William Thomas Boone.

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

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(b) Some of the solution was diluted to T Vth its original
strength, and 10 cu. cm. were then found sufficient to
neutralize 30 cu. cm. of dilute nitric acid. Find the con-
centration of HNO 3 .

(c) What dilution does the original alkali require to
make it semi-normal?

Question 34. 30.3 cu. cm. of a solution of caustic soda
were found necessary to neutralize 25 cu. cm. of N. H 2 SO 4 .
How many cubic centimetres of water must be withdrawn
from 100 cu. cm. of the alkali to make it normal?

Evidently the caustic soda is - of normal strength ;


30.3 cu. cm. contain only as much NaOH as 25 cu. cm.
N. NaOH should contain, and so 5.3 cu. cm. of water must

be withdrawn from 30.3 cu. cm. of alkali, and 5*3 x IOO
cu. cm., or 17.5 cu. cm. from 100 cu. cm.

Question 35. How is water "withdrawn" from too


weak a solution? What therefore do you consider to be
the quickest way of making the N. solution?

Question 36. Vessels containing standard caustic soda
or other alkali are usually fitted with " calcium chloride"
tubes. What is their use?

Why is it not usual to fit bottles containing alkali with
glass taps or stoppers?

Exercise 32. By means of a solution of ammonia
of unknown strength, find the amount of dilution
which concentrated hydrochloric acid has under-
gone to make ordinary dilute acid.

Directions. Bench reagents are too strong for volu-
metric work, and both the ammonia solution and strong
hydrochloric acid would lose perceptible quantities of
their dissolved gases during titration.

1. Dilute 10 cu. cm. of bench NH 4 OH to 200 cu. cm.,
and mix well.

2. Run 10 cu. cm. of the strong HC1 from a burette into
a 25o-cu.-cm. graduated flask; add water., cool well, then
fill to the mark. Label it "Strong HC1 "

25 .

3. Further dilute the dilute HC1 of bench; 50 cu. cm.

to 250 cu. cm. will do very well. Label it " [ - ".


4. Find the volume of each necessary to neutralize a

definite volume of the diluted NH 4 OH.

5. Indicator: litmus. Which indicator should not be

6. If the volumes of the first and second specially diluted
acid are respectively x cu. cm. and y cu. cm., evidently

cu. cm. of strong acid contain as much HC1 as y cu. cm.
of the dilute bench acid, and the relative strengths will

be y : 5.


Question 37. If 24 cu. cm. N. H 2 SO 4 are necessary to


neutralize 20 cu. cm. of solution of ammonia, what volume
of gaseous NH 3 , measured at S.T.P., is dissolved in i 1.
of the solution?

Solution of NFL is ^ N. strength.

.*. i 1. of solution will contain

17.034 X 2 4, or 20.4408 grm.

At S.T.P. the mol of a gas occupies 22.25 ! Since
the mol of NH 3 is 17.034 grm, 20.4408 grm. will occupy

2 44 8 x 22 . 25 i.

Exercise 33. The concentration of ammonia in
the solution ordinarily used in the laboratory is
one -fifth that of the original solution. Using

^ H 2 S0 4 find (a) the weight of NH 3 contained in

1 1. of the original ammonia solution; (b) the
volume this ammonia would occupy in the gaseous
state at S.T.P.

Directions. i. Dilute the ordinary bench solution of

NH 3 still further, so as to work well with - H 2 SO 4 . The

amount of dilution can be ascertained by a preliminary
rough measurement as given in Ex. 31.
2. Acid in burette. Which indicator?

Question 38. Why are standard solutions of ammonia
less commonly kept in laboratories than those of caustic
soda or potash?

Exercise 34. Find the solubility of calcium hy-
droxide in water at the temperature of the labora-
tory by a volumetric method. Give your answer
as the weight of Ca(OH) 2 contained in 100 eu. cm.
of solution.

Directions. i. Preparation of solution. Place some

(C906) 5


powdered slaked lime in 100 to 200 cu. cm. of water con-
tained in a flask. Cork, and allow to stand half an hour
or more with occasional shaking. Filter, and protect the
lime-water formed from contact with air as far as possible.

