1. Titration : Titration is a process by means of which the concentration (strength) of a solution determined by allowing to react with a standard solution.
2. Acidimetry : It is the process by which strength of an acid is determined by reacting with a known amount (standard solution) of base in the presence of an indicator.
3. Alkalimetry : It is the process by which strength of an is alkali determined by reacting with a known amount (standard solution) of acid in presence of an indicator.
4. Standard Solution : A solution of definite concentration is called a standard solution. That is by standard solution we mean that a known amount of solute is dissolved in a given volume of solution. As an example 10gm of NaCl dissolved in 200ml solution is a standard solution.
5. Neutralization : The process of complete reaction of hydrogen ions of an acid with the exact amount of hydroxide ions of an alkali to form alkali molecules is called neutralization.It should be noted that neutralization does not mean that the resulting solution is always neutral. i.e. PH = 7 . In fact, when the same value of equivalent amount of acid and alkali reacts, it is said neutralization process has taken place.
6. Indicators : Indicators most frequcency used in acid base titration are methyl orange and phenolphthalein. Indicators are substance by which change colour and without reacting with solution indicate the end point we mean that equivalent amounts of substances have reacted.
Indicators also indicate whether a given solution is alkaline or neutral.
2. Acidimetry : It is the process by which strength of an acid is determined by reacting with a known amount (standard solution) of base in the presence of an indicator.
3. Alkalimetry : It is the process by which strength of an is alkali determined by reacting with a known amount (standard solution) of acid in presence of an indicator.
4. Standard Solution : A solution of definite concentration is called a standard solution. That is by standard solution we mean that a known amount of solute is dissolved in a given volume of solution. As an example 10gm of NaCl dissolved in 200ml solution is a standard solution.
5. Neutralization : The process of complete reaction of hydrogen ions of an acid with the exact amount of hydroxide ions of an alkali to form alkali molecules is called neutralization.It should be noted that neutralization does not mean that the resulting solution is always neutral. i.e. PH = 7 . In fact, when the same value of equivalent amount of acid and alkali reacts, it is said neutralization process has taken place.
6. Indicators : Indicators most frequcency used in acid base titration are methyl orange and phenolphthalein. Indicators are substance by which change colour and without reacting with solution indicate the end point we mean that equivalent amounts of substances have reacted.
Indicators also indicate whether a given solution is alkaline or neutral.
7. Some commonly used indicators and their PH Range :
Indicator Colour in A Med. Colour in B. Med. Pᴴ- Range
Methyl Orange Red Yellow 3.4 – 4.7
Phenolphthalein Colourless Pink 8.2 – 10
Methyl Red Red Yellow 4.2 – 6.3
Litmus Red Blue 5.5 – 8.0
Thymol blue base Yellow Blue 8.0 – 9.6
Indicator Colour in A Med. Colour in B. Med. Pᴴ- Range
Methyl Orange Red Yellow 3.4 – 4.7
Phenolphthalein Colourless Pink 8.2 – 10
Methyl Red Red Yellow 4.2 – 6.3
Litmus Red Blue 5.5 – 8.0
Thymol blue base Yellow Blue 8.0 – 9.6
8. The choice of indicator in acid-alkali neutralization:
Acid Alkali PH-Range Indicator Solution in burette
Strong Strong 4 to 10 Ph. or M.Or. or litmus Alkali
Strong Weak 3.5 to 7 M.Or. or M.R. Acid
Weak Strong 6.5 to 10 PH Alkali
Weak Weak No Sharp None
9. Equivalent
weight of some substance
It is
hoped that students are quite familiar with definitions of equivalent weight of
acid, alkali, compound and salt. The equivalent weight of few substance are
noteworthy
Compound wt. Mol. Weight Equivalent
HCl 36.5 36.5
HNO3 63.0 63.0
H2SO4 98.0 49.0
H2C2O4.2H2O 126.0 63.0
NaOH 40.0 40.0
KOH 56.0 56.0
CaO 56.0 28.0
Na2CO3.10H2O 286.0 143.0
Na2CO3 106.0 53.0
NaHCO3 84.0 84.0
CaCO3 100.0 50.0
MgCO3 84.0 84.0
NaCl
58.5 58.5
AgCl 143.5 143.5
AgNO3 170.0 170.0
BaSO4 233 116.5
NH4Cl 53.5 53.5
NH3 17 17 10. Concentration of Strength of solution
Relative amount of the solute and the solvent present in the solution. It is expressed in different units given below.
i. Percentage strength %(v/v),%(w/v) or %(w/w)
ii. Gram per litre g/l
iii. Normality N
iv. Molarity M
v. Molality M
vi. Mole Fraction X
Relative amount of the solute and the solvent present in the solution. It is expressed in different units given below.
i. Percentage strength %(v/v),%(w/v) or %(w/w)
ii. Gram per litre g/l
iii. Normality N
iv. Molarity M
v. Molality M
vi. Mole Fraction X
i. Percentage strength :
a. % by volume (w/v) :
It means that the weight of solute
in grams dissolved per 100ml of its solution. Thus 2% NaCl solution means that
2gm NaCl is dissolved per 100ml of its solution.
