Introduction
To the student
The purpose of this workbook is to help your understanding and learning of the chemistry covered in
NCEA Level 2 Chemistry. It contains the essential material you must know.
There are seven chapters in the workbook – one for each Achievement Standard. The workbook is written
so that it does not have to be read sequentially.
Each chapter comprises explanations and examples, along with activities, which are sets of questions
designed to help you determine your understanding of the concepts covered. The content of the workbook
is based on the principle that practice makes perfect – this is especially true for mathematical calculations.
It is important to understand the language of the questions. You may be asked to:
• Describe – ie identify, name, draw, give characteristics or an account of.
• Explain – ie provide reasons for how and why.
• Discuss – ie show understanding as to how or why something occurs by linking chemistry ideas/
principles; this may require you to justify, relate, evaluate, compare and contrast and/or analyse.
At the end of most chapters there is an extra activity, called Test yourself. This activity comprises a
summary of the types of questions presented in the chapter. You can do this ‘test’ either:
• as you complete each Achievement Standard, or
• later, as a means of revising in preparation for an assessment.
Answers are provided at the end of the workbook. Most answers include an explanation and/or show full
working.
Remember: This is your workbook, for you to use in the best way to assist your
learning and understanding (and enjoyment!) of chemistry.
To the teacher
The workbook is designed to help the student work at his/her own pace, and contains the essential material
that students must know. Information has been provided to support the understanding of the principles of
chemistry.
This workbook will prove to be an excellent support for your teaching, and a complement to classroom
learning.
Acknowledgements
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Eluned FitzJohn and Jan Giffney did excellent work checkreading the book – grateful thanks are offered.
Eluned also provided many very helpful suggestions concerning the presentation of the material covered.
Thanks are also due to the typesetter, Duraid Suliman, and the editors, Helen Eastwood and Terry Bunn.
Alex Eames (Cambridge)
October 2010
Internally assessed
3 credits
Carry out qualitative analysis
Qualitative analysis
Qualitative analysis is finding out what is present in a chemical substance. (Not to be confused with quantitative
analysis – which is how much of a substance is present.) Achievement Standard 90305 (Chemistry 2.1) involves
the qualitative analysis of ions in solution. The ions that you need to know follow.
Positive ions (cations)
Ammonium
NH4+
Silver
Ag+
Sodium Barium Copper Iron(II)
Na+
Ba2+
Cu2+
Fe2+
Lead
Pb2+
Magnesium
Mg2+
Zinc
Zn2+
Iron(III) Aluminium
Fe3+
Al3+
Negative ions (anions)
Chloride
Cl –
Hydroxide
OH –
Iodide
I–
Nitrate
NO3–
Carbonate
CO32–
Sulfate
SO42–
Ionic compounds
Ionic compounds are formed by positive and negative ions combining. The formula of an ionic compound is
obtained by balancing the charges of the ions so that the overall charge of the compound is zero. The formula
represents the simplest unit or empirical formula of the compound. The term molecule should not be used for
compounds composed of ions.
Example
Sodium chloride, NaCl – formed by one positive sodium ion, Na+, balancing one negative chloride ion, Cl–.
NaCl: 1 × Na+ balances 1 × Cl–, overall charge is zero.
Zinc nitrate, Zn(NO3)2 – formed by one positive zinc ion, Zn2+, balancing two negative nitrate ions, NO3–.
Zn(NO3)2: 1 × Zn2+ balances 2 × NO3–, overall charge is zero.
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Aluminium sulfate, Al2(SO4)3 – formed by two positive aluminium ions, Al3+, balancing three negative
sulfate ions, SO42–.
Al2(SO4)3: 2 × Al3+ balances 3 × SO42–, overall charge is zero.
Activity 1A: Formulae of ionic compounds
Write the chemical formula of the following compounds:
1. Sodium carbonate
2. Zinc iodide
3. Copper hydroxide
4. Lead nitrate
5. Ammonium sulfate
6. Iron(II) chloride
7. Aluminium hydroxide
8. Iron(III) sulfate
9. Barium sulfate
10.Magnesium nitrate
AS 90305
Chemistry 2.1
2 Year 12 Chemistry Learning Workbook
Solubility rules
These solubility rules describe which ionic compounds are soluble in water and which are insoluble in water.
