Categories
Chemistry

Volumetric analysis – Titration

Introduction

Volumetric analysis is an analytical method or procedure for working out the titre or concentration of an analyte in a solution. This is done by measuring the volume of a standard solution of an appropriate reagent whose precise concentration is already known.

Preparing A Standard Solution

A standard solution is a solution in which its concentration is known.

The steps taken in preparing a standard solution are:

Determine the volume and concentration that you want to prepare.

Calculate the mass of solute needed to give the required volume and concentration.

Weigh the solute

Dissolve the solute completely dissolved in distilled water and then transfer it to a volumetric flask partially filled with distilled water.

Add distilled water to the calibration mark of the volumetric flask.

Invert the flask and shake it to make sure thorough mixing.

volumetric

Acid Base Titration

Some materials used during acid – base titration and precautions in using some of them

weighing balance

chemical balance

pipette

burette

retort stand

filter paper

funnel

white tile

standard volumetric flask

conical flask

Pipette

rinse the pipette with the solution it will be measured with e.g. base

avoid air bubbles in the pipette

make sure the mark to be read is at same level with your eye

do not blow the last drop on the burette

Burette

rinse the burette with acid or allow it to drain after rinsing

make sure the burette jar is filled

avoid air bubbles in the burette

make sure that burette is not leaking4remove the funnel befor taking your reading

avoid inconsistent burette reading

Conical flask

do not rinse with any of the solutions used in the titration but with distilled water

wash down with distilled water any drop of the solution that stick by the sides of the conical flask

Concentration of a Solution

The concentration of a solution tells you how much solute is dissolved in 1 unit volume of solution.

The volume of a solution is measured in dm³ (litres)  1 dm³ = 1000 cm³.

The amount of solute can be measured in grams or moles.

2 units of concentration used in chemistry are g dm-3 and mol dm-3

Concentration in g dm-3

Concentration is the number of moles of solute per liter of solution.

A concentration of 10 g dm-3 means there is 10 g of solute dissolved in1 dm3 of solution.

Concentration=Mass of solute (g) / Volume of solution (dm3)

Example 1:

Calculate the concentration of the solution if 28g of NaOH is dissolve in 250cm3 of water.

Answer:

Mass of solute = 28g

Volume of solvent = 250cm³ = 0.25dm³

Concentration = Mass / Volume

=28g / 0.25dm3

=112g/dm3

Concentration in mol dm-3 (Molarity)

Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution.

A concentration of 2 mol dm-3 means there are 2 moles of solute dissolved in 1 dm3 of solution.

Molarity=Mole of solute(mol) / Volume of solution(dm3)

Example 2:

What is the molarity of a solution made when water is added to 0.2 mol of CaCl2 to make 100 cm³ of solution? [RAM: Ca = 40; Cl = 35.5]

Answer:

Number of mole of solute = 0.2 mol

Volume of solvent = 100 cm³ = 0.1 dm³

Molarity = Number of Mole / Volume

= 0.2mol / 0.1dm3

=2mol/dm3

Conversion of Concentration Unit

volumetric2

The chart above shows how to convert the units of concentration from g dm-3 to mol dm-3 and vice versa.

The molar mass of the solute is equal to the relative molecular mass of the solute.

Example 3:

The concentration of a Potassium chloride solution is 14.9 g dm-3. What is the molarity ( mol dm-3) of the solution? [Relative Atomic Mass: Cl = 35.5; K = 39]

Answer:

Relative Formula Mass of Potassium Chloride (KCl)

= 39 + 35.5 = 74.5

Molar Mass of Potassium Chloride = 74.5 g/mol

Molarity of Potassium Chloride

Molarity = ConcentrationMolar Mass=14.9gdm−374.5gmol−1=0.2mol/dm3

Molarity and Number of Moles

Number of mole of solute in a solution can be calculated by using the following formula

n=MV / 1000

where

n = number of mole of solute

M = molarity of the solution

V = volume of the solution in cm3

Example 4

How many moles of zinc sulphate is present in 200cm3 of 0.1 mol dm-3 zinc sulphate solution?

Answer:

Molarity, M = 0.1 mol dm-3

Voloume, V = 200cm3

n=MV / 1000

n= (0.1)(200) / 1000 = 0.02mol

Example 5

A solution of barium hydroxide have molarity 0.1 mol dm-3. What is the concentration of the solution in g dm-3? [Relative Atomic Mass: Ba = 137; O = 16; H = 1]

Answer:

Relative Formula Mass of barium hydrokxide, Ba(OH)2

= 137 + 2(16+1) = 171

Molar Mass of Potassium Chloride = 171 g/mol

Concentration =  Molarity × Molar Mass

=0.1mol/dm3 × 171gmol−1 = 17.1gmol−1

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