TOPIC: VOLUMETRIC / QUANTITATIVE ANALYSIS 

CONTENT

• Preparation of Standard Solutions. 
• Indicators
• Calculations based on Acid-Base Titration.

 

TITRATION

There are two types of quantitative analysis namely: volumetric and gravimetric analysis.  Volumetric analysis is based on volume measurement while gravimetric analysis involves direct mass measurement.

 

Volumetric analysis is carried art using Titration.  In titration, a standard solution (one of known concentration must be using be used to react with a solution of unknown of concentration)

 

PREPARATION OF A STANDARD SOLUTION

A standard solution is a solution of which the concentration is known .A standard solution is prepared by weighing a pure solute, for instance, and dissolving it in a suitable solvent, usually water, and making up the solution to a definite volume in a volumetric flask.

 

For instance, a solution known to contain exactly 10.6g of anhydrous sodium trioxocarbonate (IV), Na2CO3, in 1 dm3 of solution is a standard solution.

Preparation of 0.1mol/dm3NaOH

         40g NaOH dissolved in 1 dm3 of the water gives 1.0mol/dm3 solution

XgNaOH will be dissolved in 1 dm3 of water to give 0.1mol/dm3

Xg = 40g x 0.1mol/dm3

1.0mol/dm3

= 4g

 

Therefore, 4g of sodium hydroxide pellet is measured, dissolved in water and made up to 1dm3 mark to obtain 0.1mol/dm3NaOH

 

Preparation of 0.1mol/dm3HCl

To prepare 0.1mol/dm3HCl, the dilution formula is used to determine the volume of the stock acid that will be measured and dissolved in water to obtain the desired concentration.

The dilution formula is C1V1 = C2V2

Where C1 = concentration of stock acid = 11.6mol/dm3 (for HCl)

V1 = volume of stock acid

C2 = desired concentration of acid = 0.1mol/dm3

V2 = volume of water = 1000cm3 (1dm3)

V1 = C2V2 = 0.1 x 1000 = 8.6cm3

        C1       11.6

Thus, 8.6cm3 of the stock acid is measured using a measuring cylinder and added to water, then made up to 1dm3 to obtained 0.1 mol/dm3HCl.

 

EVALUATION

1. Describe how to prepare 0.05mol/dm3 H2SO4
2. List the apparatuses used during volumetric analysis(titration)

 

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

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 an 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

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

• 1 : 2

1 mole of Na2CO3 reacts 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

 

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 = 25.00cm3.

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

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

 

INDICATORS FOR ACID/BASE TITRATION

Acid-base indicators are dyes that change colour when according to the pH of the medium. The table below shows some titration and their suitable indicator:

 

Acid / base Indicator

Strong acid and strong base methyl orange or phenolphthalein

Strong acid and weak base methyl orange

Weak acid and strong base phenolphthalein

Weak acid and weak base No suitable indicator

 

CONCENTRATION

The concentration of a solution is the amount of solute in a given volume of the solution.  It can be expressed as mol/dm3 or g/dm3.

 

Molar concentration

The molar concentration of a compound is one which contains one mole or the molar mass of the compound in 1dm3 of the solution. Unit of molar concentration is mol/dm3

 

Mass concentration  

The mass concentration of a compound is the mass of the compound contained in 1 dm3 of solution. The unit is g/dm3

 

Relationship between Molar concentration and Mass concentration

Concentration = number of moles  = n/V………………(i)

volume

Number of moles, n = C x V ……………………..(ii)

But, number of moles, n= m/M

Where M = molar mass and m = mass

Substituting n=m/M into …….(ii)

We have m/M = C x V

That is, m/V = C x M

But m/V = mass concentration

Therefore, mass concentration = molar concentration x molar mass = C x M

 

TITRATION REPORT

Burette reading in (cm3)	Rough	1st titration	2nd titration	3rd titration
Final burette reading	23.40	23.20	28.20	34.10
Initial burette reading	0.00	10.00	05.00	11.00
Volume of acid used (titre)	23.40	23.20	23.20	23.10

 

Average volume of acid used =  23.20 + 23.20 + 23.10

      3

= 23.17cm3

 

GENERAL EVALUATION/REVISION

• Which substance is added to the base during titration experiment?

1. What is the colour of methyl orange in a base?
2. What is the point at which the titration experiment appears to complete called?
3. State three differences between electrolytic cell and electrochemical cell
4. Calculate the standard e.m.f of a cell given that the e.m.f of Zn2+/Zn and Cu2+/Cu are -0.76V and +0.34V respectively

 

 

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1. The indicator used when titrating a weak acid against a strong base isA. methyl orange B. phenolphthalein C. methyl red D. any indicator
2. The colour of phenolphthalein in acids is   A. blue    B. red    C. colourless D. yellow
3. Which of the following formulae is direct for amount n? A. n=C/p B. n=M/m C. n=C x V D. n=C x m
4. The mass concentration of a substance can be expressed as A. mass/density B. molar concentration/molar mass     C. mass/volume D. number of moles x volume
5. At the end point there is A. a colour change     B. no change of colour C. decrease in mass     D. an increase in mass

 

SECTION B

1. Define the following terms (a) molar concentration (b) Equivalent point
2. 160cm3 of distilled water is added to 40cm3 of 0.500mol/dm3 H2SO4 solution. Determine the concentration of the diluted solution.