Compound Solubility Exceptions
Nitrate
Soluble
None
Chloride
Soluble
Silver chloride, AgCl; Lead chloride, PbCl2
Iodide
Soluble
Silver iodide, AgI; Lead iodide, PbI2
Sulfate
Soluble
Lead sulfate, PbSO4; Barium sulfate, BaSO4
Hydroxide
Insoluble
* Barium hydroxide, Ba(OH)2; Sodium hydroxide, NaOH
Carbonate
Insoluble
Sodium carbonate, Na2CO3; Ammonium carbonate, (NH4)2CO3
*Barium hydroxide is sparingly soluble.
Activity 1B: Solubility of ionic compounds
Using the table of solubility rules, identify the following compounds (listed by name or by formula) as soluble or
insoluble.
1. Lead nitrate
5. Barium sulfate
2. PbCO3
6. Al(OH)3
3. AgI
7. CuSO4
4. Magnesium chloride
8. Iron(II) sulfate
Precipitate formation
When two soluble ionic compounds are mixed, there are four ions in solution. If one of the positive ions and
one of the negative ions can combine to form an insoluble compound, then a precipitate will occur.
Example
Sodium carbonate, Na2CO3, solution added to copper sulfate, CuSO4, solution results in Na+ and CO32–, Cu2+
and SO42–, ions being present.
The table shows the possible reactions between these ions:
Cu2+
SO42–
Na+
No reaction
Soluble compound, Na2SO4
CO32–
Insoluble compound, CuCO3, formed
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No reaction
So, when a copper sulfate solution, CuSO4(aq), is mixed with a solution of sodium carbonate, Na2CO3(aq),
copper ions, Cu2+(aq), combine with carbonate ions, CO32–(aq), to form a green-blue precipitate of insoluble
copper carbonate, CuCO3(s). The solution will contain sodium ions and sulfate ions.
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Chemistry 2.1: Carry out qualitative analysis 3
Cu2+
SO42–
CuSO4(aq)
Na2CO3(aq)
solutions
mixed
Na+
CO32–
precipitation
occurs
CuCO3(s)
The sodium ions, Na+(aq), and the sulfate ions, SO42–(aq), are spectator ions since they are not involved in
the precipitation reaction. They do not form a precipitate because sodium sulfate is soluble.
In an equation:
• (aq) indicates dissolved in water.
• (g) indicates a gas.
• (s) indicates the substance is a solid.
• (l) indicates a liquid.
An accepted abbreviation of the term precipitate is ppt.
Chemical equations
The following four equations represent the same reaction – sodium carbonate, Na2CO3, solution added to
copper sulfate, CuSO4, solution. The ionic equation gives the clearest indication of the particles involved in the
reaction.
copper sulfate + sodium carbonate → copper carbonate + sodium sulfate
The word equation is:
CuSO4(aq) + Na2CO3(aq) → CuCO3(s) + Na2SO4(aq)
The formula equation is:
2+
2–
The ionic equation is: Cu (aq) + SO4 (aq) + 2Na+(aq) + CO32–(aq) → CuCO3(s) + 2Na+(aq) + SO42–(aq)
An ionic equation can be abbreviated to include only the ions that react:
Cu2+(aq) + CO32–(aq) → CuCO3(s)
The ions that are omitted (ie 2Na+(aq) and SO42–(aq)) are termed spectator ions.
Appearance of insoluble compounds
Only the following compounds listed as being insoluble can be formed as a precipitate.
Compound
Nitrate
Chloride
Iodide
Sulfate
Colour of insoluble compounds
All are soluble.
Silver chloride, white; Lead chloride, white.
Silver iodide, pale yellow; Lead iodide, yellow.
Lead sulfate, white; Barium sulfate, white.
Hydroxides of Aluminium, Lead, Magnesium and Zinc are white; Copper hydroxide, blue;
Hydroxide
Iron(II) hydroxide, green; Iron(III) hydroxide, red-brown or orange; *Silver oxide, brown.
