William Thomas Boone.

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

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and - H 2 C O 4 are mixed a result which would be ex-

pected in a reaction between the acid and an alkali. Here,
however, we are not neutralizing the acid, but breaking up
its molecules by the oxidizing power of the permanganate.
Oxalic-acid solution, therefore, made of standard strength,
either for its acid or reducing properties, contains the same
proportion of the compound.

Question 56. i 1. of a certain solution contains 0.549
grm. of manganese as potassium permanganate. 20 cu. cm.
of a ferrous solution require 24.5 cu. cm. of the perman-
ganate to transform all ferrous iron to ferric, (a) Find the
weight of ferrous iron in i 1. of the solution, (b) Express
the strength of the permanganate in the usual way, viz.:

i cu. cm. = x grm. KMnO 4
i cu. cm. = x grm. oxygen
i cu. cm. = x grm. Fe".

Question 57. How many formula weights of oxalic
acid have the same reducing effect on permanganate as
one of ferrous sulphate?


Exercise 63. Find the number of molecules of
combined water in ferrous sulphate crystals.

Directions. i. Use recently boiled and cooled distilled
water to make the solution, to which a few drops of con-
centrated H 2 SO 4 have been added to hinder oxidation.

2. Pick out clear green crystals, and grind up before
weighing- out.


3. About 4 grm. to 250 cu. cm. will work with - KMnO 4 .

4. Find the weight of FeSO 4 in that of crystals taken ;
the difference will represent the weight of water combined
with it.

5. Calculate the weight of water combined with a gramme
formula weight of FeSO 4 . The weight found divided by
the gramme molecular weight of water will give the required

Exercise 64. The solution of iron salt was ori-
ginally ferrous, (a) Find whether any ferrous iron
is still present. (&) Reduce all ferric iron to the
ferrous condition, (c) Find the weight of each kind
of iron in 1 1. of the solution.

Directions. i. To test for ferrous sail use potassium
ferricyanide (K 3 FeC 6 N 6 ), which gives Turnbull's blue (re-
sembling Prussian blue)

2 K 8 FeCy 6 + 3 FeSO 4 = Fe" 3 (FeCy 6 ) 2 + 3 K 2 SO 4 .

(Turnbull's blue)

Potassium ferricyanide gives no blue colour with ferric salt.
2. To test for ferric salt when reduction is supposed to
be complete, use potassium (or ammonium) sulphocyanate.
This reagent yields a blood-red solution with ferric com-
pounds, due to a double salt represented by the formula
Fe'"(CNS) 3 , QKCNS, 4H 2 O. In the solid state it resembles
potassium permanganate.

I2KCNS + FeCl 3 + 4H O

= Fe'"(CNS) 3 , 9 ~KCNS, 4 H 2 O + 3 KC1.

[Potassium ferrocyanide is not so suitable, as it yields

(C906) 7


Prussian blue, Fe 4 (FeCy 6 ) 3 , with ferric salts, and with
ferrous salts a white precipitate of potassium ferrous ferro-
cyanide, K.,Fe"(FeCy 6 ), which at once turns blue by reac-
tion with dissolved oxygen or with that free in the air.]

3. Reduction. Transfer a definite volume of the solution,
say 25 cu. cm., to a flask, and add a few thin pieces of
zinc free from iron, together with sufficient dilute H 2 SO 4
to dissolve them. Prevent access of air by one of the
methods given in Ex. 59. Gentle heat may be applied.
The reduction is chiefly due to the nascent hydrogen.
When the solution appears colourless, remove a drop on
a clean glass rod, and test for ferric iron by mixing it with
a drop of solution of KCNS placed on a white tile. A red
colour indicates that the process of reduction must be
carried further. It may be considered complete when only
a faint tint appears, as this may be due to the exposure
of the ferrous liquid to the air.

4. Titration. . Before titrating, note the following
points :

(i) All zinc must have been dissolved ; if this has not
been effected, undissolved pieces must be removed, or they
will reduce some of the KMnO 4 run in, and of course intro-
duce error. The ZnSO 4 does not interfere.

(ii) Ascertain whether the solution contains chloride.
If more than a trace is present, add a little MnSO 4 or
(NH 4 ) 2 SO 4 . 0.5 grm. to each 25 cu. cm. will be ample.

