Volumetric Analysis

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Welcome to class! 

In today’s class, we will be talking about volumetric analysis. Enjoy the class!

Volumetric Analysis

The volumetric analysis involves acid-base titration.

Mole ratio

Mole ratio is the ratio of the reacting species.  This determines the ratio of the acid that would react with the base.

Examples are

1. H₂SO₄  +  2NaOH          →                    Na₂SO₄ + 2H₂O

CaVa   =  ½

CbVb

2. 2HCl + Na₂CO₃        →              2NaCl +H₂O + CO₂

CaVa =    2

CbVb       1

Evaluation

1. What is volumetric analysis
2. Give the ratio of the reaction species in the following chemical reactions
3. CaCO₃ + 2 HCl   →     CaCl₂     +   H₂O    +   CO₂
4. KHCO₃ + 2HCl   →      KCl +      H₂O    +    CO₂

Calculation involving titration                                       

1. Mole ratio:

A is a solution of an acid hydrogen chloride .B is a solution of sodium trioxocarbonate(iv) containing 0.05 mole per dm³ solution A was titrated against 25cm³ of solution B, using methyl orange as an indicator during the process, the following data were obtained.

Burette reading (cm³)                   Rough           1^st                          2^nd                       3^rd

Final burette reading (cm³)        24.65              48.95              24.30              24.30

Initial burette reading (cm³)       0.00                24.65              0.00                0.00

Volume of acid used (cm³)        24.65              24.30              24.30              24.30.

1. Calculate the average titre value
2. Calculate the concentration of the acid in moldm³.
3. Calculate the concentration of the acid in g/dm³.

The equation of the reaction

NaCO₃ + 2HCl             →                    2NaCl +H₂O + CO₂

Solution

1. Average titre value = 24.30 + 24.30 + 24.30

.                                                            3

= 24.30cm³

2. Concentration of A in moldm³

from

CaVa = Na

CbVb    Nb

Ca x 24.30    =  2

0.05 x 25          1

Ca =  0.05 x 25 x 2

.              24.30

Ca = 0.103moldm³.

OR

From no  of mole  = Conc. In moldm-3  X vol/dm³

No of moles = 0.05 x  25

.                                 1000

equation of the reaction.

Na₂CO₃  + 2HCl        →          2NaCl + H₂O + CO₂

• : 2

1 mole of Na₂CO₃ react with 2 moles of HCl

:. 0.00125 mole of Na₂CO₃ will require 0.00123 x 2 of HCl

:. No of mole of A = 0.0025 mole

From conc of A in moldm-3 =  No of mole

Volume in dm3

=  0.0025       ×  10000

24.30

1000

 0.0025 x 1000

24.30.

= 0.103moldm³

3. Concentration of A in g/dm³

From:- conc in g/dm3 = conc in moldm^-3 x molar mass

Molar mass of HCl = 1 + 35.5  = 36.5 g/mol.

:. Conc in g/dm3 = 0.103 x 36.5

= 3.76g/dm³

Percentage purity and impurity

During the titration process of an impure acid or base is titrated only the pure part of either acid or base react with the base or acid.  Therefore the percentage (%) purity or impurity can be calculated.

% purity   = Conc in g/dm³ of pure solution     X     100

Conc in g/dm³ of impure solution         1

% impurity =  conc of impure – conc of pure   X  100

conc in g/dm³ of impure                 1

Mass of pure substance = Conc of pure in moldm^-3 x Molar Mass

Mass of impurity  = Conc of impure – pure

Example

A is a solution of 020mole of HCl per dm³. B is a solution of an impure sodium trioxocarbonate(iv) containing 3.0g per 250cm³.

