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An Introductory Course of Quantitative Chemical Analysis by Henry P. Talbot

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solution of the fused mass in water. The insoluble silica is filtered,
washed, and weighed, and the weight added to the weight of silica
previously obtained.]

[Note 4: Aluminium and iron are likely to be thrown down as basic
salts from hot, very dilute solutions of their chlorides, as a result
of hydrolysis. If the silica were washed only with hot water, the
solution of these chlorides remaining in the filter after the passage
of the original filtrate would gradually become so dilute as to throw
down basic salts within the pores of the filter, which would remain
with the silica. To avoid this, an acid wash-water is used until the
aluminium and iron are practically removed. The acid is then removed
by water.]

IGNITION AND TESTING OF SILICA

PROCEDURE.--Transfer the two washed filters belonging to each
determination to a platinum crucible, which need not be previously
weighed, and burn off the filter (Note 1). Ignite for thirty minutes
over the blast lamp with the cover on the crucible, and then for
periods of ten minutes, until the weight is constant.

When a constant weight has been obtained, pour into the crucible about
3 cc. of water, and then 3 cc. of hydrofluoric acid. !This must be
done in a hood with a good draft and great care must be taken not to
come into contact with the acid or to inhale its fumes (Note 2!).

If the precipitate has dissolved in this quantity of acid, add two
drops of concentrated sulphuric acid, and heat very slowly (always
under the hood) until all the liquid has evaporated, finally igniting
to redness. Cool in a desiccator, and weigh the crucible and residue.
Deduct this weight from the previous weight of crucible and impure
silica, and from the difference calculate the percentage of silica in
the sample (Note 3).

[Note 1: The silica undergoes no change during the ignition beyond the
removal of all traces of water; but Hillebrand (!loc. cit.!) has shown
that the silica holds moisture so tenaciously that prolonged ignition
over the blast lamp is necessary to remove it entirely. This finely
divided, ignited silica tends to absorb moisture, and should be
weighed quickly.]

[Note 2: Notwithstanding all precautions, the ignited precipitate of
silica is rarely wholly pure. It is tested by volatilisation of the
silica as silicon fluoride after solution in hydrofluoric acid, and,
if the analysis has been properly conducted, the residue, after
treatment with the acids and ignition, should not exceed 1 mg.

The acid produces ulceration if brought into contact with the skin,
and its fumes are excessively harmful if inhaled.]

[Note 3: The impurities are probably weighed with the original
precipitate in the form of oxides. The addition of the sulphuric
acid displaces the hydrofluoric acid, and it may be assumed that the
resulting sulphates (usually of iron or aluminium) are converted to
oxides by the final ignition.

It is obvious that unless the sulphuric and hydrofluoric acids used
are known to leave no residue on evaporation, a quantity equal to that
employed in the analysis must be evaporated and a correction applied
for any residue found.]

[Note 4: If the silicate to be analyzed is shown by a previous
qualitative examination to be completely decomposable, it may be
directly treated with hydrochloric acid, the solution evaporated to
dryness, and the silica dehydrated and further treated as described in
the case of the feldspar after fusion.

A silicate which gelatinizes on treatment with acids should be mixed
first with a little water, and the strong acid added in small portions
with stirring, otherwise the gelatinous silicic acid incloses
particles of the original silicate and prevents decomposition. The
water, by separating the particles and slightly lessening the rapidity
of action, prevents this difficulty. This procedure is one which
applies in general to the solution of fine mineral powders in acids.

If a small residue remains undecomposed by the treatment of the
silicate with acid, this may be filtered, washed, ignited and fused
with sodium carbonate and a solution of the fused mass added to the
original acid solution. This double procedure has an advantage, in
that it avoids adding so large a quantity of sodium salts as is
required for disintegration of the whole of the silicate by the fusion
method.]

PART IV

STOICHIOMETRY

The problems with which the analytical chemist has to deal are not, as
a matter of actual fact, difficult either to solve or to understand.
That they appear difficult to many students is due to the fact that,
instead of understanding the principles which underlie each of the
small number of types into which these problems may be grouped, each
problem is approached as an individual puzzle, unrelated to others
already solved or explained. This attitude of mind should be carefully
avoided.

It is obvious that ability to make the calculations necessary for
the interpretation of analytical data is no less important than the
manipulative skill required to obtain them, and that a moderate time
spent in the careful study of the solutions of the typical problems
which follow may save much later embarrassment.

1. It is often necessary to calculate what is known as a "chemical
factor," or its equivalent logarithmic value called a "log factor,"
for the conversion of the weight of a given chemical substance into an
equivalent weight of another substance. This is, in reality, a very
simple problem in proportion, making use of the atomic or molecular
weights of the substances in question which are chemically equivalent
to each other. One of the simplest cases of this sort is the
following: What is the factor for the conversion of a given weight of
barium sulphate (BaSO_{4}) into an equivalent weight of sulphur (S)?
The molecular weight of BaSO_{4} is 233.5. There is one atom of S in
the molecule and the atomic weight of S is 32.1. The chemical factor
is, therefore, 32.1/233.5, or 0.1375 and the weight of S corresponding
to a given weight of BaSO_{4} is found by multiplying the weight of
BaSO_{4} by this factor. If the problem takes the form, "What is
the factor for the conversion of a given weight of ferric oxide
(Fe_{2}O_{3}) into ferrous oxide (FeO), or of a given weight of
mangano-manganic oxide (Mn_{3}O_{4}) into manganese (Mn)?" the
principle involved is the same, but it must then be noted that, in the
first instance, each molecule of Fe_{2}O_{3} will be equivalent to two
molecules of FeO, and in the second instance that each molecule of
Mn_{3}O_{4} is equivalent to three atoms of Mn. The respective factors
then become

(2FeO/Fe_{2}O_{3}) or (143.6/159.6) and (3Mn/Mn_{3}O_{4}) or
(164.7/228.7).

It is obvious that the arithmetical processes involved in this type
of problem are extremely simple. It is only necessary to observe
carefully the chemical equivalents. It is plainly incorrect to express
the ratio of ferrous to ferric oxide as (FeO/Fe_{2}O_{3}), since each
molecule of the ferric oxide will yield two molecules of the ferrous
oxide. Mistakes of this sort are easily made and constitute one of the
most frequent sources of error.

2. A type of problem which is slightly more complicated in appearance,
but exactly comparable in principle, is the following: "What is the
factor for the conversion of a given weight of ferrous sulphate
(FeSO_{4}), used as a reducing agent against potassium permanganate,
into the equivalent weight of sodium oxalate (Na_{2}C_{2}O_{4})?" To
determine the chemical equivalents in such an instance it is necessary
to inspect the chemical reactions involved. These are:

10FeSO_{4} + 2KMnO_{4} + 8H_{2}SO_{4} --> 5Fe_{2}(SO_{4})_{3} +
K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O,

5Na_{2}C_{2}O_{4} + 2KMnO_{4} + 8H_{2}SO_{4} --> 5Na_{2}SO_{4} +
10CO_{2} + K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O.

It is evident that 10FeSO_{4} in the one case, and 5Na_{2}C_{2}O_{4}
in the other, each react with 2KMnO_{4}. These molecular
quantities are therefore equivalent, and the factor becomes
(10FeSO_{4}/5Na_{2}C_{2}O_{4}) or (2FeSO_{4}/Na_{2}C_{2}O_{4}) or
(303.8/134).

Again, let it be assumed that it is desired to determine the
factor required for the conversion of a given weight of potassium
permanganate (KMnO_{4}) into an equivalent weight of potassium
bichromate (K_{2}Cr_{2}O_{7}), each acting as an oxidizing agent
against ferrous sulphate. The reactions involved are:

10FeSO_{4} + 2KMnO_{4} + 8H_{2}SO_{4} --> 5Fe_{2}(SO_{4})_{3} +
K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O,

6FeSO_{4} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} --> 3Fe_{2}(SO_{3})_{3} +
K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + 7H_{2}O.

An inspection of these equations shows that 2KMO_{4} react with
10FeSO_{4}, while K_{2}Cr_{2}O_{7} reacts with 6FeSO_{4}. These are
not equivalent, but if the first equation is multiplied by 3 and the
second by 5 the number of molecules of FeSO_{4} is then the same in
both, and the number of molecules of KMnO_{4} and K_{2}Cr_{2}O_{7}
reacting with these 30 molecules become 6 and 5 respectively. These
are obviously chemically equivalent and the desired factor is
expressed by the fraction (6KMnO_{4}/5K_{2}Cr_{2}O_{7}) or
(948.0/1471.0).

3. It is sometimes necessary to calculate the value of solutions
according to the principles just explained, when several successive
reactions are involved. Such problems may be solved by a series of
proportions, but it is usually possible to eliminate the common
factors and solve but a single one. For example, the amount of MnO_{2}
in a sample of the mineral pyrolusite may be determined by dissolving
the mineral in hydrochloric acid, absorbing the evolved chlorine in a
solution of potassium iodide, and measuring the liberated iodine
by titration with a standard solution of sodium thiosulphate. The
reactions involved are:

MnO_{2} + 4HCl --> MnCl_{2} + 2H_{2}O + Cl_{2}
Cl_{2} + 2KI --> I_{2} + 2KCl
I_{2} + 2Na_{2}S_{2}O_{3} --> 2NaI + Na_{2}S_{4}O_{6}

Assuming that the weight of thiosulphate corresponding to the
volume of sodium thiosulphate solution used is known, what is the
corresponding weight of manganese dioxide? From the reactions given
above, the following proportions may be stated:

2Na_{2}S_{2}O_{3}:I_{2} = 316.4:253.9,

I_{2}:Cl_{2} = 253.9:71,

Cl_{2}:MnO_{2} = 71:86.9.

After canceling the common factors, there remains
2Na_{2}S_{2}O_{3}:MnO_{2} = 316.4:86.9, and the factor for the
conversion of thiosulphate into an equivalent of manganese dioxide is
86.9/316.4.

4. To calculate the volume of a reagent required for a specific
operation, it is necessary to know the exact reaction which is to be
brought about, and, as with the calculation of factors, to keep in
mind the molecular relations between the reagent and the substance
reacted upon. For example, to estimate the weight of barium chloride
necessary to precipitate the sulphur from 0.1 gram of pure pyrite
(FeS_{2}), the proportion should read

488. 120.0
2(BaCl_{2}.2H_{2}O):FeS_{2} = x:0.1,

where !x! represents the weight of the chloride required. Each of the
two atoms of sulphur will form upon oxidation a molecule of sulphuric
acid or a sulphate, which, in turn, will require a molecule of the
barium chloride for precipitation. To determine the quantity of the
barium chloride required, it is necessary to include in its molecular
weight the water of crystallization, since this is inseparable from
the chloride when it is weighed. This applies equally to other similar
instances.

If the strength of an acid is expressed in percentage by weight, due
regard must be paid to its specific gravity. For example, hydrochloric
acid (sp. gr. 1.12) contains 23.8 per cent HCl !by weight!; that is,
0.2666 gram HCl in each cubic centimeter.

