Volumetric Analysis: Acid-BaseChpt. 13

Quantitative Analysis: is analysis which involves investigating the quantities or amounts of materials present.

Gravimetric analysis (Chpt. 11) –

composition of substances

determined by careful weighing

Volumetric analysis (Chpt. 13) –

composition of substances determined

by reacting together volumes of solutions

REMEMBER:

A solution is a mixture of a solute and a solvent

A solvent: is a substance that dissolves other materials

A solute: is the substance that dissolves in the solvent

Concentration:

A concentrated solution contains a large amount of solute per litre of solution e.g. strong coffee

A dilute solution contains a small amount of solute per litre of solution e.g. weak coffee

Concentration

The concentration of a solution is the amount of

solute that is dissolved in a given volume of solution

There are several ways of expressing the concentration of a solution:

1)Percentage of solute – 3 forms

2)Parts per million (ppm)

3)Moles of solute per litre of solution (MOLARITY)

1. Percentage of solute

(This method of expressing concentration is usually used in many household solutions and in medicine)

There are 3 ways in which the percentage of solute in a solution is expressed:

a) Percentage weight per weight (w/w)

b) Percentage weight per volume (w/v)

c) Percentage volume per volume (v/v)

a) Percentage weight per weight (w/w):This is the number of grams of solute per 100g of solution e.g.

10% w/w NaCl → 10g of sodium chloride per 100g of solution

2% w/w Arnica ointment → 2g of arnica per 100g of

ointment

b) Percentage weight per volume (w/v):This is the number of grams of solute per 100cm3 of solution e.g.

10% w/v NaCl → 10g of sodium chloride per 100cm3 of solution

5% w/v NaCl → 5g of NaCl in 100cm3 of solution

*Note: usually the units are grams per litre (g/L) therefore you would have to bring it to a litre*

c) Percentage volume per volume (v/v):This is the number of cm3 of solute per 100cm3 of solution e.g.

5% v/v vinegar solution → 5cm3 of ethanoic acid per 100cm3 of vinegar

13% v/v wine solution → 13cm3 of ethanol (alcohol) per

100cm3 of wine

Complete the following table:

Concentration Unit

Example Meaning

Percentage weight per volume (w/v)

3% (w/v) NaCl solution

Percentage volume per volume (v/v)

3% (v/v) alcohol solution

Percentage weight per weight (w/w)

3% (w/w) sugar solution

*Note: DO NOT take down table

The following calculations involve working with

percentages!!!!

Make sure you understand the definitions!!!!!

Example 1:A solution contains 20g of potassium hydroxide in 1 litre of solution. Express the concentration of the solution in % (w/v).Solution:

Example 2:A bottle of vinegar contains 25cm3 ethanoic acid in 500cm3 of solution. Express the concentration of ethanoic acid in the solution in % (v/v)Solution:

Example 3:A solution contains 10g of sodium carbonate in 40g of solution. Express the concentration of the solution in % (w/w).Solution:

Example 4:The label on a bottle of wine indicates that the concentration of alcohol in the wine is 9% (v/v). What volume of alcohol is there in 250cm3 of the wine?Solution:

Try the following:

1.A sample of sea water has a mass of 1.3kg. On evaporating the water, 148g of salt was recovered from it. Express the concentration of the salt as % w/w.

2.Some illnesses can upset the salt balance in the body and it may be necessary to administer salt intravenously. The solution of salt that is injected is marked 0.85% w/v. What weight of sodium chloride is needed to make up 250cm3 of this solution?

2. Parts per million (ppm)This method of expressing the concentration of a solution is only used for very dilute solutions i.e. when dealing with very low concentrations of substances.

• This is the number of milligrams per litre (mgL-1)

*Note: 1 Litre of water has a mass of 1 million milligrams*

• So, can say 1mg/L = 1 mg per million mg = 1 ppm

• Example: the concentration of chlorine in water is 2 ppm this means there are 2 mg of chlorine per litre of water

Example 1:

1 gram = 1000 milligrams

3) Moles of solute per litre of solution (MOLARITY)

Remember:

One mole of a substance is the amount of that substance which contains 6 x 1023 particles (atoms, ions, molecules) of that substance

Mass of 1 mole of an element = Relative Atomic Mass in grams

e.g. 1 mole of Na = 23 g1 mole of Mg = 24g

The most important way of expressing the concentration of a solution is in terms of moles per litre of solution (molarity)Definitions:• The Molarity of a solution is the number of moles of solute per litre of solution

• A 1 molar solution is a solution which contains one mole of solute per litre of solution

also, - a solution which contains 2 moles of

solute in a litre of solution is said to be 2 molar (2M)

