22precise ordinary chemistry practicals final copy Copy

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MUDUKU IVAN (Bsci. Ed,MUK)

Mak.academia.edu/mudukuivan

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TABLE OF CONTENTS

1.

VOLUMETRIC ANALYSIS………...7

1.1. Details of apparatus used for titration……….7

1.2. Solutions……….8

1.3. Acid-base titration………10

1.4. Practical schedules……….11

1.4.1. Standardisation of sodium hydroxide………11

1.4.2. Standardisation of sodium carbonate……….………13

1.4.3. Determination of RAM of M in M2CO3………...…….15

1.4.4. Determination of atomic mass of X in acid………..…….17

1.4.5. Determination of the number of moles of water of crystallisation………19

1.4.6. Determination of stoichiometry of the reaction between acid HnXand sodium hydroxide……….………..20

1.4.7. Determination of basicity of acid HnX………..………22

1.4.8. Determination of value of x in acid H2C2O4.xH2O………24

1.4.9. Determination of the reaction ratio………25

1.4.10.Determination of value of R in RNH2……….…26

1.4.11.Determination of percentage purity………,..….28

1.4.12.Determination of percentage purity……….30

1.4.13.Value of n in Na2CO3.nH2O……….……31

1.4.14.Number of moles of water of crystallisation in Y.nH2O……….….33

1.4.15.Determination of composition of NaCl in a NaCl and anhydrous sodium carbonate mixture………..34

1.4.16.Worked examples………..…………35

2.

QUALITATIVE ANALYSIS……….41

2.1.

Preliminary tests………..………42

2.2.

Identification of cations……….………..45

2.3.

Activities for students………..46

2.4.

Explanations and equations……….50

2.5.

Confirmatory tests……….………..51

2.6.

Identification of anions……….………..52

2.7.

Action of hydrogen peroxide………..………55

2.8.

Displacement reactions………..………….55

2.9.

Worked examples………56

2.10.

Practical schedules………..58

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3.1. Rate of reaction in terms of product accumulation……….70

3.2. Rate of reaction in terms of disappearance of reactants……….………71

3.3. Activities for students………72

3.4. Factors that affect rate of reaction……….…75

3.4.1. Effect of temperature……….75

3.4.2. Effect of concentration………..………78

4.

THERMO-CHEMISTRY

……….82

4.1. Practical schedules ………82

4.2. Worked example 1……….………84

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DECLARATION

Appreciation goes Mr. Nagaya Henry, Mr. muhama Geofrey, Ms Naume Mr. Buyi Dick, Ms Akello, Mr. mukatabala alloycius, Mr. issat Ignatius, Mr. okitoi, Ms Abinyo Racheal, Mr Napokooli Isaac, Mr. mulegi john, Ms amole, Mr. wozemba, for the tremendous, endless effort in shaping and guiding me academic pursuit

Appreciation to my colleagues, woniala ivan, muzaki Judith, komakech, osbert, Nambozo Racheal, Abraham, Christopher woyeya for their company and guidance

Thanks to the family of Mr. massa John, my sisters suzan, junicate, joan, veron, my brothers, keney, victor, Andrew, timothy for financial and spiritual assistance.

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1.

VOLUMETRIC ANALYSIS

Volumetric analysis is the technique of finding the concentration of the solution. In this technique the solution of the known concentration is added repeatedly the one whose concentration is to be determined until there is just enough of it to neutralise the other of the unknown concentration. The method is called titration. During titration, the volume of standard solution that completely reacts with the measured volume of the solution under the analysis is recorded. The titration is repeated to obtain consistent titres. From consistent titre values, an average of volume of the standard solution is calculated. By us appropriate stoichiometry (reaction ratio) obtained from the balanced equation between the reacting substances in the experiment, the concentration of the solution under analysis can be calculated.

Volumetric analysis is performed using the following important laboratory apparatus. These include burette, pipette, conical flask, beakers, retort stand and clamp, volumetric flasks, weighing balance.

1.1.

DETAILS OF APPARATUS USED DURING

TITRATION

a) Burette.

The burette is designed to deliver variable volumes of liquids. It is graduated from 0 to 50cm3 in units of 0.10cm3. Before use, the burette is rinsed with water, then a little of solution to be put in it. It is then clamped vertically and filled carefully with solution beyond the zero mark. The meniscus of the solution is then adjusted by draining it by opening the tap. Read the lower meniscus to get the accurate volume of the solution.

b) Pipette.

A pipette is designed to deliver a fixed or definite volume of a liquid such a 50, 25, 10, 5cm3. Before use, rinse the pipette with water followed by a little of the solution. The pipette is then filled by sucking the solution into it beyond the calibration mark. Close the mouth using the first finger, then adjust the meniscus to the mark by allowing the solution to flow down slowly. Then allow the accurately measured volume of the solution to flow into the conical flask under gravity. Do not blow; touch the side of the conical flask to allow the last drops to flow into the conical flask.

c) Conical flask.

The conical flask is where the pipetted solution is put; the flask must be cleaned with water. To the solution in the conical flask is added appropriate number of drops of indicator solution given. Mainly 2-3 drops are added. The indicator solutions include methyl orange, phenolphthalein indicator, bromocresol blue, bromocresol green, starch indicator, etc.

d) Beaker.

A beaker is designed to measure the varying volumes of solutions. It should be cleaned with water before use.

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These two are used together to clamp the burette as shown below.

1.2.

SOLUTIONS



Standard solution.

A standard solution is one whose concentration is accurately known.

The term concentration refers to the amount of solute dissolved in a specific volume of a solvent. One way of expressing the concentration of a solution is to state the number of moles of solute dissolved in 1000cm3 or 1litre of the solution. When one mole of a solute is dissolved in solvent (water) to make 1000cm3_{of the solution, the solution is said to be a molar solution written as 1M or}

1mol dm-3. This is sometimes refered to as molarity.

A solution containing 5 moles of the solute in 1000cm3 or 1dm3 has a concentration of 5mol dm-3 simply written as 5M. Similarly a solution containing 0.3 moles of sulphuric acid in 1000cm3 has a concentration of 0.3M, written as 0.3M sulphuric acid.

Qn. What would be the concentration/molarity of a solution containing 5moles in 2500cm3 of solution?

... ... ... ... ... ...

a.

Concentration in grams per litre or grams per cubic decimetres

The concentration can also be expressed in grams per litre or grams per cubic decimetres written as gl-1 of gdm-3 respectively. When 4g of sodium carbonate are dissolved to make 1 litre of solution, we say that the solution has a concentration of 4gl-1 or 4gdm-3 of sodium carbonate.

Example

53.5g of sodium chloride were dissolved to make 2000cm3 of the solution, calculate the concentration of sodium chloride in grams per litre. (Na = 23, Cl = 35.5)

Solution

1000cm3 is the standard volume equivalent to 1dm3 or 1litre 1cm3 will be equal to 1 will be equal to 𝟏

𝟏𝟎𝟎𝟎 2000cm3 will be equal to 2000 X 𝟏

𝟏𝟎𝟎𝟎 = 2dm

3

2dm3 of solution contains 53.5g of NaCl 1dm3 of solution will contain 𝟓𝟑.𝟓

𝟐 g of NaCl =26.75gdm-3

Alternatively,

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R.F.M of NaCl = 23 + 35.5 = 58.5 Therefore molar mass of NaCl = 58.5g This implies that 58.5g of NaCl contain 1 mole

1g of NaCl will contain 𝟏

𝟓𝟖.𝟓 moles 53.5g of NaCl will contain 53.5 X 𝟏

𝟓𝟖.𝟓 moles = 0.9145moles 2dm3 of solution contains 0.9145moles

1dm3 of solution will contain 0.9145⁄₂ moles = 0.4573 moldm-3 = 0.4573M Therefore molarity = 0.4573M

1 mole of NaCl weighs 58.5g Therefore 0.4573moles will weigh 𝟎.𝟒𝟓𝟕𝟑 𝐗 𝟓𝟖.𝟓

𝟏 = 26.75gdm

-3

b.

