Volumetric Analysis
Answer questions
1. Distinguish between decinormal and decimolar solution.
> Following are the different between decinormal and decimolar solution:
|
Decinormal
solution |
Decimolar
solution |
|
1. The solution in which one tenth gram equivalent of solute is dissolved to prepare one litre solution. |
1. The solution in which one tenth mole of solute is dissolved to prepare one litre solution |
|
2. It is denoted by |
2. It is denoted by |
|
3. Examples:- 4 g NaoH solution dissolved to prepare 1 L solution, the strength of solution is decinormal |
3. Examples:- 9.8 g |
2.Distinguish between titration error and normality factor
>Following are the difference between titration error and
normality factor
Titration error
1.The difference between the equivalence point and the
measured end point is called titration error.
Normality factor
1.It is defined as
the ratio of observed weight of solute to the theoretical weight of solute
required to prepare a solution of desired normality
3. Differentiate between end point and equivalence point of
a reaction.
> Following are the difference between end point and
equivalence point.
End Point;
1. The point at reaction between two solutions is complete
as shown by indicator
2. It must be in acid base solution.
Equivalence Point
1. The point in a titration at which the Which one gram equivalent
of a substance completely
reacts with one
gram equivalent of another substance is called equivalence point.
2. It may or may not
in acid base solution
4. Differentiate between normality and molarity.
>Following are the difference between normality and
molarity:
Normality
The number of gram equivalent of solute is present in one
litre of it's solution is
called normality,
It is denoted by 'N'
Molarity
The number of gram mole of solute is present in one litre of
it's solution is called molarity
It is denoted by 'M'
5. Distinguish between Primary standard solution and
secondary standard solution.
> Following are the distinction between primary standard
solution and secondary
standard solution:
Primary standard solution
It is prepared by dissolving pure substance which is not
affected in its
composition for a long time.
Example: Oxalic acid, Na2CO3, AgNO3,
KCI etc.
Secondary standard solution
It is prepared by neutralizing against the primary standard
solution.
Example: NaOH, HCI, H2SO4, HNO3 ,KMnO4 etc.
6. Define the terms:
i Primary standard solution
ii. Acidimetry
i. Primary standard solution: The substances whose standard
solution can be prepared
directly by dissolving the known weight of substance in
certain volume of solution is
called primary standard substance and the solution made from
primary standard
substance is called primary standard solution. Example:
solution of oxalic acid.
ii. Acidimetry: The method of determining the strength of
acid by titrating it with the
known volume of standard alkali solution is called
acidimetry.
7.What is the importance of calculating normality factor of
solutions during
titration?
during titration. Normality factor (f) is the ratio of
observed weight of substance and
theoretical weight of substance.
Normality factor (f) = Observed weight / Theoretical weight
10. Define the terms:
i Normality factor
li. Secondary standard solution.
i. Normality factor: The ratio of actual weight of substance
taken to the theoretical weight
of substance to be taken is called normality factor. It is
denoted by symbol 'f'.
Normality factor (f) = Weight taken / Weight to be taken
ii. Secondary standard solution: The solution whose strength
is to be determined with the
help of primary standard solution is called secondary
standard solution
11. Why is crystal oxalic acid regarded as a good substance for
the preparation of primary standard solution?
> Oxalic acid is regarded as primary substance because it is
available in pure state, non toxic, high equivalent weight, non-hygroscopic and
deliquescent. So, it is used for the preparation of primary standard solution
in titration.