2. Titratton.\3se acid. Even H 2 SO 4 may be used,

as the sulphate formed is not precipitated, since it is rather
more soluble than the hydroxide.

Question 39. Find the strength of the solution of
Ca(OH) 2 . What relationship to N. does it bear?

Exercise 35. Find the percentage of Na 2 in


some soda carbonate crystals using ^ acid, and
litmus as indicator.

Directions. i. To make a solution of suitable strength.
Assume the crystals to agree approximately with the
formula Na 2 CO 3 , 10 H 2 O. Since i 1. of N. solution should
contain 143 grm. of the pure substance (as this weight
will contain the gramme equivalent of sodium), 3^ grm.
will be sufficient to make 250 cu. cm. of solution near


2. Use of weighing-bottle. Place some roughly powdered
crystals in a clean weighing-bottle (or test-tube), and find
the exact weight of the whole. Tap out a little of the
powder into the flask, avoiding any loss, until about the
necessary weight has been transferred (this may be ascer-
tained by roughly weighing at intervals) ; then find the
exact final weight. The difference between this and the
first weight gives the quantity taken.

3. Acid in burette.

4. Equation NaoO.CO,, + 2 HC1 = 2 NaCl 4- CO 2 +H 2 O



furnishes the key necessary to determine the weight of
Na 2 O in the carbonate.

Total weight of Na 2 O x IOO gives the required answer .

Weight or crystals taken

5. Use of litmus when carbon dioxide is liberated. (See
Ex. 12.) Dissolved CO 2 gives an acid reaction with litmus,
and as this gas is liberated during the titration it must
be expelled. Before final readings are taken, the liquid
should remain red after boiling for two minutes. If this
is not done, insufficient standard acid will be used.

Question 40. Do you see any objection to dropping
the carbonate into a measured quantity of acid?

Exercise 36. Of the two liquids provided one is
dilute nitric acid containing- 1.5 grin, of nitrogren
per litre, and the other a solution of potassium
carbonate. Find the weight of potassium in 1 cu.
cm. of the latter.

Directions. i. Of 63.018 grm. (the mol) of HNO 3 14.01
grm. consist of nitrogen. From this find the weight of
HNO 3 in i cu. cm. (See note on p. 152.)

2. Equation 2 HNO 3 + K 2 CO 3 = 2 KNO 3 + CO 2 + H 2 O

2 X 63 78.2

shows 39. i grm. of K = 63 grm. HNO 3 .

3. Acid in burette.

4. Use litmus for some titrations, and methyl orange
for others, and compare the readings obtained. It is not
necessary to expel the CO 2 when methyl orange is used.
(See Ex. 13.)

Question 41. 4 grm. of caustic soda were dissolved
up to i 1. when it was found that 25 cu. cm. were neutra-
lized by 22 cu. cm. of - HNO 3 . Assuming that the alkali

contained only sodium hydroxide and water, calculate the
percentage of the latter in the sample.


Note the two methods of making' the calculation :

(i) Weight of HNO 3 used = 0.0063 x 22 grm.
Equivalent weight of NaOH in 25 cu. cm.

= 0.0063 X 22 X
6 3

Weight of NaOH in i 1. = 0.0063 x 22 X |? X IOO

6 3 2 5

= 3-5 2 grm-

Weight of water in 4 grm. of caustic soda

= 0.48 grm.

Per cent weight of water in sample


= _ _ x 100 == 12 per cent.

(ii) Since " acid is used, the solution of NaOH is

evidently of decinormal strength.


Now - NaOH contains 4 grm. per litre.

/.our solution contains 4 X - (or 3.52) grm. NaOH per

Exercise 37. With seminormal acid determine
the percentage of sodium hydroxide in a sample of
caustic soda, assuming 1 that no carbonate is present.

Directions. i. Assuming sample to be mainly NaOH,
roughly calculate the weight necessary to make 250 cu. cm.

of l solution. Although this weight will give a solution

below - strength it will be sufficiently near to work with

the acid.

2. Coarsely powder up some of the solid and transfer
to a weighing-bottle with as little exposure to air as pos-
sible, then weigh the whole.


