C06 concentration of solutions and volumetric analysis

Chapter 6

LEARNING OUTCOMES

Define the term standard solution Use results from volumetric analysis to calculate

the number of moles reacting, the mole ratio in which the reactants combine and the concentrate and mass concentration of reactants

Concentration of Solutions and Volumetric Analysis

Solute and Solvent A solution is made up of two parts:

solute + solvent = solution The solute is the substance dissolved in a solution. The solvent is the substance in which the solute has

dissolved. For example, in a beaker of sugar solution, the sugar is

the solute and the water is the solvent.

Chapter 6Concentration of Solutions and Volumetric Analysis

Concentrated or dilute?

Instead of using the words “strong” and “weak” to describe coffee, we can use the terms concentrated and dilute.

A concentrated solution will contain more solute dissolved in a certain volume of solution.

A dilute solution will contain less solute dissolved in the same volume of solution.

Do you like “strong” or “weak” coffee?


Concentration of solutions

In order to standardise the volume of the solution, chemists use 1 dm3 as the unit for measurement.

The concentration of a solution is the mass of solute dissolved in 1 dm3 of the solution.

1 dm3 = 1000 cm3


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Chapter 6


Concentrations can be expressed in two ways as:

grams/dm3 or g/dm3

moles/dm3 or mol/dm3


Concentration of solution in g/dm3

► Suppose a solution of sodium chloride is made by dissolving 58.5g of the salt in 1 dm3 of the solution.

The concentration of the sodium chloride solution is equal to: 58.5 g /dm3


Concentration of solution in mol/dm3

► Since 58.5 g of sodium chloride is equal to 1 mole of the salt,

The concentration of the solution is also equal to: 1 mol/dm3 (or 1 M).

The number of moles per dm3 of a solution is also called the molarity of the solution.


Formulae

Concentration = Mass of solute in grams in g/dm3 Volume of solution in dm3

Concentration = No. of moles of solute in mol/dm3 Volume of solution in dm3

Mass of solute = Volume of solution in dm3 x Concentration in g/dm3


Worked example 1A solution of sodium chloride is made by dissolving 11.7 g of sodium chloride in 500 cm3 of the solution. Find the concentration of the solution in (a) g/dm3, (b) mol/dm3.

Solution

Volume of solution = 500 cm3 = 500 = 0.5 dm3

1000

(a) Concentration = Mass in grams Volume in dm3

= 11.7 g = 23.4 g/dm3

0.5 dm3

(b) No. of moles = 11.7 g = 11.7 = 0.2 mol Mr of NaCl 58.5

Concentration = No. of moles Volume in dm3

= 0.2 mol = 0.4 mol/dm3

0.5 dm3

Chapter 6



A solution of magnesium chloride has a concentration of 23.75 g/dm3. (a) What is the concentration of the solution in mol/dm3?(b) If 200 cm3 of the solution is evaporated to dryness, what mass of salt can be obtained?

Solution(a) Number of moles of MgCl2 in 1 dm3 = 23 g/dm3 = 23.75 = 0.25 mol

Mr of MgCl2 95

Concentration = 0.25 mol = 0.25 mol/dm3

1 dm3

(b) Mass of solute = Concentration x Volume of solution

= 23.75 g/dm3 x 200 dm3

1000 = 4.75 g

Chapter 6

Worked example 2



A solution of sulphuric acid has a concentration of 0.25 mol/dm3 (a) What is the concentration of the solution in g/dm3 ?(b) What mass of acid will be contained in 250 cm3 of the solution?

SolutionMass of H2SO4 = 0.25 mol x Mr = 0.25 x 98 g = 24.5 g

(a) Concentration = Mass in grams Volume in dm3

= 24.5 g = 24.5 g/dm3

1 dm3

(b) Mass of acid = Concentration x Volume of solution

= 24.5 g/dm3 x 250 dm3

1000 = 6.125 g

Chapter 6

Worked example 3



25 cm3 of a solution of sulphuric acid of concentration 0.400 mol/dm3 is neutralised with a solution of sodium hydroxide of concentration 0.625 mol/dm3. What is the volume of sodium hydroxide solution required?

Equation of reaction: H2SO4 + 2NaOH Na2SO4 + 2H2O

From the equation, No. of moles of H2SO4 = 1 No. of moles of NaOH 2

Vol. of H2SO4 x Conc. of H2SO4 = 1 Vol. of NaOH x Conc. of NaOH 2

0.025 dm3 x 0.400 mol/dm3 = 1Vol. of NaOH x 0.625 mol/dm3 2

Vol. of NaOH = 2 x 0.025 x 0.400 = 0.032 dm3 0.625

= 32 cm3

Solution

Chapter 6

Worked example 4



Quick check1. A solution of calcium chloride (CaCl2) contains 37 g of the salt in 250 cm3 of

the solution. Find the concentration of the solution in (a) g/dm3, (b) mol/dm3.