(iii) Sufficient H 2 SO 4 must be present, yet not in so
concentrated a form as to reduce the KMnO 4 . Use


- KMnO 4 and find the total iron present. Original fer-

rous iron must be determined by a separate titration. The
difference between the two results represents the ferric

316 grm. KMnO 4 == 558 grm. Fe".

Exercise 65. Find the effect of metallic zinc in
the form of dust on solutions of ferric salt and
potassium permanganate.


Directions. i. Shake up a little zinc dust in dilute
FeCl 3 sohition. The addition of half a dozen drops of
dilute H.,SO 4 will remove any coating of ZnO on the
particles. Continue shaking- till the liquid appears colour-
less on standing.

2. When colourless, test for ferric iron with KCNS.
("Drop test", see Ex. 64.) If the reduction is incom-
plete, continue the treatment with zinc dust.

3. Repeat i, using very dilute KMnO 4 instead of FeCl 3 .
[The above is Carnegie's method of reducing ferric salts,

and though not so perfect as that previously given, is much

Exercise 66. Find the percentage of iron in iron
alum by means of standard permanganate.

Directions. i. Solution of 2 to
3 grm. of the salt in 250 cu. cm.

will work with KMnO 4 .

2. Test for ferric iron, and re-
duce all present.

3. Reduce two or three separate
quantities of 25 cu. cm. with nascent
hydrogen (see Ex. 64), and one
quantity of about 75 cu. cm. with
zinc dust, in order to be able to
compare the two methods.

4. Since zinc reduces KMnO 4 ,
the residual dust must be removed.
Ordinary filtration would allow con-
siderable reoxidation of the iron.
A "reversed filter" is less objection-
able. Fig. 13 shows this arrange-

A double piece of filter paper is
tied tightly round the end of a
glass tube wide enough to take Fig. 13


the pipette easily. When stood vertically in the flask (or
narrow beaker), a clear liquid filters through, and may
be withdrawn in the usual way.

Exercise 67. The solution contains ferrous sul-
phate and oxalic acid mixed in proportion to their
formula weights. Find the weight of crystals of
each used in making 1 1. of the solution.

Directions. i. Since we have to find the quantity of
ferrous compound present, reduction is unnecessary.

2. Acidify each portion withdrawn for titration with


dilute H 2 SO 4 , warm to 60, and titrate with - alkali.

3. The equations

4 FeS0 4 + 2 H 2 S0 4 + O 2 = 2 Fe ,(SO 4 ) 3 + 2 H 2 O
2 H 2 C 2 O 4 + O 2 = 4 CO 2 + 2 H 2 O

show that each formula weight of oxalic acid uses twice as
much oxygen, and therefore twice as much KMnO 4 as a
formula weight of FeSO 4 . Consequently, one-third of the
KMnO 4 used will be the equivalent of the ferrous salt, and
the remaining KMnO 4 that of the oxalic acid.

Ferrous sulphate crystals are represented by FeSO 4 ,
7 H 2 O, and oxalic-acid crystals by H 2 C 2 O 4 , 2 H 2 O.

Question 58. How would you have determined the
quantity of each present, if equal weights of the crystals
had been dissolved?

Formula weight of FeSO 4 , 7 H 2 O = 278.
Formula weight of H 2 C 2 O 4 , 2 H 2 O = 126.

From equations in Ex. 67 it is evident that 63 grm. of
oxalic crystals require as much permanganate as 278 grm.
of the ferrous crystals. If x grm. of KMnO 4 is used by

the oxalic acid, 2 ? x grm. will be required by the ferrous

6 3

Then - + x = total KMnO 4 (in grammes) used.


Question 59. i 1. of ferrous sulphate solution is made
by dissolving 5 grm. of ferric oxide in dilute sulphuric
acid, reducing to the ferrous state, and adding the neces-
sary volume of water. 25 cu. cm. are found to require
35 cu. cm. of a permanganate solution to reoxidize the
ferrous salt. (a) Find the strength of the potassium
permanganate, (b) Calculate the volume of water which
must be added to or withdrawn from 750 cu. cm. of the


permanganate to make it '-.