3. Transfer to the flask (in the way described in Ex. 35)
the necessary quantity of the solid and weigh again.

4. Dissolve the solid in water, cool, fill up to the mark,
and mix well.

5. Acid in burette.

6. Indicator: methyl orange or litmus; but should any
carbonate be present boiling will be necessary with litmus.

Question 42. In working Ex. 35 a student found
21.67 P er cent of Na 2 O in the sodium carbonate crystals.
Calculate the number of molecules of water combined with
Na.,CO 3 .

Regard the crystals as made up of Na 2 O, CO 2 , and
<c(H 2 O).

21.6 parts by weight out of 100 consist of Na 2 O,

15.38 parts (i.e. 21.67 X g|) > c 2

and the remainder 63 parts ,, ,, water.

. Total weight of water in crystals __ 63
Total weight of Na 2 CO 3 37*

r> . Total weight of water in crystals
Total weight of Na 2 CO 3

x mol. of H 2 O _ a;(i8)

i formula weight of Na 2 CO 3 106

- SZ and x = 10.


1 06 37

Or, see method given in answer to Question 32 (b).

Question 43. A sample of anhydrous sodium car-
bonate is known to contain a little sulphate. 2.65 grm.
were dissolved and the solution made up to half a litre
25 cu. cm. of which required 18.5 cu. cm. of standard
acid to neutralize it; whereas it was found that 25 cu. cm.
of N. Na 2 CO 3 required 18.75 cu - cm - f tne same acid.
Find (a) the percentage of pure Na 2 CO 3 in the sample;
(b) the weight of the impure salt which should have been
taken to give strictly N. Na 2 CO 3 .


Exercise 38. The sample of acid sodium sulphate
is known to contain some sodium chloride. Find
the weight of the salt which must be used to make

a litre of y NaHS0 4 .

Directions. i. Find the weight which would be neces-
sary were the salt pure. Take a quarter of this and
dissolve up to 250 cu. cm.

2. Titrate with - alkali and find the exact weight of

NaHSO 4 in that of the salt taken; then weight of the
latter contains the necessary quantity of NaHSO 4 .

Or, suppose it is found that 25 cu. cm. of solution re-


quire 40 cu. cm. - - NaOH, or 20 cu. cm. of - - solution

of alkali. The salt contains only ^ ? of its weight of
NaHSO 4 , from which the calculation can be made.


In previous exercises the estimations have been
carried out by the " direct method", in which one
standard solution only has been used in finding the
strength of another. By using two standard solutions
one acid and one alkali greater accuracy may be
sometimes secured, and a larger number of estima-
tions become possible. In estimating the strength
of lime-water (Ex. 34) we could have added more
than sufficient standard acid to neutralize the lime,
and then found the excess by means of standard
alkali. The difference between this and the quantity
originally added represents the acid used in neutral-
izing the lime, from which the necessary calculation
can be made as before.


The strength of many neutral salt solutions may be
similarly determined. The metal may be precipitated
by the addition of excess of standard sodium car-
bonate, the precipitate filtered off, when the excess of
sodium carbonate can be found by standard acid.
The difference, as before, represents the weight con-
cerned in the reaction, from which the equivalent
weight of the salt can be calculated.

Such methods are known as " residual methods" of
analysis, or methods of " backward titration ".

Exercise 39. Determine the weight of barium
chloride crystals dissolved in 1 1. of the bench re-
agent. The formula for the crystals is BaCl 2 , 2 H 2 0.

Directions. i. To 25 cu. cm. of the reagent add a
known volume of N. Na 2 CO 3 in fair excess. Excess can
be discovered by litmus paper.

2. Boil to complete precipitation; filter off the BaCO 3
and wash the vessel, filter-paper and precipitate three
times with hot water, adding all washings to the filtrate,
which should now contain all excess of Na 2 CO 3 .

3. Cool the filtrate and make it up with distilled water
to a suitable volume (say 200 or 250 cu. cm.), and mix



4. Find the weight of Na 2 CO 3 in filtrate, using acid.

5. Weight of Na 2 CO 3 used in precipitating the barium
= weight originally taken less that found as excess. Let
it be x grm.

