guide to stpm practical 2012

1© Oxford Fajar Sdn. Bhd. (008974-T) 2012

CONTENTS

SECTION 1 Volumetric analysis

1.1 Important formulae

1.2 Acid-base titrationColour change of acid-base indicators at end pointChemical equations for acid-base titrations

1.3 Potassium manganate(VII), KMnO4, titrationBalancing half-equationsBalancing ionic equations for redox reactionsImportant half-equations for KMnO

4 titrations

Important ionic equations for KMnO4 titrations

Precautions in KMnO4 titration

1.4 Iodine-sodium thiosulphate (Na2S2O3) titrationImportant equations for iodine-sodium thiosulphate titrationsPrecautions in iodine-sodium thiosulphate titrations

SECTION 2 Physical chemistry

Thermochemistry

SECTION 3 Qualitative analysis

Table 1 Reactions with dilute hydrochloric acid Table 2 Reactions with aqueous sodium hydroxide Table 3 Reactions with aqueous ammonia Table 4 Reactions with aqueous iron(III) chloride Table 5 Reactions with silver nitrate solution Table 6 Reactions with potassium chromate(VI) solution Table 7 Reactions with potassium iodide solution Table 8 Reactions of sodium ethanoate solution Table 9 Reactions of sodium carbonate solutionTable 10 Identification of functional groups in organic compounds

GUIDE TO STPM PRACTICAL

2 © Oxford Fajar Sdn. Bhd. (008974-T) 2012

SECTION 1 Volumetric analysis

1.1 Important formulae

Concentration (in g dm–3)Concentration (in mol dm–3) = ——————————————— … (1.1) Relative molecular mass

Concentration (in dm–3) = Concentration (in mol dm–3) � Relative molecular mass … (1.2)If the reaction between X and Y to form P and Q is represented by the chemical equation:

aX + bY → cP + dQThen,

(M1V

1)

X a

————– = — … (1.3) (M

2V

2)

Y b

Example

Calculate the concentration in mol dm–3 of the following solutions:(a) F1 is hydrochloric acid of concentration 0.913 g dm–3.(b) F2 is a solution containing 3.4 g of OH– per dm3.(c) F3 is a solution containing 2.38 g MnO

4– per dm3.

(d) F4 is a solution containing 0.775 g KMnO4 per 250 cm3.

(Relative atomic mass: H, 1; O, 16.0; Cl, 35.5; K, 39.1; Mn, 54.9)

Solution

(a) Relative molecular mass of HCl = 36.5 0.913

Concentration = ———– = 0.025 mol dm–3

36.5

(b) Relative formula mass of OH– = 17.0 3.4

Concentration = ———– = 0.20 mol dm–3

17.0

(c) Relative formula mass of MnO4– = 118.9

2.38Concentration = ———– = 0.020 mol dm–3

118.9

(d) Relative molecular mass of KMnO4 = 158

0.775Concentration = ———– = 0.0049 mol per 250 cm3

158= 0.0049 � 4 = 0.0196 mol dm–3


1.2 Acid-base titration

Colour change of acid-base indicators at end point

Chemical equations for acid-base titration

aThe ratio of — in the second column is obtained by considering ba = number of moles of the first reactantb = number of moles of the second reactant

Example 1To determine the identity of X in X(OH)

2

F1 is hydrochloric acid of concentration 0.1 mol dm–3.F2 is a solution containing 6.85 g dm–3 of X(OH)

2.

25.0 cm3 of F2 required 19.85 cm3 of F1 for complete neutralisation. (a) Write the equation for the reaction between X(OH)

2 and hydrochloric acid.

(b) Calculate the concentration of X(OH)2 in solution F2.

(c) Hence calculate (i) the relative molecular mass of X(OH)2 and (ii) the relative atomic mass of X.

(d) Suggest an identity for X.

Solution

(a) X(OH)2 + 2HCl → XCl

2 + 2H

2O

(M1V

1)

X(OH)2 1(b) ——————————— = —

(M2V

2)

acid 2

1 0.1 � 19.85Concentration of X(OH)

2 = — � ——————————— = 0.0397 mol dm–3

2 25

(c) (i) Concentration (in g dm–3) = Concentration (in mol dm–3) � Relative molecular mass 6.85

Relative molecular mass = ——————— = 172.5 0.0397

(ii) 172.5 = X + (2 � 17)X = 138.5

(d) X is barium.

Solution in the burette Solution in the conical flask Indicator in the conical flask Colour change

Acid

Base

Acid

Base

Base

Acid

Base

Acid

Phenolphthalein

Phenolphthalein

Methyl orange

Methyl orange

Pink to colourless

Colourless to pink

Yellow to orange

Red to orange

Equation aRatio of — b

H+ + OH– → H2O

NaOH + HCl → NaCl + H2O

HClO4 + NaOH → NaClO

4 + H

2O

Ba(OH)2 + 2HCl → BaCl

2 + 2H

2O

H2SO

4 + 2NaOH + → Na

2SO

4 + 2H

2O

H2C

2O

4 + 2NaOH + → Na

2C

2O

4 + 2H

2O

1—1

1—1

1—1

1—2

1—2

1—2


Example 2To determine the identity of X in HXO

4

F3 is 0.10 mol dm–3 sodium hydroxide solution.F4 is an acid with molecular formula, HXO

4.