1. Calculate the

(i)   percentage purity of A

(ii)  percentage impurity of A

Va = 20.40cm³     Vb = 25.00cm³

The equation of reaction

Na₂CO₃ + 2HCl        →             2NaCl + H₂O + CO₂

(Na = 23 C= 12 O = 16 H = 1, Cl = 35.5)

Solution

CaVa  = na

CbVb    nb

 0.20 x 20.40   =  2

25 x cb              1

Cb = 0.20 x 20.40 x 1

.                25 x 2

Cb = 0.0823 moldm³

Conc in g/dm³ of pure

From

Conc in g/dm³ = Moldm³ x molar mass

Molar mass of Na₂CO³ = 2(23) + 12 + 3 (16)

Molar mass of Na₂CO₃ = 106g/mol

:. Conc in g/dm³ of pure = 0.082 x 106

= 8.692 g/dm³

Conc of impure Na₂XO₃

250 cm³ dissolve  3.0g of Na₂CO₃

1 cm3 dissolves    3.0   X 1000

.                             250

= 12.0g/dm³

1. :. % purity = Conc of pure X 1000

Conc of impure      1

=  8. 69     X  100

12              1

= 72.4%

 

% impurity  =  Conc of impure – pure  X        100

Conc of impure                   1

% impurity  =    12 – 8.6g X      100

12               1

= 27.6%

Percentage amount of water of crystallization

The water of crystallization in the wager given off when a hydrated salt is heated or exposed to the atmosphere

Hydrated salt does not contain water

Amount of water of crystallization is calculated as follows:

 Conc of anhydrous        =    molar mass of anhydrous

Conc of the hydrated           molar mass of hydrated

 

Percentage Water of Crystallization is calculated as follows:

%  water of crystallization =    Hydrated – Anhydrous    X         100

Hydrated                  1

 

Example

Solution A is a solution of hydrogen chloride acid containing 0.095 moldm₃ of solution.

B is a solution of hydrated salt Na₂CO₃. XH₂O containing 3.94g which was made up to 250cm₃ of solution with distilled water

Va = 29.00cm³, Vb = 25.00cm^3.

Calculate the

1. value of X
2. percentage of water of crystallization.

 

 

Equation of the reaction

Na₂CO₃.XH₂O  + 2HCl          →                 2NaCl + H₂O + H₂O + CO₂

Solution

1. Value of x

From

CaVa      =    Na             CaVa       =     2

CbVb           Nb                CbVb            1

 

 0.095 x 29    =  2

Cb x 25            1

Cb =  0.095 x 29 x 1

25 x 2.

Cb = 0.0550moldm³

Conc in g/dm3 of Na₂CO₃ = moldm^-3  x m.m

Molar mass of Na₂CO₃ = 2 (23) + 12 + 3(16)  = 106 g/mol

Conc in g/dm³ = 0.055 x 106  = 5.83 g/dm³

Conc in g/dm³ of hydrated:

Mass     X  1000

Volume        1

 

Conc in g/dm³  =  3.94 x 1000

250

= 15.8g/dm³

 Conc of anhydrous     =  molar mass of anhydrous

 Conc of hydrated      molar mass of hydrated.

 

 5.83     =   106

         15.76          106 x 18

(106 x 18x) 5.83  = 106 x 15.76

106 + 18x  =  106 x 15.76

5.83

106 + 18x = 286. 55

18x = 286.55 – 106

18x = 180.55

x =  180.55

18.

x = 10

The salt is Na₂CO₃.10H₂O

General evaluation

1. What is volumetric analysis
2. Name five apparatus used in volumetric analysis.
3. Define the following terms;   a. Indicator b. Buffers  c. pH scale

Weekend assignment

1. C + water give colourless solution (a) c is a soluble salt (b) c is partially dissolve in water (c) c is a filtrate (d) c is a residue
2. ____ is the apparatus used to convert vapour into a liquid during distillation. (a) conical flask (b) distillation column   (c) lie-big condenser (d) round bottom flask
3. X which fumes in moist air can be suitably stored (a) under paraffin or naphtha (b) In a white bottle (c) inside a corked conical flask (d) inside a burette.
4. The observation in bubbling SO₂ into acidified KMnO₄ solution is (a) The solution turns to green   (b) the solution becomes decolourized  (c) no visible reaction (d) the solution turns steam
5. The two substances that can give both H₂ and ZnSO₄ when added to H₂SO₄ are: (a) Magnesium and Zinc (b) Magnesium and CuO (c) Sodium and NaOH (d) iron and copper

 

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