Question 60. What weight of oxygen available for
oxidizing purposes can be obtained from i 1. of hydrogen
peroxide solution containing 40 grm. of peroxide?

Exercise 68. Find the percentage weight of
hydrogen peroxide contained in the laboratory


solution by means of ^ KMn0 4 .

Directions. i. The two highly oxidized compounds
mutually decompose each other. Equation

5 H 2 O 2 + 2 KMnO 4 + 3 H 2 SO 4

~y6~ = K 2 SO 4 + 2MnSO 4 + 5O. 2 + 8H 2 O,
shows that 316 grm. KMnO 4 = 170.1 grm. H 2 O 2 .

2. Into a weighed loo-cu.-cm. measuring -flask pom
about 10 cu. cm. of the solution, and find its weight.

3. Test the original H 2 O 2 solution for HC1. If present
in any quantity add a little MnSO 4 to the weighed per-

4. Make up to the mark with distilled water, and mix

5. To each portion titrated add about half its volume of
dilute H 2 SO 4 .

Note. The presence of organic matter will interfere with the
results. A more reliable method of estimating the peroxide is
given in Ex. 85.

Question 61. When barium peroxide is acted upon by


acid preferably phosphoric in ice water, the barium salt
is precipitated and hydrogen peroxide passes into solution.


(a} What volume of l permanganate solution will be

required to decompose the hydrogen peroxide theoretically
produced from 10 grm. of the pure barium compound?
(b) What volume would the oxygen occupy at S.T. P.?

Potassium Dichromate (K L ,Cr.,O 7 ). The use of this
reagent for oxidizing purposes was illustrated in Exs.
9 and 18. To detect the end-point of its reaction
with reducing agents, it was found necessary to use
an external indicator, as the green chromium salt
formed prevents detection of excess dichromate. But
for this one disadvantage potassium dichromate is pre-
ferable to permanganate, for the following reasons:

(i) The solid is less liable to decomposition, and
may be assumed to be in accord with its formula.

(ii) Its solution is less likely to deteriorate.

(iii) It does not react with rubber, and consequently
may be used in an ordinary Mohr's burette.

(iv) Its reactions are not interfered with by hydro-
chloric acid.

When used, however, in the estimation of iron,
reduction cannot be carried out by means of zinc; for
although zinc salts do not interfere with the reaction
between dichromate and ferrous salt, they do react
with the external indicator (potassium ferricyanide),
and make its indications of doubtful value. Mag-
nesium may be us^d instead, but this metal is expen-
sive. It is usual to employ either stannous chloride
or ammonium bisulphite to reduce any ferric iron.
(See Potassium Dichromate, p. 30.)

Question 62. What weight of potassium dichromate
\vill be required to yield 8 grm. of oxygen for oxidizing



Exercise 69. Make half a litre of ^ potassium

dichromate, and test its strength by means of a
ferrous salt. State its exact value thus:

i cu. cm. = ? grm. K 2 Cr 2 O r ; ? grm. Fe"; ? grm. O 2 .

Directions. i. Fuse some K 2 Cr 2 O 7 crystals in a porce
lain dish to drive off included water, allow to become cold,
and weigh out the necessary quantity. (See p. 30.)

2. The ferrous salt used may be either Mohr's salt or
good crystals of ferrous sulphate. The latter is preferred
by some chemists, provided H 2 SO 4 is added to the water
to hinder oxidation.

3. Equation :

6(Am,SO 4 ), FeSO 4 + K Cr O 7 + 7 H.,SO 4
= 6 (Am 2 S0 4 )" + K 2 S0 4 + 3 Fe 2 (SOJ 8 + Cr 2 (SO 4 ) 3 + 7 H 2 O.

4. Dichromate in burette; indicator (external), K 3 FeC 6 N c .

Question 63. If in the last exercise the solution of di-


chromate is found to be not exactly - -", express its relation-
ship to this value by a factor.


Assuming the Fe" solution is exactly -, and that 25 cu.
cm. require 25.2 cu. cm. of the K 2 Cr 2 O^ solution ; this latter
will be ||f X , or - X 0.992.