Calculating the amount of substance in a given volume of solution

As stated before, the concentration of solution is quoted in terms of mol dm-3 or gl-1. The quoted figure is called molarity (M).

Given a solution of sodium hydroxide with molarity of 0.5M, it means that every 1000cm3 of that solution contains 0.5moles of sodium hydroxide. If there is uniform distribution of the solute

(sodium hydroxide) in the solution and samples of different volumes of this solution are drawn from it.

Qn. How can we find the amount of sodium hydroxide in each of the drawn samples?

Let’s consider the example below,

If solution samples of volumes 5.00, 10.00, 25.00cm3 are drawn from a 0.5M solution of sodium hydroxide, calculate the number of moles of sodium hydroxide in each of the samples.

For 5.00cm3

0.5M means that 1000cm3of solution contains 0.5moles of sodium hydroxide 1.00cm3 of solution will contain 0.5

1000 moles Therefore 5.00cm3_{will contain} 0.5

1000 X 5.00 moles = 0.0025moles

For 10.00cm3

0.5M means that 1000cm3of solution contains 0.5moles of sodium hydroxide 1.00cm3 of solution will contain 0.5

1000 moles Therefore 10.00cm3 will contain 0.5

1000 X 10.00 moles = 0.005moles

For 25.00cm3

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1.00cm3 of solution will contain 0.5

1000 moles Therefore 25.00cm3 will contain 0.5

1000 X 25.00 moles = 0.0125moles

1.3.

ACID-BASE TITRATION

Acid-base titration is the titration between an acid and an alkali. The reaction taking place is in this titration is called neutralisation reaction. The reaction is represented by the equation below,

H+

(aq)+ OH-(aq) H2O(l)

The progress of the reaction is determined by adding an indicator to the solution in the conical flask. At the end point, the indicator shows a sharp colour change. Students should take care when the end point is about to be reached. Add the solution from the burette drop by drop to avoid over shooting. At this level, the indicator solutions used for titration are phenolphthalein and methyl orange indicators. They show the following colours in acid and base media.

Indicator Colour in acid medium Colour in alkali medium

Phenolphthalein Colourless pink

Methyl orange Red yellow

A.

TITRATION PROCEDURE

Given a solution for titration and a set of apparatus to be used,

a) Use a pipette to deliver appropriate volume of solution e.g 25.0, 20.0, 10.0, 5.0cm3 into a clean conical flask. The pipette must be rinsed water and the solution to be pipette.

b) Rinse the burette with some water followed by a little of the solution to be put in the burette. Fill the burette with the solution up to slightly beyond the zero mark and allow some of the solution to flow into the tip of the burette. Read the initial burette readings as V1cm3

c) Arrange the apparatus as shown below, run the solution from the burette drop wise. Use your left hand to open the burette tap and your right hand to swirl the conical flask (unless you are left handed). Stop when the indicator just changes colour. This is the end-point of the solution.

d) Read the final burette reading as V2cm3. Subtract the initial burette reading from final burette reading

to obtain the volume of solution in the burette that completely reacts with a volume of standard solution in the flask, ie V = V2 – V1.

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B. FILLING THE TABLE

Volume of pipette = 25.0cm3

Experiment 1 2 3 4

Final burette reading (cm3₎ _25.40 _27.00 _44.90 _36.80

Initial burette reading (cm3₎ _0.00 _2.20 _20.00 _12.00

Volume of solution used (cm3₎ _25.40 _24.80 _24.90 _24.80

Note the following,

 The table must be filled to two decimal places with the last figure being a zero.  The subtraction must be correct

 Take values which a close to one another to calculate the average. This enhances precision; students should know that precision does not mean accuracy. Precision is simply the closeness of the measured values to one another whereas accuracy is the degree of agreement of the measured value to the accepted or accurate value.

 The pipette volume must be indicated. Since the pipette is used to deliver a fixed volume. The volume is appropriately recorded to one decimal place; however there is no penalty for recording to two decimal places.

C.

PRACTICAL SHEDULES

a)

Practical 1

Standardisation of sodium hydroxide with 0.1M hydrochloric acid

You are provided with the following

 1 burette (50ml)

 2 conical flask

 1 pipette (20 or 25ml)

 Phenolphthalein indicator

 Solution of BA1 which is 0.1M hydrochloric acid solution

 Solution of BA2 which is sodium hydroxide solution

You are required to determine the concentration of solution of BA2 in

i. Mol dm-3 or M ii. gdm-3 or gl-1

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Pipette 25.0 0r 20.0cm3 of BA2 into a clean conical flask. Add two drops of phenolphthalein indicator and swirl. Titrate the mixture with BA1 from the burette. Repeat the titration to obtain consistent results. Record your values in the table below.

Volume of pipette used...cm3

Experiment 1 2 3 4

Final burette reading (cm3₎

Initial burette reading (cm3₎

Volume of BA1 used (cm3₎

Values used to calculate the average volume of BA1

... ... ...

Average volume of BA1

... ... ... ...

Calculate,

i. Number of moles of BA1 that reacted.

... ... ... ... ... ... ...

ii. The number of moles of BA2 that reacted.

... ... ... ... ... ...

iii. The concentration of BA2 in mol dm-3

(12)

... ... ... ... ...

iv. The concentration of BA2 that reacted in gl-1

... ... ... ... ... ... ...

Teachers comment and evaluation

... ... ...

...

b)

Practical 2

Standardisation of hydrochloric acid using sodium carbonate solution

You are provided with the following,

 BA3 which is approximately 0.2M hydrochloric acid

 BA4 which is made by dissolving 10.6g of sodium carbonate to make 1litre of solution

 Methyl orange indicator

 1 burette

 1 pipette

 2 conical flasks

 Stand and clamp

You are required to determine the concentration of BA3 in i) Mol dm-3

ii) gl-1

Procedure,

Pipette 25.0cm3_{of BA4 into a clean conical flask. Add 2 drops of methyl orange indicator}

(13)

Volume of the pipette used...cm3

Experiment 1 2 3 4

Initial burette reading (cm3₎

... ...

... ... ...

Calculate

i) number of moles of BA4 that reacted (Na = 23, C = 12, O = 16)

... ... ... ... ... ... ... ...

ii) the number of moles of BA3 that reacted

... ... ... ... ... ... ... ...

iii) the concentration of BA3 in mol dm-3

(14)

... ...

iv) the concentration of BA3 in gl-1(H = 1, Cl = 35.5)

... ... ... ... ... ... ... ... ...