Carbonates of Barium, Lead, Magnesium and Zinc are white; Copper carbonate, green/blue;
Iron(II) carbonate, grey; Silver carbonate, yellow.
Carbonate
Carbonates of iron(III) and aluminium do not exist.
*Silver oxide is the product rather than silver hydroxide because the hydroxide is unstable.
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Chemistry 2.2
Internally assessed
3 credits
AS 90306
Carry out an acid-base volumetric analysis
Volumetric analysis
Volumetric analysis involves reacting solutions of known volumes and concentrations and can be used to
determine an unknown concentration.
Acid-base titrations
0
9
10
burette
20
30
9.40 mL
10
the volume of
a burette is read
from the bottom
of the meniscus
40
pipette filler
50
wash
bottle
conical
flask
25 mL
volume mark
pipette
clamp
stand
Pipette – delivers a fixed volume (eg 20.00 mL, 25.00 mL, etc). The fixed volume of solution delivered by a
pipette is called an aliquot.
Burette – delivers a volume between 0 and 50 mL. The volume of solution delivered by the burette to reach
equivalence point is called a titre.
Conical flask – a flask with sloping sides in which a reaction occurs. The sloping sides are to avoid loss of
solutions due to splashing.
Wash-bottle – contains deionised water to rinse out glass apparatus and wash splashes from the inside of the
conical flask.
Titration – volumetric analysis using burettes and pipettes to accurately determine the concentrations or the
amounts of substances.
Equivalence point – when the amount of acid/base has been neutralised by the base/acid.
End point – the point in the titration at which the indicator changes colour.
Indicator – dye that changes colour with a change in pH.
Concordant – consistent; within small (acceptable) differences of other values.
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16 Year 12 Chemistry Learning Workbook
Example
Indicator
Phenolphthalein
Methyl orange
Common indicators used, and
their colours in acid and base,
are shown in the table.
Colour in acid
Colourless
Red
Colour in base
Pink
Yellow/orange
Practising the titration technique should enable you to get concordant (agreeing) titres. At least three concordant
results within 0.2 mL of each other are necessary for an accurate titration (ie a titration of an ‘Excellence’ standard).
This titre value must also be within 0.2 mL of the expected titre for ‘Excellence’.
Titration procedure
• Wash all glass apparatus to be used thoroughly with tap water to remove any soluble impurities.
• Rinse all glass apparatus with deionised water to remove any soluble impurities in tap water.
• Rinse a clean pipette with the solution it is to contain – this is to remove the water clinging to the inside of
the pipette, which otherwise would dilute the solution by an unknown factor.
• Rinse a clean burette with the solution it is to contain – to remove the water clinging to the inside of the
burette, which otherwise would dilute the solution by an unknown factor.
• Using the pipette, transfer an aliquot (eg 25.00 mL (titres to 4 sf)) of the solution into the conical flask.
• Add 2 or 3 drops of indicator to the conical flask.
• Carry out a rough titration – add 1 mL portions of the solution from the burette, with constant swirling of
the conical flask, until the indicator changes colour. This titre gives you an estimate of the volume required
for complete reaction. Read the burette to 2 decimal places.
• Repeat the titration with constant swirling of the flask, but add 1 mL less than was used in the rough
titration. Then, carefully add one or two drops of solution at a time from the burette until a colour change is
observed. Swirl the conical flask after each addition of the solution from the burette and use the deionised
water from the wash-bottle to rinse any solution on the side of the flask into acid-base mixture.
• Repeat the above step until at least three concordant results are obtained.
Start of titration
pipette
25
conical
flask
Step 1
Solution is pipetted
into a conical flask.
drops of
indicator
mL 0
initial reading
mL 0
10
10
burette 20
20
End of titration
final reading
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Step 2
Indicator is added.
30
30
40
40
50
50
Step 3
Solution of unknown
concentration is
added slowly from
the burette.
Step 4
Titration continues until sufficient
solution is added to just change
colour of the indicator.