Then weight of crystals dissolved in 25 cu. cm. = x X
as shown by equation

BaCl 2 , 2 H 2 O + Na 2 CO 3 = BaCO 3 + 2 NaCl + 2 H 2 O.
244.3 io6

6. Indicator : methyl orange.


N.B. The calculation may be made in the usual way;


but when solutions are used strictly N. or - it can be


much shortened, as seen by the following example.
The bench reagent is about N. strength.
Suppose a student's records are

Volume of BaCl 2 solution taken 25 cu. cm.
Volume of N. Na 2 CO 3 taken ... 40 ,,
Alkaline filtrate diluted to ... 200 ,,

25 cu. cm. of diluted filtrate required 17.5 cu. cm. ' acid.


The total excess of Na 2 CO 3 will be that found in

(17.5 cu. cm. x X ?),

10 257

or 14 cu. cm. of N. Na 2 CO 3 .

Volume of N. Na 2 CO 3 used to precipitate the
barium in 25 cu. cm. of solution

= (40 14) cu. cm.,

and weight of Na 2 CO 3 so used

= 26 X o. 106 grm.

.'. weight of crystals dissolved in i 1.

10 25

Exercise 40. A solution of hydrochloric acid
provided is approximately normal. Find its exact
value by means of calcite (CaC0 3 ).

Directions. i. (The calcite should be quite transparent
and colourless; otherwise you had better use pure dried
precipitated calcium carbonate.) Grind a portion to fine
powder and weigh out an exact quantity (about i grm.)
from a weighing-bottle into a 25o-cu.-cm. graduated flask.

2. Use 50 cu. cm. of the acid for solution. This should
leave ample excess of acid.


3. Add water to the mark and mix well.


4. Find excess of acid with - alkali.


5. Indicator: methyl orange.

6. Equation

CaCO 3 + 2 HC1 = CaCl 2 + CO 2 + H 2 O.

TOO. i 7 2 -9

Note. If the acid is run into the alkali a precipitate of Ca(OH) 2
at first appears, but dissolves before the end-point is reached.

Exercise 41. Find the concentration of calcium
sulphate in your bench reagent by a volumetric
method, and confirm your result gravimetrieally.

Directions. i. Precipitate the calcium in 100 cu. cm. of
solution, using excess of N. Na 2 CO 3 . Complete precipita-
tion by boiling for 3 min.

2. Filter into 25o-cu.-cm. measuring-flask, washing pre-
cipitate, &c., well, and adding washings to the filtrate.
(See Ex. 39.)

3. Cool, add water to mark, and mix well.

4. Find excess of Na 2 CO 3 with suitable acid. What
indicator will you use?

5. Equation to show the proportions of CaSO 4 and
Na 2 CO 3 which react:

CaSO 4 + Na 2 CO 3 = CaCO 3 + Na 2 SO 4 .

136.1 1 06

6. Gravimetric Method. Evaporate a suitable quantity
50 to 100 cu. cm. of the solution to dryness on a water
bath, in a dish previously weighed. Then heat in an air
oven kept at 120 to 140 till the weight after cooling in a
desiccator is constant.



Exercise 42. Titrate equal volumes of j^ Na 2 C0 3


with ^ acid, using methyl orange alternately with
phenolphthalein as indicators.

Directions. i. Acid in burette.

2. Find the number of cubic centimetres of acid used
when the colour change occurs with each indicator. What
do you infer?

3. When the pink colour due to the phenolphthalein
disappears add one drop of methyl orange, and continue
adding the acid till neutralization is shown.

If the acid and alkali have been properly standardized,
equal volumes should have been used when methyl orange
shows the pink colour; but with phenolphthalein the solu-
tion should become colourless when half as much acid has
been run in. The colour change of methyl orange occurs
when acid and alkali are in the proportion shown by the

Na 2 C0 3 + 2 HC1 = 2 NaCl + CO 2 + H 2 O,

and that of phenolphthalein when in this proportion:
Na 2 CO 3 + HC1 = NaHCO 8 + NaCl.

In other words, phenolphthalein is not sensitive to the
bicarbonate. (See Exs. 13 and 14.)

It will also be noted that the presence of phenolphtha-
lein does not interfere with the indications of methyl orange
when no free alkali or normal carbonate is present. (See
also notes on these indicators, pp. 139, 141.)

Exercise 43. Estimate the percentage of sodium
hydroxide and sodium carbonate present in the
given sample of "soda ash", assuming that these
are the only compounds present soluble in water.