Teacher’s comment and evaluation

... ... ... ...

c)

practical 3

Determination of relative atomic mass of metal M in M

2

CO

3

You are provided with the following solutions

 BA1 which is 0.2M hydrochloric acid solution

 BA2 which is a salt solution made by dissolving 10.6g 0f M2CO3 in 1litre of solution

You are required to determine the relative atomic mass of metal M in M2CO3

Procedure,

Pipette 25ml of BA2 into cleans a conical flask. Add two drops of methyl orange indicator to it and swirl. Titrate the mixture with BA1 from the burette. Repeat the titration to obtain consistent titres. Record your results in the table below,

Volume...cm3

Experiment 1 2 3 4

Initial burette reading (cm3)

(15)

... ...

The average volume of BA1

... ... ...

Calculate,

a) i) The number of moles of BA1 that reacted

... ... ... ... ... ... iii)write the equation of reaction between BA1 and BA2

... ...

iv)write the mole ratio of the reaction

... ...

b) The number of moles of BA2 that reacted

... ... ... ... ... ... ...

c) Calculate the molarity of BA2

... ... ... ... ... ... ...

d) Calculate the relative formula mass of M2CO3

(16)

e) Determine the relative atomic mass of M( C = 12, O = 16)

... ... ... ... ... ... ... ...

Teachers comment and evaluation

... ... ...

...

d)

Practical 4

Determination of atomic mass of X in an acid

BA3 which is a solution containing 1.7g of OH- _{ions per litre}

BA4 which is a dibasic acid H2X containing 9.8gl-1

You are required to determine the atomic mass of X in the acid (H = 1, O = 16)

Procedure,

Pipette 25 or 20ml of BA3 into a clean conical flask. Add 3 drops of phenolphthalein indicator. Titrate the mixture with BA4 from the burette. Repeat the titration to obtain consistent results. Record your values in the table below,

Initial burette reading (cm3)

Volume of BA4 used(cm3₎

... ...

(17)

Write the ionic equation for the reaction between BA3 and BA4

... ...

i) Calculate the molarity of BA3

... ... ... ... ... ... ...

i) Calculate the number of moles of BA3 that reacted

... ... ... ... ... ...

ii) Calculate the number of moles of BA4 that reacted

... ... ... ... ... ...

iii)Determine the molarity of BA4

... ... ... ... ... ...

iv)Calculate the relative atomic mass of X in the acid

... ... ... ... ... ...

Teachers comment and evaluation

(18)

... ...

e)

Practical 5

Determination of the number of moles of water of crystallisation

BA5 which is a solution containing 12.6g of dibasic acid Y.nH2O per litre of solution

BA6 which is made by dissolving 8g of sodium hydroxide to make 1 litre of the solution (1 mole of Y reacts with 2 moles of sodium hydroxide, molar mass of Y = 90)

Procedure,

Pipette 25 or 20ml of BA5 into a clean conical flask. Add 3 drops of phenolphthalein indicator, then titrate with BA6 from the burette. Repeat the titration to obtain consistent results. Record your values in the table below.

Initial burette reading (cm3)

... ...

... ... ...

a) Calculate,

i) Molarity of sodium hydroxide.

... ... ... ... ... ...

(19)

... ... ... ... ... ...

iii)The number of moles of Y.nH2O that reacted

... ... ... ... ... ...

iv)The molarity of Y.nH2O

... ... ... ... ... ...

v) Determine the value of n in Y.nH2O

... ... ... ... ... ...

Teachers comment and evaluation

... ... ... ...

f)

Practical 6

Determination of the stoichiometry of the reaction between the acid H

n

A and

sodium hydroxide

 BA1 which is 0.3M sodium hydroxide solution

 Ba2 which is 0.2M solution of acid HnA

(20)

Pipette 25 or 20ml of BA1 into a clean conical flask. Add 2 drops of methyl orange indicator. Titrate the mixture with BA2 from the burette. Repeat the titration to obtain consistent results. Record your values in the table below,

Experiment 1 2 3 4

... ...

... ... ...

a) Calculate the,

i) Number of moles of sodium hydroxide that reacted.

... ... ... ... ... ...

ii) The number of moles of HnAthat reacted.

... ... ... ... ... ...

b) Determine the mole ratio of the acid HnA to sodium hydroxide.

(21)

... ...

c) The value of n in HnA

... ... ... ... ... ...

Write the equation for the reaction.

... ...

Teachers comment and evaluation

... ... ... ...

g)

Practical 7

Determining the basicity of an acid H

n

X

 BA1 which is 0.2 moles per litre of HnX

 BA2 which is 0.5M solution of sodium hydroxide. You are required to determine the basicity, n of the acid

Procedure,

Pipette 25 or 20ml of BA2 into a clean conical flask. Add 3 drops of methyl orange indicator. Titrate the mixture with BA1 from the burette. Repeat the procedure to obtain consistent titres. Record the results in the table below,

Final burette reading (cm3)

(22)

... ...

... ... ...

a) Calculate the,

I. Number of moles of sodium hydroxide that reacted

... ... ... ... ... ...

II. The number of moles of HnX that reacted

... ... ... ... ... ...

III. Determine the mole ratio of the acid HnX to sodium hydroxide

... ... ... ... ... ...

IV. The basicity, n of the acid HnX

... ... ... ... ... ...

Write an ionic equation of the reaction between the acid and base (sodium hydroxide)

... ...

Teachers comment and evaluation

(23)

h)

Practical 8

You are provided with the following:

BA1 which is a solution containing 12.6gl- of a dibasic acid of formula H2C2O4.xH2O.

BA2 which is a 0.1M sodium hydroxide solution. You are required to determine the value of x

Procedure:

Pipette 20cm3 ( or 25cm3) of BA2 into a clean conical flask and add 2-3 drops of methyl orange

indicator. Titrate this mixture with BA1 from the burette until the colour changes from yellow to orange. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

Volume of pipette used………..cm3

Experiment 1 2 3

Final burette reading (cm3) Initial burette reading (cm3₎

Volume of BA1 used (cm3)

Volumes of BA1 used for calculating the average………cm3

Average volume of BA1 used……….. cm3

Questions:

(a)Calculate the:-

(i) number of moles of BA2 reacted.

……… ……… ……… ………

(ii) number of moles of BA1 that reacted with BA2.

……… ……… ………...

(iii) The concentration in moll- of BA1 .

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……… ……… ………

(iv) The R.M.M of the acid and hence the value of x.

……… ……… ……… ……… ……… ……… ……… …

i)

practical 9

BA1 which is a solution of acid HnX of concentration 0.625M.

BA2 which is a 0.5M sodium hydroxide.

You are required to determine the reaction ratio for BA1 and BA2

Procedure:

Pipette 20cm3 (or 25cm3) of sodium hydroxide solution into a clean conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this mixture with BA1 from the burette until the pink colour changes to colourless. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

Experiment 1 2 3

Final burette reading (cm3)

Initial burette reading (cm3)

(25)

Questions:

(a)Calculate the:-

(i) Number of moles of BA1 reacted.

……… ……… ……… ……… ………

(ii) Number of moles of BA2 reacted.

……… ……… ……… ……… ………

(iii) The value of n (basicity of HnX).

……… ……… ……… ……… ………

(iv) The mole ratio.

……… ……… ……… ………

j)

practical 10

You are to determine the value of R in RH2 BA1 is an acid of formula RH2 and was prepared by

dissolving 1.125g of RH2 in 250cm3 of distilled water.

