The document provides calculations to determine the molar and normal concentrations of a 200 g BaCl2 solution in 500 mL. It first calculates the molar mass of BaCl2 as 208 g/mol and its equivalent mass as 104 g/mol. It then calculates the molar concentration as 1.92 mol/L and the normal concentration as 3.85 mol/L. The document provides similar calculations for FeCl2 and NaOH solutions to determine their normal concentrations based on given masses and volumes. It also calculates the mass of ZnCl2, H2SO4, and H3PO4 needed to make solutions with given normal concentrations and volumes.

1. Определите молярную и нормальную концентрацию раствора BaCl2 объемом 500 мл и массой соли 200 г.
Дано:
m(ВаCl2) = 200 г
V(solution) = 500 мл
Решение:
1. Расчетные формулы:
2. Расчет молярной массы и молярной массы эквивалента:
М(ВаCl2) = 137 + 35, 5 × 2 = 208 г/моль
Эm = f × М.m.
f (ВаCl2) = 1
2
Эm = 1
2 × 208 = 104
3. Расчет молярной и нормальной концентрации:
𝑀 =
200
208 × 0,5
= 1,92 моль/л
𝑁 =
200
104 ×0,5
= 3,85 моль экв/л
Найти:
См - ?
СN- ?
M (СM)=
𝑚вещества
𝑀.𝑚.вещества × 𝑉раствора
N (CN)=
𝑚вещества
Э𝑚вещества
× 𝑉раствора

2. Вычислите нормальность раствора, содержащего 35 г FeCl2 в 500 мл раствора.
Дано:
m(FeCl2) = 35 г
V(раствора) = 500 мл
Найти:
СN - ?
Решение:
1. Расчетная формула: СN =
𝑚в
−
ва
Эм × 𝑉раствора
2. Расчет эквивалентной массы FeCl2:
M.m. FeCl2 = 55,8 + (35,5) • 2 = 126,8
3. Расчет нормальной концентрации:
СN=
35
63,4 × 0,5
= 1,1M
Эм = М.м. · f Эм = 126,8 · ½ = 63,4

3. Гидроксид натрия массой 21,4 г растворили в воде. Объем раствора довели до 300 мл. Определите нормальную
концентрацию полученного раствора.
Дано:
m(NaOH) = 21,4 г
V(раствора) = 300 мл
Найти:
СN - ?
Решение:
1. Расчетная формула:
2. Расчет эквивалентной массы NaOH:
M.m. NaOH = 23 + 16 + 1= 40
3. Расчет нормальности раствора:
СN =
𝑚в
−
ва
Эм × 𝑉раствора
Эм = М.м. · f Эм = 40 · 1 = 40
СN=
21,4
40 × 0,3
= 1,8M

4. Какая масса хлорида цинка потребуется для приготовления 500 мл раствора этой соли. и с концентрацией 1
моль/л.
Дано:
СN = 1 (моль экв/ л)
V(раствора) = 500 мл
Найти:
m (ZnCl2) - ?
Решение:
1. Расчетная формула:
2. Расчет эквивалентной массы ZnCl2 :
M.m. ZnCl2 = 65,4 + (35,5) • 2 = 136,4
3. Расчет нормальности раствора ZnCl2
1 =
𝑚
68,2 ×0,5
;
𝑚 = 1 × 68,2 × 0,5 = 34,1 г
СN =
𝑚в
−
ва
Эм × 𝑉раствора
Эм = М.м. · f
Эм = 136,4 · ½ = 68,2

5. Какая масса серной кислоты потребуется для приготовления 2 литров 2-нормального раствора?
Дано:
СN H2SO4 = 2 N
V(раствора) = 2 л
Найти:
mH2SO4 - ?
Решение:
1. Расчетная формула:
2. Расчет эквивалентной массы H2SO4:
M.m.H2SO4 = (1) • 2 + 32 + (16) • 4 = 98
3. Расчет массы навески H2SO4
𝑚𝐻2𝑆𝑂4 = 2 × 49 × 2 = 196 г
𝑚𝐻2𝑆𝑂4 = СN × Эм × 𝑉раствора
СN =
𝑚в
−
ва
Эм × 𝑉раствора
Эм = М.м. · f
Эм = 98 · ½ = 49

6. Какая масса ортофосфорной кислоты потребуется для приготовления 2 л 0,1 Н раствора?
Дано:
NH3PO4 = 0,1N
V(раствора) = 2 л
Найти:
mH3PO4 - ?
Решение:
1. Расчетная формула:
2. Расчет эквивалентной массы H3PO4 :
M.m. H3PO4 = (1) • 3 + 31 + (16) • 4 = 98
3. Расчет массы навески H3PO4
m = 0,1 • 32,67 • 2 = 6,53 g
𝑚вещества = СN × Эм × 𝑉раствора
СN =
𝑚в
−
ва
Эм × 𝑉раствора
Эм = М.м. · f
Эм = 98 · 1/3 = 32,67

