# VOLUMETRIC ANALYSIS

## Volumetric Analysis

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

1. 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 indicator during the process, the following data were obtained.

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

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

1. 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

1. 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

Water of crystallization in the wager given off when an 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           =    moalr mass of anhydrous

Conc of 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 = 25.00cm3.

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

Practical Chemistry by Makanjuola pages 1-15.

New School Chemistry by Osei Yaw Ababio pages 165 – 183

Practical Chemistry for Schools and Colleges pages 100 – 170

## GENERAL EVALUATION

1. What is volumetric analysis
2. Name five  apparatus used in volumeric 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 filterate (d) c is a residue
2. ____ is the apparatus use to convert vapor into liquid during distillation. (a) conical flask (b) distillation column   (c) lie-big condenser (d) round bottom flask
3. X which fumes in most 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

## THEORY

1. State what would observe on
1. mixing Zinc dust with CuSO4 solution
2. adding concentrated HNO3 to freshly prepared FeSO4 solution
1. A salt sample was suspected to be either Na2CO3 or NaHCO3. A student who was required to identify it, tested a portion for solubility in water and for effects on litmus paper.
1. What was the observation in each case?
2. State the reason why the student’s procedure was unsuitable.
3. Describe briefly how you would have identified the salt.

Anions

QUALITATIVE ANALYSIS

METALS

PRELIMINARY PREPARATION

METALLIC BONDING

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