BA2 is a n aqueous solution of sodium carbonate prepared by dissolving 2.65g of it in 250cm3_of

distilled water.

(26)

Pipette 20 or 25 cm3 of BA1 into a conical flask add 1 – 2 drops of phenolphthalein indicator and titrate into it BA2 from a burette. Repeat the experiment until consistent results are obtained.

Results:

Capacity of pipette used ……….cm3

Experiment Number 1 2 3 Final burette readings cm3

Initial burette reading cm3

Volume of BA2 used cm3

Best 2 titre values to use to determine average volume of BA2

used……….

used………..

Questions:

(a) Calculate the

(i) Molarity of BA2

……… ……… ……… ……… ……… ………

(ii) Number of moles of BA2 that reacted

……… ……… ……… ………

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……… ……… ………

(iv) Relative formula mass of RH2

……… ……… ………

(v) Value of R in RH2

……… ……… ………

k)

practical 11

BA1 which is a solution of an impure acid H2X of concentration 2gdm-3.

BA2 which is a solution of sodium hydrogen carbonate made by dissolving 4.2g in 1dm-3.You are required to determine the percentage purity of BA1

Procedure:

(28)

Results:

Average volume of BA1 used………..… cm3

Questions:

(a)Calculate the:-

……… ……… ……… ……….

(ii) Number of moles of BA1reacted.

……… ……… ……… ……….

(iii) Molarity of BA1.

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……… ………

(iv) The percentage impurity of BA1 ( X = 88g)

……… ……… ……… ……… ……… ……… ………

l)

practical 12

BA3 which is a 0.04M hydrochloric acid solution.

BA4 which is a solution made by dissolving 3.6g of Na2CO3.10H2O to make 250cm3 of a solution.

You are required to determine the percentage impurity of BA4

Procedure:

Pipette 20cm3 (or 25cm3) of BA4into a clean conical flask and add 2-3 drops of methyl orange indicator. Titrate this mixture with BA3 from the burette until the colour changes from yellow to orange. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

(30)

Questions:

(a) Write an equation for the reaction.

……… (b) Calculate:-

(i) Moles of BA3 that reacted.

……… ……… ……… ………

(ii) The moles of Na2CO3.10H2O in 250cm3 of a solution.

……… ……… ……… ……… ………

(Iii) The percentage purity of Na2CO3.10H2O.

……… ……… ……… ……… ……… ……… ……… ……… ( Na = 23, C = 12, O = 16, H = 1)

m)

practical 13

BA3 which is a 0.2M solution of a monobasic acid.

BA4 which is a solution made by dissolving 11.6g of Na2CO3.nH2O to make a litre of a solution.

You are required to determine the value of n

Procedure:

(31)

Results:

Initial burette reading (cm3) Volume of BA3 used (cm3₎

Questions:

(a)Calculate:-

(i) Moles of BA3 that reacted.

……… ……… ……… ……… ………

(ii) Moles of Na2CO3.10H2O in 1000cm3 of a solution.

……… ……… ……… ……… ………..

(Iii) The formula mass of BA4 and hence the value of n.

(32)

n)

practical 13

BA1 which is a solution containing 6.2g of Y.nH2O in one litre ( Y = 106, 1 mole of Y reacts with

2 moles of hydrochloric acid).

BA2 which is a 0.1M hydrochloric acid.

You are required to determine the number of moles of water of crystallization in BA1

Procedure:

Pipette 20cm3 ( or 25cm3) of BA1 solution into a clean conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this mixture with BA2 from the burette until the pink colour changes to colourless. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

Initial burette reading (cm3) Volume of BA2 used (cm3)

Questions:

(a)Calculate the:-

……… ……… ……… ……… ………

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……… ……… ……… ……… ………

(iii) Concentration of BA1 in moles per litre.

……… ……… ……… ……… ………

(iv) The value of n in Y.nH2O.

……… ……… ………

o)

practical 14

BA1, which is a solution containing 12.5gl- of a mixture of anhydrous sodium carbonate and sodium

chloride.

BA2, which is a 0.1M hydrochloric acid solution.

You are required to determine the composition of sodium chloride in BA1

Procedure:

Pipette 20cm3 (or 25cm3) of BA1into a clean conical flask and add 2-3 drops of methyl orange indicator. Titrate this mixture with BA2 from the burette until the colour changes from yellow to orange. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results: Volume of pipette used………..cm3

(34)

(a)Calculate the:-

……… ……… ……… ………

(ii) Number of moles of BA1 that reacted with BA2.

……… ……… ……… ……… ………

(iii) The concentration in moles per litre of BA1.

……… ……… ……… ………..

(iv) The percentage of sodium chloride in BA1.

……… ……… ……… ……… ……… ……… ………

D.

WORKED EXAMPLES

WORKED EXAMPLE 1

BA1 which is a solution of 0.1M calcium hydroxide.BA2 which is sulphuric acid. You are required to determine the concentration of BA2

(35)

Pipette 20cm3 (or 25cm3) of BA1 into a clean conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this mixture with BA2 until the pink color changes to colorless. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

Volume of pipette used: 20cm3

Final burette reading (cm3) 8.30 16.30 24.40 Initial burette reading (cm3) 0.00 8.30 16.30 Volume of BA2 used (cm3) 8.30 8.00 8.10

NB: The table must be filled in two decimal places, accurately and not in pencil.

Volumes of BA2 used for calculating the average; 8.00cm3 and 8.10cm3

Average volume of BA2 used 8.00 + 8.10 = 16.10 = 8.05cm3 2 2

Questions:

(a)Calculate the:-

(I) number of moles of calcium hydroxide in BA1 that reacted.

1000cm3 of solution contained 0.1moles of calcium hydroxide 20.0cm3 of solution contained 0.1 x 20.0 moles of calcium hydroxide

1000

= 0.002moles of calcium hydroxide

(ii) Number of moles of BA2 reacted.

Ca(OH)2(aq) + H2SO4 (aq) CaSO4 (aq) + 2H2O (l)

From the equation;

1 mole of calcium hydroxide reacted with 1 mole of sulphuric acid.

0.002moles of calcium hydroxide reacted with 1/1X 0.002 moles of sulphuric acid.

= 0.002moles

(iii) Concentration of BA2 in moles per litre.

8.05cm3 of solution contained 0.002moles of sulphuric acid. 1000cm3 of solution contained 0.002 x 1000 moles of sulphuric acid.

8.05

= 0.25 moles per litre.

(36)

(iv) Concentration of BA2 in grams per litre. (S = 32, O = 16, H = 1)

RFM of H2SO4 = (1X2 )+ (32X1) + (16X4) = 98

1 mole of HCl weighs 98g 0.25 moles of HCl weigh (98x 0.25)g

=24.5 g/l

NB: Using a formula anywhere leads to loss of marks.

WORKED EXAMPLE 2

BA1, which is a solution made by dissolving 4.8g of metal hydroxide MOH in one litre.

BA2, which is a 0.0625 moll- sulphuric acid.