5. It is sometimes desirable to avoid the manipulation required for
the separation of the constituents of a mixture of substances by
making what is called an "indirect analysis." For example, in the
analysis of silicate rocks, the sodium and potassium present may be
obtained in the form of their chlorides and weighed together. If the
weight of such a mixture is known, and also the percentage of chlorine
present, it is possible to calculate the amount of each chloride in
the mixture. Let it be assumed that the weight of the mixed chlorides
is 0.15 gram, and that it contains 53 per cent of chlorine.

The simplest solution of such a problem is reached through algebraic
methods. The weight of chlorine is evidently 0.15 x 0.53, or 0.0795
gram. Let x represent the weight of sodium chloride present and y
that of potassium chloride. The molecular weight of NaCl is 58.5 and
that of KCl is 74.6. The atomic weight of chlorine is 35.5. Then

x + y = 0.15
(35.5/58.5)x + (35.5/74.6)y = 0.00795

Solving these equations for x shows the weight of NaCl to be 0.0625
gram. The weight of KCl is found by subtracting this from 0.15.

The above is one of the most common types of indirect analyses. Others
are more complex but they can be reduced to algebraic expressions and
solved by their aid. It should, however, be noted that the results
obtained by these indirect methods cannot be depended upon for high
accuracy, since slight errors in the determination of the common
constituent, as chlorine in the above mixture, will cause considerable
variations in the values found for the components. They should not be
employed when direct methods are applicable, if accuracy is essential.

PROBLEMS

(The reactions necessary for the solution of these problems are either
stated with the problem or may be found in the earlier text. In the
calculations from which the answers are derived, the atomic weights
given on page 195 have been employed, using, however, only the first
decimal but increasing this by 1 when the second decimal is 5 or
above. Thus, 39.1 has been taken as the atomic weight of potassium,
32.1 for sulphur, etc. This has been done merely to secure uniformity
of treatment, and the student should remember that it is always well
to take into account the degree of accuracy desired in a particular
instance in determining the number of decimal places to retain.
Four-place logarithms were employed in the calculations. Where four
figures are given in the answer, the last figure may vary by one or
(rarely) by two units, according to the method by which the problem is
solved.)

VOLUMETRIC ANALYSIS

1. How many grams of pure potassium hydroxide are required for exactly
1 liter of normal alkali solution?

!Answer!: 56.1 grams.

2. Calculate the equivalent in grams (a) of sulphuric acid as an acid;
(b) of hydrochloric acid as an acid; (c) of oxalic acid as an acid;
(d) of nitric acid as an acid.

!Answers!: (a) 49.05; (b) 36.5; (c) 63; (d) 63.

3. Calculate the equivalent in grams of (a) potassium hydroxide;
(b) of sodium carbonate; (c) of barium hydroxide; (d) of sodium
bicarbonate when titrated with an acid.

!Answers!: (a) 56.1; (b) 53.8; (c) 85.7; (d) 84.

4. What is the equivalent in grams of Na_{2}HPO_{4} (a) as a
phosphate; (b) as a sodium salt?

!Answers!: (a) 47.33; (b) 71.0.

5. A sample of aqueous hydrochloric acid has a specific gravity
of 1.12 and contains 23.81 per cent hydrochloric acid by weight.
Calculate the grams and the milliequivalents of hydrochloric acid
(HCl) in each cubic centimeter of the aqueous acid.

!Answers!: 0.2667 gram; 7.307 milliequivalents.

6. How many cubic centimeters of hydrochloric acid (sp. gr. 1.20
containing 39.80 per cent HCl by weight) are required to furnish 36.45
grams of the gaseous compound?

!Answer!: 76.33 cc.

7. A given solution contains 0.1063 equivalents of hydrochloric acid
in 976 cc. What is its normal value?

!Answer!: 0.1089 N.

8. In standardizing a hydrochloric acid solution it is found that
47.26 cc. of hydrochloric acid are exactly equivalent to 1.216 grams
of pure sodium carbonate, using methyl orange as an indicator. What is
the normal value of the hydrochloric acid?

!Answer!: 0.4855 N.

9. Convert 42.75 cc. of 0.5162 normal hydrochloric acid to the
equivalent volume of normal hydrochloric acid.

!Answer!: 22.07 cc.

10. A solution containing 25.27 cc. of 0.1065 normal hydrochloric acid
is added to one containing 92.21 cc. of 0.5431 normal sulphuric acid
and 50 cc. of exactly normal potassium hydroxide added from a pipette.
Is the solution acid or alkaline? How many cubic centimeters of
0.1 normal acid or alkali must be added to exactly neutralize the
solution?

!Answer!: 27.6 cc. alkali (solution is acid).

11. By experiment the normal value of a sulphuric acid solution is
found to be 0.5172. Of this acid 39.65 cc. are exactly equivalent to
21.74 cc. of a standard alkali solution. What is the normal value of
the alkali?

!Answer!: 0.9432 N.

12. A solution of sulphuric acid is standardized against a sample of
calcium carbonate which has been previously accurately analyzed and
found to contain 92.44% CaCO_{3} and no other basic material. The
sample weighing 0.7423 gram was titrated by adding an excess of acid
(42.42 cc.) and titrating the excess with sodium hydroxide solution
(11.22 cc.). 1 cc. of acid is equivalent to 0.9976 cc. of sodium
hydroxide. Calculate the normal value of each.

!Answers!: Acid 0.4398 N; alkali 0.4409 N.

13. Given five 10 cc. portions of 0.1 normal hydrochloric acid, (a)
how many grams of silver chloride will be precipitated by a portion
when an excess of silver nitrate is added? (b) how many grams of pure
anhydrous sodium carbonate (Na_{2}CO_{3}) will be neutralized by a
portion of it? (c) how many grams of silver will there be in the
silver chloride formed when an excess of silver nitrate is added to a
portion? (d) how many grams of iron will be dissolved to FeCl_{2} by a
portion of it? (e) how many grams of magnesium chloride will be formed
and how many grams of carbon dioxide liberated when an excess of
magnesium carbonate is treated with a portion of the acid?

!Answers!: (a) 0.1434; (b) 0.053; (c) 0.1079; (d) 0.0279; (e) 0.04765,
and 0.022.

14. If 30.00 grams of potassium tetroxalate
(KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O) are dissolved and the solution
diluted to exactly 1 liter, and 40 cc. are neutralized with 20 cc.
of a potassium carbonate solution, what is the normal value of the
carbonate solution?

!Answer!: 0.7084 N.

15. How many cubic centimeters of 0.3 normal sulphuric acid will be
required to neutralize (a) 30 cc. of 0.5 normal potassium hydroxide;
(b) to neutralize 30 cc. of 0.5 normal barium hydroxide; (c) to
neutralize 20 cc. of a solution containing 10.02 grams of potassium
bicarbonate per 100 cc.; (d) to give a precipitate of barium sulphate
weighing 0.4320 gram?

!Answers!: (a) 50 cc.; (b) 50 cc.; (c) 66.73 cc.; (d) 12.33 cc.

16. It is desired to dilute a solution of sulphuric acid of which 1
cc. is equivalent to 0.1027 gram of pure sodium carbonate to make it
exactly 1.250 normal. 700 cc. of the solution are available. To what
volume must it be diluted?

!Answer!: 1084 cc.

17. Given the following data: 1 cc. of NaOH = 1.117 cc. HCl. The HCl
is 0.4876 N. How much water must be added to 100 cc. of the alkali to
make it exactly 0.5 N.?

!Answer!: 9.0 cc.

18. What is the normal value of a sulphuric acid solution which has a
specific gravity of 1.839 and contains 95% H_{2}SO_{4} by weight?

!Answer!: 35.61 N.

19. A sample of Rochelle Salt (KNaC_{4}H_{4}O_{6}.4H_{2}O), after
ignition in platinum to convert it to the double carbonate, is
titrated with sulphuric acid, using methyl orange as an indicator.
From the following data calculate the percentage purity of the sample:

Wt. sample = 0.9500 gram
H_{2}SO_{4} used = 43.65 cc.
NaOH used = 1.72 cc.
1 cc. H_{2}SO_{4} = 1.064 cc. NaOH
Normal value NaOH = 0.1321 N.

!Answer!: 87.72 cc.

20. One gram of a mixture of 50% sodium carbonate and 50% potassium
carbonate is dissolved in water, and 17.36 cc. of 1.075 N acid is
added. Is the resulting solution acid or alkaline? How many cubic
centimeters of 1.075 N acid or alkali will have to be added to make
the solution exactly neutral?

!Answers!: Acid; 1.86 cc. alkali.

21. In preparing an alkaline solution for use in volumetric work, an
analyst, because of shortage of chemicals, mixed exactly 46.32 grams
of pure KOH and 27.64 grams of pure NaOH, and after dissolving in
water, diluted the solution to exactly one liter. How many cubic
centimeters of 1.022 N hydrochloric acid are necessary to neutralize
50 cc. of the basic solution?

!Answer!: 74.18 cc.

22. One gram of crude ammonium salt is treated with strong potassium
hydroxide solution. The ammonia liberated is distilled and collected
in 50 cc. of 0.5 N acid and the excess titrated with 1.55 cc. of 0.5 N
sodium hydroxide. Calculate the percentage of NH_{3} in the sample.

!Answer!: 41.17%.

23. In titrating solutions of alkali carbonates in the presence of
phenolphthalein, the color change takes place when the carbonate has
been converted to bicarbonate. In the presence of methyl orange, the
color change takes place only when the carbonate has been completely
neutralized. From the following data, calculate the percentages of
Na_{2}CO_{3} and NaOH in an impure mixture. Weight of sample, 1.500
grams; HCl (0.5 N) required for phenolphthalein end-point, 28.85 cc.;
HCl (0.5 N) required to complete the titration after adding methyl
orange, 23.85 cc.

!Answers!: 6.67% NaOH; 84.28% Na_{2}CO_{3}.

24. A sample of sodium carbonate containing sodium hydroxide weighs
1.179 grams. It is titrated with 0.30 N hydrochloric acid, using
phenolphthalein in cold solution as an indicator and becomes colorless
after the addition of 48.16 cc. Methyl orange is added and 24.08 cc.
are needed for complete neutralization. What is the percentage of NaOH
and Na_{2}CO_{3}?

!Answers!: 24.50% NaOH; 64.92% Na_{2}CO_{3}.

25. From the following data, calculate the percentages of Na_{2}CO_{3}
and NaHCO_{3} in an impure mixture. Weight of sample 1.000 gram;
volume of 0.25 N hydrochloric acid required for phenolphthalein
end-point, 26.40 cc.; after adding an excess of acid and boiling out
the carbon dioxide, the total volume of 0.25 N hydrochloric acid
required for phenolphthalein end-point, 67.10 cc.

!Answer!: 69.95% Na_{2}CO_{3}; 30.02% NaHCO_{3}.

26. In the analysis of a one-gram sample of soda ash, what must be the
normality of the acid in order that the number of cubic centimeters of
acid used shall represent the percentage of carbon dioxide present?

!Answer!: 0.4544 gram.

27. What weight of pearl ash must be taken for analysis in order that
the number of cubic centimeters of 0.5 N acid used may be equal to one
third the percentage of K_{2}CO_{3}?

!Answer!: 1.152 grams.