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Chemistry 2.2: Carry out an acid-base volumetric analysis 17
Activity 2A: Titration procedure
1.The following is an account of a titration procedure. Complete the account using the following words or
numerals (note that some words are used twice):
burette, colourless, concordant, conical, conical, end point, mL, pink, pipette, swirling, titration,
titre, wash-bottle, 0.2
25.00 mL of sodium hydroxide solution, measured by using a
flask. Three drops of phenolphthalein indicator were added and the colour became
Dilute hydrochloric acid was added from a
The
recorded as 9.85
using a
clean as possible. A fresh 25.00 mL sample of the base was placed in the flask and the
, were placed in a
with constant
was reached when the indicator changed to
(units required). The
.
of the conical flask.
. The
was
flask was washed out with tap water and then,
, the flask was rinsed with deionised water to make the inside of the flask as
repeated to get
results. At least three titres should be within
of each other.
2. Explain each of the following operations in the titration procedure.
a. Why all glass apparatus is washed thoroughly with tap water.
b. Why all glass apparatus is then rinsed with deionised water.
c. Why the pipette and the burette are rinsed out with the solutions they are to contain.
d. Rough titration.
(digits required) mL
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Standard solution and standardising solutions
A standard solution is one for which the concentration is known accurately – to the limits of the equipment
being used.
To prepare a standard solution, a primary standard must be used. A primary standard is a chemical that is
pure, and unaffected by the atmosphere, and soluble in water; ie it is:
• Not oxidised by oxygen in the air.
• Not deliquescent or hygroscopic (ie does not absorb water from the atmosphere).
• Not efflorescent (ie does not lose water of crystallisation to the atmosphere).
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18 Year 12 Chemistry Learning Workbook
A standard solution can be used to standardise (ie find the accurate concentration of) another solution.
Example
Titration experiment
Standardising a solution of hydrochloric acid using a standard solution of sodium hydroxide
You are provided with:
•
•
•
•
250 mL of a 0.0950 mol L–1 sodium hydroxide solution.
250 mL of a solution of hydrochloric acid of unknown concentration.
The apparatus required for a titration procedure.
Phenolphthalein indicator.
Procedure
• Using a pipette, place 25.00 mL of sodium hydroxide solution in a conical flask.
• Add 2 or 3 drops of phenolphthalein indicator – the solution becomes pink.
• Using a burette, add hydrochloric acid, 1 mL at a time, with constant swirling of the flask, until the colour
of the solution is colourless.
• Repeat the titration to the point where one drop of acid produces a colour change from pink to colourless.
• Obtain three concordant results (within 0.2 mL of each other).
Titration results (pipette size = 25 mL)
Titration
1
2
3
4
Average titre (volume HCl) =
Burette reading (mL)
Initial
Final
0.60
23.04
23.04
44.52
1.20
22.74
21.16
42.68
Titre (mL)
22.44
21.48
21.54
21.52
21.48 + 21.54 + 21.52 64.54
=
= 21.51 mL
3
3
Calculation
Equation for the reaction is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l), which indicates a 1 : 1 mol ratio
between the reactants.
Amount of sodium hydroxide, n(NaOH) = c × V = 0.0950 × 0.0250 = 0.00238 = 2.38 × 10–3 mol
Amount of hydrochloric acid, n(HCl) = 2.38 × 10–3 mol
2.38 × 10 –3
n
Concentration of hydrochloric acid, c(HCl) = =
= 0.111 mol L–1,
0.02151
V
or 1.11 × 10–1 mol L–1
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46 Year 12 Chemistry Learning Workbook
Preparing a primary standard solution
• Accurately weigh the solute (ie primary standard reagent) into a small, clean, dry glass container (eg small
beaker). Record this mass.
• Clean a volumetric flask by washing thoroughly with tap water, and then rinsing with deionised/distilled water.
• Using a wash bottle, add just enough deionised/distilled water to the beaker to dissolve the weighed solute.
• Transfer the solution to the volumetric flask using a funnel.
• Partially fill the volumetric flask with deionised water, stopper, and shake to mix thoroughly.
• Remove stopper and let any solution run down from the top of the neck of the flask. Add deionised water
to fill the volumetric flask to the volume mark. Ensure the bottom of the meniscus sits on the volume mark.
