Ace Ahead Chemistry (CD-Rom)1st(17.2.11)
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Transcript of Ace Ahead Chemistry (CD-Rom)1st(17.2.11)
1© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
PRACTICAL GUIDE
2 © Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
3© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
4 © Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
5© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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)
6 © Oxford Fajar Sdn. Bhd. (008974-T) 2011
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.
7© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
8 © Oxford Fajar Sdn. Bhd. (008974-T) 2011
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+
9© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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.
10 © Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
11© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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
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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.
13© Oxford Fajar Sdn. Bhd. (008974-T) 2011
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)
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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
Inference Explanation
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
Inference Explanation
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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
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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.
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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