% by volume =
Weight of solute (in gm) × 100
Volume of solution (in ml)
b. % by weight (w/w) :
It express the weight of solute in grams
dissolved per 100gm of its solution. As an examples 5% Na2CO3 by weight means
that 5gm of Na2CO3 is displayed per 100gm of its solution.
% by weight =
Weight of solute ×100
Volume of solution
ii. Gram per litre (g/l) :
The term itself implies the weight
of solute in grams dissolved per litre or 100ml of its solution.
Gram per litre =
Weight of solute (in gm)
Volume of solution (in l)
Weight of solute (in gm) × 1000
Volume of solution (in ml)
iii. Normality :
It is the number of gram equivalents
of solute dissolved per litre solution. The numerical value is following by the
letter N.
Thus 0.5 N solution means that the
normality is 0.5, It than follows that 0.5 gram equivalent of solute is present
in 1 L of solution.
Normal solution (1N) : it is the
solution containing one gram equivalent of solute dissolved per litre of its
solution.
Semi normal solution (N/2) : It is
the equation containing half of the gram equivalent of solute per litre
solution.W =
N(normality) × E(Eq. wt) ×V(ml)
1000
iv. Molarity :
It is the number of g-molecules
(moles) of solute dissolved per litre its solution. As shown for normality,
molarity, M =
W ×1000
V × MW
Where Mw is the molecular wt. of
solute.
Molar solution (1M) : It is the solution
containing one gm- molecule of the solute dissolved per litre solution.
v. Molality :
Molality is the number moles of the
solute dissolved per 1 Kg or 1000gm of solvent.
If ‘a’ gm of solute of molecular
weight MW is dissolved in “b” gm of solvent, then by definition, Molality =
a × 1000
MW × b
vi. Mole Fraction :
Mole fraction of a component in a
solution is the ratio of the number of moles of that component and the number
of moles of all the components in solution.
If a solution is prepared by
dissolving nx mole of solute in ny mole of solvent, then
Xsolute = nx/(nx
+ ny)
Xsolvent = ny/(nx
+ ny)
Obviously; Xsolute+ Xsolvent
= 1.0
11. Important :
- The most common mode of expressing
concentration of a solution is molarity.
- Since mass does not change with
rise in temperature, the value of molality, and mole fraction remains the same.
- As there in increase in volume by
increase of temperature, the value of molarity and normality decreases when
temperature is increased.
- When a certain volume of a
standard solution is diluted (i.e. the volume is increased by addition of
water) the concentration decreases.
- Normality and molarity of a
solution is related as
Normality = molarity × mol. wt./eq.
wt.
12. Normality factor :
It is the ratio of the weight of
solute taken and the wt. of the solute required to prepare a solution of a
particular strength.
Suppose 1.3gm of Na2CO3
is dissolved in 250 ml of a solution to prepare N/10 Na2CO3.By
calculation the weight of Na2CO3 required = (250×53)/(10×1000)
= 1.325 gm
Hence, by definition normality
factor of N/10 Na2CO3.
Solution = 1.3/1.325= 0.98
The strength of solution =N/10 , (f = 0.98)
13. Basic principles of calculation :
1. When two solutions containing different
solutes are able to different solutes are mixed and are to react together, it
is the solutes present that react with themselves, water just speeds up the
reaction.
2. Solutes completely react and products
there by formed in a given reaction are always in proportion to their
gm-equivalent.
3. If V1ml of a solution of
strength N1 react completely with V2 ml of another
solutes of strength N2, then their number gm-eq. of solute is
V1× N1 = V2×
N2
4. Strength of a mixture : when different
solution of the same nature and of given volume and strength are mixed, the
principles (2) still holds
That is VmNm = V1N1+
V2N2+ V3N3+……………………….
When all solution are acids or
alkalies, The subscript m stands for mixture, When acids and alkalies are mixed
VmNm = V1N1+
V2N2 _ V1N1+ V2N2 ……………………
Acids Alkalies
5. Acids and alkalies of the same strength in
terms strength in terms of normality require the equal volume for neutralization.
Important formulae:
1. Strength in percentage = (gm/lit)/10.
2. Required amount (w) = (NEV)/1000.
3. Normality (N) = [(w/w)% × SP. gr × 10
]/[eq. weight.]
4. Molarity (M) = [(w/w)% × SP. gr × 10 ]/[mol. weight.]
5. Normality (N) = [(w/v)% × 10 ]/[eq.
weight.]
6. Molarity (M) = [(w/v)% × 10 ]/[mol.
weight.]
7. N1V1 = N2V2
8. S1V1 = S2V2(only
for dilution)
9. NV = N1V1+ N2V2+
N3V3
10. M1V1 × Basicity = M2V2
× Acidity.
0 comments:
Post a Comment