**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

- 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

**⇓ Fully Funded Scholarships ⇓**

**⇒ Fully Funded Scholarships for Students & Teachers**

**⇒ USA Fully Funded Scholarships for Internatioal Students**

**⇒ Canada Fully Funded Scholarships for International Students**

**⇒ Germany Fully Funded Scholarships for Students**

**⇒ European Scholarships for International Students **

**⇒ US Fully Funded Scholarships for Africans**

**⇒ UK Fully Funded Scholarships for International Students**

**⇒ Australia Fully Funded Scholarships for International Students**

**⇒ US Scholarships for Nigerian Students**

We built School Portal NG to specially serve as an alternative means to provide free education for the less-privileged children or children born to low-income earners, adults who otherwise cannot afford to finance their way through school.

Building & maintaining an elearning portal is very expensive, but this portal is free. Help to keep this learning portal free by donating or supporting us. We accept donations, grants, sponsorships & support to help reach out to more children. Thank you so much. Click here to donate