THE MOLE, CONCENTRATION AND VOLUMETRIC ANALYSIS



VOLUMETRIC ANALYSIS

We have already established that the number of moles of a substance can be obtained by dividing the mass of a substance by the mass of one mole of it.

Often in chemistry chemical reactions are carried out in solution and we need to know the number of moles of the substances present so that the reaction will take place in the correct reacting quantities as determined by the equation for the reaction.

Concentration means the number of moles dissolved in 1000 cm3 (1 dm3) of a solvent, usually water.

For example if we dissolve 58.5g of sodium chloride in 1dm3 of water then we have a solution whose concentration is 1mole per dm3 (1mol dm-3)

If we dissolve 29.25g of sodium chloride in 1dm3 of water, what would be the concentration in mol dm-3?

If we dissolved 5.85g of sodium chloride in 1 dm3 of water what would be the concentration?

If we dissolve 5.85g of sodium chloride in 100cm3 of water what would be the concentration?

If we dissolve 58.5g of sodium chloride in 2dm3 of water what would be the concentration?

So, knowing the number of moles of a substance dissolved in a given volume we can scale up or down to find the number of moles present in 1 dm3, which is the concentration.

Sometimes we might express concentration in terms of the mass of a substance dissolved in 1 dm3. For example, the concentration of a solution of sodium chloride is 0.25 moldm-3, what is its concentration in terms of mass?

1 mole of sodium chloride is 58.5g, therefore 0.25 mole is 14. 625g therefore the mass concentration is 14.625g dm-3

The concentration of a solution of sulphuric acid (H2 SO4 is 0.1 moldm-3, what is its mass concentration?

Often, given the concentration of a solution, we may want to determine the number of moles present in a given volume of the solution.

A solution of sulphuric acid has a concentration of 0.5 moldm-3 how many moles are present in 25cm3 ?

How many moles are present in 1000cm3?

How many moles are present in 1 cm3?

So, how many moles are present in 25 cm3 ?

Use this method to answer the following questions:

A solution of nitric acid has a concentration of 0.1 moldm-3 calculate the number of moles present in 15 cm3

A solution of sodium hydroxide (NaOH) has a mass concentration of 2.0g dm-3, calculate the number of moles present in 25 cm3

How many moles of a 0.75 mol dm-3 solution of calcium hydroxide will be present in 35 cm3 of solution?

We often need to perform these types of calculations when we carry out reactions in solution, particularly if we want to calculate the concentration of one of the reactants or products.

Often we would perform these reactions via a titration in a process known as volumetric analysis.

In this process a solution of a reactant, A, whose concentration is usually known, is added from an instrument called a burette (see below) to an accurately measured volume of another reactant, B whose concentration is usually unknown. Some kind of indicator previously added to B or another measuring instrument immersed in B tells us when enough of A has been added to completely react with B.

An indicator is a solution whose colour changes when enough of one reactant has been added to completely react with the other reactant.

The entire process is shown below:

a) Accurate delivery of a volume of B to a conical flask using a pipette

Pipette containing B

[pic] Conical flask

[pic]

b) Accurate delivery of reactant A to reactant B to which a few drops of indicator have been added

Reactant A in a burette

[pic]

[pic]

Reactant B + indicator

Supposing we wanted to determine the concentration of B by reacting it with A, whose concentration was known.

• The volume of A needed to react with B would be determined.

• The number of moles of A needed to react with B would be worked out.

• The equation for the reaction would be consulted to see in what mole ratio A reacts with B.

• The number of moles of B would be worked out.

• The concentration of B would be determined.

We can represent this process by a U- turn calculation chart. The U- turn template for calculations is a series of operation boxes arranged like a flow chart in the shape of a letter U. The rules of use of the template are:

• The top left or right hand box contains the information you have been given.

• The operation boxes that follow down a leg of the U are operations you perform on the information you have been given.

• The operation box at the bottom of the U contains the equation and mole ratio of the substances you are interested in.

• The operation boxes on the opposite leg of the U help you to work out the mass or moles of the substance you need to produce an answer for.

We can use the U- turn chart to perform volumetric calculations for example complete the U- turn chart for the following reaction.