You are required to determine the relative atomic mass of M in MOH

Procedure:

Pipette 20cm3_{(or 25cm}3_{) of}_BA1_{into a clean conical flask and add 2-3 drops of phenolphthalein}

indicator. Titrate this mixture with BA2 until the pink color changes to colorless. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

Volume of pipette used; 25.0cm3

Experiment 1 2 3 4

Final burette reading (cm3) 24.10 48.10 24.00 48.10 Initial burette reading (cm3) 0.00 24.10 0.00 24.00 Volume of BA2 used (cm3) 24.10 24.00 24.00 24.10

Volumes of BA2 used for calculating the average. 24.00cm3and 24.00cm3

Average volume of BA2 used 24.00 + 24.00 = 48.00 = 24.00cm3 2 2

Questions:

(a) Calculate the:-

1000cm3 of solution contained 0.0625 moles of sulphuric acid 24.00cm3 of solution contained 0.0625 x 24.00 moles of sulphuric acid

1000

= 0.0015moles of sulphuric acid

(ii) number of moles of MOH in BA1 reacted.

2MOH (aq) + H2SO4(aq) M2SO4(aq) + 2H2O (l)

From the equation;

1 mole of sulphuric acid reacted with 2 mole of MOH

(37)

= 0.003 mole of MOH

(iii) molarity of BA1 .

25.0cm3 of solution contained 0.003moles of hydrochloric acid. 1000cm3_{of solution contained 0.003 x 1000 moles of hydrochloric acid.}

25.0

= 0.12M

(iv) The formula mass of MOH.

0.12moles of MOH were contained in 4.8g 1 moles of MOH was contained in (4.8/0.12)g

= 40g

(V) The R.A.M of M (O = 16, H = 1)

Relative formula mass of MOH = (mx1) + (16x1) + (1x1) = 40 Then; m+17 = 40

M = 23.

WORKED EXAMPLE 3

You are provided with the following;

GA1, which was a solution made by dissolving 12.6g of a monobasic acid HNOX to make 1 litre of

solution.

GA2, which is a 0.2M sodium hydroxide solution.

You are required to determine the value of X in HNOX

Procedure:

Pipette 25cm3 (or 20cm3) of GA2 into a clean conical flask and add 2-3 drops of phenolphthalein

indicator. Titrate this mixture with GA1 until the pink colour changes to colourless. Repeat the titrations two more times for consistent results and enter your results in the table below.

Results:

Volume of pipette used = 25.00cm3

Experiment 1 2 3 4

Final burette reading (cm3) 25.20 25.10 25.00 25.00

Initial burette reading (cm3) 0.00 0.00 0.00 0.00 Volume of GA2 used (cm3) 25.20 25.10 25.00 25.00

Volumes of GA2 used for calculating the average; 25.00cm3_{and 25.00cm}3

Average volume of GA2 used 25.00 + 25.00 = 50.00 = 25.00cm3 2 2

(38)

(a) Write an ionic equation for the reaction between the acid in GA1 and sodium hydroxide in GA2

H+(aq) + OH- (aq) H2O (l)

(b) Calculate;

(i) Moles of sodium hydroxide in GA2 that reacted with the acid 1000cm3 _{of solution contained 0.2moles of sodium hydroxide}

25.00cm3 of solution contained 0.2 x 25.00 moles of sodium hydroxide 1000

= 0.005moles of sodium hydroxide

(ii) Number of moles of HNOX in GA1 reacted.

From the equation;

1 mole of sodium hydroxide reacted with 1 mole of MOH

0.0015mole of sodium hydroxide reacted with (1/1x 0.005) mole of MOH

=0.005 mole of MOH

(iii) Molarity of GA1.

25.0cm3 _{of solution contained 0.005}_{moles of hydrochloric acid.}

1000cm3 of solution contained 0.005 x 1000 moles of hydrochloric acid. 25.0

= 0.2M

(iv) The formula mass ofHNOX

0.2moles of HNOX were contained in 12.6g

1 moles of HNOX was contained in (12.6/0.2)g

= 63g

(V) the value of X ( O = 16, H = 1, N=14)

Formolar mass of MOH = (1x1) + (14x1) + (Xx16) =40 Then; 16X+15 = 63

16X = 48.

X = 3

WORKED EXAMPLE 4

BA1, which is a solution containing 10g of solid Y.10H2O per 500cm3 (Y = 202)

BA1, which is 0.1M hydrochloric acid solution

(39)

You are required to determine the percentage purity of solid Y.10H2O

Procedure:

Pipette 25 (or 20) cm3 of BA1 into a clean conical flask. Then add 2 – 3 drops of methyl orange indicator and titrate with solution BA2 until you obtain a persistent colour change. Repeat the titration 2 – 3 times to obtain consistent results. Enter your results in the table below.

Results:

Volume of pipette used = 25cm3

Final burette reading/cm3 15.40 30.90 46.30

Initial burette reading/cm3 0.00 15.40 30.90

Volume of BA2 used/cm3 15.40 15.50 15.40

Titre values used for calculating average volume of BA2; 15.50cm3_{and 15.40cm}3

Average volume of BA2 = 15.45cm3

Questions:

(a) Calculate the

(i) Moles of hydrochloric acid in BA2 that reacted 1000cm3 _{of solution contained 0.1}_{moles of calcium hydroxide}

15.45cm3 of solution contained 0.1 x 15.45 moles of calcium hydroxide 1000

= 0.001545moles of calcium hydroxide

(ii) Moles of Y in BA1 that reacted with hydrochloric acid. From the mole ratio;

2 moles of the acid reacted with 1 mole of Y

0.001545 moles of the acid reacted with (0.001545/2) moles of Y = 0.000772 moles of Y

(iii) Molarity of BA1 and hence the percentage purity of the solid, Y.10H2O (4 marks)

25.0cm3 of solution contained 0.000772moles of hydrochloric acid. 1000cm3 of solution contained 0.000772 x 1000 moles of hydrochloric acid.

25.0

= 0.03088M

RFM of Y.10H2O = 202 + 180 = 382

1 mole of Y.10H2O weighs 382g

0.03088 moles of Y.10H2O weigh (382x 0.03088) g

(40)

500cm3 of BA1 contained 10g of Y.10H2O

1000cm3 of BA1 contained 10 x 1000 g of Y.10H2O

500 = 20g/l

Therefore, the percentage purity of Y.10H2O in BA1 = 11.7 x 100 % 20

= 58.5%

NB: Percentage impurity = 100 – (percentage purity)

2.

QUALITATIVE ANALYSIS

This section forms question number two of chemistry paper 545/3 or paper 545/4. It is mainly based on the chemistry of salts which is learnt in senior two. Students are therefore advised to revise the chemistry of salts in order to present quality work in this section. Safety is the most important aspect for students of chemistry in the laboratory; therefore students should take caution and responsibility. Students should revise all rules and regulations of the laboratory before performing practical.

Qualitative analysis requires a student to carry out a number of sample tests on one or more substances mixed, the observation a recorded from which deduction about the substance are made. The common apparatus used in this case include,

 Test tube rack, test tubes, boiling tubes, spatula, Bunsen burner, test tube holder, glass rod, filter paper, filter funnel, wash bottle, a set of reagent bottles

The assessment in this section is based on the quality of the information provided by the student. The following points should be noted while making a practical report in this section.

 Make clear and true observations  Precisely write observation

 Make your deductions logically from the observation  A wrong observation cannot lead to a correct deduction.