28. What weight of cream of tartar must have been taken for analysis
in order to have obtained 97.60% KHC_{4}H_{4}O_{6} in an analysis
involving the following data: NaOH used = 30.06 cc.; H_{2}SO_{4}
solution used = 0.50 cc.; 1 cc. H_{2}SO_{4} sol. = 0.0255 gram
CaCO_{3}; 1 cc. H_{2}SO_{4} sol. = 1.02 cc. NaOH sol.?

!Answer!: 2.846 grams.

29. Calculate the percentage of potassium oxide in an impure sample of
potassium carbonate from the following data: Weight of sample = 1.00
gram; HCl sol. used = 55.90 cc.; NaOH sol. used = 0.42 cc.; 1 cc. NaOH
sol. = 0.008473 gram of KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O; 2 cc.
HCl sol. = 5 cc. NaOH sol.

!Answer!: 65.68%.

30. Calculate the percentage purity of a sample of calcite
(CaCO_{3}) from the following data: (Standardization); Weight of
H_{2}C_{2}O_{4}.2H_{2}O = 0.2460 gram; NaOH solution used = 41.03
cc.; HCl solution used = 0.63; 1 cc. NaOH solution = 1.190 cc. HCl
solution. (Analysis); Weight of sample 0.1200 gram; HCl used = 36.38
cc.; NaOH used = 6.20 cc.

!Answer!: 97.97%.

31. It is desired to dilute a solution of hydrochloric acid to exactly
0.05 N. The following data are given: 44.97 cc. of the hydrochloric
acid are equivalent to 43.76 cc. of the NaOH solution. The NaOH
is standardized against a pure potassium tetroxalate
(KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O) weighing 0.2162 gram and
requires 49.14 cc. How many cc. of water must be added to 1000 cc. of
the aqueous hydrochloric acid?

!Answer!: 11 cc.

32. How many cubic centimeters of 3 N phosphoric acid must be added to
300 cc. of 0.4 N phosphoric acid in order that the resulting solution
may be 0.6 N?

!Answer!: 25 cc.

33. To oxidize the iron in 1 gram of
FeSO_{4}(NH_{4})_{2}SO_{4}.6H_{2}O (mol. wgt. 392) requires 3 cc. of
a given solution of HNO_{3}. What is the normality of the nitric
acid when used as an acid? 6FeSO_{4} + 2HNO_{3} + 2H_{2}SO_{4} =
3Fe_{2}(SO_{4})_{3} + 2NO + 4H_{2}O.

!Answer!: 0.2835 N.

34. The same volume of carbon dioxide at the same temperature and the
same pressure is liberated from a 1 gram sample of dolomite, by adding
an excess of hydrochloric acid, as can be liberated by the addition of
35 cc. of 0.5 N hydrochloric acid to an excess of any pure or impure
carbonate. Calculate the percentage of CO_{2} in the dolomite.

!Answer!: 38.5%.

35. How many cubic centimeters of sulphuric acid (sp. gr. 1.84,
containing 96% H_{2}SO_{4} by weight) will be required to displace the
chloride in the calcium chloride formed by the action of 100 cc. of
0.1072 N hydrochloric acid on an excess of calcium carbonate, and how
many grams of CaSO_{4} will be formed?

!Answers!: 0.298 cc.; 0.7300 gram.

36. Potassium hydroxide which has been exposed to the air is found on
analysis to contain 7.62% water, 2.38% K_{2}CO_{3}. and 90% KOH. What
weight of residue will be obtained if one gram of this sample is added
to 46 cc. of normal hydrochloric acid and the resulting solution,
after exact neutralization with 1.070 N potassium hydroxide solution,
is evaporated to dryness?

!Answer!: 3.47 grams.

37. A chemist received four different solutions, with the statement
that they contained either pure NaOH; pure Na_{2}CO_{3}; pure
NaHCO_{3}, or mixtures of these substances. From the following data
identify them:

Sample I. On adding phenolphthalein to a solution of the substance, it
gave no color to the solution.

Sample II. On titrating with standard acid, it required 15.26 cc. for
a change in color, using phenolphthalein, and 17.90 cc. additional,
using methyl orange as an indicator.

Sample III. The sample was titrated with hydrochloric acid until the
pink of phenolphthalein disappeared, and on the addition of methyl
orange the solution was colored pink.

Sample IV. On titrating with hydrochloric acid, using phenolphthalein,
15.00 cc. were required. A new sample of the same weight required
exactly 30 cc. of the same acid for neutralization, using methyl
orange.

!Answers!: (a) NaHCO_{3}; (b) NaHCO_{3}+Na_{2}CO_{3}; (c)NaOH; (d)
Na_{2}CO_{3}.

38. In the analysis of a sample of KHC_{4}H_{4}O_{6} the following
data are obtained: Weight sample = 0.4732 gram. NaOH solution used =
24.97 cc. 3.00 cc. NaOH = 1 cc. of H_{3}PO_{4} solution of which 1
cc. will precipitate 0.01227 gram of magnesium as MgNH_{4}PO_{4}.
Calculate the percentage of KHC_{4}H_{4}O_{6}.

!Answer!: 88.67%.

39. A one-gram sample of sodium hydroxide which has been exposed to
the air for some time, is dissolved in water and diluted to exactly
500 cc. One hundred cubic centimeters of the solution, when titrated
with 0.1062 N hydrochloric acid, using methyl orange as an indicator,
requires 38.60 cc. for complete neutralization. Barium chloride in
excess is added to a second portion of 100 cc. of the solution, which
is diluted to exactly 250 cc., allowed to stand and filtered. Two
hundred cubic centimeters of this filtrate require 29.62 cc. of 0.1062
N hydrochloric acid for neutralization, using phenolphthalein as an
indicator. Calculate percentage of NaOH, Na_{2}CO_{3}, and H_{2}O.

!Answers!: 78.63% NaOH; 4.45% Na_{2}CO_{3}; 16.92% H_{2}O.

40. A sodium hydroxide solution (made from solid NaOH which has been
exposed to the air) was titrated against a standard acid using methyl
orange as an indicator, and was found to be exactly 0.1 N. This
solution was used in the analysis of a material sold at 2 cents per
pound per cent of an acid constituent A, and always mixed so that
it was supposed to contain 15% of A, on the basis of the analyst's
report. Owing to the carelessness of the analyst's assistant, the
sodium hydroxide solution was used with phenolphthalein as an
indicator in cold solution in making the analyses. The concern
manufacturing this material sells 600 tons per year, and when the
mistake was discovered it was estimated that at the end of a year
the error in the use of indicators would either cost them or their
customers $6000. Who would lose and why? Assuming the impure NaOH used
originally in making the titrating solution consisted of NaOH and
Na_{2}CO_{3} only, what per cent of each was present?

!Answers!: Customer lost; 3.94% Na_{2}CO_{3}; 96.06% NaOH.

41. In the standardization of a K_{2}Cr_{2}O_{7} solution against iron
wire, 99.85% pure, 42.42 cc. of the solution were added. The weight of
the wire used was 0.22 gram. 3.27 cc. of a ferrous sulphate solution
having a normal value as a reducing agent of 0.1011 were added
to complete the titration. Calculate the normal value of the
K_{2}Cr_{2}O_{7}.

!Answer!: 0.1006 N.

42. What weight of iron ore containing 56.2% Fe should be taken to
standardize an approximately 0.1 N oxidizing solution, if not more
than 47 cc. are to be used?

!Answer!: 0.4667 gram.

43. One tenth gram of iron wire, 99.78% pure, is dissolved in
hydrochloric acid and the iron oxidized completely with bromine water.
How many grams of stannous chloride are there in a liter of solution
if it requires 9.47 cc. to just reduce the iron in the above? What
is the normal value of the stannous chloride solution as a reducing
agent?

!Answer!: 17.92 grams; 0.1888 N.

44. One gram of an oxide of iron is fused with potassium acid sulphate
and the fusion dissolved in acid. The iron is reduced with stannous
chloride, mercuric chloride is added, and the iron titrated with a
normal K_{2}Cr_{2}O_{7} solution. 12.94 cc. were used. What is the
formula of the oxide, FeO, Fe_{2}O_{3}, or Fe_{3}O_{4}?

!Answer!: Fe_{3}O_{4}.

45. If an element has 98 for its atomic weight, and after reduction
with stannous chloride could be oxidized by bichromate to a state
corresponding to an XO_{4}^{-} anion, compute the oxide, or valence,
corresponding to the reduced state from the following data: 0.3266
gram of the pure element, after being dissolved, was reduced with
stannous chloride and oxidized by 40 cc. of K_{2}Cr_{2}O_{7}, of which
one cc. = 0.1960 gram of FeSO_{4}(NH_{4})_{2}SO_{4}.6H_{2}O.

!Answer!: Monovalent.

46. Determine the percentage of iron in a sample of limonite from the
following data: Sample = 0.5000 gram. KMnO_{4} used = 50 cc. 1 cc.
KMnO_{4} = 0.005317 gram Fe. FeSO_{4} used = 6 cc. 1 cc. FeSO_{4} =
0.009200 gram FeO.

!Answer!: 44.60%.

47. If 1 gram of a silicate yields 0.5000 gram of Fe_{2}O_{3} and
Al_{2}O_{3} and the iron present requires 25 cc. of 0.2 N KMnO_{4},
calculate the percentage of FeO and Al_{2}O_{3} in the sample.

!Answer!: 35.89% FeO; 10.03% Al_{2}O_{3}.

48. A sample of magnesia limestone has the following composition:
Silica, 3.00%; ferric oxide and alumina, 0.20%; calcium oxide, 33.10%;
magnesium oxide, 20.70%; carbon dioxide, 43.00%. In manufacturing lime
from the above the carbon dioxide is reduced to 3.00%. How many cubic
centimeters of normal KMnO_{4} will be required to determine the
calcium oxide volumetrically in a 1 gram sample of the lime?

!Answer!: 20.08 cc.

49. If 100 cc. of potassium bichromate solution (10 gram
K_{2}Cr_{2}O_{7} per liter), 5 cc. of 6 N sulphuric acid, and 75 cc.
of ferrous sulphate solution (80 grams FeSO_{4}.7H_{2}O per liter) are
mixed, and the resulting solution titrated with 0.2121 N KMnO_{4}, how
many cubic centimeters of the KMnO_{4} solution will be required to
oxidize the iron?

!Answer!: 5.70 cc.

50. If a 0.5000 gram sample of limonite containing 59.50 per cent
Fe_{2}O_{3} requires 40 cc. of KMnO_{4} to oxidize the iron, what
is the value of 1 cc. of the permanganate in terms of (a) Fe, (b)
H_{2}C_{2}O_{4}.2H_{2}O?

!Answers!: (a) 0.005189 gram; (b) 0.005859 gram.

51. A sample of pyrolusite weighing 0.6000 gram is treated with 0.9000
gram of oxalic acid. The excess oxalic acid requires 23.95 cc. of
permanganate (1 cc. = 0.03038 gram FeSO_{4}.7H_{2}O). What is the
percentage of MnO_{2}, in the sample?

!Answer!: 84.47%.

52. A solution contains 50 grams of
KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O per liter. What is the normal
value of the solution (a) as an acid, and (b) as a reducing agent?