25.0 cm3 of F4 required 21.00 cm3 of F3 for complete neutralisation. (a) Write the equation for the reaction between HXO

4 and NaOH.

(b) Calculate the concentration of HXO4 in solution F4.

(c) Hence calculate (i) the relative molecular mass of X(OH)2 and (ii) the relative atomic mass of X.

(The concentration of HXO4 is 8.44 g dm–3)

(d) Suggest an identity for X.

Solution

(a) HXO4 + NaOH → NaXO

4 + H

2O

(M1V

1)

HXO4 1(b) ——————————— = —

(M2V

2)

NaOH 1

0.1 � 21.00 Concentration of HXO

4 = ——————————— = 0.084 mol dm–3

25

(c) (i) Concentration (in g dm–3) = Concentration (in mol dm–3) � Relative molecular mass 8.44

Relative molecular mass = —————— = 100.5 0.084

(ii) 100.5 = 1 + X + (4 � 16)X = 35.5

(d) X is chlorine.

Example 3 You are asked to determine the accurate concentration of a monoprotic acid, HX in F5 solution from the following experiment. The rough concentration of HX is about 1.0 mol dm–3.By means of a pipette, 50.0 cm3 of F5 is transferred into a 250 cm3 volumetric flask. Distilled water is then added and make up to the mark on the volumetric flask. The solution is labelled as F6 solution.F7 is prepared by dissolving 2.65 g of anhydrous sodium carbonate in250 cm3 solution. 25.0 cm3 of F7 required 20.95 cm3 of F6 for complete neutralisation using methyl orange as indicator.(a) Calculate the concentration of sodium carbonate in F7 solution.(b) Write the chemical equation for the reaction between sodium carbonate and HX.(c) Calculate the concentration of F6.(d) Hence, calculate the accurate concentration of HX in F5.

Solution

(a) Relative molecular mass of Na2CO

3 = 106

2.65 � 4Concentration of Na

2CO

3 = ———————— = 0.1 mol dm–3

106

(b) 2HX + Na2CO

3 → 2NaX + CO

2 + H

2O

(M1V

1)

HX 2

(c) ——————————— = — (M

2V

2)

Na2CO3 1

2 � 0.1 � 25.0Concentration of HX in F6 = ———————————— = 0.239 mol dm–3

20.95 250(d) Concentration of HX in F5 = 0.239 � ——— = 1.20 mol dm–3

50

NaOH solution is not used to standardise acid because it is a deliquescent solid. Thus, it is difficult to find the accurate mass of NaOH as it absorbs the moisture from the air (deliquescent) during weighing.

Take NoteTake Note


1.3 Potassium manganate(VII), KMnO4, titration

Balancing half-equations

1. A half-equation is a chemical equation containing electrons.

Fe2+ → Fe3+ + e– … oxidation reactionCl

2 + 2e– → 2Cl– … reduction reaction

2. The following steps are used to balance half-equations for redox reactions.Step 1 : Determine the oxidation number of the atom that undergoes oxidation or reduction.Step 2 : Balance the half-equation in terms of the total charge and the number of atoms on

both sides of the equation.Step 3 : Calculate the number of oxygen atoms on both sides of the equation and add water (if

necessary) on the side of equation that has insufficient number of oxygen atoms.

Example 1Balance the equation : MnO

4– + H+ → Mn2+

Step 1 The oxidation number of Mn decreases from +7 to +2. Hence, 5e– must be added to the left-hand side of the equation.

MnO4– + H+ + 5e– → Mn2+ … not balanced

Step 2 Balance the total charge and number of atoms on both sides of the equation

MnO4– + 8H+ + 5e– → Mn2+ + 4H

2O … balanced

Total charge on LHS of equation = –1 + 8 � (+1) + 5 � (–1) = +2Total charge on RHS of equation = +2Total number of atoms on LHS of equation = 1Mn + 4O + 8HTotal number of atoms on RHS of equation = 1Mn + 4O + 8HHence, the half-equation is a balanced equation.

Example 2Balance the equation : XO

2– → XO

3–

Step 1 The oxidation number of X increases from +3 to +5. Hence, 2e– must be added to the right-hand side of the equation.

XO2– → XO

3– + 2e– … not balanced

Steps 2 and 3

XO2– + H

2O → XO

3– + 2e– + 2H+ … balanced

Total charge on LHS of equation = –1 Total charge on RHS of equation = (–1) + 2 � (–1) + 2 � (+1) = –1Total number of atoms on LHS of equation = 1X + 2H + 3OTotal number of atoms on RHS of equation = 1X + 3O + 2HHence, the half-equation is a balanced equation.

Balancing ionic equations for redox reactions

Step 1 : Write the half-equations for the oxidation and reduction reactions.Step 2 : Combine the half-equations to get an ionic equation that does not contain electrons.

Example 3Balance the equation for the reaction between HCOO– and MnO

4–.

