts of alkalis are usually colourless.

The composition of litmus is unknown: but if we
may apply an explanation of its behaviour similar to
the above, red is the colour of its molecules, and blue
that of its anions; and since its neutral solution is
purple, we must assume that a considerable proportion
of its molecules are dissociated to provide sufficient
anions to modify the red due to the undissociated


molecules. Litmus thus appears to dissociate more
freely than phenolphthalein ; or, in other words, to
possess a somewhat stronger acid character. It will
thus be seen that although the indicators used in
acidimetry and alkalimetry show small tendency to
dissociate, there is considerable difference between
them in this respect; and in making a selection, the
character of the acid or alkali to be titrated must be
taken into account. For instance, an indicator com-
parable in strength with the acid used in a titration
will require too great an excess of acid to overcome
the tendency of the indicator to dissociate and cause
the disappearance of any colour due to its anions.
Again, a very weak acid indicator used with a weak
base will produce a salt which can be hydrolysed by
water; and its small amount of ionization will be
masked by its acid molecules in spite of a fair excess
of base. This is seen to occur when phenolphthalein
is used in titrating ammonia. Strong acids like picric
acid, although coloured, are useless as indicators; for
their dissociation is so great that the colour seen is
due to their anions, just as when their salts dis-

Many chemists, however notably Hantzsch do
not accept the explanation of the behaviour of indi-
cators as due to dissociation. They consider the
changes in colour to be due to rearrangements of the
atoms within the molecules.

Phenolphthalein results when phenol and the anhy-
dride of phthalic acid are gently heated with strong
sulphuric acid. Its composition is well known. For
use as an indicator it is generally dissolved in alcohol.
The solution is colourless, and shows no change when
mixed with water or acid, but sharply becomes red
when the slightest trace of alkali is added. It is an


excellent indicator for estimating the weaker acids
especially organic when titrated with caustic soda
or potash; but it is useless if the acid is very weak,
for then the salt formed with the indicator will dis-
sociate and show a pink colour due to its anions
before the acid being titrated is neutralized. Neither
may it be used with ammonia or other weak bases,
nor in the presence of compounds of ammonia, as
under these circumstances the colour-change is not
sharp, and too great excess of base is required.

Normal carbonates of alkalis produce the same red
colour as the alkalis themselves, but acid carbonates
do not. The point when the former are wholly con-
verted into the latter may be thus readily determined.
This is supposed to be due to dissociation of the
normal salt, which in the presence of water becomes
free alkali and acid salt. The acid salt dissociates
scarcely at all. Since the decomposition of a car-
bonate charges the solution with carbonic acid, which
is more prone to dissociate than the indicator, the
carbon dioxide must be boiled out before assuming
that the reaction is complete. The disappearance of
the red colour on standing may be due to absorption
of carbon dioxide from the air; or, as Sutton suggests,
because the tendency of the indicator salt to ionize is
gradually overcome by its tendency to react with
water when acid molecules are again formed.

Litmus. As a general indicator for strong in-
organic acids and bases, litmus is best known and
most generally useful ; but for the titration of weak
acids or bases its indications are unreliable. Amongst
the weak acids must be included phosphoric, arsenic,
and sulphurous acids. It is an aqueous solution of
colouring matter extracted from certain lichens.
Litmus test-paper is prepared by soaking paper in


the solution made weakly acid, alkaline, or neutral.
That made with glazed paper is most sensitive. As
in the case of phenolphthalein, so with litmus: carbon
dioxide set free during a titration must be expelled
before assuming that the end-point of the reaction is

Methyl orange or Tropceolin D is the sodium salt of
the sulphonic acid of an azo dye. Its behaviour sug-
gests that its acid is considerably stronger than those
contained in phenolphthalein and litmus. Orange-
coloured in the solid state, it dissolves to a yellow
solution in water, and turns red on acidifying. It is
an excellent indicator in alkalimetry, being thoroughly
reliable even when weak bases are titrated; but it is
useless for estimating organic or other weak acids.
The fact that its colour is uninfluenced by carbonic
acid makes it specially useful in titrations where
carbon dioxide is liberated, as boiling off becomes un-
necessary. Boric acid, too, makes no difference to its
colour; hence it can be used in the titration of borax
with mineral acid. The colour-change is best fol-
lowed when the tint is light.

The behaviour of methyl orange may be thus ex-
plained: the acid being moderately strong, its sodium
salt dissociates, and the yellow colour is due to its
anions. The same colour is seen when the acid itself
is dissolved, and points to a considerable dissociation
of its molecules. The pink colour due to these does
not appear until a sufficient quantity of hydrions has
been introduced to overcome this ionizing tendency.
Hence its uselessness for the titration of weak acids.
A buff colour suggests a mixture of molecules and
anions. The tendency of methyl orange to dissociate
is too great to be affected by many organic acids and
some inorganic acids notably carbonic and boric;


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Online LibraryWilliam Thomas BooneA complete course of volumetric analysis for middle and higher forms of schools → online text (page 9 of 11)