Question 64. What weight of potassium dichromate
is equivalent as an oxidizer to i grm. of potassium per-

Exercise 70. A sample of spathic iron ore pro-
vided consists mainly of ferrous carbonate. By

means of ^ dichromate, find the percentage of
ferrous iron present.

Directions. i. Dissolve an accurately weighed quantity
of about i grm. of the ore in dilute acid, and make up


to 100 cu. cm. with recently boiled and cooled distilled
water ("boiled out" water), and mix well. See Ex. 59
for means of preventing oxidation. This is especially
necessary if HC1 is used.

2. Let insoluble matter settle, and neglect it as it
occupies so little volume.

3. Equation shows 4.9 grin. K. 2 Cr 2 O r = 5.58 grm. Fe".


Question 65. It is found that 25 cu. cm. - ferrous


solution require respectively 17.5 cu. cm. of KMnO 4 solu-
tion and 18.2 cu. cm. of K. 2 Cr> 2 O 7 solution to convert the
ferrous iron to ferric. How many grammes of available
oxygen are there in i 1. of each solution?

Question 66. On titrating a dichromate solution with


- FeSO 4 , it was found that 25 cu. cm. of the latter re-

quired 20 cu. cm. of the former to oxidize the iron. What
weight of CrO 3 was present in i 1.?

Exercise 71. Find the percentage of iron in a
sample of hematite, using ^ K,Cr,0 7 .

Directions. i. Finely powder the sample; dry it, and
weigh out about i grm. by difference, and dissolve in
small quantity of concentrated HC1 diluted with equal
volume of H. 2 O.

2. Proceed by method A or B.

Method A. Reduction of iron -with ammonium bisul-
phite (N H 4 HSO 3 ). [ Made by bubbling SO. 2 through strong
NH 4 OH, contained in a test-tube standing in water, till
any crystals which may form redissolve and the liquid
smells strongly of SO 2 .]

(i) Neutralize as nearly as possible the acid iron solution
with AmOH.

(ii) Add 5 to 10 cu. cm. of the NH 4 HSO 3 solution, and


(iii) Mix 10 cu. cm. concentrated H 2 SO 4 with 10 cu. cm.
H 2 O, add it to the solution, and continue boiling till all SO 2
has been expelled. (20 minutes.)

(iv) Cool quickly, and make up to 100 cu. cm. with
boiled out and cooled water. This solution should work

with - 1 K 2 Cr 2 O 7 .

(v) Indicator : freshly-made solution of K 8 FeC 6 N 6 .

Method B. Reduction of iron with good SnCl 2 .

(i) Cool and make up the iron solution to 100 cu. cm.
This will contain a fair excess of HC1, which is necessary
in this method.

(ii) Boil up each quantity of ferric solution withdrawn,
and add SnCl 2 solution, one drop only at a time, till the
red colour just disappears, i.e. till the ferric iron is reduced.
(Look at a piece of white paper through the liquid.) A
very slight excess of SnCl 2 will probably now be present,
which must be oxidized to prevent it reducing- any K 2 Cr 2 O 7 .

(iii) Cool somewhat, and add a few drops of HgCl 2 solu-
tion (2 cu. cm. should be plenty), when some cloudiness
due to HgCl should be seen. This precipitate of mer-
curous chloride has no action on K 2 Cr 2 O 7 .

(iv) Titrate as soon as possible.

(v) Indicator: freshly-made solution of K.,FeC N .

3. Equation : 6 FeCl 2 + K 2 Cr 2 O 7 + I4HC1

= 6 FeCl 3 + 2 KC1 + 2 CrCl 3 + 7 H 2 O.

Why cannot zinc be used for reduction purposes in the
above exercise?

Question 67. You are asked to find the strength of

some potassium permanganate which is about -, and are


provided with acid potassium oxalate (C 2 O 4 HK). How

much of the latter salt would you require to make i 1. of -


strength? This solution may also be used for standardiz-


ing alkali. Would you assume it to be also for this


purpose? If not, state its relationship to strength.


Question 68 Two solutions containing respectively
potassium permanganate and ferrous iron are roughly
decinormal. Being provided with pure solid potassium
dichromate, how would you find the exact strength of
the permanganate solution?