!Answers!: (a) 0.5903 N; (b) 0.7872 N.

53. In the analysis of an iron ore containing 60% Fe_{2}O_{3}, a
sample weighing 0.5000 gram is taken and the iron is reduced with
sulphurous acid. On account of failure to boil out all the excess
SO_{2}, 38.60 cubic centimeters of 0.1 N KMnO_{4} were required to
titrate the solution. What was the error, percentage error, and what
weight of sulphur dioxide was in the solution?

!Answers!: (a) 1.60%; (b) 2.67%; (c) 0.00322 gram.

54. From the following data, calculate the ratio of the nitric acid as
an oxidizing agent to the tetroxalate solution as a reducing agent:
1 cc. HNO_{3} = 1.246 cc. NaOH solution; 1 cc. NaOH = 1.743 cc.
KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O solution; Normal value NaOH =
0.12.

!Answer!: 4.885.

55. Given the following data: 25 cc. of a hydrochloric acid, when
standardized gravimetrically as silver chloride, yields a precipitate
weighing 0.5465 gram. 24.35 cc. of the hydrochloric acid are exactly
equivalent to 30.17 cc. of KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O
solution. How much water must be added to a liter of the oxalate
solution to make it exactly 0.025 N as a reducing agent?

!Answer!: 5564 cc.

56. Ten grams of a mixture of pure potassium tetroxalate
(KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O) and pure oxalic acid
(H_{2}C_{2}O_{4}.2H_{2}O) are dissolved in water and diluted to
exactly 1000 cc. The normal value of the oxalate solution when used as
an acid is 0.1315. Calculate the ratio of tetroxalate to oxalate used
in making up the solution and the normal value of the solution as a
reducing agent.

!Answers!: 2:1; 0.1577 N.

57. A student standardized a solution of NaOH and one of KMnO_{4}
against pure KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O and found the former
to be 0.07500 N as an alkali and the latter exactly 0.1 N as an
oxidizing agent. By coincidence, exactly 47.26 cc. were used in each
standardization. Find the ratio of the oxalate used in the
NaOH standardization to the oxalate used in the permanganate
standardization.

!Answer!: 1:1.

58. A sample of apatite weighing 0.60 gram is analyzed for its
phosphoric anhydride content. If the phosphate is precipitated as
(NH_{4})_{3}PO_{4}.12MoO_{3}, and the precipitate (after solution and
reduction of the MoO_{3} to Mo_{24}O_{37}), requires 100 cc. of normal
KMnO_{4} to oxidize it back to MoO_{3}, what is the percentage of
P_{2}O_{5}?

!Answer!: 33.81%.

59. In the analysis of a sample of steel weighing 1.881 grams the
phosphorus was precipitated with ammonium molybdate and the yellow
precipitate was dissolved, reduced and titrated with KMnO_{4}. If the
sample contained 0.025 per cent P and 6.01 cc. of KMnO_{4} were used,
to what oxide was the molybdenum reduced? 1 cc. KMnO_{4} = 0.007188
gram Na_{2}C_{2}O_{4}.

!Answer!: Mo_{4}O_{5}.

60. What is the value of 1 cc. of an iodine solution (1 cc. equivalent
to 0.0300 gram Na_{2}S_{2}O_{3}) in terms of As_{2}O_{3}?

!Answer!: 0.009385 gram.

61. 48 cc. of a solution of sodium thiosulphate are required to
titrate the iodine liberated from an excess of potassium iodide
solution by 0.3000 gram of pure KIO_{3}. (KIO_{3} + 5KI + 3H_{2}SO_{4}
= 3K_{2}SO_{4} + 3I_{2} + 3H_{2}O.) What is the normal strength of the
sodium thiosulphate and the value of 1 cc. of it in terms of iodine?

!Answers!: 0.1753 N; 0.02224 gram.

62. One thousand cubic centimeters of 0.1079 N sodium thiosulphate
solution is allowed to stand. One per cent by weight of the
thiosulphate is decomposed by the carbonic acid present in the
solution. To what volume must the solution be diluted to make it
exactly 0.1 N as a reducing agent? (Na_{2}S_{2}O_{3} + 2H_{2}CO_{3} =
H_{2}SO_{3} + 2NaHCO_{3} + S.)

!Answer!: 1090 cc.

63. An analyzed sample of stibnite containing 70.05% Sb is given for
analysis. A student titrates it with a solution of iodine of which 1
cc. is equivalent to 0.004950 gram of As_{2}O_{3}. Due to an error on
his part in standardization, the student's analysis shows the sample
to contain 70.32% Sb. Calculate the true normal value of the iodine
solution, and the percentage error in the analysis.

!Answers!: 0.1000 N; 0.39%.

64. A sample of pyrolusite weighing 0.5000 gram is treated with an
excess of hydrochloric acid, the liberated chlorine is passed into
potassium iodide and the liberated iodine is titrated with sodium
thiosulphate solution (49.66 grams of pure Na_{2}S_{2}O_{3}.5H_{2}O
per liter). If 38.72 cc. are required, what volume of 0.25 normal
permanganate solution will be required in an indirect determination
in which a similar sample is reduced with 0.9012 gram
H_{2}C_{2}O_{4}.2H_{2}O and the excess oxalic acid titrated?

!Answer!: 26.22 cc.

65. In the determination of sulphur in steel by evolving the sulphur
as hydrogen sulphide, precipitating cadmium sulphide by passing the
liberated hydrogen sulphide through ammoniacal cadmium chloride
solution, and decomposing the CdS with acid in the presence of a
measured amount of standard iodine, the following data are obtained:
Sample, 5.027 grams; cc. Na_{2}S_{2}O_{3} sol. = 12.68; cc. Iodine
sol. = 15.59; 1 cc. Iodine sol. = 1.086 cc. Na_{2}S_{2}O_{3} sol.; 1
cc. Na_{2}S_{2}O_{3}= 0.005044 gram Cu. Calculate the percentage of
sulphur. (H_{2}S + I_{2} = 2HI + S.)

!Answer!: 0.107%.

66. Given the following data, calculate the percentage of iron in
a sample of crude ferric chloride weighing 1.000 gram. The iodine
liberated by the reaction 2FeCl_{3}+ 2HI = 2HCl + 2FeCl_{2} + I_{2} is
reduced by the addition of 50 cc. of sodium thiosulphate solution and
the excess thiosulphate is titrated with standard iodine and requires
7.85 cc. 45 cc. I_{2} solution = 45.95 cc. Na_{2}S_{2}O_{3} solution;
45 cc. As_{2}O_{3} solution = 45.27 cc. I_{2} solution. 1 cc. arsenite
solution = 0.005160 gram As_{2}O_{3}.

!Answer!: 23.77%.

67. Sulphide sulphur was determined in a sample of reduced barium
sulphate by the evolution method, in which the sulphur was evolved as
hydrogen sulphide and was passed into CdCl_{2} solution, the acidified
precipitate being titrated with iodine and thiosulphate. Sample, 5.076
grams; cc. I_{2} = 20.83; cc. Na_{2}S_{2}O_{3} = 12.37; 43.45 cc.
Na_{2}S_{2}O_{3} = 43.42 cc. I_{2}; 8.06 cc. KMnO_{4} = 44.66 cc.
Na_{2}S_{2}O_{3}; 28.87 cc. KMnO_{4} = 0.2004 gram Na_{2}C_{2}O_{4}.
Calculate the percentage of sulphide sulphur in the sample.

!Answer!: 0.050%.

68. What weight of pyrolusite containing 89.21% MnO_{2} will oxidize
the same amount of oxalic acid as 37.12 cc. of a permanganate
solution, of which 1 cc. will liberate 0.0175 gram of I_{2} from KI?

!Answer!: 0.2493 gram.

69. A sample of pyrolusite weighs 0.2400 gram and is 92.50% pure
MnO_{2}. The iodine liberated from KI by the manganese dioxide is
sufficient to react with 46.24 cc. of Na_{2}S_{2}O_{3} sol. What is
the normal value of the thiosulphate?

!Answer!:: 0.1105 N.

70. In the volumetric analysis of silver coin (90% Ag), using a
0.5000 gram sample, what is the least normal value that a potassium
thiocyanate solution may have and not require more than 50 cc. of
solution in the analysis?

!Answer!: 0.08339 N.

71. A mixture of pure lithium chloride and barium bromide weighing
0.6 gram is treated with 45.15 cubic centimeters of 0.2017 N silver
nitrate, and the excess titrated with 25 cc. of 0.1 N KSCN solution,
using ferric alum as an indicator. Calculate the percentage of bromine
in the sample.

!Answer!: 40.11%.

72. A mixture of the chlorides of sodium and potassium from 0.5000
gram of a feldspar weighs 0.1500 gram, and after solution in water
requires 22.71 cc. of 0.1012 N silver nitrate for the precipitation of
the chloride ions. What are the percentages of Na_{2}O and K_{2}O in
the feldspar?

!Answer!: 8.24% Na_{2}O; 9.14% K_{2}O.

GRAVIMETRIC ANALYSIS

73. Calculate (a) the grams of silver in one gram of silver chloride;
(b) the grams of carbon dioxide liberated by the addition of an excess
of acid to one gram of calcium carbonate; (c) the grams of MgCl_{2}
necessary to precipitate 1 gram of MgNH_{4}PO_{4}.

!Answers!: (a) 0.7526; (b) 0.4397; (c) 0.6940.

74. Calculate the chemical factor for (a) Sn in SnO_{2}; (b) MgO
in Mg_{2}P_{2}O_{7}; (c) P_{2}O_{5} in Mg_{2}P_{2}O_{7}; (d) Fe in
Fe_{2}O_{3}; (e) SO_{4} in BaSO_{4}.

!Answers!: (a) 0.7879; (b) 0.3620; (c) 0.6378; (d) 0.6990; (e) 0.4115.

75. Calculate the log factor for (a) Pb in PbCrO_{4}; (b) Cr_{2}O_{3}
in PbCrO_{4}; (c) Pb in PbO_{2} and (d) CaO in CaC_{2}O_{4}.

!Answers!: (a) 9.8069-10, (b) 9.3713-10; (c) 9.9376-10; (d) 9.6415-10.

76. How many grams of Mn_{3}O_{4} can be obtained from 1 gram of
MnO_{2}?

!Answer!: 0.8774 gram.

77. If a sample of silver coin weighing 0.2500 gram gives a
precipitate of AgCl weighing 0.2991 gram, what weight of AgI could
have been obtained from the same weight of sample, and what is the
percentage of silver in the coin?

!Answers!: 0.4898 gr.; 90.05%.

78. How many cubic centimeters of hydrochloric acid (sp. gr. 1.13
containing 25.75% HCl by weight) are required to exactly neutralize
25 cc. of ammonium hydroxide (sp. gr. .90 containing 28.33% NH_{3} by
weight)?

!Answer!: 47.03 cc.

79. How many cubic centimeters of ammonium hydroxide solution (sp. gr.
0.96 containing 9.91% NH_{3} by weight) are required to precipitate
the aluminium as aluminium hydroxide from a two-gram sample of alum
(KAl(SO_{4})_{2}.12H_{2}O)? What will be the weight of the ignited
precipitate?