2. 500 cm3 of a solution of sodium nitrate contains 14.7 g of the salt. (a) Find the concentration of the solution in mol/dm3. (b) If 100 cm3 of the solution is evaporated, how much salt can be obtained?

3. A solution of magnesium sulphate has a concentration of 0.25 mol/dm3. (a) What is the concentration of the solution in g/dm3? (b) What mass of magnesium sulphate is contained in 250 cm3 of the solution?

4. A solution of nitric acid has an unknown concentration. 25.0 cm3 of the acid is completely neutralised by 22.5 cm3 of potassium hydroxide solution of concentration 0.485 mol/dm3. What is the concentration of the nitric acid?

Solution


Solution to Quick check

1. (a) Concentration = 37 g = 148 g/ dm3

0.25 dm3

(b) No. of moles = 37 = 0.333 mol 111

Concentration = 0.333 = 1.33 mol/dm3 0.25 dm3

2. (a) No. of moles = 14.7 = 0.173 mol 85

Concentration = 0.173 = 0.346 mol/dm3

0.5 (b) Mass of salt = 0.1 x 0.346 x 85 = 2.94 g

3. (a) Concentration = (0.25 x 120) mol x 1 dm3

= 30 g/dm3

(b) Mass of magnesium sulphate = 0.250 x 30 = 7.5 g

4. Equation: HNO3 + KOH KNO3 + H2ONo. of moles of nitric acid = 1No. of moles of KOH 1 25.0 cm3 x Conc. of acid = 1

22.5 cm3 x 0.485 mol/dm3 1Conc. of nitric acid = 0.437 mol/dm3


Return

1. http://www.ausetute.com.au/concsols.html

2. http://dl.clackamas.edu/ch105-04/tableof.htm

3. http://en.wikipedia.org/wiki/Concentration

To Learn more about Concentrations of Solutions, click on the links below!


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Introduction Volumetric Analysis or VA is a method of finding out

the quantity of substance present in a solid or solution. It usually involves titrating a known solution, called a

standard solution, with an unknown solution. Based on the equation of reaction, calculations are

then made to find out the concentration of the unknown solution.


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Using a pipette A pipette is used to deliver an exact volume, usually

25.0 cm3 of solution into a conical flask. The solution in the titrating flask is called the titrate. Before using a pipette, it should be washed with tap

water, then rinsed with distilled water and finally with the liquid it is to be filled.

For safety reasons, a pipette filler is used to suck up the solution.

To use the pipette filler, first fit it to the top of the pipette, as shown in the diagram.

Squeeze valve 1 with right index finger and thumb and squeeze the bulb with the left palm to expel all the air in the bulb.

Then place the tip of the pipette below the surface of the liquid to be sucked up, and squeeze valve 2 to suck up the liquid.


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Using a pipette When the liquid rises to a level higher than the

mark, remove the tip of the pipette from the liquid.

Gently squeeze valve 3 to release the liquid slowly until the meniscus of the liquid is exactly at the mark of the pipette.

Now place the tip of the pipette into the titration flask, and squeeze valve 3 to release all the liquid into the flask.

When all the liquid in the pipette has run out, touch the tip of the pipette on the inside of the flask so that only a drop of liquid is left inside the tip of the pipette.


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Using a burette A burette is used to contain and measure the volume

of the liquid, called the titrant used in the titration. Before using a burette, it should be washed first with

tap water, then rinsed with distilled water and finally with the liquid (titrant) it is to be filled.

The liquid (titrant) in the burette must be released slowly, a few drops at a time, into the titration flask.

The readings must be taken accurate to 0.1 cm3. E.g. 24.0 cm3, not 24 cm3.

Make sure that the clip of the burette is tight and the liquid is not leaking.

Also make sure that the burette jet is filled with liquid, it must not contain any air bubbles.


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Using a burette The burette should be clamped

to the retort stand in a vertical position so that the reading will be accurate.

When reading the burette, the eye must be horizontal to the bottom of meniscus to avoid parallax error. (See diagram).


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Other tips on safety and accuracy When filling or reading the burette, it should be lowered to a

suitable height. Do not attempt to read it by climbing onto a stool. Make sure that the tip of the pipette is always kept below the

surface of the liquid when it is being filled, otherwise air bubbles will get into the pipette.

After filling a burette, the small funnel should be removed from the top of the burette, otherwise drops of liquid may run down into the burette during a titration and affect the reading.

The titration flask should be placed on a white tile or paper so that the colour of the indicator can be seen easily.

Use the wash bottle to wash down the insides of the conical flask towards the end of the titration.