12. how could you convert 500cc of 2MH2SO4
into
i. Gram/liter ii. Normality
> Given,
volume of H2SO4 = 500cc
Molecular weight of H2SO4 = 98
Equivalent weight of H2SO4 = 49
Molarity of H2SO4 = 2M
i.Gram/L = Molarity X Molecular weight = 2 X 98 = 196 g/L
ii.Normality = Gram/L
by Equivalent weight = 196/49 = 4N
13. Which one has higher concentration and why ?
a. 80g / liter NaOH solution
and 3M NaOH solution.
b. 5.3g / liter Na2CO3
and N/10 Na2CO3 solution.
a. > Given,
g/liter of NaOH= 80 g/litre
Equivalent weight of NaOH = 40 = Molecular weight
we know that,
Normality = g/liter by Equivalent weight = 80/40 = 2N of NaOH
Given,
Molarity of NaOH = 3M = 3N of NaOH
Hence, Concentration of 3M NaOH is higher than 80 g/liter NaOH
solution.
b. > Given,
g/liter of Na2CO3 = 5.3 g/liter
Molecular weight of Na2CO3 = 106
Equivalent weight of Na2CO3 = 53
We know that,
Normality = g/liter by Equivalent weight = 5.3/53 = 0.1 N of Na2CO3
Given, the normality of Na2CO3 = N/10
= 0.1 N
Hence, the concentration of both 5.3 g/liter Na2CO3 and N/10 Na2CO3
are same.
15. A sample of Na2CO3 weighing 0.53 g is
added to 101 ml of 0.1 N H2SO4 solution. Will the resulting solution be
acidic, basic or neutral ?
> Given,
Weight of Na2CO3 = 0.53g
Equivalent weight of Na2CO3 = 53
From Normality solution,
1000 ml of Na2CO3 of 1 N requires 53 g of Na2CO3.
100 ml of Na2CO3 of 0.1 N requires 0.53 g of
Na2CO3
For complete neutralization
100 ml of Na2CO3 of 0.1 N = 100 ml of H2SO4
of 0.1 N
We know that,
Equal volume of acid neutralizes equal volume of base, if
acid and base have the same normality. So, 100 ml oh H2SO4
of 0.1 N neutralizes 100 ml of Na2CO3 of
0.1 N. Thus, 1 ml. of acid is left unreacted and the resulting solution must be
acidic.
16. Calculate the normality and molarity of 5% of NaOH solution.
> Given, weight of NaOH means 5g of NaoH is dissolve in 100 cc
of solution
Given;
Weight of NaOH=5 g
Volume of solution = 100 cc.
Equivalent weight = Molecular weight of NaOH = 40
We know that,
Normality (N) = No. of g equivalent weight of solute / Volume of
solution in litre = weight of solute/ Equivalent weight by Volume of solution
in litre = 5/40 // 100/1000
=5/40×10 =
5/4 = 1.25 N
The equivalent weight = Molecular weight of NaOH. So,
normality is equal to normality.
N = M = 1.25
Hence, Normality (N) = Molarity (M) = 1.25
17. What mass of 90% pure CaCO3 is required to
neutralize 2 liter deci-normal solution of HCL?
> Given,
2 liter 0.1 N HCL = 0.2 liter 1N HCL = 200 ml 1N HCL
Volume of HCL = 200 ml
Normality oh HCL = 1N
Equivalent weight of CaCO3 = 100/2 = 50
Weight of CaCO3 (W) =?
We know that,
W = NEV/1000 = 1×50×200 /1000 =10 g
Let, X g of 90% = 10g
Or, X 90/100 = 10
Or, X = 10×100/90 = 11.11 g
Hence, 11.11 g of 90% pure CaCO3 is required to
neutralize 2 liter of deci-normal solution of HCL.
18. how many moles of H2SO4 are required to
neutralize 4 litres of 2N NaOH solution ?
> Given,
4 liters of 2N NaOH = 8 liters of 1N NaOH
Volume of NaOH = 8 liter = 8000 ml.
Normality of NaOH = 1N
Equivalent weight of H2SO4 =49
Weight required for H2SO4 (W) = ?
We know that,
W = NEV/1000 = 1×49×8000 / 1000 = 392g
Molecular weight of H2SO4 = 98 g
So, no. of moles of H2SO4 = Given weight/
Molecular weight = 392/98 = 4 mol.
Hence, 4 mol of H2SO4 are required to
neutralize 4 liters of 2N NaOH solution.