Step 1 Write the half-equations for HCOO– and MnO4–

HCOO– → H+ + CO2 + 2e– … oxidation (1)

MnO4– + 8H+ + 5e– → Mn2+ + 4H

2O … reduction (2)


Step 2 Combine the half-equations into the ionic equation

5 � equation (1) gives5HCOO– → 5H+ + 5CO

2 + 10e– … (3)

2 � equation (2) gives 2MnO

4– + 16H+ + 10e– → 2Mn2+ + 8H

2O … (4)

Adding equations (3) and (4) gives5HCOO– + 2MnO

4– + 11H+ → 5CO

2 + 2Mn2+ + 8H

2O

Important half-equations for KMnO4 titrations

MnO4– + 8H+ + 5e– → Mn2+ + 4H

2O

Fe2+ → Fe3+ + e–

C2O

42– → 2CO

2 + 2e–

H2O

2 → O

2 + 2H+ + 2e–

NO2– + H

2O → NO

3– + 2H+ + 2e–

Important ionic equations for KMnO4 titrations

Note: a = number of moles for the first reactant, b = number of moles for the second reactant

Ionic equation aRatio of — b

5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H

2O

5C2O

42– + 2MnO

4– + 16H+ → 2Mn2+ + 10CO

2 + 8H

2O

5H2O

2 + 2MnO

4– + 6H+ → 5O

2 + 2Mn2+ + 8H

2O

5NO2– + 2MnO

4– + 6H+ → 5NO

3– + 2Mn2+ + 3H

2O

5FeC2O

4 + 3MnO

4– + 24H+ → 5Fe3+ + 3Mn2+ + 12H

2O + 10CO

2

5—1

5—2

5—2

5—2

5—3

Precautions in KMnO4 titration

1. KMnO4 titration is used to determine the concentration of iron(II) salt, ethanedioate (also

called oxalate, C2O

42–), hydrogen peroxide and nitrites.

2. Solutions used for KMnO4 titrations must be acidified. In neutral or alkaline solution, brown

MnO2 is precipitated and detection of end point is difficult.

MnO4– + 2H

2O + 3e– → MnO

2 + 4OH–

3. Sulphuric acid is always used for acidification. Hydrochloric acid and nitric acid must not be used for acidification.

4. KMnO4 titrations involving Fe2+, H

2O

2 and NO

2– should be carried out at room temperature.

5. KMnO4 titration involving C

2O

42– should be carried out at temperatures above 70 °C. If the

titration is carried out at lower temperatures, the redox reaction is incomplete, precipitation of MnO

2 may occur and the result obtained is inaccurate.

6. KMnO4 solution should be added from the burette slowly. If it is added quickly, a brown

suspension of MnO2 is formed.

7. No external indicator is required for KMnO4 titration because KMnO

4 can act as it own indicator.

The end point is reached when one drop of KMnO4 solution produces a light pink colour in the

reaction mixture which should last for more than 35 seconds.


Example 4 To find the mass of FeSO

4.7H

2O in 250 cm3 solution

A sample of iron(II) sulphate crystals, FeSO4.7H

2O, was dissolved in dilute sulphuric acid and

made up to 250 cm3 with distilled water in a standard flask. 25.0 cm3 of this solution needed 25.80 cm3 of KMnO

4 for complete reaction.

(a) Write the ionic equation for the reaction between iron(II) sulphate and KMnO4.

(b) If the concentration of KMnO4 solution is 3.16 g dm–3, calculate the concentration (in g dm–3) of

(i) FeSO4, (ii) FeSO

4.7H

2O.

(c) What was the mass of FeSO4.7H

2O dissolved in 250 cm3 solution?

Solution

(a) 5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H

2O

(b) (i) Step 1 Calculate the concentration of KMnO4 in mol dm–3

Relative molecular mass of KMnO4 = 158

3.16Concentration of KMnO

4 = ——— = 0.020 mol dm–3

158

Step 2 Calculate the concentration of Fe2+

(M1V

1)

Fe2+ 5—————————— = —

(M2V

2)

KMnO4 1

0.020 � 25.8Concentration of Fe2+ = 5 � —————————————————————————— = 0.103 mol dm–3

25Relative molecular mass of FeSO

4 = 152

Concentration of FeSO4

= 0.103 � 152 = 15.7 g dm–3

(ii) Relative molecular mass of FeSO4.7H

2O = 278

Concentration of FeSO4.7H

2O = 0.103 � 278 = 28.6 g dm–3

(c) Mass of FeSO4.7H

2O dissolved in 250 cm3 solution

28.6= ————— = 7.15 g 4

Example 5 To find the concentration of Na

2C

2O

4

F1 is a solution containing oxalic acid, H2C

2O

4 and sodium oxalate, Na

2C

2O

4.

F2 is 0.10 mol dm–3 sodium hydroxide solution.

F3 is 0.024 mol dm–3 KMnO4 solution.

25.0 cm3 of F1 required 18.50 cm3 of F2 for complete neutralisation.25.0 cm3 of F1 required 33.60 cm3 of F3 for complete redox reaction.(a) Calculate the concentration (in g dm–3) of H

2C

2O

4 in F1 solution

(b) Calculate the concentration (in mol dm–3) of C2O

42– ions in F1 solution.