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Use of Indicators

IndicatorIndicator Colour in Colour in acidsacids

Colour at end Colour at end pointpoint

Colour in alkalisColour in alkalis

Methyl orangeMethyl orange redred orangeorange yellowyellow

Screened methyl orangeScreened methyl orange redred greygrey greengreen

LitmusLitmus redred purplepurple blueblue

PhenolphthaleinPhenolphthalein colourlesscolourless pinkpink redred


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In a normal titration, candidates are usually advised to carry out at least one rough and two accurate titrations.

You should record your readings in a table like this.

Chapter 6

Titration readings

In general, you should carry out as many titrations as needed to obtain two or more consistent volumes.

If no consistent volumes are obtained, the average value should be calculated.

Titration numberTitration number 11 22 33 44

Final burette readingFinal burette reading/cm/cm33 25.225.2 24.824.8 33.333.3 24.924.9

Initial burette readingInitial burette reading/cm/cm33 0.00.0 0.00.0 7.47.4 0.10.1

Volume of NaOH usedVolume of NaOH used /cm /cm33 25.225.2 24.824.8 25.925.9 24.824.8

Best titration results (√)Best titration results (√)


√√ √√

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Chapter 6

Suppose that in an experiment, you are asked to find the concentration of a solution of sulphuric acid by titrating 25.0 cm3 of the acid against a standard solution of sodium hydroxide of concentration 0.100 mol/dm3, using phenolphthalein as an indicator.

First set up the apparatus as shown in the diagram and then carry out the titration, repeating it as many times as necessary to obtain a set of consistent results.

Titration of a known acid with an alkali


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Mean volume of sodium hydroxide used = 24.8 cm3

Chapter 6

Titration numberTitration number 11 22 33 44

Final burette readingFinal burette reading/cm/cm33 25.225.2 24.824.8 33.333.3 24.924.9

Initial burette readingInitial burette reading/cm/cm33 0.00.0 0.00.0 7.47.4 0.10.1

Volume of NaOH usedVolume of NaOH used /cm /cm33 25.225.2 24.824.8 25.925.9 24.824.8

Best titration results (√)Best titration results (√) √√ √√

Suppose the following readings are obtained:

Results


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Titration of a known acid with an alkali You are then asked to calculate the concentration of the sulphuric acid from

your results. The equation for the reaction is:

H2SO4 + 2NaOH Na2SO4 + 2H2O From the equation,

No. of moles of H2SO4 = 1No. of moles of NaOH 2Vol. of H2SO4 x Conc. of H2SO4 = 1Vol. of NaOH x Conc. of NaOH 2

25.0 x Conc. of H2SO4 = 1 24.8 x 0.100 mol/dm3 2

Therefore, Conc. of H2SO4 = 1 x 24.8 x 0.100 mol/dm3

2 x 25.0 = 0.0496 mol/dm3


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Va x Ma = x Vb x Mb y

where Ma, Mb are the concentrations of the acid and base and Va, Vb are the volumes of the acid and base used in the titration.

Chapter 6

In general if x moles of an acid reacts with y moles of a base, then No. of moles of acid = x

No. of moles of base y Vol. of acid x Conc. of acid = x

Vol. of base x Conc. of base y

Hence, it can be shown that :

Acid-base titration


2828

Aim: You are provided with a solution containing 5.00 g/dm3 of the acid H3XO4. You are to find the relative molecular mass of the acid by titrating 25.0 cm3 portions of the acid with the standard (0.100 mol/dm3) sodium hydroxide solution, and hence find the relative atomic mass of element X.

The equation for the reaction is:H3XO4 + 2NaOH Na2HXO4 + 2H2O

Titration No. 1 2 3 4

Final reading/ cm3 25.4 25.5 25.6 35.8

Initial reading/ cm3 0.0 0.0 0.0 10.0

Volume of NaOH/ cm3 25.4 25.5 25.6 25.8

Average volume of NaOH used = 25.5 cm3

Results:

Chapter 6

Titration of an unknown acid with an alkali


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From the equation,No. of moles of H3XO4 = 1No. of moles of NaOH 2Vol. of H3XO4 x Conc. of H3XO4 = 1Vol. of NaOH x Conc. of NaOH 2

25.0 x Conc. of H3XO4 = 1 25.5 x 0.100 mol/dm3 2Therefore, conc. of H3XO4 = 1 x 25.5 x 0.100 2 x 25.0 = 0.0510 mol/dm3

Since 1 dm3 of the acid contains 5.00 g of the acid, therefore 0.0510 x Mr of H3XO4 = 5.00 g Mr of H3XO4 = 5.00 = 98.0

0.0510 Calculate the relative atomic mass of X:

1x3 + X + 16x4 = 98 X = 98 – 67 = 31

Chapter 6

Titration of unknown acid with an alkali


3030

25.0 cm3 portions of hydrogen peroxide solution (H2O2) was titrated with standard (0.020 mol/dm3) potassium manganate(VII) solution.