20 cm3 of a solution of sulphuric acid whose concentration was 0.2 mol dm-3 was required to completely neutralise 25 cm3 of a solution of sodium hydroxide. Calculate the concentration of sodium hydroxide:

The U- turn calculation can be extended to have as many steps as you want, for example

Use a similar U- turn template as the one above but fill it in yourself for the following problem:

30cm3 of aqueous ammonia was neutralised by 14 cm3 of a solution of sulphuric acid whose concentration was 0.2 mol dm-3. Calculate the concentration of the ammonia solution

We will use U- turn calculations to solve other types of chemical problems to do with the mole as well. We will meet these in the next chapter.

HOMEWORK

Use the U- turn template above to calculate the following:

The concentration of:

1) A solution of calcium hydroxide, when 20 cm3 of it was neutralised by 20 cm3 of a solution of hydrochloric acid whose concentration was 0.5 moldm-3.

2) A solution of sulphuric acid, when 25 cm3 of it was used to neutralise a 15 cm3 of a solution of sodium hydroxide whose concentration was 0.4 mol dm-3.

3) Consider the following reaction when a solution of 2.0mol dm-3 sodium hydroxide was reacted with hydrochloric acid. Hydrochloric acid was added from a burette to 25 cm3 of the sodium hydroxide in 1 cm3 increments and the temperature measured after each 1 cm3 had been added. The results are shown below:

|VOLUME OF HCl / cm3 | TEMPERATURE/C |

|1.0 |22.0 |

|2.0 |26.5 |

|3.0 |30.1 |

|4.0 |33.9 |

|5.0 |37.2 |

|6.0 |35.3 |

|7.0 |32.3 |

|8.0 |29.1 |

|9.0 |26.0 |

|10.0 |23.2 |

|11.0 |20.1 |

a) Use a graphical method to determine the volume of HCl required to neutralise the NaOH.

b) Use a U- turn calculation and template to determine the concentration of the HCl.

4) A small corrie (upland lake) had a pH of 3 as a result of acid deposition which can be regarded as H2SO4 . The concentration of hydrogen ions was 10-3 mol dm-3.

• What would be the concentration of the sulphuric acid?

• The lake is approximately circular of diameter 200m and has an average depth of 12 m.

• In order to restore the lake to pH 7 it has been decided to drop calcium hydroxide from a light aircraft to neutralise the sulphuric acid.

• Write a balanced equation for the reaction.

• Use a U- turn calculation and template to calculate the mass of calcium hydroxide that would need to be used to restore the lake to pH 7.

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Concentration of A is known

Volume of A added to completely react with B is determined

Number of moles of A added is worked out

Equation consulted

nA + y B( zC

Ratio of A:B = n:y

Work out the number of moles of B = y/n x number of moles of A

This number of moles is present in the volume of B

Concentration of B worked out

___moles NaOH present in ___cm3 NaOH used in the titration.

Volume of sulphuric acid added to completely react with sodium hydroxide =___ cm3

Number of moles of NaOH reacted = ___x __moles H2SO4; this is

___moles

Number of moles of sulphuric acid present in this volume = ___moles

Equation consulted

H2SO4 + 2NaOH( Na2 SO4 + 2H2O

Ratio = 1mole H2SO4 : __ mole NaOH

Concentration of sulphuric acid = ___ mol dm-3

Therefore the concentration of NaOH = ___mol dm-3

Therefore the number of moles NaOH in 1 cm3= ____moles

Concentration of sulphuric acid = ___ mol dm-3

___moles NaOH present in 25 cm3 NaOH used in the titration.

Number of moles H2SO4 present in 1 cm3 = ____ moles

Number of moles of NaOH reacted = ___x __moles H2SO4; this is

___moles

Number of moles of sulphuric acid present in volume added (20cm3) = _____ moles

Equation consulted

H2SO4 + 2NaOH( Na2 SO4 + 2H2O

Ratio = 1mole H2SO4 : __ mole NaOH

Number of moles NaOH in 1000 cm3 = ______moles. So, concentration = ___mol dm-3

Equation consulted:

H2SO4 + 2NH3( (NH4)2 SO4

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