(c) Hence, calculate the concentration (in g dm–3) of Na2C

2O

4 in F1 solution.

Solution

(a) H2C

2O

4 + 2NaOH → Na

2C

2O

4 + 2H

2O

(M1V

1)

H2C2O4 1———————————— = —

(M2V

2)

NaOH 2

1 0.1 � 18.5Concentration of H

2C

2O

4 = — � —————————— = 0.037 mol dm–3

2 25

Relative molecular mass of H2C

2O

4 = 90

Concentration of H2C

2O

4 = 0.037 � 90 = 3.33 g dm–3

(b) 5C2O

42– + 2MnO

4– + 16H+ → 2Mn2+ + 10CO

2 + 8H

2O

(M1V

1)

C2O42– 5

———————————— = — (M

2V

2)

KMnO4 2


5 0.024 � 33.6Concentration of C

2O

42– = — � ————————————— = 0.0806 mol dm–3

2 25

(c) Total concentration of C2O

42– = 0.0806 mol dm–3

Concentration of H2C

2O

4 = 0.037 mol dm–3

Concentration of Na2C

2O

4 = 0.0806 – 0.037 = 0.0436 mol dm–3

Relative molecular mass of Na2C

2O

4 = 134

Concentration of Na2C

2O

4 = 0.0436 � 134 = 5.84 g dm–3

Example 6 Acidified hydroxyammonium ion, NH

3OH+, reduces iron(III) ion, Fe3+, to iron(II) ion, Fe2+. In the

following experiment, you are asked to determine the chemical equation for the reaction between hydroxyammonium ion and iron(III) ion. F4 is a solution prepared by boiling 1.56 g of hydroxyammonium sulphate, NH

3OH+HSO

4– ,with

excess iron(III) ammonium sulphate and dilute sulphuric acid. The reaction mixture is then made up to 250 cm3 with distilled water. F5 is potassium manganate(VII) solution containing 1.58 g of KMnO

4 per 500 cm3.

In a titration experiment, 25.0 cm3 of F4 required 24.4 cm3 of F5 for complete reaction.(a) Calculate the concentration of (i) NH

3OH+HSO

4–, (ii) KMnO

4 in mol dm–3.

(b) Calculate the concentration (in mol dm-3) of Fe2+ ions produced in F4.(c) Hence, calculate the number of moles of Fe3+ that react with 1 mol of NH

3OH+.

(d) The half-equation for the oxidation of hydroxylamine to nitrogen is:

2NH2OH → N

2O + H

2O + 4H+ + 4e–

Hence, write a balanced redox equation between NH3OH+ ions and Fe3+ ions.

Solution

(a) (i) Relative molecular mass of NH3OH+HSO

4– = 131

1.56 � 4Concentration = ———————— = 0.0476 mol dm–3

131

(ii) Relative molecular mass of KMnO4 = 158

1.58 � 2Concentration = ———————— = 0.020 mol dm–3

158

(b) 5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H

2O

(M1V

1)

Fe2+

5———————————— = —

(M2V

2)

MnO4– 1

5 � 0.02 � 24.4Concentration of Fe2+ = ———————————————— = 0.0976 mol dm–3

25

(c) Fe3+ + e– → Fe2+

1 mol of Fe2+ is produced from 1 mol of Fe3+.

Number of moles of Fe3+ that have reacted with NH3OH+ = 0.0976

NH3OH+ : Fe3+ = 0.0476 : 0.0976 = 1 : 2

(d) 2NH2OH → N

2O + H

2O + 4H+ + 4e–

Fe3+ + e– → Fe2+

2NH2OH + 4Fe3+ → 4Fe2+ + N

2O + H

2O + 4H+


1.4 Iodine-sodium thiosulphate (Na2S2O3) titration

1. The half-equations for the redox reaction between iodine and thiosulphate are

2S2O

32– → S

4O

62– + 2e– … (1)

I2 + 2e– → 2I– … (2)

2. The ionic equation for the reaction between sodium thiosulphate and iodine is

2S2O

32– + I

2 → S

4O

62– + 2I– … (3)

This reaction is used to determine the concentration of iodine by titration against sodium thiosulphate. In some cases, the reaction is used to determine the concentration of an oxidising agent. For example, the oxidising agent is added to potassium iodide solution. The iodine liberated is titrated against sodium thiosulphate. From this, the concentration of the oxidising agent can be deduced indirectly.

3. Iodine has a low solubility in water but dissolves readily in potassium iodide solution.

I2(s) + KI(aq) I

3–(aq) + K+(aq)

In most direct titrations with iodine, a solution of iodine in potassium iodide is used. For simplicity, equation (3) is commonly used for calculation.

Important equations for iodine-sodium thiosulphate titration

Note: a = number of moles of the first reactant; b = number of moles for thiosulphate

Ionic equation aRatio of — b

2S2O

32– + I

2 → S

4O

62– + 2I–

In the following reactions, iodine produced by the reaction is titrated against sodium thiosulphate solution.