Method /. By the use of methyl orange and phenol-

Directions. i. Assume sample to be wholly Na 2 CO 3 ,
and make 250 cu. cm. of about - or - strength. Use


weighing -bottle. About what weight will you take?
(See note on p. 152.)

2. Mix thoroughly, then allow the insoluble matter to
subside before titrating.

3. N. acid will probably be found suitable to work with ;
if not, it must be diluted.

4. Titrations. First, use phenolphthalein, which will
show when all NaOH and half the Na 2 C(X have been
neutralized ; then add methyl orange, which will show
when the remaining half of Na 2 CO 3 has been neutralized.
The acid equivalent of this portion is thus directly read,
and the total Na 2 CO 3 can then be easily calculated.

If preferred, an entirely fresh titratioii may be made
with methyl orange as indicator, when the excess of acid
used over that used with the phenolphthalein will be the
equivalent of half the Na 2 CO 3 in the solution.

From the total quantity of acid used subtract that
required by the Na 2 CO 3 the remainder has been used
by the NaOH present.

Method II. By using barium chloride.

Directions. i. Make solution as in Method I.

2. Find volume of N. acid necessary to neutralize both
hydroxide and carbonate with methyl orange as indi-

3. Precipitate all carbonate in 50 cu. cm. of the solution
by adding excess of BaCl 2 . (See Ex. 39.)

4. Filter off the BaCO 3 , wash the precipitate, paper, and
beaker well, adding washings to filtrate. If hot water is
used, cool and make up to 100 or 200 cu. cm., and mix
well. Or, the precipitate need not be filtered off. In this
case make the solution up to 100 or 200 cu. cm. at once,


and after allowing- the BaCO 3 to subside, neglect it, as it
occupies very little volume.

5. Titrate the clear liquid with - acid. The quantity

used will be equivalent to the NaOH in solution, and can
thus be directly determined.

6. Weight of Na 2 CO 3 will be the equivalent of the acid
used in 2 less that used in 5.

It might appear that in the above exercise Ba(OH).,
should also be precipitated by reaction with the NaOH
thus: BaCl 2 + 2NaOH = Ba(OH) 2 + 2NaCl. But from
such dilute solutions as we have used the small amount of
Ba(OH) 2 formed remains dissolved. Solutions of Ba(OH) 2
can be made of one-fourth N. strength.

If soda ash contains free caustic soda, is its solution
likely to contain bicarbonate as well?


Exercise 44. Determine the temporary hardness
of spring- water by Hehner's method, and express
your result in " degrees of hardness ".

Directions. i. Assume temporary hardness as wholly
due to CaH 2 (CO 3 ) 2 .

2. " Degrees of hardness" are given as the number of
grammes of CaCO 3 found in 100,000 grm. of water. As-
sume i cu. cm. weighs i grm.


3. Titrate 100 cu. cm. at a time with - H 2 SO 4 (50 cu. cm.

N. 5

- diluted to 250 cu. cm.), using methyl orange as indi-
cator. From the weight of H 2 SO 4 used, that of CaCO 3 in
the water can be found.


4. Practically i cu. cm. - H 2 SO 4 = o.ooi grm. H 2 SO 4 .

Therefore each cubic centimetre used will indicate o.ooi grm.

CaCO 3 . If in the above exercise 13.5 cu. cm. - H 2 SO 4 are


used, the presence of 0.0135 grm. CaCO 3 in 100 cu. cm. of
the water- is indicated; or 13.5 grm. in 100,000 grm. of
water. The water will be said to possess 13.5 degrees
of temporary hardness.

Question 44. I add excess of sodium carbonate to
some hard water known to contain both calcium carbonate
and calcium sulphate in solution and boil. State what

Question 45. It was found that 0.53 grm. of sodium
carbonate was necessary to react with the salts contained
in a certain quantity of hard water. Assuming that the
hardness was wholly due to calcium sulphate, (a) find the
weight of this salt present in the water, (b) What quantity
of water was used if it was found to possess 6.8 degrees
of permanent hardness?

Exercise 45. Determine the "permanent" hard-
ness of the same sample of water, and express
your result in "degrees of hardness".