• Stopper the flask, and invert it a number of times to mix well. Label the flask with name and concentration
of solution.
dissolved
solute
small beaker
electronic balance
reads zero after tare
button pressed
solute
added
funnel
wash-bottle
deionised
water
readout gives mass
of solute
Step 1 Weigh the solute.
Step 2
Transfer dissolved solute completely
to the volumetric flask.
volume mark
bottom of meniscus
stopper flask
Step 3
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shake thoroughly
Partially fill flask with deionised water,
stopper, and shake thoroughly.
volumetric flask
Fill the flask to the volume mark.
Step 4 Ensure the bottom of the meniscus
sits on the volume mark. Mix
thoroughly. Label the solution.
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Chemistry 2.3: Solve simple quantitative chemical problems 47
Example
To prepare 250 mL standard sodium carbonate solution of concentration approximately 0.1 mol L–1
First, find out the mass of Na2CO3 required for a 0.100 mol L–1 solution.
M(Na2CO3) = 2 × 23.0 + 12.0 + 3 × 16.0 = 106 g mol–1
n(Na2CO3) in 250 mL of 0.100 mol L–1 solution = c × V = 0.100 × 0.250 = 0.0250 (or 2.5 × 10–2) mol
m(Na2CO3) in 250 mL of 0.100 mol L–1 solution = n × M = 0.0250 × 106 = 2.65 g
Then weigh out approximately 2.65 g Na2CO3.
For example, if 2.58 g of Na2CO3 was weighed out (this is close enough to the value of 2.65 g) and made up
to 250 mL in a volumetric flask, the concentration of the standard solution can be calculated:
m
2.58
n(Na2CO3)= =
= 0.0243 mol
M
106
n
0.0243
c(Na2CO3)= =
= 0.0972 mol L–1
V
0.250
Activity 3H: Preparing a primary standard solution
1. Calculate the masses of solute needed to prepare the following solutions.
a. 100 mL of 0.0500 mol L–1 sodium carbonate, Na2CO3, solution.
b. 500 mL of 0.0100 mol L–1 sodium carbonate, Na2CO3, solution.
2. Calculate the concentration of the standard solution prepared by:
a.Dissolving 3.14 g of sodium carbonate, Na2CO3, in water and making the volume up to 250 mL in a
volumetric flask.
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b.Dissolving 1.26 g oxalic acid crystals, H2C2O4.2H2O, in water and making the volume up to 500 mL in a
volumetric flask.
c.Dissolving 24.5 g of sucrose, C12H22O11 , and making the volume up to 250 mL in a volumetric flask.
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Answers 2.1
Achievement Standard 90305
b. i.
ii. Chloride, Cl–
Sulfate, SO42–
iii. Iodide, I–
iv. Hydroxide, OH– or Carbonate, CO32–
Ba2+(aq) + SO42–(aq) → BaSO4(s) or Ba2+(aq) + CO32–(aq) → BaCO3(s)
ii. Ag+(aq) + I–(aq) → AgI(s)
Activity 1F: Complex ions (page 9)
Activity 1A: Formulae of ionic compounds (page 1)
1. complex
1. Na2CO3
2. ZnI2
3. Cu(OH)2
4. Pb(NO3)2
5. (NH4)2SO4
6. FeCl2
7. Al(OH)3
8. Fe2(SO4)3
9. BaSO4
10. Mg(NO3)2
2. ammonia molecule, NH3 / hydroxide ion, OH– / thiocyanate ion, SCN–
Activity 1G: Confirming the identity of metals by the formation of complex ions (page 10)
1. a. Aluminium hydroxide, Al(OH)3 , Lead hydroxide, Pb(OH)2 , Zinc hydroxide, Zn(OH)2
Activity 1B: Solubility of ionic compounds (page 2)
1. Soluble. 2. Insoluble. 3. Insoluble. 4. Soluble. 5. Insoluble. 6. Insoluble. 7. Soluble. 8. Soluble.
Pb(OH)2(s) + 2OH–(aq) → [Pb(OH)4]2–(aq)
Activity 1C: Forming precipitates (page 4)
Zn(OH)2(s) + 2OH–(aq) → [Zn(OH)4]2–(aq)
b. Al(OH)3(s) + OH–(aq) → [Al(OH)4]– (aq)
ii. Chloride, Cl–.