H2O

2 + 2H+ + 2I–

→ 2H

2O + I

2

(H2O

2 ≡ I

2 ≡ 2S

2O

32–)

IO3– + 6H+ + 5I– → 3H

2O + 3I

2

(IO3– ≡ 3I

2 ≡ 6S

2O

32–)

2Cu2+ + 4I– → Cu

2I

2 + I

2

(2Cu2+ ≡ I

2 ≡ 2S

2O

32–)

1—2

1—2

1—6

1—1

Precautions in iodine-sodium thiosulphate titrations

1. Iodine is a volatile substance. Hence the iodine solution prepared should be titrated immediately and quickly to avoid loss of iodine due to evaporation.

2. In I2 – S

2O

32– titration, starch is used as an indicator. The starch solution should not be added

at the beginning of the titration where there is a high concentration of iodine. This is because iodine is adsorbed onto the starch molecule and may remain adsorbed even at the end point.

3. The starch solution should be added towards the end of the titration when the reaction mixture turns pale yellow. The reaction mixture in the conical flask should be shaken vigorously but carefully as the titration proceeds.

4. The starch solution will produce a dark blue colour with iodine and when the end point is reached, the solution turns colourless abruptly.

5. When the solution in the conical flask is left aside after titration, the solution becomes blue again. Ignore this, because iodine is formed due to the atmospheric oxidation of excess potassium iodide in the reaction mixture.

6. Starch solution is unstable and should be prepared fresh for each titration.


Example 1To determine the concentration of iodine solutionF1 contains 20.0 g of Na

2S

2O

3.5H

2O per dm3.

F2 is an iodine solution.25.0 cm3 of F2 required 22.5 cm3 of F1 for complete reaction.Calculate (a) the concentration (in mol dm–3) of iodine in solution F2, (b) the mass of iodine required to prepare 200 cm3 of F2.

Solution

(a) Relative molecular mass of Na2S

2O

3.5H

2O = 248

20.0Concentration of Na

2S

2O

3.5H

2O = ———— = 0.0806 mol dm–3

248

2S2O

32– + I

2 → S

4O

62– + 2I–

(M1V

1)

I2 1

———————————— = — (M

2V

2)

S2O32– 2

1 0.0806 � 22.5Concentration of I

2 = — � ————————— = 0.0363 mol dm–3

2 25.0

(b) Relative molecular mass of I2 = 254

200Mass of iodine = (0.0363 � 254) � ———— = 1.84 g

1000

Example 2 To determine the percentage purity of hydrated sodium sulphite, Na

2SO

3.7H

2O

F3 is a solution containing 11.1 g of hydrated sodium sulphite, Na2SO

3.7H

2O per dm3.

F4 is 0.050 mol dm–3 iodine.F5 is 0.025 mol dm–3 sodium thiosulphate solution.25.0 cm3 of F4 is mixed with 25.0 cm3 of F3. The resulting solution required 18.90 cm3 of F5 for complete reaction. Sulphite ion reacts with iodine as represented by the equation

SO32– + I

2 + H

2O SO

42– + 2HI

SolutionStep 1 Calculate excess iodine 0.025 � 18.90

Number of moles of S2O

32– that reacted with excess I

2 = —————————————— = 4.725 � 10–4

10002S

2O

32– + I

2 → S

4O

62– + 2I–

1Number of moles of excess I

2 = — � 4.725 � 10–4 = 2.36 � 10–4

2

Step 2 Calculate amount of iodine reacted with SO32–

0.050 � 25Number of moles of I

2 added = ——————————— = 1.25 � 10–3

1000

Number of moles of I2 reacted = 1.25 � 10–3 – 2.36 � 10–4 = 1.014 � 10–3

Step 3 Calculate mass of pure Na2SO

3.7H

2O

SO32– + I

2 + H

2O SO

42– + 2HI

Number of moles of pure Na2SO

3.7H

2O = 1.014 � 10–3

Relative molecular mass of Na2SO

3.7H

2O = 252

Mass of Na2SO

3.7H

2O in 25.0 cm3 solution = (1.014 � 10–3) � 252 = 0.256 g

1000Mass of Na

2SO

3.7H

2O in 1000 cm3 solution = 0.256 � ————— = 10.24 g

25


Step 4 Calculate percentage purity of Na2SO

3.7H

2O

10.24% purity = —————— � 100 = 92.3

11.1

Alternative method

Step 1 Calculate the volume of iodine that did not react with the sulphite ion

2S2O

32– + I

2 → S

4O

62– + 2I–

(M1V

1)

I2 1————————————————— = —

(M2V

2)

S2O32– 2

1 0.025 � 18.9Volume of I

2 = — � —————————————

2 0.05

= 4.73 cm3

Step 2 Calculate the volume of iodine that reacted with the sulphite ion25.0 – 4.73 = 20.3 cm3

Step 3 Calculate the concentration of Na2SO

3.7H

2O

SO32– + I

2 + H

2O SO

42– + 2HI

(M1V

1)

SO32– 1———————————————— = —

(M2V

2)

I2 1

0.050 � 20.3Concentration of SO

32– = ————————————— = 0.0406

25

Relative molecular mass of Na2SO

3.7H

2O = 252

Concentration of Na2SO

3.7H

2O = 0.0406 � 252 = 10.23 g dm–3

Step 4 Calculate the percentage purity of hydrated sodium sulphite

10.23% purity = ——————— � 100 = 92.2 %

11.1


SECTION 2 Physical chemistry

Thermochemistry

Important formula

∆H = mc∆Twhere m = mass of solution, c = specific heat capacity (4.2 J g–1 K–1) ∆T = maximum rise or fall in temperature

Example 1 To determine the heat of reaction between a metal hydrogen carbonate and dilute hydrochloric acid.F1 is 1.0 mol dm–3 hydrochloric acid.F2 is a metal hydrogen carbonate (MHCO

3).