!Answers!: 2.26 cc.; 0.2154 gram.

80. What volume of nitric acid (sp. gr. 1.05 containing 9.0%
HNO_{3} by weight) is required to oxidize the iron in one gram of
FeSO_{4}.7H_{2}O in the presence of sulphuric acid? 6FeSO_{4} +
2HNO_{3} + 3H_{2}SO_{4} = 3Fe_{2}(SO_{4})_{3} + 2NO + 4H_{2}O.

!Answer!: 0.80 cc.

81. If 0.7530 gram of ferric nitrate (Fe(NO_{3})_{3}.9H_{2}O) is
dissolved in water and 1.37 cc. of HCl (sp. gr. 1.11 containing 21.92%
HCl by weight) is added, how many cubic centimeters of ammonia (sp.
gr. 0.96 containing 9.91% NH_{3} by weight) are required to neutralize
the acid and precipitate the iron as ferric hydroxide?

!Answer!: 2.63 cc.

82. To a suspension of 0.3100 gram of Al(OH)_{3} in water are added
13.00 cc. of aqueous ammonia (sp. gr. 0.90 containing 28.4% NH_{3} by
weight). How many cubic centimeters of sulphuric acid (sp. gr. 1.18
containing 24.7% H_{2}SO_{4} by weight) must be added to the mixture
in order to bring the aluminium into solution?

!Answer!: 34.8 cc.

83. How many cubic centimeters of sulphurous acid (sp. gr. 1.04
containing 75 grams SO_{2} per liter) are required to reduce the
iron in 1 gram of ferric alum (KFe(SO_{4})_{2}.12H_{2}O)?
Fe_{2}(SO_{4})_{3} + SO_{2} + 2H_{2}O = 2FeSO_{4} + 2H_{2}SO_{4}.

!Answer!: 0.85 cc.

84. How many cubic centimeters of a solution of potassium bichromate
containing 26.30 grams of K_{2}Cr_{2}O_{7} per liter must be taken
in order to yield 0.6033 gram of Cr_{2}O_{3} after reduction and
precipitation of the chromium?

K_{2}Cr_{2}O_{7} + 3SO_{2} + H_{2}SO_{4} = K_{2}SO_{4} +
Cr_{2}(SO_{4})_{3} + H_{2}O.

!Answer!: 44.39 cc.

85. How many cubic centimeters of ammonium hydroxide (sp. gr. 0.946
containing 13.88% NH_{3} by weight) are required to precipitate
the iron as Fe(OH)_{3} from a sample of pure
FeSO_{4}.(NH_{4})_{2}SO_{4}.6H_{2}O, which requires 0.34 cc. of nitric
acid (sp. gr. 1.350 containing 55.79% HNO_{3} by weight) for oxidation
of the iron? (See problem No. 80 for reaction.)

!Answer!: 4.74 cc.

86. In the analysis of an iron ore by solution, oxidation and
precipitation of the iron as Fe(OH)_{3}, what weight of sample must be
taken for analysis so that each one hundredth of a gram of the ignited
precipitate of Fe_{2}O_{3} shall represent one tenth of one per cent
of iron?

!Answer!: 6.99 grams.

87. What weight in grams of impure ferrous ammonium sulphate should
be taken for analysis so that the number of centigrams of BaSO_{4}
obtained will represent five times the percentage of sulphur in the
sample?

!Answer!: 0.6870 gram.

88. What weight of magnetite must be taken for analysis in order that,
after precipitating and igniting all the iron to Fe_{2}O_{3}, the
percentage of Fe_{2}O_{4} in the sample may be found by multiplying
the weight in grams of the ignited precipitate by 100?

!Answer!: 0.9665 gram.

89. After oxidizing the arsenic in 0.5000 gram of pure As_{2}S_{3} to
arsenic acid, it is precipitated with "magnesia mixture" (MgCl_{2} +
2NH_{4}Cl). If exactly 12.6 cc. of the mixture are required, how many
grams of MgCl_{2} per liter does the solution contain? H_{3}AsO_{4} +
MgCl_{2} + 3NH_{4}OH = MgNH_{4}AsO_{4} + 2NH_{4}Cl + 3H_{2}O.

!Answer!: 30.71 grams.

90. A sample is prepared for student analysis by mixing pure apatite
(Ca_{3}(PO_{4})_{2}.CaCl_{2}) with an inert material. If 1 gram of
the sample gives 0.4013 gram of Mg_{2}P_{2}O_{7}, how many cubic
centimeters of ammonium oxalate solution (containing 40 grams of
(NH_{4})_{2}C_{2}O_{4}.H_{2}O per liter) would be required to
precipitate the calcium from the same weight of sample?

!Answer!: 25.60 cc.

91. If 0.6742 gram of a mixture of pure magnesium carbonate and pure
calcium carbonate, when treated with an excess of hydrochloric acid,
yields 0.3117 gram of carbon dioxide, calculate the percentage of
magnesium oxide and of calcium oxide in the sample.

!Answers!: 13.22% MgO; 40.54% CaO. 92. The calcium in a sample of
dolomite weighing 0.9380 gram is precipitated as calcium oxalate and
ignited to calcium oxide. What volume of gas, measured over water
at 20 deg.C. and 765 mm. pressure, is given off during ignition, if the
resulting oxide weighs 0.2606 gram? (G.M.V. = 22.4 liters; V.P. water
at 20 deg.C. = 17.4 mm.)

!Answer!: 227 cc.

93. A limestone is found to contain 93.05% CaCO_{3}, and 5.16 %
MgCO_{3}. Calculate the weight of CaO obtainable from 3 tons of the
limestone, assuming complete conversion to oxide. What weight of
Mg_{2}P_{2}O_{7} could be obtained from a 3-gram sample of the
limestone?

!Answers!: 1.565 tons; 0.2044 gram.

94. A sample of dolomite is analyzed for calcium by precipitating
as the oxalate and igniting the precipitate. The ignited product is
assumed to be CaO and the analyst reports 29.50% Ca in the sample.
Owing to insufficient ignition, the product actually contained 8% of
its weight of CaCO_{3}. What is the correct percentage of calcium in
the sample, and what is the percentage error?

!Answers!: 28.46%; 3.65% error.

95. What weight of impure calcite (CaCO_{3}) should be taken for
analysis so that the volume in cubic centimeters of CO_{2} obtained by
treating with acid, measured dry at 18 deg.C. and 763 mm., shall equal the
percentage of CaO in the sample?

!Answer!: 0.2359 gram.

96. How many cubic centimeters of HNO_{3} (sp. gr. 1.13 containing
21.0% HNO_{3} by weight) are required to dissolve 5 grams of brass,
containing 0.61% Pb, 24.39% Zn, and 75% Cu, assuming reduction of the
nitric acid to NO by each constituent? What fraction of this volume of
acid is used for oxidation?

!Answers!: 55.06 cc.; 25%.

97. What weight of metallic copper will be deposited from a cupric
salt solution by a current of 1.5 amperes during a period of 45
minutes, assuming 100% current efficiency? (1 Faraday = 96,500
coulombs.)

!Answer!: 1.335 grams.

98. In the electrolysis of a 0.8000 gram sample of brass, there is
obtained 0.0030 gram of PbO_{2}, and a deposit of metallic copper
exactly equal in weight to the ignited precipitate of Zn_{2}P_{2}O_{7}
subsequently obtained from the solution. What is the percentage
composition of the brass?

!Answers!: 69.75% Cu; 29.92% Zn; 0.33% Pb.

99. A sample of brass (68.90% Cu; 1.10% Pb and 30.00% Zn) weighing
0.9400 gram is dissolved in nitric acid. The lead is determined by
weighing as PbSO_{4}, the copper by electrolysis and the zinc by
precipitation with (NH_{4})_{2}HPO_{4} in a neutral solution.

(a) Calculate the cubic centimeters of nitric acid (sp. gr. 1.42
containing 69.90% HNO_{3} by weight) required to just dissolve the
brass, assuming reduction to NO.

!Answer!: 2.48 cc.

(b) Calculate the cubic centimeters of sulphuric acid (sp. gr. 1.84
containing 94% H_{2}SO_{4} by weight) to displace the nitric acid.

!Answer!: 0.83 cc.

(c) Calculate the weight of PbSO_{4}.

!Answer!: 0.0152 gram.

(d) The clean electrode weighs 10.9640 grams. Calculate the weight
after the copper has been deposited.

!Answer!: 11.6116 grams.

(e) Calculate the grams of (NH_{4})_{2}HPO_{4} required to precipitate
the zinc as ZnNH_{4}PO_{4}.

!Answer!: 0.5705 gram.

(f) Calculate the weight of ignited Zn_{2}P_{2}O_{7}.

!Answer!: 0.6573 gram.

100. If in the analysis of a brass containing 28.00% zinc an error is
made in weighing a 2.5 gram portion by which 0.001 gram too much is
weighed out, what percentage error in the zinc determination would
result? What volume of a solution of sodium hydrogen phosphate,
containing 90 grams of Na_{2}HPO_{4}.12H_{2}O per liter, would be
required to precipitate the zinc as ZnNH_{4}PO_{4} and what weight of
precipitate would be obtained?

!Answers!: (a) 0.04% error; (b) 39.97 cc.; (c) 1.909 grams.

101. A sample of magnesium carbonate, contaminated with SiO_{2} as its
only impurity, weighs 0.5000 gram and loses 0.1000 gram on ignition.
What volume of disodium phosphate solution (containing 90 grams
Na_{2}HPO_{4}.12H_{2}O per liter) will be required to precipitate the
magnesium as magnesium ammonium phosphate?

!Answer!: 9.07 cc.

102. 2.62 cubic centimeters of nitric acid (sp. gr. 1.42 containing
69.80% HNO_{2} by weight) are required to just dissolve a sample
of brass containing 69.27% Cu; 0.05% Pb; 0.07% Fe; and 30.61% Zn.
Assuming the acid used as oxidizing agent was reduced to NO in every
case, calculate the weight of the brass and the cubic centimeters of
acid used as acid.

!Answer!: 0.992 gram; 1.97 cc.

103. One gram of a mixture of silver chloride and silver bromide is
found to contain 0.6635 gram of silver. What is the percentage of
bromine?

!Answer!: 21.30%.

104. A precipitate of silver chloride and silver bromide weighs 0.8132
gram. On heating in a current of chlorine, the silver bromide is
converted to silver chloride, and the mixture loses 0.1450 gram
in weight. Calculate the percentage of chlorine in the original
precipitate.

!Answer!: 6.13%.

105. A sample of feldspar weighing 1.000 gram is fused and the silica
determined. The weight of silica is 0.6460 gram. This is fused with 4
grams of sodium carbonate. How many grams of the carbonate actually
combined with the silica in fusion, and what was the loss in weight
due to carbon dioxide during the fusion?

!Answers!: 1.135 grams; 0.4715 gram.

106. A mixture of barium oxide and calcium oxide weighing 2.2120 grams
is transformed into mixed sulphates, weighing 5.023 grams. Calculate
the grams of calcium oxide and barium oxide in the mixture.

!Answers!: 1.824 grams CaO; 0.3877 gram BaO.