1. a. i.
Silver nitrate (or Silver sulfate), Sodium iodide (or any soluble iodide).
ii. Pale yellow.
2. a. i.
b. i.
Iron(II) sulfate (or chloride or iodide), Sodium (or Ammonium) carbonate.
ii. Grey.
c. i.
Zinc nitrate (or sulfate, or iodide, or chloride), Sodium (or Ammonium or Barium) hydroxide.
ii. White.
Cu(OH)2(s) + 4NH3(aq) → [Cu(NH3)4]2+(aq) + 2OH–(aq)
d. i.
Magnesium nitrate (or sulfate, or iodide, or chloride), Sodium (or Ammonium) carbonate.
ii. White.
Ag2O(s) + 4NH3(aq) + H2O(l) → 2[Ag(NH3)2]+(aq) + 2OH–(aq)
e. i.
Barium nitrate (or chloride or hydroxide), sodium sulfate (or any soluble sulfate).
ii. White.
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl–(aq)
f. i.
Iron(III) sulfate (or nitrate, or chloride, or iodide), Sodium (or Ammonium or Barium) hydroxide. ii. Red-brown.
g. i.
Barium nitrate (or chloride, or iodide, or hydroxide), Sodium (or Ammonium) carbonate.
2. a. Yes, ppt forms – it is Mg(OH)2.
d. Yes, ppt forms – it is FeCO3.
g. No ppt forms.
ii. White.
b. No ppt forms.
c. Yes, ppt forms – it is BaSO4.
e. Yes, ppt forms – it is Cu(OH)2.
f. Yes, ppt forms – it is AgCl.
2. a. Al3+(aq) + 3OH–(aq) → Al(OH)3(s)
b. AlI3, Al(NO3)3
3. a. 2OH–
b. 3OH–
4. a. Ppt is iron(II) hydroxide, Fe(OH)2
Ionic equation is Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)
b. Ppt is iron(II) hydroxide, Fe(OH)2
Ionic equation is Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)
c. Ppt is lead sulfate, PbSO4
Ionic equation is Pb2+(aq) + SO42–(aq) → PbSO4(s)
d. Ppt is aluminium hydroxide, Al(OH)3
Ionic equation is Al3+(aq) + 3OH–(aq) → Al(OH)3(s)
e. Ppt is aluminium hydroxide, Al(OH)3
Ionic equation is Al3+(aq) + 3OH–(aq) → Al(OH)3(s)
f. Ppt is silver chloride, AgCl
Ionic equation is Ag+(aq) + Cl–(aq) → AgCl(s)
g. Ppt is copper hydroxide, Cu(OH)2
Ionic equation is Cu (aq) + 2OH (aq) → Cu(OH)2(s)
Sodium Na+, Ammonium NH4+
iv. Copper Cu2+
b. i.
2+
ii. Aluminium Al3+, Lead Pb2+, Zinc Zn2+ c.White ppt (with NaOH ppt is barium hydroxide, Ba(OH)2 ; with H2SO4 ppt is barium sulfate, BaSO4 ); ionic equation is
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iii. Zinc Zn2+
Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s)
Mg (aq) + 2OH (aq) → Mg(OH)2(s)
Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s)
–
With NH3(aq) – blue ppt of copper hydroxide, Cu(OH)2. Ionic equation is Cu2+ + 2OH– → Cu(OH)2
Complex ion equation is Cu(OH)2 + 4NH3 → [Cu(NH3)4]2+ + 2OH–
c. With NaOH – white ppt of lead hydroxide, Pb(OH)2. Ionic equation is Pb2+ + 2OH– → Pb(OH)2
Complex ion equation is Pb(OH)2 + 2OH– → [Pb(OH)4 ]2–
With NH3(aq) – white ppt of lead hydroxide, Pb(OH)2. Ionic equation is Pb2+ + 2OH– → Pb(OH)2
With H2SO4 – white ppt, lead sulfate, PbSO4. Ionic equation is Pb2+ + SO42– → PbSO4
d. With NaOH – white ppt of zinc hydroxide, Zn(OH)2. Ionic equation is Zn2+ + 2OH– → Zn(OH)2
With excess NaOH – Zn(OH)2 + 2OH– → [Zn(OH)4 ]2–
With NH3(aq) – white ppt of zinc hydroxide, Zn(OH)2. Ionic equation is Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s)
With excess NH3 – Zn(OH)2 + 4NH3 → [Zn(NH3)4]2+ + 2OH–
Ionic equation is Ba
2+
+ SO4
2–
b. Pale yellow ppt of silver iodide, AgI.