50.0 cm3 of F1 is placed in a polystyrene cup. 4.7 g of F2 is added to F1. The mixture is constantly stirred and the temperature recorded.

1. The density of solution is usually assumed to be 1.0 g cm–3.

Hence, the mass of solution = volume of solution used.

2. The mass of solution does not include the mass of solid added to water. For example, when 1.0 g of NaOH is added to 100 cm3 of water, the mass of solution is 100 g and not 101 g.

Take NoteTake Note

Temperature of F1 before mixing

Lowest temperature reached after mixing

Temperature change

30.0 °C

25.0 °C

5.0 °C

(a) Calculate the amount of heat absorbed in the experiment.(b) Calculate the number of moles of HCl used.(c) Calculate the number of moles of MHCO

3 used.

(d) Calculate the number of moles of MHCO3 reacted.

(The relative molecular mass of the metal hydrogen carbonate = 100)(e) Calculate the enthalpy change for the reaction.

Solution

(a) ∆H = mc∆T= 50.0 � 4.2 � 5.0 = 1050 J = 1.05 kJ

1.0 � 50.0(b) Number of moles of HCl used = —————————— 1000

= 0.05

4.7(c) Number of moles of MHCO

3 used = ———

100

= 0.047

(d) MHCO3(aq) + HCl(aq) → MCl(aq) + H

2O(l) + CO

2(g); ∆H (+)ve

HCl is in excess. Number of moles of MHCO

3 reacted = 0.047

1.05 (e) ∆H = ————— = 22.3 kJ mol–1

0.047

Example 2To determine the partition of an organic acid, HOOC(CH

2)

nCOOH between water and ether.

F3 is a solution of an organic acid, HOOC(CH2)

nCOOH.

F4 is 0.1 mol dm–3 NaOH solution.F5 is 0.01 mol dm–3 NaOH solution.40.0 cm3 of F3 is added to a bottle, followed by 60.0 cm3 of ether. The bottle is tightly closed and shaken vigorously. After 30 minutes, 10.0 cm3 of the ether layer is pipetted out and titrated with F5. Then, 10.0 cm3 of aqueous layer is pipetted out and titrated with F4.


The experiment is repeated using (a) 50.0 cm3 of ether and 50.0 cm3 of F3.(b) 40.0 cm3 of ether and 60.0 cm3 of F3.

Results

Experiment

1

2

3

Volume of solution usedVolume of F4 (V

1) Volume of F5 (V

2)

40 cm3 F3+

60 cm3 ether

50 cm3 F3+

50 cm3 ether

60 cm3 F3+

40 cm3 ether

12.5 cm3

15.5 cm3

18.8 cm3

18.25 cm3

23.4 cm3

28.1 cm3

Titration results

(a) Plot a graph of V1 (on the y-axis) against V

2 (on the x-axis).

(b) Based on the graph, determine the partition coefficient for the organic acid between water and ether.

Solution

Concentration of acid in aqueous layer(b) K = ——————————————————————————————— Concentration of acid in ether layer

Concentration of acid is proportional to the number of moles of alkali used.

V1 (20 – 10)

—— = ——————— = 0.67 V

2 (30 – 15)

V1 � 0.1

∴ K = ———————— V

2 � 0.01

0.67 � 0.1= ————————

0.01

= 6.67

20

10

0 10 20 30

15

10

Solution

V1 (cm3)

V2 (cm3)


SECTION 3 Qualitative analysis

Table 1 Reactions with dilute hydrochloric acid

(I) Gas released

(II) Precipitate formed

Table 2 Reactions with aqueous sodium hydroxide

(I) Gas liberated

(II) Precipitate formed

Gas

CO2

SO2

NO2

CH3COOH

Colour OdourAction on blue litmus paper Specific test

Inference

Observation

Colourless

Colourless

Brown

Colourless

Odourless

Pungent

Pungent

Vinegar

Little effect

Blue → red

Blue → red

Blue → red

Limewater turns milky

• Acidified K2Cr

2O

7

Orange → Green• Acidified KMnO

4

Decolorisation

–

–

CO32– or HCO

3– present

SO32– or S

2O

32– present

NO2– present

CH3COO– present

Observation

White precipitate, soluble in concentrated HCl

White precipitate, turns grey on exposure to sunlight

Yellow precipitate

White precipitate, dissolves when solution is heated. On cooling, yellow crystals formed

