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Saturday 4 March 2017

VOLUMETRIC ANALYSIS

Standard Volumetric Apparatus
The Concept of Volumetric Analysis
Explain the concept of volumetric analysis
           

Volumetric analysis is a quantitative analysis involving the measurement of different solutions. These solutions are made to react completely and the completion of the reaction is indicated by certain substances called indicators. The quantitative composition of the solution is then determined.
Important steps of volumetric analysis include:
Weighing;
Preparation of the solution;
Titration; and
Calculation
In volumetric analysis, we deal with volumes of solutions. That is why this quantitative determination of solutions of substances is called volumetric analysis.
The amount of a substance present in a solution is given in terms of its volume and its concentration. The volume of a solution is usually given in litres (dm3). The concentration of a solution is given in moles per litre (mol/dm3) or grams per litre (g/dm3).
Volumetric analysis is a means of finding the concentration of an unknown solution. For example, the concentration of an unknown solution of an acid can be found if it is reacted with a standard solution of an alkali.
A standard solution is one whose concentration is well known and does not change with time.
In volumetric analysis, the reaction is carried out in a carefully controlled way. The volumes are measured accurately using a pipette and burette.
The method is to add a solution of one reactant to the solution of another reactant until the reaction is complete. When the reaction is complete, we say the end-point has been reached. If the reactants are acids and bases, completion (end-point) is determined by the change in colour of an acid-base indicator. The method is called titration. In other reactions, completion is determined by a colour change of reactant(s). The concentration of one of the reactant solutions must be known in order to be able to find the concentration of unknown solution.
Significance of Volumetric Analysis
Volumetric analysis is used to quantify the amount of substances present in solutions by analytical procedure, which involves precise measurements of volumes of solutions and masses of solids.
Volumetric analysis helps in the determination of the accurate volumes and concentrations of the reacting substances, often solutions.
Volumetric analysis (titration) helps in the preparations of standard solutions.
Volumetric analysis knowledge helps in the standardization of acids and bases.
Volumetric Apparatus
Use volumetric apparatus
We have seen that volumetric analysis involves determinations of quantities of substances, usually acids and alkalis, present in volumes of solutions. This is usually done by using measuring apparatus.
Apparatus used in volumetric analysis is based on volume measurements and since the analysis demands high accuracy, the apparatus has to be calibrated with the highest possible accuracy. It is for this reason that all apparatus for volumetric analysis are specifically for this and not other purposes.
Apparatus used for volumetric analysis include, burette, pipette, burette stand, white tile, conical flask, filter funnel, reagent bottle, watch glass, beaker, measuring cylinder and measuring flask (or volumetric flask). For approximate measurements, measuring cylinders may be used. For accurate measurements of volumes, volumetric flasks are used.
Burette
This is a long glass tube with a narrow lower part, which is fitted with a tap that controls the amount of solution let out of the burette. This instrument is calibrated from 0 to 50 cm3.Before measuring the solution, rinse the burette with distilled water, then with the solution it is going to hold. It has to be filled to the tip and all gas bubbles removed. Thus, the burette is an apparatus used for transferring the solution to the titration vessel (normally a flask).

Pipette
This apparatus has a wider middle part with narrow parts at either ends. The upper narrow part has a mark which marks the volume of all the space below it. If, say, the pipette is one that is marked 25 c

TOPIC 3 MOLE CONCEPT

          MOLE CONCEPT
The Mole as a Unit of Measurement
The Mole with Other Units of Measurements
Compare the mole with other units of measurements
When carrying out an experiment, a chemist cannot weigh out a single atom, ion, electron, proton or molecule of a substance. These particles are simply very small. A counting unit that is useful in practical chemistry must be used.
The standard unit is called one  mole of the substance. One mole of each of these different substances contains the same number of the particles (atoms, molecules, ions, electrons, protons, neutrons, etc). That number per mole has been worked by several different experimental methods and is found to be 6.0 × 1023. The value 6.0 × 1023 is called Avogadro’s constant or Avogadro’s number and is abbreviated as L. It is named after the nineteenth-century Italian chemist, Amedeo Avogadro.
The value 6.0 × 1023 is obtained through the following relationship.The mass of one atom of carbon-12 is 1.993 × 10-23g. Then, the number of atoms present in 12g of carbon-12 is derived as follows:
1 atom = 1.993 × 10-23g
X atoms = 12g

\ X = 6.0 × 1023 atoms.
Therefore, the number of atoms in 12g of carbon-12 and hence the number of particles in a mole are 6.02 × 1023 atoms.
Hence, Avogadro’s number is the number of atoms in exactly 12g of carbon-12 isotope.One mole of any substance contains as many as many elementary particles as the Avogadro’s number (constant).
So, from the above explanation, the mole can be defined as the amount of a substance that contains as many elementary particles as the number of atoms present in 12g of carbon-12 isotope.
Substance Formula Relative formula mass, Mr Mass of one mole (molar mass) This mass (1 mole) contains
Carbon C 12 12g 6.0 × 1023 carbon atoms
Iron Fe 56 56g 6.0 × 1023 iron atoms
Hydrogen H2 2 × 1 = 2 2g 6.0 × 1023 molecules
Oxygen O2 2 × 16 = 32 32 6.0 × 1023 molecules
Water H2O (2×1) + 16 = 18 18g 6.0 × 1023 formula units
Magnesium oxide MgO 24 + 16 = 40 40g 6.0 × 1023 formula units
Calcium carbonate CaCO3 40+12+(3×16) = 100 100g 6.0 × 1023 formula units
Silicon oxide SiO2 28 + (2 × 16) = 60 60g 6.0 × 1023 formula units
Fe3+ Fe3+ 56 56g 6.0 × 1023 iron(III) ions
Cl- Cl- 35.5 35.5g 6.0 × 1023 chloride ions
e- e- - - 6.0 × 1023 electrons