→ BaSO4
2. White ppt of silver chloride, AgCl, forms.
–
ii. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)
b. With NaOH – blue ppt of copper hydroxide, Cu(OH)2. Ionic equation is Cu2+ + 2OH– → Cu(OH)2
1. a. White ppt of barium sulfate, BaSO4 .
–
–
Complex ion equation is Fe3+ + SCN– → [FeSCN]2+
Activity 1I: Using the anion flow chart (page 13)
v. Barium Ba2+, Lead Pb2+
Ba (aq) + 2OH (aq) → Ba(OH)2(s) Ba(OH)2 is formed if [OH (aq)] is moderate rather than dilute.
b. Green ppt of iron(II) hydroxide, Fe(OH)2 , ionic equation is Fe2+ + 2OH– → Fe(OH)2
–
2+
Al (aq) + 3OH (aq) → Al(OH)3(s)
2+
1. a. White ppt of magnesium hydroxide, Mg(OH)2, ionic equation is Mg2+ + 2OH– → Mg(OH)2
Activity 1E: Identifying ions (page 8)
Activity 1H: Using the cation flow chart (page 11)
Answers
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b. Pb(NO3)2
3+
b. Zn(OH)2(s) + 4NH3(aq) → [Zn(NH3)4]2+(aq) + 2OH–(aq)
2. a. Orange ppt of iron(III) hydroxide, Fe(OH)3. Ionic equation is Fe3+ + 3OH– → Fe(OH)3
1. a. Lead hydroxide, white.
Zinc, Zn2+; Copper, Cu2+; Silver, Ag+.
Ba2+ + 2OH– → Ba(OH)2 , Ba2+ + SO42– → BaSO4
Activity 1D: Ionic equations for precipitation reactions (page 6)
1. a. i.
3. lone pair
Ionic equation is Ag+ + Cl– → AgCl
Complex ion equation is AgCl + 2NH3 → [Ag(NH3)2]+ + Cl–
Ionic equation is Ag+ + I– → AgI
217 Year 12 Chemistry Learning Workbook
2. a. i.
ii. 18.40 mL rejected – not within 0.2 mL of the other titres.
18.12 + 18.08 + 17.92
= 18.04 mL.
3
Anion is SO42–
iii. 18.04 mL. Working:
With H2SO4 is Ba2+ + SO42– → BaSO4
iv. Amount of sodium carbonate, n(Na2CO3) = c × V = 0.0920 × 0.020 = 0.00184 mol.
Cation is Zn2+
With NaOH
Amount of hydrochloric acid, n(HCl) = 2 × 0.00184 = 0.00368 mol
Ionic equation is Zn2+ + 2OH– → Zn(OH)2
Concentration of hydrochloric acid, c(HCl) =
Complex ion equation is Zn(OH)2 + 2OH– → [Zn(OH)4]2– (aq)
With NH3(aq)
ii. 21.76 is rejected – not within 0.2 mL of the other titres.
Ionic equation is Zn2+ + 2OH– → Zn(OH)2
iii.
Complex ion equation is Zn(OH)2 + 4NH3 → [Zn(NH3)4]2+(aq) + 2OH–
iv. n(CH3COOH) : n(HCl) = 1 : 1
Achievement Standard 90306
2. a. To clean the flask of all soluble chemicals other than those present in tap water.
b. To remove the impurities that may be present in tap water.
c. To ensure the solution is not diluted by an unknown volume of water inside the glassware.
d. An estimate of the volume of solution required to be delivered from the burette to reach the end point.
Activity 2B: Titration calculations (page 20)
1. A solution whose concentration is known precisely.
2. a.22.70 mL. All acceptable titres must be within 0.2 mL of each other and within 0.2 mL of the expected result for
'Excellence'.