APPENDIX

ELECTROLYTIC DISSOCIATION THEORY

The following brief statements concerning the ionic theory and a few
of its applications are intended for reference in connection with the
explanations which are given in the Notes accompanying the various
procedures. The reader who desires a more extended discussion of the
fundamental theory and its uses is referred to such books as Talbot
and Blanchard's !Electrolytic Dissociation Theory! (Macmillan
Company), or Alexander Smith's !Introduction to General Inorganic
Chemistry! (Century Company).

The !electrolytic dissociation theory!, as propounded by Arrhenius in
1887, assumes that acids, bases, and salts (that is, electrolytes)
in aqueous solution are dissociated to a greater or less extent into
!ions!. These ions are assumed to be electrically charged atoms or
groups of atoms, as, for example, H^{+} and Br^{-} from hydrobromic
acid, Na^{+} and OH^{-} from sodium hydroxide, 2NH_{4}^{+} and
SO_{4}^{--} from ammonium sulphate. The unit charge is that which is
dissociated with a hydrogen ion. Those upon other ions vary in sign
and number according to the chemical character and valence of the
atoms or radicals of which the ions are composed. In any solution the
aggregate of the positive charges upon the positive ions (!cations!)
must always balance the aggregate negative charges upon the negative
ions (!anions!).

It is assumed that the Na^{+} ion, for example, differs from the
sodium atom in behavior because of the very considerable electrical
charge which it carries and which, as just stated, must, in an
electrically neutral solution, be balanced by a corresponding negative
charge on some other ion. When an electric current is passed through a
solution of an electrolyte the ions move with and convey the current,
and when the cations come into contact with the negatively charged
cathode they lose their charges, and the resulting electrically
neutral atoms (or radicals) are liberated as such, or else enter at
once into chemical reaction with the components of the solution.

Two ions of identically the same composition but with different
electrical charges may exhibit widely different properties. For
example, the ion MnO_{4}^{-} from permanganates yields a purple-red
solution and differs in its chemical behavior from the ion
MnO_{4}^{--} from manganates, the solutions of which are green.

The chemical changes upon which the procedures of analytical chemistry
depend are almost exclusively those in which the reacting substances
are electrolytes, and analytical chemistry is, therefore, essentially
the chemistry of the ions. The percentage dissociation of the same
electrolyte tends to increase with increasing dilution of its
solution, although not in direct proportion. The percentage
dissociation of different electrolytes in solutions of equivalent
concentrations (such, for example, as normal solutions) varies widely,
as is indicated in the following tables, in which approximate figures
are given for tenth-normal solutions at a temperature of about 18 deg.C.

ACIDS
=========================================================================
|
SUBSTANCE | PERCENTAGE DISSOCIATION IN
| 0.1 EQUIVALENT SOLUTION
_____________________________________________|___________________________
|
HCl, HBr, HI, HNO_{3} | 90
|
HClO_{3}, HClO_{4}, HMnO_{4} | 90
|
H_{2}SO_{4} <--> H^{+} + HSO_{4}^{-} | 90
|
H_{2}C_{2}O_{4} <--> H^{+} + HC_{2}O_{4}^{-} | 50
|
H_{2}SO_{3} <--> H^{+} + HSO{_}3^{-} | 20
|
H_{3}PO_{4} <--> H^{+} + H_{2}PO_{4}^{-} | 27
|
H_{2}PO_{4}^{-} <--> H^{+} + HPO_{4}^{--} | 0.2
|
H_{3}AsO_{4} <--> H^{+} + H_{2}AsO_{4}^{-} | 20
|
HF | 9
|
HC_{2}H_{3}O_{2} | 1.4
|
H_{2}CO_{3} <--> H^{+} + HCO_{3}^{-} | 0.12
|
H_{2}S <--> H^{+} + HS^{-} | 0.05
|
HCN | 0.01
|
=========================================================================

BASES
=========================================================================
|
SUBSTANCE | PERCENTAGE DISSOCIATION IN
| 0.1 EQUIVALENT SOLUTION
_____________________________________________|___________________________
|
KOH, NaOH | 86
|
Ba(OH)_{2} | 75
|
NH_{4}OH | 1.4
|
=========================================================================

SALTS
=========================================================================
|
TYPE OF SALT | PERCENTAGE DISSOCIATION IN
| 0.1 EQUIVALENT SOLUTION
_____________________________________________|___________________________
|
R^{+}R^{-} | 86
|
R^{++}(R^{-})_{2} | 72
|
(R^{+})_{2}R^{--} | 72
|
R^{++}R^{--} | 45
|
=========================================================================

The percentage dissociation is determined by studying the electrical
conductivity of the solutions and by other physico-chemical methods,
and the following general statements summarize the results:

!Salts!, as a class, are largely dissociated in aqueous solution.

!Acids! yield H^{+} ions in water solution, and the comparative
!strength!, that is, the activity, of acids is proportional to the
concentration of the H^{+} ions and is measured by the percentage
dissociation in solutions of equivalent concentration. The common
mineral acids are largely dissociated and therefore give a relatively
high concentration of H^{+} ions, and are commonly known as "strong
acids." The organic acids, on the other hand, belong generally to the
group of "weak acids."

!Bases! yield OH^{-} ions in water solution, and the comparative
strength of the bases is measured by their relative dissociation in
solutions of equivalent concentration. Ammonium hydroxide is a weak
base, as shown in the table above, while the hydroxides of sodium and
potassium exhibit strongly basic properties.

Ionic reactions are all, to a greater or less degree, !reversible
reactions!. A typical example of an easily reversible reaction is that
representing the changes in ionization which an electrolyte such as
acetic acid undergoes on dilution or concentration of its solutions,
!i.e.!, HC_{2}H_{3}O_{2} <--> H^{+} + C_{2}H_{3}O_{2}^{-}. As was
stated above, the ionization increases with dilution, the reaction
then proceeding from left to right, while concentration of the
solution occasions a partial reassociation of the ions, and the
reaction proceeds from right to left. To understand the principle
underlying these changes it is necessary to consider first the
conditions which prevail when a solution of acetic acid, which has
been stirred until it is of uniform concentration throughout, has come
to a constant temperature. A careful study of such solutions has shown
that there is a definite state of equilibrium between the constituents
of the solution; that is, there is a definite relation between the
undissociated acetic acid and its ions, which is characteristic for
the prevailing conditions. It is not, however, assumed that this is a
condition of static equilibrium, but rather that there is continual
dissociation and association, as represented by the opposing
reactions, the apparent condition of rest resulting from the fact that
the amount of change in one direction during a given time is exactly
equal to that in the opposite direction. A quantitative study of
the amount of undissociated acid, and of H^{+} ions and
C_{2}H_{3}O_{2}^{-} ions actually to be found in a large number of
solutions of acetic acid of varying dilution (assuming them to be in
a condition of equilibrium at a common temperature), has shown that
there is always a definite relation between these three quantities
which may be expressed thus:

(!Conc'n H^{+} x Conc'n C_{2}H_{3}O_{2}^{-})/Conc'n HC_{2}H_{3}O_{2} =
Constant!.

In other words, there is always a definite and constant ratio between
the product of the concentrations of the ions and the concentration of
the undissociated acid when conditions of equilibrium prevail.

It has been found, further, that a similar statement may be made
regarding all reversible reactions, which may be expressed in general
terms thus: The rate of chemical change is proportional to the product
of the concentrations of the substances taking part in the reaction;
or, if conditions of equilibrium are considered in which, as stated,
the rate of change in opposite directions is assumed to be equal, then
the product of the concentrations of the substances entering into
the reaction stands in a constant ratio to the product of the
concentrations of the resulting substances, as given in the expression
above for the solutions of acetic acid. This principle is called the
!Law of Mass Action!.

It should be borne in mind that the expression above for acetic acid
applies to a wide range of dilutions, provided the temperature remains
constant. If the temperature changes the value of the constant changes
somewhat, but is again uniform for different dilutions at that
temperature. The following data are given for temperatures of about
18 deg.C.[1]

==========================================================================
| | | |
MOLAL | FRACTION | MOLAL CONCENTRA- | MOLAL CONCENTRA- | VALUE OF
CONCENTRATION | IONIZED | TION OF H^{+} AND| TION OF UNDIS- | CONSTANT
CONSTANT | | ACETATE^{-} IONS | SOCIATED ACID |
______________|__________|__________________|__________________|__________
| | | |
1.0 | .004 | .004 | .996 | .0000161
| | | |
0.1 | .013 | .0013 | .0987 | .0000171
| | | |
0.01 | .0407 | .000407 | .009593 | .0000172
| | | |
===========================================================================

[Footnote 1: Alexander Smith, !General Inorganic Chemistry!, p. 579.]

The molal concentrations given in the table refer to fractions of a
gram-molecule per liter of the undissociated acid, and to fractions of
the corresponding quantities of H^{+} and C_{2}H_{3}O_{2}^{-} ions
per liter which would result from the complete dissociation of a
gram-molecule of acetic acid. The values calculated for the constant
are subject to some variation on account of experimental errors in
determining the percentage ionized in each case, but the approximate
agreement between the values found for molal and centimolal (one
hundredfold dilution) is significant.

The figures given also illustrate the general principle, that the
!relative! ionization of an electrolyte increases with the dilution of
its solution. If we consider what happens during the (usually) brief
period of dilution of the solution from molal to 0.1 molal, for
example, it will be seen that on the addition of water the conditions
of concentration which led to equality in the rate of change, and
hence to equilibrium in the molal solution, cease to exist; and since
the dissociating tendency increases with dilution, as just stated,
it is true at the first instant after the addition of water that the
concentration of the undissociated acid is too great to be
permanent under the new conditions of dilution, and the reaction,
HC_{2}H_{3}O_{2} <--> H^{+} + C_{2}H_{3}O_{2}^{-}, will proceed from
left to right with great rapidity until the respective concentrations
adjust themselves to the new conditions.

That which is true of this reaction is also true of all reversible
reactions, namely, that any change of conditions which occasions
an increase or a decrease in concentration of one or more of the
components causes the reaction to proceed in one direction or the
other until a new state of equilibrium is established. This principle
is constantly applied throughout the discussion of the applications
of the ionic theory in analytical chemistry, and it should be clearly
understood that whenever an existing state of equilibrium is disturbed
as a result of changes of dilution or temperature, or as a consequence
of chemical changes which bring into action any of the constituents of
the solution, thus altering their concentrations, there is always a
tendency to re-establish this equilibrium in accordance with the law.
Thus, if a base is added to the solution of acetic acid the H^{+} ions
then unite with the OH^{-} ions from the base to form undissociated
water. The concentration of the H^{+} ions is thus diminished, and
more of the acid dissociates in an attempt to restore equilbrium,
until finally practically all the acid is dissociated and neutralized.

Similar conditions prevail when, for example, silver ions react with
chloride ions, or barium ions react with sulphate ions. In the former
case the dissociation reaction of the silver nitrate is AgNO_{3} <-->
Ag^{+} + NO_{3}^{-}, and as soon as the Ag^{+} ions unite with the
Cl^{-} ions the concentration of the former is diminished, more of the
AgNO_{3} dissociates, and this process goes on until the Ag^{+} ions
are practically all removed from the solution, if the Cl^{-} ions are
present in sufficient quantity.