Qualitative analysis at this level mainly deals with the chemistry of few selected non-transitional metals cations and some transitional metal cations and some anions.

a)

Non- transition cations

Salts of non-transition metals are usually white or colourless and the generally dissolves in water or dilute acids to form colourless solutions. Cations considered in this case are,

o Zinc ions, Zn2+

(41)

o Calcium ions, Ca2+

o Magnesium ions, Mg2+

o Barium ions, Ba2+

o Ammonium ions, NH4+

b)

Transition metal cations

Salts of transition metals exhibit characteristic colours and they usually dissolves in water or dilute acids to form coloured solutions. The cations considered in this case include,

o Copper (II) ions, Cu2+

o Iron (II) ions, Fe2+

o Iron (III) ions, Fe3+

c)

Anions considered at this level include,

o Carbonate ions, CO3

2-o Sulphate ions, SO4 2-o Chloride ions, Cl

-o Nitrate ions, NO3 -o Sulphite ions, SO3

2-Others include,

o Bromide ions, Br-

o Iodide ions, I

-o Nitrite ions, NO2

2.1.

PRELIMINARY TESTS

a)

Appearance,

Note the physical properties of the substance provided, the physical properties considered at this level include colour, texture, smell etc. Physical properties give crucial information about the presence of particular ions. This is possible because ions/cations or anions have unique physical properties, as shown in table below.

Observation deduction

White solid (crystalline or pawderly), if crystalline it may be a hydrated salt

Non-transition metal cations present

Therefore Ca2+_{, Mg}2+_{, Ba}2+_{, Al}3+_{, Pb}2+_{, Zn}2+_,

NH4+ ions probably present.

Blue crystals Cu2+ ions present

Green solid Fe2+ ions present

(42)

b. Action of heat,

Heat two spatula endful of the substance in a dry hard glass tube initially slowly and then strongly until there is no further change. Identify the gases evolved, their smell, colour, action on damp litmus paper, action on the splint, and usual confirmatory tests for gases. Observe the residue when hot and when cold.

Consider the table below,

Observation Deduction

Colourless liquid condenses on sides of boiling tube, liquid turns anhydrous copper (II) sulphate from white to blue. Or turns cobalt (II) chloride from blue to pink.

Water of crystallisation

∴ Hydrated salt, HCO3-, HSO4-, OH-

probably present

White sublimate, colourless gas with pungent smell, gas turns damp red litmus paper blue, gas forms dense white fumes with conc.HCl

NH3 gas evolved,

∴ NH4+ ions present

Colourless gas evolved, gas turns moist blue litmus paper red, gas turns lime water milky

CO2 gas evolved

∴ CO32-, HCO3- probably present

Colourless gas liberated, gas is neutral to litmus paper, gas relights a glowing splint.

O2 gas evolved

∴ NO3- probably present

Cracking noise, brown fumes of a gas evolved, gas has irritating smell, gas turns damp blue litmus paper red, gas relights a glowing splint.

Mixture of NO2 and O2 gases

evolved.

∴ NO3- present.

Pale yellow fumes of gas evolved, gas turns moist blue litmus paper red and bleaches, gas turns KI solution from colourless to brown.

Note. Most chlorides usually impart the characteristic colour of the metal in the Bunsen flame. The pale yellow colour may not be observed

Cl2 gas given off

∴ Cl-_{probably present}

Misty white fumes of a gas, gas has pungent irritating smell, gas turns moist blue litmus paper red, gas forms dense white fumes with ammonia

(43)

This is also true for chlorides, here the chlorides should be heated with conc. Acid

Colourless gas evolved, gas has irritating smell, gas turns moist blue litmus paper red and bleaches it, gas turns purple acidified potassium permanganate

colourless and turns acidified dichromate solution from orange to green

SO2 gas evolved

∴ SO42- probably present

White fumes with irritating smell, turns moist blue litmus paper red, turns barium nitrate milky.

SO3 gas given off

∴ SO42-, HSO4- probably present.

Colour of the residue

c.

Solubility test.

The solvent used to dissolve solutes is water. For practical work, distilled water should be used for making solutions for qualitative analysis.

Here the knowledge of solubility of salts is crucial. Consider the summary of the solubilities of common salts in water in the table below.

Salt Solubility

Carbonates All are insoluble except potassium, sodium, and ammonium carbonates

Chlorides All are soluble except silver and lead chlorides. Lead chloride is soluble in hot water.

Observation Deduction

Residue is yellow hot, white cold Residue is ZnO

∴ Zn2+ _present

Residue is brown hot, yellow cold Residue is PbO

∴ Pb2+_present

Residue turns from black to brown Fe2O3 is formed

∴ Fe2+_present

Residue is black hot, black cold Residue is CuO

(44)

Nitrates All are soluble

Sulphates All are soluble except barium and lead sulphates, calcium sulphate is only sparingly soluble.

Note.

 If a salt is soluble, state the colour of the solution formed.

 For a mixture of salts where one is soluble and the other is insoluble, you will be required to filter. In this case, describe the colour of the residue and the filtrate.

 Where you are require to wash the residue, do it at least two times to ensure that the particles of the soluble salt are dissolved. Allow to settle and decant the solution before you analyse the residue as required.

 Where the salt is insoluble in water, you will be required to dissolve in cold dilute acids or warm dilute acid. To a spatula endfull of the substance in the test tube add dilute nitric acid or hydrochloric acid or sulphuric acid.

 When effervescence occurs, identify the gas evolved by their common tests as described before. Note the colour of the resultant solution.

Colour of the solutions

Observation Deduction

Dissolves to form a colourless solution

Non-transitional metal cations ie Mg2+, Ca2+, Ba2+, Al3+, Pb2+, Zn2+_{, NH}

4+ probably present.

Dissolves to form a blue solution Cu2+ present Dissolves to form a green solution Fe2+ present Dissolves to form a brown

solution, the solution turns blue litmus paper red

Fe3+ present

2.2

IDENTIFICATION OF CATIONS

There are two categories of cations considered here, these are

1) Non-transition metal cations, these are non-coloured and their solutions are colourless. They include Ca2+, Mg2+, Ba2+, Al3+, Pb2+, Zn2+or NH4+ ions

2) Transition metal cations, these are coloured and their solutions are also coloured. Refer to colour of the solutions described before. These include Cu2+, Fe2+, Fe3+ which are blue, green, and brown respectively.

(45)

Activity

 Solutions containing Mg2+, Ca2+, Ba2+, Al3+, Pb2+, Zn2+, NH4+, Cu2+, Fe2+, Fe3+ ions

 Test tubes

 2M sodium hydroxide solution

 2M ammonium solution Procedure,

 To 2cm3 of each of the solutions in a separate test tubes, add sodium hydroxide solution drop wise until excess. Record your observations in the table 1 below

 Repeat the procedure with 2M ammonia solution and record your observation in table 2

a.

Effect of sodium hydroxide

Table 1

cations Observation

Ca2+

Mg2+ Ba2+

Al3+

Pb2+

Zn2+

NH4+

Cu2+

Fe2+

Fe3+

b.

Effect of ammonia solution

Table 2

(46)

Questions,

1. List the ions that form white ppt with sodium hydroxide.

... ... ... ...

2. List the ions that form white ppt with sodium hydroxide insoluble in excess.

... ... ... ...

3. List the ions that form white ppt with sodium hydroxide soluble in excess.

... ... ... ...

4. Which ions form coloured ppt with sodium hydroxide insoluble in excess, state the respective colours of the ions.

... ... ... ...

5. Which ions form coloured ppt with sodium hydroxide soluble in excess, specify respective colours of the ions for both ppt and solutions.