- a solution which contains 0.5 moles of solute in a litre of solution is said to be 0.5 molar (0.5M)

• Symbols Used: - M - mol/L or mol L-1

Remember:

No. of Moles of = Mass of Substance Substance Molar Mass

Concentration Examples:

-1 mol/L NaOH = 40g NaOH (Mr NaOH = 40) per litre of solution

- 2mol/L NaOH = 80g NaOH per litre of solution

- 0.5mol/L NaOH = 20g NaOH per litre of solution

- 0.1 M (decimolar) NaOH = 4g NaOH per litre of solution

Complete the following:

1M H2SO4 =

0.5M H2SO4 =

3M H2SO4 =

Calculations Involving Molarity

Three types:

1. Converting Molarity to Grams per Litre

2. Converting Grams per Litre(Volume) to Molarity

3. Calculation of number of Moles from Molarity and Volume

1. Converting Molarity to Grams per Litre

Example 1:What is the concentration in g/L of a 0.1 M H2SO4

Solution?

Example 2:How many grams of NaCl per litre are present in a solutionmarked 0.25 M NaCl.

Example 3:Calculate the concentration in grams per litre ofbench dilute sulphuric acid whose concentration Is 1.5mol/L

Concentration in g/L = Molar Mass x Molarity

2. Converting Grams per Litre(Volume) to Molarity

Molarity = Grams per Litre Molar Mass

Example 1: What is the molarity of a NaOH solution containing 4g of NaOH per litre?Solution:

Example 2:What is the molarity of a solution that contains 3.68g of NaOH per litre of solution?

Example 3:Calculate the concentration in moles per litre of a solution containing 45 grams of sulphuric acid per 250cm3 of solution.

3. Calculation of number of moles from molarity and volume

No. of moles = Volume(L) x Molarity

Example 1:How many moles are there in 250cm3 of 0.1 M HCl?Solution:

Example 2:How many moles of NaOH are present in 25cm3 of 0.55M NaOH

Example 3:How many moles of hydrochloric acid are present in 30cm3 of 0.2M HCl

Example 4:What mass of sodium hydroxide is contained in 25cm3 of a 0.75M solution of sodium hydroxide?

Example 5:What volume of 0.15M sodium hydroxide solution will contain 5 grams of sodium hydroxide?

Balanced Chemical EquationsA balanced equation tells you the amounts of substances that react together and the amounts of products formed.

Consider the following balanced equation for the reaction between hydrogen gas and oxygen gas to form water:

2H2 + O2 → 2H2O

This equation can be interpreted in a number of ways.

In terms of molecules:

2 molecules 1 molecules 1 molecule

In terms of Avogadro’s number of molecules:

2 x 6 x 1023 1 x 6 x 1023 2 x 6 x 1023

molecules molecules molecules

*REMEMBER: the amount of a substance which contains the Avogadro number of particles is called a mole of that substance

2H2 + O2 → 2H2O

2H2 + O2 → 2H2O

In terms of Moles of a substance:

2 moles 1 mole 2 moles

Further examples:

a) 2Mg + O2 → 2MgO 2 moles 1 mole 2 moles

b) CaCO3 → CaO + CO2 1 mole 1 mole 1 mole

c) CH4 + 2O2 → CO2 + 2H2O 1 mole 2 moles 1 mole 2 moles

2H2 + O2 → 2H2O

Reactions between a solution and a solid

In a number of chemical reactions solids react with solutions. You may be asked to calculate the mass of metal which reacts with a given volume of acid

Example 1:What mass of magnesium will react with 50cm3 of 0.5M H2SO4. The balanced equation for the reaction is:

Mg + H2SO4 → MgSO4 + H2

Example 1 Solution:

Example 2:Sodium carbonate, Na2 CO3, reacts with dilute hydrochloric acid according to the equation:

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

What volume of hydrochloric acid of concentration 0.75M would be needed to neutralise 7.5g of anhydrous sodium carbonate?

Example 2 Solution:

Concentration of Solutions

1 mole/L1 mole/500cm3

1 mole/250cm3

1 mole/100cm3

If in each volumetric flask one mole of solute is dissolved then as the volume becomes smaller, the concentration increases.

In the case of a coloured solution , as the concentration increases, the intensity of the colour also increases (see diagram pg. 148 book)

Dilution of SolutionsTo save space in our prep room we buy solutions in concentrated form, i.e. 18M HCl (18 mol L-1). We call these stock solutions

Definition:The process of adding more solvent to a solution is called dilution.