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
22.32 + 22.44 + 22.48
= 22.41 mL.
3
3. a. 19.10 mL, 19.22 mL, 19.28 mL, 19.25 mL.
19.10 mL + 19.22 mL +19.28 mL + 19.25 mL
= 19.21.
4
b. 19.21 mL. Working:
c. No. All are in the acceptable range of 0.2 mL of each other and the expected value.
4. a. i.
18.16 mL, 18.24 mL, 17.92 mL, 18.32 mL.
ii. 17.92 mL – not within 0.2 mL of the other titres.
iii. 18.24 mL. Working:
iv.Equation for the reaction is: HCl(aq) + NH3(aq) → NH4Cl(aq), which indicates a 1 : 1 mol ratio between the
18.16 + 18.24 + 18.32
= 18.24 mL.
3
reactants.
Amount of hydrochloric acid, n(HCl) = c × V = 0.113 × 0.01824 = 0.00206 mol.
Amount of ammonia solution, n(NH3) = 0.00206 mol.
Concentration of ammonia solution, c(NH3)2 =
b. i.
18.12 mL, 18.08 mL, 17.92 mL, 18.40 mL.
Amount of ethanoic acid, n(CH3COOH) = c × V = 0.108 × 0.025 = 0.00270 mol.
Amount of sodium hydroxide, n(NaOH) = 0.00270 mol.
n
0.00270
=
= 0.126 mol L–1.
0.02135
V
Activity 2C: Titration procedure and calculations (page 23)
1.pipette, conical, pink, burette, swirling, end point, colourless, titre, mL, conical, wash-bottle, titration, concordant, 0.2
b. 22.41 mL. Working:
21.32 + 21.44 + 21.28
= 21.35 mL.
3
Concentration of sodium hydroxide, c(NaOH) =
Activity 2A: Titration procedure (page 17)
21.76 mL, 21.32 mL, 21.44 mL, 21.28 mL.
Answers 2.2
c. i.
0.00368
n
=
= 0.204 mol L–1.
V
0.01804
0.00206
n
=
= 0.103 mol L–1, or 1.03 × 10–1 mol L–1.
0.020
V
1.
a.
The substance must be:
• In a pure state.
• Soluble in water.
• Unaffected by the atmosphere.
E
L
P
M
A
S
S
E
G
A
P
b. M(Na2CO3) = (2 × 23) + 12 + (3 × 16) = 106 g mol–1
n(Na2CO3) =
c(Na2CO3) =
m(Na2CO3)
M(Na2CO3)
0.0130
0.250
=
1.38
106
= 0.0130 mol
= 0.0521 mol L–1
2. a. • A
ll glass apparatus (burette, conical flask, pipette) is washed thoroughly with tap water and then rinsed with
deionised water.
• The 25.00 mL pipette is rinsed with a small volume of the sodium carbonate solution. The solution is drawn
into the pipette using a pipette filler.
• Three drops of methyl orange indicator are added to the conical flask.
• The burette is rinsed with a small volume of the hydrochloric acid and then filled to a suitable level with the
solution of hydrochloric acid.
• The acid is delivered from the burette until the solution in the flask undergoes a distinct colour change. The
initial titration is the rough titration.
• The procedure is repeated with clean apparatus to within 1 mL of the rough titration value, and then the acid is
added one or two drops at a time until the colour change is observed.
• The procedure is repeated until concordant results (at least three titres within 0.2 mL) are obtained.
b. Orange to red.
c. i.
The third titration (17.70 mL) is outside the 0.2 mL range of the other values and is considered to be inaccurate.
18.24 + 18.42 + 18.38
= 18.35 mL
3
ii. Average titre =
iii. n(Na2CO3) = c(Na2CO3) × V(Na2CO3) = 0.0521 × 0.0250 = 0.00130 mol
n(HCl) : n(Na2CO3) = 2 : 1
n(HCl) = 2 × 0.00130 = 0.00260 mol
c(HCl) =
n(HCl) 0.00260
=
= 0.142 mol L–1
V(HCl) 0.01835
218 Year 12 Chemistry Learning Workbook
Activity 1J: Detective work (page 13)