7. direct titration
Back titration
substitution titration
Classification
Titrimetric
Methods
according to
the method of titration

8. direct titration
the reaction of the interaction of the test substance (A) with the reagent under the conditions
of the analysis must be specific;
the interaction between the substance and the titrant must occur stoichiometrically (i.e.
proceed strictly according to the reaction equation), with a clear fixation of the end point of
the titration;
the reaction must proceed at a sufficient rate and be practically irreversible;
the equilibrium constant must be high enough, the reaction must go to the end, so that the
errors that arise from the incomplete reaction do not exceed the allowable values;
the solution should be free of substances that interfere with the course of the main reaction or
the fixation of the end point of the titration.
А + Т(titrant) → product

9. Back titration
slow forward reaction;
there is no corresponding indicator;
possible loss of the analyte due to its volatility.
А + Т1(excess) → product 1 + T(remainder)
T(remainder) + Т2 → product2
For example, the concentration of an NH3 solution is determined by back titration due to its volatility. An excess of
titrated hydrochloric acid solution is added to the ammonia solution. Unreacted hydrochloric acid is titrated with sodium
hydroxide solution..
Back acid-base titration, methyl orange indicator.
NH4OH + HClexcess. → NH4Cl + H2O
HCl + NaOH → NaCl + H2O

10. substitution titration
analyte does not interact with this titrant;
the interaction of the analyte and the titrant leads to the formation of a mixture of several
products, the quantitative ratio of which is not constant;
titration reaction is not stoichiometrically;
there is no corresponding indicator;
analyte is unstable.
А + В(reagent) → А1(deputy)
А1(deputy) + Т → product

11. For example, potassium dichromate interacts with Na2S2O3 titrant nonstoichiometrically.
Therefore, reagent KI is added to a solution of the test substance K2Cr2O7, as a result of which
an equivalent amount of iodine is formed, which is then titrated with a standard solution of
sodium thiosulfate.
K2Cr2O7 + 6KI + 7H2SO4 → 3I2 + 4K2SO4 + Cr2(SO4)3 + 7H2O
I2 + 2Na2S2O3 → Na2S4O6 + 2NaI
substitution titration

12. (direct and surrogate titration)
(back titration)
Calculation formulas in titrimetric analysis
 The method of individual measurements :
% =
𝑽𝑻 × 𝑻 𝑻 𝑶
× КП × 𝟏𝟎𝟎
𝒎
% =
(𝑽𝑻𝟏 − 𝑽𝑻𝟐) × 𝑻 𝑻 𝑶
× КП × 𝟏𝟎𝟎
𝒎
 Pipetting method :
% =
𝑽𝑻 × 𝑻 𝑻 𝑶
× КП × 𝟏𝟎𝟎 × 𝑽м. к.
𝒎 × 𝑽п
% =
𝑽𝑻𝟏 − 𝑽𝑻𝟐 × 𝑻 𝑻 𝑶
× КП × 𝟏𝟎𝟎 × 𝑽м. к
𝒎 × 𝑽п
(direct and surrogate titration)
(back titration)
𝑻 𝑻 𝑶
=
𝑬м × С𝑵
𝟏𝟎𝟎𝟎
Ем – equivalent mass of the substance under investigation; CN – normal titrant concentration.

13. Calculate the weight of glutamic acid (M.m. = 147.13), if 20.06 ml of 0.1 N sodium hydroxide was used for titration by the
method of direct alkalimetry, Cf = 1.0000. The percentage content of the active substance is 99.1 %
given:
М.м. = 147,13
V(titrant) = 20,06 ml
CN = 0,1N
Cf = 1,0000
% = 99,1%
Find :
msubstunce- ?
The solution:
1. Calculation formula :
2. Reaction equation :
3. Calculation of the titer of the titrant by the determining substance:
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100
𝑚𝑠𝑢𝑏𝑠𝑡𝑢𝑛𝑐𝑒
HOOC
COOH
N
H2
+ NaOH
HOOC
COONa
N
H2
+ H2O
𝑇 𝑇 𝑂
=
𝐸м × С𝑁
1000
𝑇 𝑇 𝑂
=
147,13 × 0,1
1000
= 0,0147 г/мл
𝑚 =
𝑉𝑇 × 𝐶𝑓 × 𝑇 𝑇 𝑂
× 100
%
4. Calculation of the mass of the active substance:
𝑚 =
20,06 × 1,0000 × 0,0147 × 100
99,1
m = 0,297 г