... ... ... ...

Ca2+

Mg2+ Ba2+

Al3+ Pb2+ Zn2+

NH4+

Cu2+ Fe2+

(47)

6. List the ions that form white ppt with ammonium hydroxide.

... ... ... ...

7. List the ions that form white ppt with ammonium hydroxide insoluble in excess.

... ... ... ...

8. List the ions that form white ppt with ammonium hydroxide soluble in excess.

... ... ... ...

9. Which ions form coloured ppt with ammonium hydroxide insoluble in excess, state the respective colours of the ions.

... ... ... ...

10.Which ions form coloured ppt with ammonium hydroxide soluble in excess, specify respective colours of the ions for both ppt and solutions.

... ... ... ...

b) Compare the observations with sodium hydroxide and ammonia solutions, show differences and similarities.

... ... ... ... ...

(48)

Now compare your results with the tables below,

Effect of sodium hydroxide

Cations Observation

Ca2+ _{White ppt, insoluble in excess sodium hydroxide}

Mg2+ White ppt, insoluble in excess sodium hydroxide Ba2+ White ppt, insoluble in excess sodium hydroxide

Al3+ White ppt, soluble in excess, forming a colourless solution Pb2+ White ppt, soluble in excess, forming a colourless solution Zn2+ White ppt, soluble in excess, forming a colourless solution

NH4+ Solution remains colourless, colourless gas evolved, gas turns moist red

litmus paper blue, gas forms dense white fumes with conc. HCl Cu2+ Blue ppt, insoluble in excess sodium hydroxide

Fe2+ _{Green ppt, insoluble in excess, turns brown on standing}

Fe3+ Brown ppt, insoluble in excess sodium hydroxide

Effect of ammonia solution

NOTE

 NH4+, K+ and Na+ ions do not form precipitates with both sodium hydroxide and ammonia

solution.

 When a solution containing NH4+ ions and sodium hydroxide is warmed, ammonia gas is

evolved. Refer to the tests for ammonia

Cations Observation

Ca2+ White ppt, soluble in excess, forming a colourless solution Mg2+ White ppt, insoluble in excess ammonium hydroxide Ba2+ White ppt, insoluble in excess ammonium hydroxide

Al3+ White ppt, insoluble in excess ammonium hydroxide solution Pb2+ White ppt, insoluble in excess ammonium hydroxide

Zn2+ _{White ppt, soluble in excess, forming a colourless solution}

NH4+ No observable change, solution remains colourless.

(49)

2.3

Explanation and equations

The cations of the salt solutions react with sodium hydroxide or ammonia solutions to form insoluble metal hydroxides which are observed as precipitates. The alkalis provides the hydroxide ions, OH- to the cations.

Generally,

Mn+ (aq) + nOH-(aq) M(OH)n(s)

Where M is the metal, n is the valance of the metal

Mg2+

(aq) + 2OH-(aq) Mg(OH)2(s)

(White ppt)

Ca2+(aq) + 2OH-(aq) Ca(OH)2(s)

(White ppt)

Zn2+(aq) + 2OH-(aq) Zn(OH)2(s)

(White ppt)

Pb2+(aq) + 2OH-(aq) Pb(OH)2(s)

(White ppt)

Al3+(aq) + 3OH-(aq) Al(OH)3(s)

(White ppt)

Cu2+(aq) + 2OH-(aq) Cu(OH)2(s)

(blue ppt)

Fe2+(aq) + 2OH-(aq) Fe(OH)2(s)

(green ppt)

Fe3+(aq) + 3OH-(aq) Fe(OH)3(s)

(brown ppt)

In excess sodium hydroxide, the white ppt of zinc hydroxide, lead hydroxide, aluminium hydroxide dissolves forming colourless solutions of zincate, plumbate, aluminate complex ions respectively.

Zn(OH)2(s) + 2OH-(aq) [Zn(OH)4]2-(aq)

(50)

Pb(OH)2(s) + 2OH-(aq) [Pb(OH)4]2-(aq)

(plumbate ions)

In excess ammonia solution, zinc (II) hydroxide dissolves to form colourless solution of tetra ammine Zinc (II) ions whereas copper (II) hydroxide dissolves to form a deep blue solution of tetra ammine copper (II) ions as below

Zn(OH)2(s) + NH3(aq) [Zn(NH3)4]2+(aq) + 2 OH-(aq)

(tetra ammine Zinc (II) ions)

Cu(OH)2(s) + NH3(aq) [Cu(NH3)4]2+(aq) + 2 OH-(aq)

(tetra ammine copper (II) ions)

2,4 CONFIRMATORY TESTS

a) Zinc (II) ions, Zn2+

To a solution of zinc ions, add ammonia solution drop wise until excess

Observation,

White precipitate, soluble in excess ammonia forming a colourless solution

b) Lead (II) ions, Pb2+

Note, the test here can be used to distinguish Pb2+_{from Al}3+ _ions

I) Effect of dilute sulphuric acid

A white ppt is formed with Pb2+ according to the equation. Pb2+(aq) + SO42-(aq) PbSO4(s)

No observable change with Al3+_{, solution remains colourless}

II) With dilute hydrochloric acid

Add few drops of dilute hydrochloric acid to the solution of suspected ions

Observation,

White ppt formed, soluble on warming and reappears on cooling. Pb2+

(aq) + 2Cl-(aq) PbCl2(s)

No observable change with Al3+, solution remains colourless

III) With potassium iodide solution

To a solution of suspected ions is added 3 drops of KI solution

Observation,

Yellow ppt is formed

Pb2+(aq) + 2l-(aq) Pbl2(s)

IV) With potassium dichromate solution

(51)

Observation,

Yellow ppt is formed

Pb2+(aq) + CrO42-(aq) PbCrO4(s)

c) Copper (II) ions, Cu2+

I) potassium iodide solution

To solution of unknown in test tube, add equal volume of KI

Observation,

White ppt formed in a brown solution (you can filter to clearly see the white ppt, or add sodium thiosulphate solution to it, the brown colour is discharged leaving the white ppt)

II) potassium hexacynoferrate (II) solution

To a solution of suspected ions, add 4 drops of potassium hexacynoferrate (II) solution

Observation,

Brown ppt is formed

d) Iron (II) ions, Fe2+

To the solution of suspected ions, add 3 drops of potassium hexacynoferrate (III) solution

Observation,

A dark blue ppt is formed

e) Iron (III) ions, Fe3+

To the solution of Fe3+ ions add potassium hexacynoferrate (II) solution

Observation,

Dark blue ppt is formed Or,

Add Aluminium thiocyanate solution

Observation,

Deep red solution is formed.

2.4

IDENTIFICATION OF ANIONS

1. Carbonate ions, CO3

2-a) Effect of heat.

On heating strongly, carbonates decompose producing a colourless gas, gas turns moist blue litmus paper red and lime water milky. Carbonates of group one and ammonium carbonate,

MCO3(s) MO(s) + CO2(g)

(52)

Carbonates react with dilute acids with effervescence of a colourless gas, gas turns moist blue litmus paper red and lime water milky. Solution is colourless.

Generally,

MCO3(s) + 2HX (aq) MX2(aq) + H2O(l) + CO2(g)

c) Reaction with barium chloride or nitrate solution.