Methods involving Indirect Oxidation


Many examples of oxidation by halogens will be
familiar to you. In each case the oxygen is derived
from water through its reaction with the halogen.
When an aqueous solution of chlorine is exposed to
direct sunlight a small quantity of oxygen is actually
set free, and may be collected, though the greater
part goes to form chloric acid.

8 C1 2 + 8 H 2 O = O 2 + 14 HC1 + 2 HC1O 3 .

Exercise 72. Examine the power of each halogen
to oxidize different substances.

Directions. The following experiments will serve as
examples :

1. Test some solution of FeSO 4 for ferric salt, and
repeat after adding a few drops of chlorine water.

2. To i cu. cm. of SnCl., solution add bromine water in
slight excess, i.e. till the solution remains permanently red.
Test for SnCl, with HgCl 2 .


3. Test some H 2 SO, solution for H 2 SO 4 . Repeat after
adding" a little solution of iodine, and contrast the results.

Of these three halogens, that most generally useful
for making standard solutions is iodine, for the fol-
lowing reasons:

(i) It is more easily purified than either of the
others, (ii) Being solid, it is most easily handled and
weighed out. (iii) Excess used in a titration can be
most readily detected, partly by its colour, and particu-
larly by the starch test. Although not very soluble in
water, it dissolves readily in the presence of potassium
iodide, which does not interfere with its reactions.

Exercise 73. Make 500 cu. cm. of approximately
deci-normal solution of iodine.

Directions. i. To Purify the Iodine. If the purity of
your sample is doubtful, powder up rather more than you
require, and heat it in a porcelain dish on a water-bath for
15 min. Next, mix it thoroughly with about T \yth its weight
of pure KI well ground up; cover the mixture with a well-
fitting inverted funnel or dish, and heat on a sand-bath till
the iodine has sublimed into the inverted vessel. The sub-
limed iodine may be assumed to be pure.

2. Weighing out. The equivalent of iodine being 126.92


the same as its atomic weight, ?; 1. of - solution should


contain 6.346 grm. But the volatility of iodine renders it
doubtful whether the difference between two weighings
represents exactly the weight dissolved; and so it is usual
to take rather more than the theoretical quantity. Put
about 6\ grm. into a 5oo-cu.-cm. measuring-flask.

3. Solution. Half-fill the flask \vith water, and add
10 grm. of pure KI or more if required to complete the
solution of the iodine. Add water to the mark, and mix

4. Preservation of Solution. Keep in a dark place in a
stoppered bottle, properly labelled.


Standard Sodium Thiosulphate. It was seen in
Ex. 16 that the colour due to iodine disappeared on the
addition of sodium thiosulphate. The reaction between
these two substances is expressed by the equation

I 2 + 2 Na 2 S 2 O 8 = Na.,S 4 O 6 + 2 Nal.
253-8 316.3

253.84 grm. of iodine reacting with 316.28 grm. of
the thiosulphate (or with 496.44 grm. of the penta-
hydrate crystals, Na 2 S 2 O 3 , 5 H 2 O) produce the tetra-
thionate and sodium iodide; and since free iodine is
so readily detected, the end-point of the reaction can
be easily determined. By the combined use of these
reagents a much larger number of estimations become

The weight of thiosulphate crystals which corre-
spond with 126.92 grm. of iodine is 248.22 grm.,
hence decinormal thiosulphate will require 24.82 grm.
of crystals containing 15.81 grm. of Na 2 S 2 O 3 to the
litre. It will be noticed that this weight of crystals
contains two equivalent weights of sodium ; but just
as decinormal solutions of permanganate and di-
chromate contain that weight per litre which can
yield T Vth of the gramme-equivalent weight of oxygen,
so a litre of decinormal thiosulphate must contain
that weight of the salt which can react with ^th of
the gramme-equivalent of iodine. (See also p. 97.)

Exercise 74. Make \ 1. of deeinormal thiosulphate,
and with it determine the strength of the iodine
solution made in last exercise.

Directions. i. Assume the crystals to be pure. This
will be ascertained in a later exercise. Crush up the
crystals (Na 2 S 2 O 3 , 5 H 2 O), and dry the powder well be-
tween filter paper. Accurately weigh out 12.41 grm.,
dissolve in water, and make up to 500 cu. cm.