For the sake of accuracy it should be stated that the mass law cannot
be rigidly applied to solutions of those electrolytes which are
largely dissociated. While the explanation of the deviation from
quantitative exactness in these cases is not known, the law is still
of marked service in developing analytical methods along more logical
lines than was formerly practicable. It has not seemed wise to qualify
each statement made in the Notes to indicate this lack of quantitative
exactness. The student should recognize its existence, however, and
will realize its significance better as his knowledge of physical
chemistry increases.

If we apply the mass law to the case of a substance of small
solubility, such as the compounds usually precipitated in quantitative
analysis, we derive what is known as the !solubility product!, as
follows: Taking silver chloride as an example, and remembering that it
is not absolutely insoluble in water, the equilibrium expression for
its solution is:

(!Conc'n Ag^{+} x Conc'n Cl^{-})/Conc'n AgCl = Constant!.

But such a solution of silver chloride which is in contact with the
solid precipitate must be saturated for the existing temperature, and
the quantity of undissociated AgCl in the solution is definite and
constant for that temperature. Since it is a constant, it may be
eliminated, and the expression becomes !Conc'n Ag^{+} x Conc'n
Cl^{-} = Constant!, and this is known as the solubility product. No
precipitation of a specific substance will occur until the product of
the concentrations of its ions in a solution exceeds the solubility
product for that substance; whenever that product is exceeded
precipitation must follow.

It will readily be seen that if a substance which yields an ion in
common with the precipitated compound is added to such a solution as
has just been described, the concentration of that ion is
increased, and as a result the concentration of the other ion must
proportionately decrease, which can only occur through the formation
of some of the undissociated compound which must separate from the
already saturated solution. This explains why the addition of an
excess of the precipitant is often advantageous in quantitative
procedures. Such a case is discussed at length in Note 2 on page 113.

Similarly, the ionization of a specific substance in solution tends to
diminish on the addition of another substance with a common ion, as,
for instance, the addition of hydrochloric acid to a solution
of hydrogen sulphide. Hydrogen sulphide is a weak acid, and the
concentration of the hydrogen ions in its aqueous solutions is very
small. The equilibrium in such a solution may be represented as:

(!(Conc'n H^{+})^{2} x Conc'n S^{--})/Conc'n H_{2}S = Constant!, and a
marked increase in the concentration of the H^{+} ions, such as would
result from the addition of even a small amount of the highly ionized
hydrochloric acid, displaces the point of equilibrium and some of the
S^{--} ions unite with H^{+} ions to form undissociated H_{2}S. This
is of much importance in studying the reactions in which hydrogen
sulphide is employed, as in qualitative analysis. By a parallel course
of reasoning it will be seen that the addition of a salt of a weak
acid or base to solutions of that acid or base make it, in effect,
still weaker because they decrease its percentage ionization.

To understand the changes which occur when solids are dissolved where
chemical action is involved, it should be remembered that no substance
is completely insoluble in water, and that those products of a
chemical change which are least dissociated will first form. Consider,
for example, the action of hydrochloric acid upon magnesium hydroxide.
The minute quantity of dissolved hydroxide dissociates thus:
Mg(OH)_{2} <--> Mg^{++} + 2OH^{-}. When the acid is introduced,
the H^{+} ions of the acid unite with the OH^{-} ions to form
undissociated water. The concentration of the OH^{-} ions is thus
diminished, more Mg(OH)_{2} dissociates, the solution is no longer
saturated with the undissociated compound, and more of the solid
dissolves. This process repeats itself with great rapidity until, if
sufficient acid is present, the solid passes completely into solution.

Exactly the same sort of process takes place if calcium oxalate, for
example, is dissolved in hydrochloric acid. The C_{2}O_{4}^{--} ions
unite with the H^{+} ions to form undissociated oxalic acid, the acid
being less dissociated than normally in the presence of the H^{+} ions
from the hydrochloric acid (see statements regarding hydrogen sulphide
above). As the undissociated oxalic acid forms, the concentration of
the C_{2}O_{4}^{--} ions lessens and more CaC_{2}O_{4} dissolves,
as described for the Mg(OH)_{2} above. Numerous instances of the
applications of these principles are given in the Notes.

Water itself is slightly dissociated, and although the resulting H^{+}
and OH^{-} ions are present only in minute concentrations (1 mol. of
dissociated water in 10^{7} liters), yet under some conditions they
may give rise to important consequences. The term !hydrolysis! is
applied to the changes which result from the reaction of these ions.
Any salt which is derived from a weak base or a weak acid (or both)
is subject to hydrolytic action. Potassium cyanide, for example, when
dissolved in water gives an alkaline solution because some of the
H^{+} ions from the water unite with CN^{-} ions to form (HCN), which
is a very weak acid, and is but very slightly dissociated. Potassium
hydroxide, which might form from the OH^{-} ions, is so largely
dissociated that the OH^{-} ions remain as such in the solution. The
union of the H^{+} ions with the CN^{-} ions to form the undissociated
HCN diminishes the concentration of the H^{+} ions, and more water
dissociates (H_{2}O <--> H^{+} + OH^{-}) to restore the equilibrium.
It is clear, however, that there must be a gradual accumulation of
OH^{-} ions in the solution as a result of these changes, causing the
solution to exhibit an alkaline reaction, and also that ultimately the
further dissociation of the water will be checked by the presence of
these ions, just as the dissociation of the H_{2}S was lessened by the
addition of HCl.

An exactly opposite result follows the solution of such a salt as
Al_{2}(SO_{4})_{3} in water. In this case the acid is strong and the
base weak, and the OH^{-} ions form the little dissociated Al(OH)_{3},
while the H^{+} ions remain as such in the solution, sulphuric acid
being extensively dissociated. The solution exhibits an acid reaction.

Such hydrolytic processes as the above are of great importance in
analytical chemistry, especially in the understanding of the action of
indicators in volumetric analysis. (See page 32.)

The impelling force which causes an element to pass from the atomic
to the ionic condition is termed !electrolytic solution pressure!, or
ionization tension. This force may be measured in terms of electrical
potential, and the table below shows the relative values for a number
of elements.

In general, an element with a greater solution pressure tends to cause
the deposition of an element of less solution pressure when placed in
a solution of its salt, as, for instance, when a strip of zinc or
iron is placed in a solution of a copper salt, with the resulting
precipitation of metallic copper.

Hydrogen is included in the table, and its position should be noted
with reference to the other common elements. For a more extended
discussion of this topic the student should refer to other treatises.

POTENTIAL SERIES OF THE METALS

__________________________________________________________________
| | |
| POTENTIAL | | POTENTIAL
| IN VOLTS | | IN VOLTS
_____________________|___________|____________________|___________
| | |
Sodium Na^{+} | +2.44 | Lead Pb^{++} | -0.13
Calcium Ca^{++} | | Hydrogen H^{+} | -0.28
Magnesium Mg^{++} | | Bismuth Bi^{+++}|
Aluminum A1^{+++} | +1.00 | Antimony | -0.75
Manganese Mn^{++} | | Arsenic |
Zinc Zn^{++} | +0.49 | Copper Cu^{++} | -0.61
Cadmium Cd^{++} | +0.14 | Mercury Hg^{+} | -1.03
Iron Fe^{++} | +0.063 | Silver Ag^{+} | -1.05
Cobalt Co^{++} | -0.045 | Platinum |
Nickel Ni^{++} | -0.049 | Gold |
Tin Sn^{++} | -0.085(?) | |
_____________________|___________|____________________|__________

THE FOLDING OF A FILTER PAPER

If a filter paper is folded along its diameter, and again folded along
the radius at right angles to the original fold, a cone is formed on
opening, the angle of which is 60 deg.. Funnels for analytical use are
supposed to have the same angle, but are rarely accurate. It is
possible, however, with care, to fit a filter thus folded into a
funnel in such a way as to prevent air from passing down between the
paper and the funnel to break the column of liquid in the stem,
which aids greatly, by its gentle suction, in promoting the rate of
filtration.

Such a filter has, however, the disadvantage that there are three
thicknesses of paper back of half of its filtering surface, as a
consequence of which one half of a precipitate washes or drains more
slowly. Much time may be saved in the aggregate by learning to fold a
filter in such a way as to improve its effective filtering surface.
The directions which follow, though apparently complicated on first
reading, are easily applied and easily remembered. Use a 6-inch filter
for practice. Place four dots on the filter, two each on diameters
which are at right angles to each other. Then proceed as follows:
(1) Fold the filter evenly across one of the diameters, creasing it
carefully; (2) open the paper, turn it over, rotate it 90 deg. to the
right, bring the edges together and crease along the other diameter;
(3) open, and rotate 45 deg. to the right, bring edges together, and
crease evenly; (4) open, and rotate 90 deg. to the right, and crease
evenly; (5) open, turn the filter over, rotate 22-(1/2) deg. to the right,
and crease evenly; (6) open, rotate 45 deg. to the right and crease
evenly; (7) open, rotate 45 deg. to the right and crease evenly; (8) open,
rotate 45 deg. to the right and crease evenly; (9) open the filter, and,
starting with one of the dots between thumb and forefinger of the
right hand, fold the second crease to the left over on it, and do
the same with each of the other dots. Place it, thus folded, in the
funnel, moisten it, and fit to the side of the funnel. The filter will
then have four short segments where there are three thicknesses
and four where there is one thickness, but the latter are evenly
distributed around its circumference, thus greatly aiding the passage
of liquids through the paper and hastening both filtration and washing
of the whole contents of the filter.

!SAMPLE PAGES FOR LABORATORY RECORDS!

!Page A!

Date

CALIBRATION OF BURETTE No.

___________________________________________________________________________
| | | |
BURETTE | DIFFERENCE | OBSERVED | DIFFERENCE | CALCULATED
READINGS | | WEIGHTS | | CORRECTION
_______________|______________|______________|______________|______________
0.02 | | 16.27 | |
10.12 | 10.10 | 26.35 | 10.08 | -.02
20.09 | 9.97 | 36.26 | 9.91 | -.06
30.16 | 10.07 | 46.34 | 10.08 | +.01
40.19 | 10.03 | 56.31 | 9.97 | -.06
50.00 | 9.81 | 66.17 | 9.86 | +.05
_______________|______________|______________|______________|______________

These data to be obtained in duplicate for each burette.

!Page B!