Introduction:Oxidising agents can be titrated with reducing agents. Hydrogen peroxide is a reducing agent and can be titrated against acidified potassium manganate(VII), an oxidising agent.

No indicator is required for this titration as potassium manganate(VII) solution is purple in colour and is decolourised by the hydrogen peroxide solution when the reaction is complete.

Chapter 6

Titration of hydrogen peroxide with potassium manganate(VII) solution


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Volume of KMnO4 used = 25.2 cm3

A. Calculate the number of moles of KMnO4 used. No. of moles of KMnO4 = Volume in dm3 x Conc. = 25.2 dm3 x 0.020 mol/dm3

1000 = 0.000504 mol

Chapter 6

Results:Titration No. 1 2 3

Final reading/ cm3 25.1 25.2 25.2

Initial reading/ cm3 0.0 0.0 0.0

Volume of KMnO4/ cm3 25.1 25.2 25.2



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B. If 1 mole of KMnO4 reacts with 2.5 moles of H2O2, (a) Calculate the number of moles of H2O2 that react with the KMnO4. (b) Find the concentration of H2O2 solution. (a) No. of moles of H2O2 = 2.5 x 0.000504 mol = 0.00126 mol (b) Concentration = 0.00126 mol = 0.0504 mol/dm3

0.025 dm3

C. If 2 moles of H2O2 decompose during the reaction to give 1 mole of oxygen, calculate the volume of oxygen given off during the titration.No. of moles of O2 given off = 1 x 0.00126 mol 2 = 0.00063 molTherefore, Volume of O2 = 0.00063 x 24000 cm3 = 15.1 cm3

Chapter 6



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1. After washing the pipette, it should be rinsed with ________.(A) distilled water (B) the titrate(C) the titrant (D) tap water

2. After washing the titration flask, it should be rinsed with ________. (A) distilled water (B) the titrate(C) the titrant (D) tap water

3. After washing the burette, it should be rinsed with ________. (A) distilled water (B) the titrate(C) the titrant (D) tap water

4. A titration flask contains 25.0 cm3 of sodium hydroxide and a few drops of phenolphthalein as indicator. It is titrated against hydrochloric acid contained in a burette. What colour change would you observe when the end point is reached?(A) colourless to light pink (B) light pink to colourless(C) red to colourless (D) blue to pink

Solution

Chapter 6

Quick Check


3434

25.0 cm3 samples of sodium hydroxide solution are titrated against hydrochloric acid which has a concentration of 0.225 mol/dm3.

The results obtained are shown in the table below.


Final burette reading/ cm3 24.4 48.9 23.6 48.0

Initial burette reading/ cm3 0.0 24.4 0.0 23.6

Volume of HCl/ cm3

Best titration result (√)

Chapter 6

Solution

Quick Check


(a) Complete the table above.

(b) Calculate the concentration of the sodium hydroxide solution.

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1. http://www.tele.ed.nom.br/buret.html

2. http://www.chem.ubc.ca/courseware/154/tutorials/exp6A/

To learn more about titration, click on the links below!


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1. After washing the pipette, it should be rinsed with(A) distilled water (B) the titrate(C) the titrant (D) tap water

2. After washing the titration flask, it should be rinsed with(A) distilled water (B) the titrate(C) the titrant (D) tap water

3. After washing the burette, it should be rinsed with(A) distilled water (B) the titrate(C) the titrant (D) tap water

4. A titration flask contains 25.0 cm3 of sodium hydroxide and a few drops of phenolphthalein as indicator. It is titrated against hydrochloric acid contained in a burette. What colour change would you observe when the end point is reached?(A) colourless to light pink (B) light pink to colourless(C) red to colourless (D) blue to pink

Return

Chapter 6



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5.

Average volume of HCl used = 24.4 cm3

Equation: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

25.0 x Conc. of NaOH = 124.4 x 0.225 mol/dm3 1Conc. of NaOH = 24.4 x 0.225 mol/dm3

25.0

= 0.220 mol/dm3

Chapter 6

Return



Final burette reading/ cm3 24.4 48.9 23.6 48.0

Initial burette reading/ cm3 0.0 24.4 0.0 23.6

Volume of HCl/ cm3 24.4 24.5 23.6 24.4

Best titration result (√) √ √


ReferencesReferences

Chemistry for CSEC Examinations by Mike Taylor and Tania Chung

Longman Chemistry for CSEC by Jim Clark and Ray Oliver

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Science

C06 concentration of solutions and volumetric analysis