2. Thiosulphate in burette.

3. Indicator: starch solution. (See Ex. 13.)

This is best added towards the end of the reaction ; that
is, when only sufficient iodine is left to produce a faint
yellowish colour. Shake up after each addition of Na 2 S 2 O 3 .
The reaction is complete when the blue colour is just de-


4. Either dilute your iodine solution to - strength, or


find the factor expressing its relationship to this value.
For example

If 25 cu. cm. of I solution = 26.5 cu. cm. Na 2 S 2 O 3 ,


the 25 cu. cm. require dilution to 26.5 cu. cm., and so on.


Or, its strength may be given as T - x 1.06 I.J.

Use of Sodium Bicarbonate (NaHCO 3 ). In the
exercise just completed, the conversion of thiosulphate
into the more highly oxidized tetrathionate was accom-
plished by the splitting off of a portion of the positive
radical, which then united with iodine. But in many
estimations oxygen is actually supplied to the reduc-
ing substance from the water present, whose hydrogen
then combines with iodine to form hydriodic acid.
The reaction, however, is reversible, and may be

H 2 O + I 2 ^ 2 HI + O.

To prevent the backward reaction, the hydriodic
acid is neutralized by a suitable alkali an acid car-
bonate being found most convenient. Caustic alkali
cannot be used, because it reacts with iodine as well
as with the acid.

2 NaOH + I 2 = NalO + Nal + H 2 O.
3NaIO = NaIO 3 + 2 Nal.


Normal alkali carbonates are also unsuitable, as
they suffer some hydrolysis to caustic soda and acid
carbonate, when the former can react with iodine as
shown above.

Na,CO 3 + H 2 O = NaOH + NaHCO 3 .

Even the bicarbonates hydrolyse to some slight
extent, but unless present in great excess their use
introduces practically no error.

Exercise 75. Prepare some 3; solution of arseni-


ous oxide and titrate it against your 1( J iodine.

Directions. i. Unless sure of its purity, resublime suffi-
cient As 4 O 6 from a crucible or dish, up into an inverted
vessel (a funnel does well) placed over it. This should be
done in a fume closet.

2. Arsenious compounds oxidize to the arsenic state in
presence of iodine and water, as expressed in the equa-

As 4 O 6 + 4 I 2 + 4 H 2 O = 2As 2 O 5 + 8HI,
395.8 1015.4 1023.4

from which it will be evident that -^ mol of As 4 O 6 should

produce a litre of - solution. 250 cu. cm. will be suffi-

cient for this and later exercises. Time will be saved if
you accurately weigh out (by difference) approximately
this quantity (about 1.5 grm.) into the graduated flask.

3. Solution of As 4 O 6 . Probably your best method will
be to dissolve it in the least quantity of NaOH with gentle
heat, then to prevent reaction of the alkali with iodine
neutralize with dilute HC1. Cool, and add water to the
mark. Preserve the solution in a bottle properly labelled.

4. To each 25 cu. cm. withdrawn for titration, add 0.75
to i.o grm. of powdered NaHCO 3 or 10 to 12 cu. cm. of


cold saturated solution, which contains about one-twelfth
its weight of the acid carbonate.

5. Iodine solution in burette (glass stopcock).

6. Indicator: i cu. cm. starch solution for each quantity
titrated. The end-point will be reached when the blue
colour persists for five minutes after thorough mixing.

Assuming your As 4 O 6 solution to be accurately deci-
normal, how does the strength of your iodine solution
compare with the previous determination?

Note on the quantity of NaHCO^ used. The equation
8 [HI + NaHC0 3 = Nal + CO 2 + H 2 O]
1023.4 672

combined with that in (2) above indicates that for 396 grm. of
As 4 O 6 , 672 grm. of NaHCO 3 are necessary to neutralize the HI
formed. 25 cu. cm. of solution prepared in above exercise contain
approximately 0.15 grm. As 4 O 6 , and will require theoretically
about 0.25 grm. NaHCO 3 to neutralize the HI formed; so that
the addition of (0.75 to i) grm. to each quantity will allow a
fair margin of excess, yet not so great as to introduce error.

Question 69. How would you proceed to estimate an
arsenate by means of iodine?

1 2 3 4 6 8 9 10 11

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