19. Define the term
i. Semi normal
solution ii. Alkalimetry
> Semi normal solution :- If the half gram equivalent of solute
is present in one liter of solution then it is called semi normal solution.
It is denoted by N/2
> Alkalimetry :- The method of determining the strength of
alkali by titrating it with the known volume of standard acid solution is
called alkalimetry .
20. Define secondary standard solution with a suitable example.
> The solution made from secondary standard substance is called
secondary standard solution.
Example : HCL solution.
21. Write an example of redox titration. Why is it called so ?
> Example of redox reaction is
Here, KMnO4 is reduced to MnSO4 and C2H2O4
is oxidized to CO2 and H2O. So, this reaction is called
redox titration.
22. What is the normality of 20 cc of 2 M phosphoric acid (H3PO4)
?
> Molarity of H3PO4 = 2 M
Basicity of H3PO4 = 3
Normality of 20 cc of H3PO4 = ?
We know that,
Normality = Molarity × basicity of acid = 2×3 = 6 N
So, the normality of 20 cc of H3PO4 is
6N.
23. What is normality ? How is it related with molarity ?
> Normality (N) :- If the equivalent weight of a substance is expressed
in gram is called gram equivalent. It is defined as the number of gram
equivalent of solute present in one liter of its solution. It is denoted by
(N).
Normality (N) = No. of gram equivalent weight of solute / Volume
of solution in liter = weight of solute in g/ Equivalent weight of solute × 1000/volume of solution in ml.
Normality is related with molarity as ;
Normality (N) = Molarity (M) × basicity of acid or acidity of base.
24. What do you mean by equivalent weight of an element ?
> The equivalent weight of an element is that parts by weight
of it which combines with or displace from a compound 1.008 part by weight of
hydrogen or 8 parts by weight of oxygen or 35.5 part by weight of chlorine.
25. What are the requisites for a substance to be a primary
standard ?
> Following are the requisites for a substance to be a primary
standard.
a. They should have high molecular and equivalent weight.
b. They should be easily available and highly purity.
c. They should be non-volatile, non-toxic, non-hygroscopic in
nature.
d. They should be highly soluble in water.
26. Define decinormal solution.
> A solution containing 1/10th of gram equivalent of
solute present in one liter of solution is known as decinormal solutions.
27. Are all standard solutions, primary standard
solutions or not? Give reason. 1g of a
divalent metal was dissolved in 25 mL of 2N H2SO4 (f
= 1.01). The excess acid required 15.1mL of 1N NaOH (f = 0.8) for complete neutralization. Find the atomic weight of the
metal.
> Not, all satandard solutions are not primary standard solution. To be a
primary
standard, a substance must fulfill the following criteria:
i. The substance must be easy to obtain and purity
ii. The substance should not be hygroscopic or efflorescent of deliquescent.
iii. The composition should not change during storage or weighing.
iv. The substance should be readily soluble under the employed condition
v. The substance should have high molecular mass.
Numerical:
Given,
The weight of metal =1g
The valency of metal =2
For acid:
Volume of acid (H2SO4) = 25 mL
Normality of acid =2N (f=1.01)
We know that
The gram equivalent of H2SO4 =Volume ×Normality /1000=25 x 2 x 1.01/1000=0.0505
For base:
Volume of base (NaOH) = 15.1 mL
Normality of acid = 1N (f=0.8)
We know that
The gram equivalent of NaOH=Volume x Normality/1000= 15.1 x 1 x 0.8/1000=0.01208
For Metal:
Weight of metal
The gram equivalent of metal =weight of metal / Equivalent weight of
metal=1/Equivalent of metal
We know that
The gram equivalent of metal =The gram equivalent of H2SO4 - The gram
equivalent of NaOH
1/Equivalent weight of metal = (0.0505 -0.01208) = 0.03842
Or, Equivalent weight of metal = 26.02
Hence, the atomic weight of metal= Equivalent weight × Valency = 26.02 ×2
= 52.05 amu.
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