Carbonates form white ppt of barium carbonate, when little of dilute acid is added; the ppt dissolves in the acid forming colourless solution.

Ba2+(aq) + CO32-(aq) BaCO3(s)

d) Reaction with silver nitrate solution

A white ppt of silver carbonate is formed. The ppt dissolves in aqueous ammonia or dilute nitric acid forming colourless solution.

Ag+(aq) + CO32-(aq) Ag2CO3(s)

e) Reaction with lead nitrate or lead ethanoate solutions

White ppt of lead carbonate is formed which dissolves in dilute acid forming colourless solution.

Pb2+(aq) + CO32-(aq) PbCO3(s)

2. Test for chloride ions, Cl

-a) Reaction with silver nitrate solution.

To the solution of the chloride, add dilute nitric acid followed by silver nitrate solution.

A white ppt is of silver chloride is formed, Ag+(aq) + Cl-(aq) AgCl(s)

b) Reaction with concentrated sulphuric acid

When solid chlorides are warmed with concentrated sulphuric acids, a colourless gas with irritating smell, turns moist blue litmus red and forms dense white fumes with ammonia is evolved.

SO42-(aq) + Cl-(s) HSO4-(aq) + HCl(g)

3. Test for sulphate ions, SO42-

a) Add dilute HCl or HNO3 acid followed by barium chloride or barium nitrate

respectively to a solution of SO42- ions.

A white ppt is formed, insoluble in dilute HCl or HNO3 acid

Ba2+(aq) + SO42-(aq) BaSO4(s)

b) Add 3 drops of lead nitrate solution and warm White ppt formed

Pb2+(aq) + SO42-(aq) PbSO4(s)

4. Test for sulphite ions, SO3

(53)

A white ppt is formed, soluble in dilute hydrochloric acid or nitric acid forming a colourless solution.

Ba2+(aq) + SO32-(aq) BaSO3(s)

BaSO3(s) + 2H+(aq) Ba2+(aq) + SO2(g) + H2O(l)

b) Add lead nitrate solution to the solution of suspected ions

A white ppt is formed which dissolves in dilute nitric acid forming a colourless solution.

Pb2+(aq) + SO32-(aq) PbSO3(s)

PbSO3(s) + 2H+(aq) Pb2+(aq) + SO2(g) + H2O(l)

5. Test for nitrates

Add an equal volume of freshly prepared iron (II) sulphate solution to the suspected solution of the nitrate, tilt the test tube as shown in the figure below, then add conc. Sulphuric acid to the mixture along the sides of the test tube.

Diagram

Observation,

A brown ring is formed between the two layers of solutions

Explanation,

Conc. Sulphuric acid is denser than the layer containing the suspected nitrate ion and iron (II) sulphate solution.

Conc. Sulphuric acid reacts with the nitrate to produce nitric acid. The nitric acid is reduced to nitrogen (II) oxide by iron (II) sulphate. The nitrogen (II) oxide formed reacts with unreacted iron (II) sulphate to form a brown nitroso-iron (II) sulphate which appears as a brown ring

FeSO4(aq) + NO(g) FeSO4.NO(aq)

Note,

(54)

2.5ACTION OF HYDROGEN PEROXIDE, H2O2

This test applies when the solution contains ions which can be oxidised; such as iron (II) ions.

Test Observation Deduction

To the test solution, add 5 drops of hydrogen peroxide and warm

Effervescence occurs and a colourless gas which relights a glowing splint is evolved. Blue solution turns brown.

- Colourless gas is oxygen. .- Fe 2+ present

( Fe2+ ions are oxidised to Fe3+ ions)

2.6

DISPLACEMENT OF IONS DEPENDINING ON THEIR POSITION IN

REACTIVITY SERIES

Zinc and iron can be used to displace copper from its salt solution.

Test Observation Deduction

To 2cm3 of copper (II) Sulphate solution, add a spatula end ful of zinc powder , shake and allow to cool

- the solution changes from blue to colourless

- A brown solid is formed.

Cu2+ present

Brown solid is copper.

Explanation:

Zinc is more reactive and displaces copper (II) ions Cu2+ from its salt solution to form brown copper solid and itself is oxidised of colourless zinc ions

i.e. Zn( S) + CuSO4 (aq) ZnSO4(aq) + Cu(S).

(blue) ( colourless)

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WORKED EXAMPLES

Example 1

You are provided with substance Z which contains one cation and two anions. Carry out the following tests to identify the cation and anions in Z. Identify any gases evolved.

Test Observation Deduction

Heat a spatula end-full of Z in a dry test tube until there is no further change

A colourless gas evolved, gas turns moist blue litmus paper red, gas turns lime water milky

Residue is yellow hot, white cold

Gas is CO2

Hence CO32-, HCO3-,

present

Residue is ZnO Hence Zn2+ present

Dissolve two spatula end-full of Z in distilled water, filter and keep both the filtrate and residue.

Partially dissolves forming a colourless solution and white residue

Zn2+, Pb2+, Al3+,Ba2+, Mg2+, Ca2+ probably present

Divide the filtrate into four equal portions,

To the first portion, add sodium hydroxide solution drop wise until excess

White ppt, dissolves in excess forming a colourless solution

Zn2+, Pb2+, Al3+, probably present

To the second portion, add ammonia solution drop wise until excess

White ppt, soluble in excess

forming a colourless soltion Zn2+ confirmed present

To the third portion, add 3 drops of lead nitrate solution

White ppt formed SO42-, Cl- probably

present Carry out a test of your own to

confirm the anion in Z

To the fourth portion, add dilute nitric acid followed by barium nitrate solution

White ppt formed SO42- confirmed

present

Dissolve the residue in dilute hydrochloric acid and divide the resultant solution into two equal portions.

Dissolves with

effervescence evolving a colourless gas, gas turns moist blue litmus paper red, gas turns lime water milky A colourless solution formed

Gas is CO2

Hence CO32-

confirmed present

To the first portion, add sodium hydroxide solution drop wise until excess

white ppt formed Zn2+, Pb2+, Al3+, probably present

To the second portion, add ammonia solution drop wise until excess

White ppt, soluble in excess ammonia solution forming a colourless solution

Zn2+ confirmed present

Identify the (i) cation in Z Zn2+

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EXAMPLE 2

1. You are provided with substance X which contains one cation and two anions. Carry out the following tests to identify the cation and anions present in X.

TEST OBSERVATION DEDUCTION

(a)To 2 spatula endfuls of X add about 15cm3 of water and shake well. Filter and keep both the residue and the filtrate.

X is white powderly crystalline solid, partially soluble forming white residue and colorless filtrate.

Al3+, Zn2+,Pb2+ probably present

CO32-, NO3-,SO42-, Cl

-probably present

(b)Divide the filtrate into four portions.

(i)To the first portion add dilute sodium hydroxide solution drop wise until in excess.

A white ppt soluble in excess forming colorless solution

Al3+_{, Zn}2+_orPb2+_probably

present

(ii)To the second portion add dilute aqueous ammonia drop wise until in excess.

A white ppt insoluble in excess Al3 +,Pb2+ probably present.

(iii)To the third portion add drops of potassium iodide solution.

A yellow ppt formed Pb2+confirmed present.

(iv)To the fourth portion add copper fillings followed conc. sulphuric acid and boil.

A white ppt was formed on addition of the acid. On boiling, a brown gas which turned moist blue litmus red