Pb2+ present

Ag+ present

S2O

32– present

Pb2+ present

PbCl2 formed. Solubility in

concentrated HCl due to formation of H

2PbCl

4

AgCl formed. On exposure to sunlight, Ag deposited

Sulphur formed

PbCl2 formed. PbCl

2 is insoluble in

water but dissolves readily in hot water

Inference Explanation

Observation

Red litmus → blue

White fumes with concentrated HCl vapour

NH4+ present NH

3 liberated

NH3 + HCl → NH

4Cl


Observation

White precipitate insoluble in excess NaOH

White precipitate soluble in excess NaOH

White precipitate turns rapidly to brown; insoluble in excess NaOH

Brown precipitate, insoluble in excess NaOH

Dirty green precipitate, insoluble in excess NaOH

Green precipitate, insoluble in excess NaOH

Blue precipitate, insoluble in excess NaOH

Bluish-green precipitate, soluble in excess NaOH to form a green solution

Mg2+, Ca2+ or Ba2+

present

Pb2+, Zn2+ or Al3+ present

Mn2+ present

Fe3+ present

Fe2+ present

Ni2+ present

Cu2+ present

Cr3+ present

The basic hydroxides, M(OH)2 are formed

The amphoteric hydroxides are formed

Mn(OH)2 is formed; oxidised by air to

Mn(OH)3

The basic hydroxide, Fe(OH)3 is formed

The basic hydroxide, Fe(OH)2 is formed,

Some Fe(OH)2 is oxidised to brown Fe(OH)3

The basic hydroxide, Ni(OH)2 is formed

The basic hydroxide, Cu(OH)2 is formed

The amphoteric hydroxide, Cr(OH)3 is

formed



Table 3 Reactions with aqueous ammoniaNote: The asterisk (*) shows the cation that will form a precipitate with NH

3(aq) but the precipitate

will dissolve when NH4Cl is added.

1. Ammonia is a weak base. In aqueous solution, it undergoes partial dissociation to form NH4

+ and OH– ions.NH3(aq) + H2O(l) NH4

+(aq) + OH–(aq)2. Ca2+(aq) forms a

precipitate with NaOH(aq) but it does not form a precipitate with NH3(aq). This is because the concentration of OH– ions in aqueous ammonia is very low. The ionic product of Ca(OH)2 for ammonia is lower than the solubility product (Ksp) of Ca(OH)2. Hence, precipitation of Ca(OH)2 cannot occur.

3. Some metal hydroxides, such as Mg(OH)2, dissolve in NH4Cl, because of the common ion effect. In the presence of NH4Cl, the concentration of OH– is decreased due to the common ion, NH4

+. The metal hydroxide dissolves because the ionic product is less than the solubility product.

Take NoteTake NoteObservation

White precipitate, insoluble in excess NH

3(aq)

White precipitate, soluble in excess NH

3(aq)

White precipitate, turns rapidly to brown colour; insoluble in excess NH

3(aq)

Dirty green precipitate, insoluble in excess NH

3(aq)

Brown precipitate, insoluble in excess NH

3(aq)

Bluish-green precipitate, insoluble in excess NH

3(aq)

Green precipitate, soluble in excess NH

3(aq) to form

light blue solution

Blue precipitate, soluble in excess NH

3(aq) to form

dark blue solution

Inference

Pb2+, *Mg2+, Al3+ present

*Zn2+ present

*Mn2+ present

*Fe2+ present

Fe3+ present

Cr3+ present

*Ni2+ present

*Cu2+ present

Explanation

The metal hydroxides are formed

Zn(OH)2 is soluble in

NH3 due to the formation

of [Zn(NH3)

4]2+

Mn(OH)2 is oxidised

easily to Mn(OH)3

Fe(OH)2 formed and

oxidised by air to brown Fe(OH)

3

Fe(OH)3 formed

Cr(OH)3 formed

Ni(OH)2 formed which

dissolves in NH3(aq) to

form the complex ion [Ni(NH

3)

6]2+

Cu(OH)2 formed which

dissolves in NH3(aq) to

form the complex ion, [Cu(NH

3)

4]2+

Table 4 Reactions with aqueous iron(III) chloride, FeCl3(aq)

(I) Formation of precipitate

Observation

Buff-coloured (yellowish-brown) precipitate

Brown precipitate; CO2 liberated

Inference

C6H

5COO– (benzoate) present

CO32– present

Explanation

Formation of iron(III) benzoate, (C

6H

5COO)

3Fe

Formation of iron(III) carbonate; CO

2 liberated because FeCl

3(aq)

undergoes hydrolysis to produce H+ ions

(II) Colour change without precipitation

Observation

Solution turns red; Brown precipitate formed on heating

Inference

CH3COO– or HCOO– present

Explanation

(CH3COO)

3Fe or (HCOO)

3Fe

formed. On heating, (CH3COO)

3Fe

or (HCOO)3Fe undergoes

hydrolysis to form the basic salt, e.g. (CH

3COO)Fe(OH)