Directions. i. Assume all permanent hardness to be
due to CaSO 4 .


2. To 100 cu. cm. of the water add 20 cu. cm. - Na 2 CO 3 ,

and evaporate to dryness on a steam bath. (Platinum dish
is preferable for this.)

3. Dissolve as much as possible of the residue in distilled
water and filter. Wash all traces of Na 2 CO 3 both from
dish and filter paper, and add washings to filtrate.


4. Estimate the Na 2 CO 3 in the filtrate with - ' H 2 SO 4 .
This will be the excess Na L ,CO 3 used.

5. Indicator: methyl orange.

6. Weight of Na. 2 Co 3 originally taken (i.e. 0.0053 X 20
grm.), less that found in excess, will have been used to
precipitate the CaSO 4 .

Note. The carbonate precipitated will be \$$ of the weight of
Na 2 CO. ? used up ; and it is usual to give as the number of degrees
of permanent hardness the number of grammes of CaCO 3 which
would be precipitated from 100,000 cu. cm. of the water, in the


way stated instead of the number of grammes of CaSO 4 existing
in this volume.

Question 46. Some "soda water" containing nothing
but carbon dioxide dissolved in water was passed into a
known volume of standardized caustic soda, the latter
being in excess. The solution then contained either caustic
soda and normal carbonate, or the latter salt and acid
carbonate. Can you suggest a volumetric method of
estimating the carbon dioxide in i cu. cm. of the "soda


Suppose I wish to displace all the ammonia contained in
10 grm of ammonium nitrate ; what weight (a) of sodium
hydroxide, (b) of caustic soda containing 70 per cent of
NaOH, will be theoretically required?

Equation NH 4 NO 3 + NaOH = NaNO 3 + NH 3 + H 2 O
80. i 40

shows that practically 80 grm. of the salt require 40 grm.
of NaOH; so that 10 grm. will require (a) 5 grm. NaOH,

(b) 5 x -^? or 7| grm. of the caustic soda.

Exercise 46. Find the percentage of ammonia
in some sal-ammoniac crystals (fairly pure).

Directions. i. Use the "indirect method" since it is
known that the sample is not very impure. Dissolve
about a gramme in a small quantity of water contained in
a small ordinary flask, and determine whether the solution
is neutral. If found to be acid the amount of standard
alkali necessary to neutralize it must be found and allowed
for in your calculations.

2. Calculate the volume of N. NaOH necessary to expel
the NH 3 from the weight of NH 4 C1 taken. Equation

NH 4 C1 + NaOH = NaCl + NH 3 + H 2 O

- i



shows that 53^ grm. require 40 grm. NaOH, or i grm.
requires about grm. NaOH, which will be contained in
20 cu. cm. N. NaOH. About three times this volume,
therefore, should be ample both to expel all NH 3 and leave
sufficient excess when diluted to 250 cu. cm. to be accu-

rately determined by - acid.

3. Add the N. NaOH noting" exact volume taken to
the solution of NH 4 C1, and boil till all NH 3 has been
expelled. This may be determined by holding a damp
litmus or turmeric paper in the steam.

4. Cool, transfer to a suitable measuring-flask, together
with rinsings, add water to the mark, and mix well.

5. Determine excess of NaOH with suitable standard
acid. The difference between this and that originally
taken will be the equivalent of the NH 3 or NH 4 C1 in the
quantity of salt taken.

Note the following methods of calculation :

A. Weight of crystals taken ... ... i grm.

Solution found to be neutral.

Quantity of N. NaOH added to expel NH 3

50 cu. cm. or ... ... ... ... 2 grm.

Residual alkali diluted to ... ... 250 cu. cm.

Volume taken at each titration ... 25 ,,

Average volume - H 2 SO 4 required to

neutralize the alkali ... ... ... 31.3 ,,

Weight of H 2 SO 4 used for each titration

(0.0049 X 31.3) grm.

Total H 2 SO 4 used for excess of NaOH

= (0.0049 x 31.3 X 10) grm. H 2 SO 4 .

1 2 4 6 7 8 9 10 11

Online LibraryWilliam Thomas BooneA complete course of volumetric analysis for middle and higher forms of schools → online text (page 4 of 11)