A typical dilution involves determining how much water must be added to an amount of stock solution to achieve a solution of the desired concentration

When a solution is diluted, more solvent is added but the quantity of solute is unchanged:

Moles of solute = Moles of solute

before dilution after dilution

Since the volume of the solution increases and the number of moles present remains the same, then the concentration of the solution must decrease

*Note: diluting a coloured solution results in a lightening of the colour of the solution i.e. colour intensity is proportional to concentration

Calculation of the Effect of Dilution on Concentration

MolarityDil x VolumeDil = MolarityConc. x VolumeConc.

M1 x V1 = M2 x V2

Example 1:If 20cm3 of a 3M hydrochloric acid solution is diluted to a volume of 1 L with water, what is the concentration of the diluted acid?Solution:

Example 2:What volume of a 2M sodium hydroxide solution is needed to make up 100 cm3 of a 0.1 M sodium hydroxide solution

Student Questions:Question 1: If 12cm3 of a 0.1M sodium hydroxide solution is diluted to a volume of 500cm3 with water, what is the concentration of the diluted solution?

Question 2:What volume of 1M NaOH solution is needed to make 300cm3 of 0.05M solution?

Standard SolutionsDefinition:A standard solution is a solution whose concentration is accurately known

e.g. a solution containing 10 grams of NaCl per litre is a standard solution

In the determination of the concentration of an acid a standard solution of an alkali is used and to determine the concentration of an alkali a standard acid would be used. However, before any determinations can be made a starting accurately standardised solution is required – from which to find the exact concentration of other solutions

A standard solution is prepared by weighing out a sample of solute, transferring it completely to a volumetric flask, and adding enough solvent (usually deionised water) to bring the volume up to the mark on the neck of the flask.

Due to the fact that many substances can not be obtained in a high degree of purity standard solutions of common laboratory acids and bases cannot be prepared directly e.g.:

- cannot weigh out 1 mole of sulphuric acid as it absorbs moisture from the air- cannot weigh out 1 mole of nitric acid as it is volatile- cannot weigh out 1 mole of iodine as it sublimes at room temperature

In order to make up standard solutions substances which can be obtained in a highly pure state and which are stable in air are required

Primary StandardDefinition:A primary standard is a substance which can be obtained in a stable, pure and soluble solid form so that it can be weighed out and dissolved in water to give a solution of accurately known concentration

Primary Standard Solution = Pure 100% Soluble Stable once made up

Examples of Primary Standards:

- Anhydrous sodium carbonate Na2CO3 - Sodium Chloride NaCl- Potassium Dichromate K2Cr2O7

Mandatory Expt. 13.1:

To prepare a standard solution of sodium carbonate

Note: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each step

Volumetric Analysis - TitrationsDefinitions:Standardise: means to find the concentration of a solution using titration

A Titration: is a laboratory procedure where a measured volume of one solution is added to a known volume of another solution until the reaction is complete.

Equivalence Point (End Point): the stage when the two solutions just react completely with each other

Theory regarding apparatus and method involved in carrying out a titration on handout

Mandatory Expt. 13.2:

To use a standard solution of sodium carbonate to standardise a given

hydrochloric acid solution

Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each stepNote 2: You must be able to carry out calculations on your results – see notes to follow*Need to calculate concentration of HCl in mol\L & g/L

Volumetric Analysis Calculations

1.Calculating the unknown concentration of a solution from titration data

2.Calculating the relative molecular mass and the amount of water of crystallisation in a compound from titration data.

1. Calculating the unknown concentration of a solution from titration dataIn straightforward titration calculations, where only the unknown concentration is required, the following formula can be used:

VA x MA = VB x MB

nA nB

VA = volume (cm3) of acid used VB = volume (cm3) of base usedMA = concentration of acid MB = concentration of base nA = no. of moles of acid in nB = no. of moles of base in balanced eqn for rxn balanced eqn for rxn

Example 1:In a titration, 25cm3 of a 0.05M sodium carbonate solution required 22cm3 of a hydrochloric acid solution for complete neutralisation. Calculate the concentration of the hydrochloric acid solution. The equation for the reaction is:

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g)

Example 1 Solution:

Example 2:Hydrochloric acid and sodium hydroxide react according to the equation:

HCl + NaOH NaCl + H2O

25cm3 of a sodium hydroxide solution was titrated against a 0.2M HCl solution. The average titration figure was 23.5cm3. Calculate the concentration of the sodium hydroxide solution in:

a) mol/Lb) g/L

Example 2 Solution:

Using results from your experiment calculate the concentration of the given hydrochloric acid solution in mol/L and g/L

*Note:

The first titration you performed was a rough titration which gave you an idea of where the end point is and so this result should be neglected.

The remaining two titration results should agree within 0.1cm3 of each other. The average of these results should be used in your calculation of the concentration of HCl.