2


Table 5 Reactions with silver nitrate solution

Observation

Solution turns brownish-red; black deposit may also formed

Solution turns deep purple/violet

Inference

I– present

Phenol, C6H

5OH, or derivatives

of phenol, e.g. HOC6H

4COOH

present

Explanation

FeCl3 reduces I– to I

2

Formation of complex ion

Observation

White precipitate, insoluble in HNO

3 but soluble in dilute

ammonia solution

Cream-coloured precipitate; insoluble in HNO

3; insoluble in

dilute ammonia solution

Yellow precipitate; insoluble in HNO

3; insoluble in dilute

ammonia solution

White precipitate, soluble in HNO

3

White precipitate, soluble in NH

3 and hot water

Inference

Cl– present

Br– present

I– present

SO32–, NO

2– , C

2O

42–, CH

3COO–

or HCOO– present

C6H

5COO– present

Explanation

AgCl formed. Dissolves in NH

3(aq) because of complex ion

formation, [Ag(NH3)

2]+(aq)

AgBr formed

AgI formed

Ag+ ions react with these anions to form insoluble salts. Insoluble salts of weak acids are soluble in HNO

3

C6H

5COOAg is insoluble in cold

water but soluble in hot water

Table 6 Reactions with potassium chromate(VI) solution

Observation

Yellow precipitate, soluble in HNO

3

Yellow precipitate, soluble in HCl

Solution turns green

Inference

Ba2+ or Pb2+ present

Ba2+ present

Reducing agent, such as SO32–

or NO2– present

Explanation

BaCrO4 or PbCrO

4 precipitated

BaCrO4 precipitated

Reducing agent reduces CrO42– to

Cr3+ (green)

Table 7 Reactions with potassium iodide solution

NO2– (nitrite) can act as

an oxidising agent or a reducing agent depending on the reagent added to it.

Take NoteTake NoteObservation

Pale yellow precipitate in brown solution

Yellow precipitate, soluble in hot water. Yellow crystals formed on cooling

Solution turns brown; black deposit formed

Inference

Cu2+ present

Pb2+ present

Oxidising agent present

Explanation

2Cu2+ + 4KI →2CuI(s) + I

2 + 4K+

PbI2 formed. PbI

2 is

soluble in hot water but insoluble in cold water

Oxidising agents such as Fe3+, CrO

42–, Cr

2O

72– or

NO2– oxidises KI to I

2.


Table 8 Reactions of sodium ethanoate, CH3COONa

Observation

Solution turns red. On heating, a brown precipitate formed

No visible change. On heating, white precipitate formed

Inference

Fe3+ present

Al3+ present

Explanation

3CH3COO–(aq) + Fe3+(aq) → (CH

3COO)

3Fe

red solutionOn heating,(CH

3COO)

3Fe + 2H

2O → (CH

3COO)Fe(OH)

2 + 2CH

3COOH

brown precipitate

3CH3COO–(aq) + Al3+(aq) → (CH

3COO)

3Al

colourless

(CH3COO)

3Al + 2H

2O → (CH

3COO)Al(OH)

2 + 2CH

3COOH

white precipitate

Table 9 Reactions of sodium carbonate solution

Table 10 Identification of functional groups in organic compounds

Reagent

Cl2 water

Sodium chlorate(I), NaClO

Br2 water

NaNO2 in HCl at 5 °C

followed by phenol

Alcohol + a few drops of concentrated H

2SO

4, then heat

White precipitate

Solution turns purple

(a) Decolourisation

(b) Decolourisation and white precipitate formed

Red dye

Sweet-smelling odour

C6H

5OH or C

6H

5NH

2

present

C6H

5NH

2 present

Unsaturated organic compound present

C6H

5OH or C

6H

5NH

2

present

C6H

5NH

2 present

RCOOH present

Precipitation of2,4,6–trichlorophenol or2,4,6–trichlorophenylamine

Formation of organic complex

C = C + Br2 →

� �

– C – C – � � Br Br

colourless

Precipitation of2,4,6–trichlorophenol or 2,4,6–trichlorophenylamine

Diazotisation followed by coupling reaction to form coloured dye

Observation Inference Explanation

Observation

Effervescence. Gas turns limewater milky

Pungent gas liberated.Gas turns red litmus blue

White precipitate formed

White precipitate, turns brown

Brown precipitate, gasliberated turns limewatermilky

Green precipitate

Inference

Acid, such as C6H

5COOH

or H2C

2O

4 or acid salt,

such as HSO4– present

NH3 given off. NH

4+

present

Ba2+, Ca2+, Mg2+, Pb2+ orZn2+ present

Mn2+ present

Fe3+ present

Fe2+, Cr3+ or Ni2+ present

Explanation

2H+ + CO32– → H

2O + CO

2

Na2CO

3 is a base. A base will react with

NH4+ to give NH

3(g)

These cations react with CO32–(aq) to form

insoluble metal carbonates

MnCO3 precipitate is oxidised to Mn(III) salt

Brown precipitate is Fe(OH)3.

Fe2(CO

3)

3 is unstable and undergoes rapid

hydrolysis to form Fe(OH)3. Fe3+(aq) is

acidic and reacts with CO32– to give CO

2

These cations react with CO32–(aq) to form

insoluble metal carbonates

guide to stpm practical 2012

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relative atomic

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dilute sulphuric

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