# Volumetric and Quantitative Analysis

In qualitative analysis, the elements and compounds present in a given substance are identified, whereas in quantitative analysis, the amount of elements or compounds present in a substance are identified.
There are two approaches to quantitative analysis:
Volumetric analysis which has to do with the determination of volume of solution
Gravimetric analysis which is associated with the determination of mass of substances.
Volumetric analysis is carried out by titration. In this process, a solution from a vessel is added to another solution in a second vessel until the reaction is completed. This is determined by a colour change in the resulting solution or the added indicator. Also here, a standard solution is added to a solution of unknown concentration which is later determined.
A Standard solution is a solution whose concentration is accurately known.
Concentration: This is the amount of a substance in moles or grams dissolved in 1dm3 of solution. It may either be;
Molar Concentration: This is the amount of a substance in moles dissolved in 1dm3 of solution.
Molar Concentration = Amount of substance in moles/Volume in dm3
The unit is mol/dm3
Mass Concentration: This is the amount of a substance in grammes dissolved in 1dm3 of solution.
Mass Concentration = Mass of substance in grammes/Volume in dm3
The unit is g/dm3
Volumetric analysis can be;
Acid against base or trioxocarbonates
Oxidising agent against reducing agent
A substance against another to give a precipitate.

Acid-base Titration
This is a neutralization reaction between acids and bases. Equipment or materials needed in titration include:
Burette
Pipette
Funnel
White tile
Beaker
Retort stand, etc
Relationship between Molar Mass, Molar Concentration and Mass Concentration
Mass Concentration = Molar concentration x Molar mass
Using their units: g/dm3 = mol/dm3 x g/mol
Examples
Calculate the mass concentration of 3g of a substance in 400cm3 of solution.
Solution
Mass conc = mass (g)/ Volume (dm3)
1000cm3 = 1dm3
Therefore, 400cm3 = 1/1000 x400
= 0.4dm3
3/0.4 = 7.5dm3
Mole ratio
This is simply the coefficient of elements or compounds in a balanced equation
Given the equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Mole ratio: 1 : 2
Calculations on Acid- base titration
The concentration of acid or base can be determined using the formular:
CAVA/CBVB = na/nb
Where CA – Molar Concentration of acid in mol/dm3
CB – Molar Concentration of base in mol/dm3
VA – Volume of acid used in cm3
VB – Volume of pipette or volume of base used in cm3
na – Mole ratio of acid
nb – mole ratio of base
Example
A is a solution containing 0.5mol/dm3 HCl. B is a solution of Na2CO3. 27.5cm3 of A requires 25cm3 portion of B for complete neutralization. Calculate
Concentration of B in mol/dm3
Comcentration of B in g/dm3
Volume of CO2 liberated at stp during the titration. (Na = 23, C = 12, O = 16, H = 1, Cl = 35.5).
Sol ution
Equation for the reaction
2HCl + Na2CO3 → 2NaCl + H2O = CO2
Mole ratio: 2 : 1
CA – 0.5mol/dm3
VA – = 27.5cm3
CB – ?
VB – 25cm3
na = 2
nb = 1
Using CAVA/CBVB = na/nb
CB = CAVAnb/VBna
CB = 0.5mol/dm3 x 27.5cm3 x 1/ 25cm3 x 2
CB = 0.275mol/dm3

Mass Concn of B = Molar concentration of B x Molar mass of B
Molar mass of B – Na2CO3 = (23×2) + 12 + (16 x 3)
= 106g/mol
Molar concn of B = 0.275mol/dm3
Therefore mass concn of B = 0.275mol/dm3 x 106g/mol
= 29.15g/dm3
Na2CO3 → CO2
1mol : 1mol
0.275mole of Na2CO3 gives 22.4dm3 of CO3
Molar Volume = Volume/Amount
Volume = Molar Volume x amount
= 22.4dm3 x 0.275
= 6.16dm3