Date

DETERMINATION OF COMPARATIVE STRENGTH HCl vs. NaOH

___________________________________________________________________________
| |
DETERMINATION | I | II
_________________________|________________________|________________________
| |
| Corrected | Corrected
Final Reading HCl | 48.17 48.08 | 43.20 43.14
Initial Reading HCl | 0.12 .12 | .17 .17
| ----- ----- | ----- -----
| 47.96 | 42.97
| |
| Corrected | Corrected
Final Reading HCl | 46.36 46.29 | 40.51 40.37
Initial Reading HCl | 1.75 1.75 | .50 .50
| ----- ----- | ----- -----
| 44.54 | 39.87
| |
log cc. NaOH | 1.6468 | 1.6008
colog cc. HCl | 8.3192 | 8.3668
| ------ | ------
| 9.9680 - 10 | 9.9676 - 10
1 cc. HCl | .9290 cc. NaOH | .9282 cc. NaOH
Mean | .9286 |
_________________________|________________________|________________________

Signed

!Page C!
Date

STANDARDIZATION OF HYDROCHLORIC ACID
=====================================================================
| |
Weight sample and tube| 9.1793 | 8.1731
| 8.1731 | 6.9187
| ------ | ------
Weight sample | 1.0062 | 1.2544
| |
Final Reading HCl | 39.97 39.83 | 49.90 49.77
Initial Reading HCl | .00 .00 | .04 .04
| ----- ----- | ----- -----
| 39.83 | 49.73
| |
Final Reading NaOH | .26 .26 | .67 .67
Initial Reading NaOH | .12 .12 | .36 .36
| --- --- | --- ---
| .14 | .31
| |
| .14 | .31
Corrected cc. HCl | 39.83 - ----- = 39.68 | 49.73 - ----- = 49.40
| .9286 | .9286
| |
log sample | 0.0025 | 0.0983
colog cc | 8.4014 - 10 | 8.3063 - 10
colog milli equivalent| 1.2757 | 1.2757
| ------ | ------
| 9.6796 - 10 | 9.6803 - 10
| |
Normal value HCl | .4782 | .4789
Mean | .4786 |
| |
=====================================================================

Signed

!Page D!
Date

DETERMINATION OF CHLORINE IN CHLORIDE, SAMPLE No.
=====================================================================
| |
Weight sample and tube| 16.1721 | 15.9976
| 15.9976 | 15.7117
| ------- | -------
Weight sample | .1745 | .2859
| |
Weight crucible | |
+ precipitate | 14.4496 | 15.6915
Constant weights | 14.4487 | 15.6915
| 14.4485 |
| |
Weight crucible | 14.2216 | 15.3196
Constant weight | 14.2216 | 15.3194
| |
Weight AgCl | .2269 | .3721
| |
log Cl | 1.5496 | 1.5496
log weight AgCl | 9.3558 - 10 | 9.5706 - 10
log 100 | 2.0000 | 2.0000
colog AgCl | 7.8438 - 10 | 7.7438 - 10
colog sample | 0.7583 | 0.5438
| ------- | -------
| 1.5075 | 1.5078
| |
Cl in sample No. | 32.18% | 32.20%
| |
=====================================================================

Signed

STRENGTH OF REAGENTS

The concentrations given in this table are those suggested for use
in the procedures described in the foregoing pages. It is obvious,
however, that an exact adherence to these quantities is not essential.

Approx. Approx.
Grams relation relation
per to normal to molal
liter. solution solution

Ammonium oxalate, (NH_{4})_{2}C_{2}O_{4}.H_{2}O 40 0.5N 0.25
Barium chloride, BaCl_{2}.2H_{2}O 25 0.2N 0.1
Magnesium ammonium chloride (of MgCl_{2}) 71 1.5N 0.75
Mercuric chloride, HgCl_{2} 45 0.33N 0.66
Potassium hydroxide, KOH (sp. gr. 1.27) 480
Potassium thiocyanate, KSCN 5 0.05N 0.55
Silver nitrate, AgNO_{3} 21 0.125N 0.125
Sodium hydroxide, NaOH 100 2.5N 2.5
Sodium carbonate. Na_{2}CO_{3} 159 3N 1.5
Sodium phosphate, Na_{2}HPO_{4}.12H_{2}O 90 0.5N or 0.75N 0.25

Stannous chloride, SnCl_{2}, made by saturating hydrochloric acid (sp.
gr. 1.2) with tin, diluting with an equal volume of water, and adding
a slight excess of acid from time to time. A strip of metallic tin is
kept in the bottle.

A solution of ammonium molybdate is best prepared as follows: Stir
100 grams of molybdic acid (MoO_{3}) into 400 cc. of cold, distilled
water. Add 80 cc. of concentrated ammonium hydroxide (sp. gr. 0.90).
Filter, and pour the filtrate slowly, with constant stirring, into a
mixture of 400 cc. concentrated nitric acid (sp. gr. 1.42) and 600
cc. of water. Add to the mixture about 0.05 gram of microcosmic salt.
Filter, after allowing the whole to stand for 24 hours.

The following data regarding the common acids and aqueous ammonia
are based upon percentages given in the Standard Tables of the
Manufacturing Chemists' Association of the United States [!J.S.C.I.!,
24 (1905), 787-790]. All gravities are taken at 15.5 deg.C. and compared
with water at the same temperature.

Aqueous ammonia (sp. gr. 0.96) contains 9.91 per cent NH_{3} by
weight, and corresponds to a 5.6 N and 5.6 molal solution.

Aqueous ammonia (sp. gr. 0.90) contains 28.52 per cent NH_{3} by
weight, and corresponds to a 5.6 N and 5.6 molal solution.

Hydrochloric acid (sp. gr. 1.12) contains 23.81 per cent HCl by
weight, and corresponds to a 7.3 N and 7.3 molal solution.

Hydrochloric acid (sp. gr. 1.20) contains 39.80 per cent HCl by
weight, and corresponds to a 13.1 N and 13.1 molal solution.

Nitric acid (sp. gr. 1.20) contains 32.25 per cent HNO_{3} by weight,
and corresponds to a 6.1 N and 6.1 molal solution:

Nitric acid (sp. gr. 1.42) contains 69.96 per cent HNO_{3} by weight,
and corresponds to a 15.8 N and 15.8 molal solution.

Sulphuric acid (sp. gr. 1.8354) contains 93.19 per cent H_{2}SO_{4} by
weight, and corresponds to a 34.8 N or 17.4 molal solution.

Sulphuric acid (sp. gr. 1.18) contains 24.74 per cent H_{2}SO_{4} by
weight, and corresponds to a 5.9 N or 2.95 molal solution.

The term !normal! (N), as used above, has the same significance as
in volumetric analyses. The molal solution is assumed to contain one
molecular weight in grams in a liter of solution.

DENSITIES AND VOLUMES OF WATER AT TEMPERATURES FROM 15-30 deg.C.

Temperature Density. Volume.
Centigrade.

4 deg. 1.000000 1.000000
15 deg. 0.999126 1.000874
16 deg. 0.998970 1.001031
17 deg. 0.998801 1.001200
18 deg. 0.998622 1.001380
19 deg. 0.998432 1.001571
20 deg. 0.998230 1.001773
21 deg. 0.998019 1.001985
22 deg. 0.997797 1.002208
23 deg. 0.997565 1.002441
24 deg. 0.997323 1.002685
25 deg. 0.997071 1.002938
26 deg. 0.996810 1.003201
27 deg. 0.996539 1.003473
28 deg. 0.996259 1.003755
29 deg. 0.995971 1.004046
30 deg. 0.995673 1.004346

Authority: Landolt, Boernstein, and Meyerhoffer's !Tabellen!, third
edition.

CORRECTIONS FOR CHANGE OF TEMPERATURE OF STANDARD SOLUTIONS

The values below are average values computed from data relating to a
considerable number of solutions. They are sufficiently accurate for
use in chemical analyses, except in the comparatively few cases
where the highest attainable accuracy is demanded in chemical
investigations. The expansion coefficients should then be carefully
determined for the solutions employed. For a compilation of the
existing data, consult Landolt, Boernstein, and Meyerhoffer's
!Tabellen!, third edition.

Corrections for 1 cc.
Concentration. of solution between
15 deg. and 35 deg.C.

Normal .00029
0.5 Normal .00025
0.1 Normal or more dilute solutions .00020

The volume of solution used should be multiplied by the values given,
and that product multiplied by the number of degrees which the
temperature of the solution varies from the standard temperature
selected for the laboratory. The total correction thus found is
subtracted from the observed burette reading if the temperature is
higher than the standard, or added, if it is lower. Corrections are
not usually necessary for variations of temperature of 2 deg.C. or less.

INTERNATIONAL ATOMIC WEIGHTS

==========================================================
| | |
| 1920 | | 1920
_________________|_________|___________________|__________
| | |
Aluminium Al | 27.1 | Molybdenum Mo | 96.0
Antimony Sb | 120.2 | Neodymium Nd | 144.3
Argon A | 39.9 | Neon Ne | 20.2
Arsenic As | 74.96 | Nickel Ni | 58.68
Barium Ba | 137.37 | Nitrogen N | 14.008
Bismuth Bi | 208.0 | Osmium Os | 190.9
Boron B | 11.0 | Oxygen O | 16.00
Bromine Br | 79.92 | Palladium Pd | 106.7
Cadmium Cd | 112.40 | Phosphorus P | 31.04
Caesium Cs | 132.81 | Platinum Pt | 195.2
Calcium Ca | 40.07 | Potassium K | 39.10
Carbon C | 12.005 | Praseodymium Pr | 140.9
Cerium Ce | 140.25 | Radium Ra | 226.0
Chlorine Cl | 35.46 | Rhodium Rh | 102.9
Chromium Cr | 52.0 | Rubidium Rb | 85.45
Cobalt Co | 58.97 | Ruthenium Ru | 101.7
Columbium Cb | 93.1 | Samarium Sm | 150.4
Copper Cu | 63.57 | Scandium Sc | 44.1
Dysprosium Dy | 162.5 | Selenium Se | 79.2
Erbium Er | 167.7 | Silicon Si | 28.3
Europium Eu | 152.0 | Silver Ag | 107.88
Fluorine Fl | 19.0 | Sodium Na | 23.00
Gadolinium Gd | 157.3 | Strontium Sr | 87.63
Gallium Ga | 69.9 | Sulphur S | 32.06
Germanium Ge | 72.5 | Tantalum Ta | 181.5
Glucinum Gl | 9.1 | Tellurium Te | 127.5
Gold Au | 197.2 | Terbium Tb | 159.2
Helium He | 4.00 | Thallium Tl | 204.0
Hydrogen H | 1.008 | Thorium Th | 232.4
Indium In | 114.8 | Thulium Tm | 168.5
Iodine I | 126.92 | Tin Sn | 118.7
Iridium Ir | 193.1 | Titanium Ti | 48.1
Iron Fe | 55.84 | Tungsten W | 184.0
Krypton Kr | 82.92 | Uranium U | 238.2
Lanthanum La | 139.0 | Vanadium V | 51.0
Lead Pb | 207.2 | Xenon Xe | 130.2
Lithium Li | 6.94 | Ytterbium Yb | 173.5
Lutecium Lu | 175.0 | Yttrium Y | 88.7
Magnesium Mg | 24.32 | Zinc Zn | 65.37
Manganese Mn | 54.93 | Zirconium Zr | 90.6
Mercury Hg | 200.6 | |
==========================================================

INDEX

Acidimetry
Acid solutions, normal
standard
Acids, definition of
Acids, weak, action of other acids on
action of salts on
Accuracy demanded
Alkalimetry
Alkali solutions, normal
standard
Alumina, determination of in stibnite
Ammonium nitrate, acid
Analytical chemistry, subdivisions of
Antimony, determination of, in stibnite
Apatite, analysis of
Asbestos filters
Atomic weights, table of

Balances, essential features of
use and care of
Barium sulphate, determination of sulphur in

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