14. Calculate the percentage content of diphenhydramine (M.m. = 291.82) in the substance, if 10.49 ml of 0.1N perchloric acid
solution was used for the titration of the weight (0.2976 g). Cf = 1.0018. The volume of the titrant in the control study is 0.36 ml.
given:
М.м. = 291,82
m = 0,2976 g.
V(titrant) = 10,49 ml
CN = 0,1N
Cf = 1,0018
Vc = 0,36 ml.
Find:
%- ?
The solution:
1. Calculation formula:
2. Reaction equation:
3. Calculation of the titer of the titrant by the determining substance:
% =
(𝑉𝑇 − 𝑉𝑐) × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100
𝑚𝑠𝑢𝑏𝑠𝑡𝑢𝑛𝑐𝑒
𝑇 𝑇 𝑂
=
𝐸м × С𝑁
1000
𝑇 𝑇 𝑂
=
291,82 × 0,1
1000
= 0,0292 г/мл
4. Calculation of the percentage content of the active substance:
% = 99,5%
CH O
H5C6
H5C6
CH2 CH2 N
CH3
CH3
HCl
2 + 2HClO4
(CH3COO)2Hg
CH O
H5C6
H5C6
CH2 CH2 N
CH3
CH3
2 HClO4 + HgCl2 + CH3COOH
% =
(𝑉𝑇 − 𝑉𝑐) × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100
𝑚𝑠𝑢𝑏𝑠𝑡𝑢𝑛𝑐𝑒
% =
(10,49 − 0,36) × 0,0292 × 1,0018 × 100
0,2976

15. Calculate the weight of potassium permanganate (M.m. = 158.04), if 23.68 ml of 0.1N sodium thiosulfate (Cf = 1.0000) were
used for titration by the method of indirect iodometry; the percentage content in the substance is 99.8%. The volume of the
measuring flask is 100 ml, the volume of the pipette is 25 ml.
given :
М.м. = 158,04
V(titrant) = 23,68 ml
CN = 0,1N
Cf = 1,0000
% = 99,8%
Vmf. = 100 ml
Vp = 25 ml
Find :
msubstunce - ?
The solution:
1. Calculation formula:
2. Reaction equation:
3. Calculation of the titer of the titrant by the determining substance :
𝑇 𝑇 𝑂
=
𝐸м × С𝑁
1000
4. Calculation of the suspended mass of the active substance :
m = 0,3037 г
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100 × 𝑉𝑚. 𝑓.
𝑚 × 𝑉𝑝
KMnO4 + Na2S2O3 + H2SO4 → MnSO4 + Na2SO4 + K2SO4 + H2O
MnO4
- + 8H+ + 5e → Mn2+ + 4H2O
𝐸м = М. м. × 𝑓 𝐸м = 158,04 × 1
5 Eм = 31,61
𝑇 𝑇 𝑂
=
31,61 × 0,1
1000
𝑇 𝑇 𝑂
= 0,0032 g/ml
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100 × 𝑉𝑚. 𝑓.
𝑚 × 𝑉𝑝
99,8 =
23,68 × 0,0032 × 1,0000 × 100 × 100.
𝑚 × 25

16. Calculate the percentage content of potassium chloride (M.m. = 77.56) in the substance, if 13.02 ml of 0.1 N silver nitrate
solution (Cf = 1.0100) was used for the titration of a weight of 0.9850 g. The volume of the measuring flask is 50 ml, the volume
of the pipette is 5 ml.
given :
М.м. = 77,56
m = 0,9850 g.
V(titrant) = 13,2 ml
CN = 0,1N
Cf = 1,0100
Vm.f. = 50 ml
Vp. = 5 ml
Find:
%- ?
The solution:
1. Calculation formula:
2. Reaction equation:
3. Calculation of the titer of the titrant by the determining substance:
𝑇 𝑇 𝑂
=
𝐸м × С𝑁
1000
𝑇 𝑇 𝑂
=
77,56 × 0,1
1000
4. Calculation of the percentage content of the active substance:
% = 99,5%
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100 × 𝑉𝑚. 𝑓.
𝑚 × 𝑉𝑝
KCl + AgNO3 → AgCl↓ + KNO3
𝑇 𝑇 𝑂
= 0,0077 g/ml
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100 × 𝑉𝑚. 𝑓.
𝑚 × 𝑉𝑝
% =
13,2 × 0,0077 × 1,0100 × 100 × 50.
0,9850 × 5

17. Calculate the volume of 0.1N sodium edetate (Cf = 0.9998), which will be used for the titration of 0.7422 g of calcium gluconate
(M.m. = 448.4), if the percentage content in the substance is 99.7%.
given :
М.м. = 448,4
m = 0,7422 g.
CN = 0,1N
Cf = 0,9998
% = 99,7%
Find:
VT - ?
The solution:
1. Calculation formula:
2. Calculation of the titer of the titrant by the determining substance:
𝑇 𝑇 𝑂
=
𝐸м × С𝑁
1000 𝑇 𝑇 𝑂
=
448,4 × 0,1
1000
3. Calculation of the percentage content of the active substance:
V = 16,67 ml
𝑇 𝑇 𝑂
= 0,04484 g/ml
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100
𝑚
% =
𝑉𝑇 × 𝑇 𝑇 𝑂
× 𝐶𝑓 × 100
𝑚
99,7 =
𝑉𝑇 × 0,04484 × 0,9998 × 100
0,7422