**Volumetric Analysis**

Volumetric analysis involves acid base titration.

Table of Contents

**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

- H
_{2}SO_{4}+ 2NaOH Na_{2}SO_{4}+ 2H_{2}O

__CaVa __ = ½

CbVb

- 2HCl + Na
_{2}CO_{3 }2NaCl +H_{2}O + CO_{2}

__CaVa__ = __2__

CbVb 1

**EVALUATION**

- What is volumetric analysis
- Give the ratio of the reaction species in the following chemical reactions
- CaCO
_{3}+ 2 HCl CaCl_{2 }+ H_{2}O + CO_{2} - KHCO
_{3}+ 2HCl KCl + H_{2}O + CO_{2}

** **

**Calculation Involving Titration **

- Mole Ratio

A is a solution of an acid **hydrogen** chloride .B is a solution of sodium trioxocarbonate(iv) containing 0.05 mole per dm^{3} solution A was titrated against 25cm^{3} of solution B, using methyl orange as indicator during the process, the following data were obtained.

Burette reading (cm^{3}) Rough 1^{st } 2^{nd } 3^{rd}

Final burette reading (cm^{3}) 24.65 48.95 24.30 24.30

Initial burette reading (cm^{3}) 0.00 24.65 0.00 0.00

Volume of acid used (cm^{3}) 24.65 24.30 24.30 24.30.

- Calculate the average titre
**value** - Calculate the concentration of the acid in moldm
^{3}. - Calculate the concentration of the acid in g/dm
^{3}.

The equation of the reaction

NaCO_{3} + 2HCl 2NaCl +H_{2}O + CO_{2}

Solution

- Average titre
**value**=__24.30 + 24.30 + 24.30__

3

= 24.30cm^{3}

- Concentration of A in moldm
^{3}

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 = 0.103moldm^{3}.

OR

From no of mole = Conc. In moldm-3 X vol/dm^{3}

No of moles = 0.05 x __ 25__

1000

equation of the reaction.

Na_{2}CO_{3} + 2HCl 2NaCl + H_{2}O + CO_{2}

- : 2

1 mole of Na_{2}CO_{3} react with 2 moles of HCl

:. 0.00125 mole of Na_{2}CO_{3} 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.

= 0.103moldm^{3}

- Concentration of A in g/dm
^{3}

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/dm^{3}

** **

**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/dm ^{3 }of pure solution __ X

__100__

Conc in g/dm^{3} of impure solution 1

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

conc in g/dm^{3} 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 dm^{3}. B is a solution of an impure sodium trioxocarbonate(iv) containing 3.0g per 250cm^{3}.

- Calculate the

(i) percentage purity of A

(ii) percentage impurity of A

Va = 20.40cm^{3} Vb = 25.00cm^{3}

The equation of reaction

Na_{2}CO_{3} + 2HCl 2NaCl + H_{2}O + CO_{2}

(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 moldm^{3}

Conc in g/dm^{3} of pure

From

Conc in g/dm^{3} = Moldm^{3} x molar mass

Molar mass of Na_{2}CO^{3} = 2(23) + 12 + 3 (16)

Molar mass of Na_{2}CO_{3} = 106g/mol

:. Conc in g/dm^{3 }of pure = 0.082 x 106

= 8.692 g/dm^{3}

Conc of impure Na_{2}XO_{3}

250 cm^{3} dissolve 3.0g of Na_{2}CO_{3}

1 cm3 dissolves __ 3.0 __ X 1000

250

= 12.0g/dm^{3}

- :. % 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 moldm_{3} of solution.

B is a solution of hydrated salt Na_{2}CO_{3}. XH_{2}O containing 3.94g which was made up to 250cm_{3} of solution with distilled water

Va = 29.00cm^{3}, Vb = 25.00cm^{3.}

Calculate the

**value**of X- percentage of water of crystallization.

Equation of the reaction

Na_{2}CO_{3}.XH_{2}O + 2HCl 2NaCl + H_{2}O + H_{2}O + CO_{2}

Solution

**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 = 0.0550moldm^{3}

Conc in g/dm3 of Na_{2}CO_{3} = moldm^{-3 } x m.m

Molar mass of Na_{2}CO_{3} = 2 (23) + 12 + 3(16) = 106 g/mol

Conc in g/dm^{3} = 0.055 x 106 = 5.83 g/dm^{3}

Conc in g/dm^{3} of hydrated:

__Mass __ X __ 1000__

Volume 1

Conc in g/dm^{3} = __ 3.94 x 1000__

250

= 15.8g/dm^{3}

__ 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 Na_{2}CO_{3}.10H_{2}O

**READING ASSIGNMENT **

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**

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

**WEEKEND ASSIGNMENT**

- 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
- ____ 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
- 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.
- The observation in bubbling SO
_{2 }into acidified KMnO_{4}solution is (a) The solution turns to green (b) the solution becomes decolourized (c) no visible reaction (d) the solution turns steam - The two substances that can give both H
_{2}and ZnSO_{4}when added to H_{2}SO_{4}are: (a) Magnesium and Zinc (b) Magnesium and CuO (c) Sodium and NaOH (d) iron and copper

**THEORY**

- State what would observe on

- mixing Zinc dust with CuSO
_{4}solution - adding concentrated HNO
_{3 }to freshly prepared FeSO_{4}solution

- A salt sample was suspected to be either Na
_{2}CO_{3}or NaHCO_{3}. A student who was required to identify it, tested a portion for solubility in water and for effects on litmus paper.

- What was the observation in each case?
- State the reason why the student’s procedure was unsuitable.
- Describe briefly how you would have identified the salt.

See also