Example 3:1.45g of sodium carbonate was dissolved in water and the solution was made up to 250cm3 in a volumetric flask. 25cm3 of this solution were titrated against a solution of hydrochloric acid using methyl orange as indicator. One rough and two accurate titrations were performed. The titration results and the equation for the reaction are given below. What is the concentration of the HCl solution in

a) mol/L b) g/L

Titration 1 2 3

Volume Acid (cm3) 19.8 19.5 19.6

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g)

Example 3 Solution:

Example 4:10.0g of impure sodium hydroxide were weighed out, dissolved in water and solution made up to 250cm3 in a volumetric flask. 25cm3 of this solution, on being titrated with 1.1M HCl, required 21.8cm3 of the acid for neutralisation. Calculate the % purity of the original sodium hydroxide.

Example 4 Solution:

Definition Secondary Standard: Make up a solution and then standardise this solution using a primary standard. This secondary standard can then be used to standardise other solutions e.g. HCl standardised and then used to standardise NaOH

Mandatory Expt. 13.3 (Ordinary Level):

A Hydrochloric Acid/Sodium Hydroxide titration and the use of this titration in making the salt

sodium chloride

Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each stepNote 2: You must be able to carry out calculations on your results

*Calculate concentration of NaOH in mol/L and g/L

More Difficult ProblemsIn a calculation involving a standard solution where more than just the unknown concentration is required, or where a solid is one of the reactants an alternative method from first principles should be used.

Example 1: In a titration, 25cm3 of a 0.12M NaOH solution required 24cm3 of a H2SO4 solution for complete neutralisation. Calculate:

i) the number of moles of NaOH consumedii) the number of moles of H2SO4 consumediii) the concentration of the H2SO4

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H20(l)

Example 1 Solution:

Example 2:What mass of magnesium will react with 20cm3 of a 0.09M hydrochloric acid solution? The equation for the reaction is:

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

Example 2 Solution:

Applications of Acid-Base Titrations

Mand. Expt.13.4: To determine the percentage of ethanoic acid in vinegar

Mand. Expt. 13.5: To determine the percentage of water of crystallisation in hydrated sodium carbonate (washing soda)

Mandatory Expt. 13.4:

To determine the percentage of ethanoic acid (acetic acid) in vinegar

Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each stepNote 3: You must understand the need for and the use of the dilution factorNote 2: You must be able to carry out calculations on your results

*Calculate concentration of ethanoic acid in the original vinegar in mol/L, g/L and % w/v

Example 1: A sample of vinegar was diluted from 25cm3 to 250cm3 with water. In a titration, 25cm3 of a 0.1M NaOH solution required 30cm3 of the diluted vinegar for complete neutralisation. Calculate the concentration of ethanoic acid (CH3 COOH) in the vinegar in:

i) mol L-1

ii) g/Liii) %w/v

The equation for the reaction is

CH3 COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

Example 1 Solution:

2. Calculating the amount of water of crystallisation in a compound and the relative molecular mass of the compound from the titration data

Definition:• Water of crystallisation: is water that is chemically bound in the compound, which gives rise to the crystalline form.

• Crystals that contain water of crystallisation are said to be hydrated.

Mandatory Expt. 13.5:

To determine the percentage of water of crystallisation in hydrated sodium carbonate (washing soda)

• Hydrated sodium carbonate has the formula Na2CO3.xH2O ( where x = no. of molecules of water of crystallisation present)

• The purpose of this experiment is to determine a value for x

Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each stepNote 2: You must be able to carry out calculations on your results

*Calculate the molar mass of hydrated sodium carbonate, the percentage of water of crystallisation and the value of X in the formula Na2CO3.xH2O

Mandatory Expt. 13.5:

To determine the percentage of water of crystallisation in hydrated sodium carbonate (washing soda)

Mandatory Expt. 13.5:NOTE:

• Washing soda is composed of large translucent crystals of Na2CO3.10H2O (form of hydrated sodium carbonate)

• It is important to realise that hydrated sodium carbonate gradually looses water of crystallisation over a period of time.

• Thus, if washing soda is used a lower value than the expected value of 10 for x in Na2CO3.xH2O will be obtained.

Example 1:Crystals of hydrated sodium carbonate (Na2CO3.xH2O) of mass 3.15g were dissolved in water and made up to 250cm3 in a volumetric flask. 25cm3 of this solution required 15cm3 of a 0.15M HCl solution for complete neutralisation. The equation for the reaction is:

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g)

Find the:i) concentration of the sodium carbonate solutionii) the molar mass of the sodium carbonate solutioniii) the value of x ion the formulaiv) the % of water of crystallisation in the

hydrated sodium carbonate

Na2CO3.xH2O

Example 1 Solution:

Student Questions:

Book: pg. 174 No’s 13.15 & 3.16