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Brown's Chemistry Advanced A Level Notes - Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5
Part
5.0 Acids, Bases and Salts – Revision and A level update
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This page summarises, ideally what you know, or
SHOULD NOW KNOW about acids and bases, from your course prior to doing an A level chemistry course.
Sub-index for
Part 5
5.1
Equilibria:
Lewis and Bronsted-Lowry acid-base theories
5.2
Self-ionisation of water and pH scale
5.3
Strong acids - examples and pH calculations
5.4
Weak acids - examples & pH, Ka and pKa calculations
5.5
Strong bases - examples and pH calculations
5.6
Weak bases - examples and pH, Kb and pKb calculations
5.0
Basic notes and equations on acids, bases, salts,
uses of
acid-base titrations - upgrade from GCSE!
5.0
Acids, Bases and Salts – An Advanced A level chemistry update
5.0.1 Revision of some basics
-
Alkaline solutions have a
pH of over 7 and the higher the pH the stronger is the alkali. Weak
alkalis (soluble bases) like ammonia give a pH of 10–11 but strong alkalis
(soluble bases) like sodium hydroxide give a pH of 13–14.
They
give blue–purple–violet colour with universal indicator or litmus paper.
-
NEUTRALISATION usually
involves mixing an acid (pH <7) with a
base
or alkali (pH > 7) which react to form a neutral
salt solution of pH7.
-
THE
IONIC THEORY of ACIDS and
ALKALIS and a few technical terms:
-
The majority of liquid water
consists of covalent H2O molecules, but there
are trace quantities of H+ and OH– ions from the
self–ionisation of water, BUT they
are of equal concentration and so water is neutral.
-
In acid solutions
there are more H+ ions than OH– ions.
-
In alkaline solution
there are more OH– ions than H+ ions. -
When alkalis and acids
react, the 'general word' and 'molecular formula' equation might be
for NEUTRALISATION ...
-
BUT
the ionic equation for ANY neutralisation is -
hydrogen ion +
hydroxide ion
==> water -
H+(aq)
+ OH–(aq)
==>
H2O(l) -
because all
acids form hydrogen ions in water and all alkalis (soluble
bases) form hydroxide ions in water. -
and, in this case,
the remaining ions e.g. Na+(aq) and
Cl–(aq)
become the salt crystals NaCl(s)
on evaporating the
water.
-
NOTE: Its much
cheaper to produce sodium chloride 'salt' by evaporating seawater!
BASES e.g. oxides,
hydroxides and carbonates, are substances that react and neutralise
acids to form salts and water.
-
Bases
which are soluble in water are called alkalis
e.g. NaOH sodium
hydroxide, KOH potassium hydroxide or Ca(OH)2 calcium
hydroxide.
-
Bases which are water
insoluble include CuO copper(II) oxide, MgO magnesium oxide.
After
a neutralisation, the
salt
solutions
consist of a mixture of positive and negative
ions (and their names are in the salt name!) e.g.
sodium chloride (NaCl) is a mixture of Na+
and Cl– ions, calcium chloride (CaCl2)
is a mix of Ca2+ and Cl– ions;
magnesium nitrate (Mg(NO3)2) is a mix of Mg2+
and NO3– ions, aluminium sulphate (Al2(SO4)3)
consists of Al3+ and SO42–
ions etc.
Bronsted–Lowry
Theory
-
The main concept of
the advanced Bronsted–Lowry theory is ...
-
a Bronsted–Lowry
acid is defined as a proton donor,
-
and a Bronsted–Lowry
base is defined as a proton acceptor.
-
Incidentally water
is a neutral oxide because its pH is 7
-
However water is
an amphoteric oxide i.e. it reacts as both a proton acceptor and a
proton donator.
-
e.g. water
acting as a base – proton acceptor with a stronger acid like the
hydrogen chloride gas
-
e.g. water
acting as an acid – proton donor with a weak BUT stronger base like
the alkaline gas ammonia
5.0.2
Some
important reactions of Acids
Acids
are neutralised by reaction with metals, oxides, hydroxides or carbonates
to form salts and other products.
Apart from metals (which is
an electron loss/gain redox reaction), the other reactants
listed above are considered as bases (meaning they react by
accepting a proton from an acid). Water soluble bases are known as alkalis.
The reaction between acids
and bases like oxides, hydroxides and carbonates are called neutralisation
reactions.
-
metal + acid
==>
a
salt
+ hydrogen
-
e.g. zinc +
hydrochloric acid ==> zinc chloride + hydrogen
-
Zn(s) + 2HCl(aq)
==> ZnCl2(aq)
+ H2(g)
-
Its the same equation
for many other Group 2 and Transition metals e.g. Mg, Ca and Fe, Co,
Ni
-
magnesium + sulphuric acid
==> magnesium sulphate +
hydrogen
-
Mg(s) +
H2SO4(aq)
==> MgSO4(aq)
+ H2(g)
-
Note 1: sulphuric
acid gives sulphate salt and hydrogen,
-
Note 2: Nitric
acid (HNO3) doesn't usually form hydrogen with a
metal, instead you get nasty
brown fumes of nitrogen dioxide! but you still get
the metal nitrate salt
-
alkali (soluble base )
+ acid ==> salt + water
(the 'classic' neutralisation reaction)
-
e.g.
sodium hydroxide + hydrochloric acid
==>
sodium chloride + water
-
metal hydroxide + acid
==>
a salt + water
-
e.g.
sodium hydroxide + sulphuric acid ==> sodium sulphate +
water
-
2NaOH(aq) + H2SO4(aq)
==> Na2SO4(aq)
+ 2H2O(l)
-
Its the same equation
for any Group 1 Alkali Metal hydroxide e.g. LiOH, KOH etc.
-
or potassium hydroxide + nitric acid
==> potassium nitrate +
water
-
NaOH(aq) + HNO3(aq)
==> NaNO3(aq)
+ 2H2O(l)
-
insoluble base + acid ==> salt + water
-
(note:
oxides that react with acids to form salts are known as 'basic
oxides')
-
e.g.
metal oxide +
acid ==> salt + water
-
e.g. copper(II) oxide +
sulphuric acid ==> copper(II) sulphate
+
water
-
CuO(s) +
H2SO4(aq) ==> CuSO4(aq) +
H2O(l)
-
calcium
hydroxide + hydrochloric acid ==> calcium chloride + water
-
metal carbonate or
hydrogencarbonate + acid ==> a salt + water
+ carbon dioxide
-
e.g.
calcium carbonate + nitric acid ==> calcium nitrate +
water + carbon dioxide
-
CaCO3(s) + 2HNO3(aq)
==>
Ca(NO3)2(aq) + H2O(l) + CO2 (g)
-
Its the same equation
for many other Group 2 and Transition metals e.g. Mg, Sr and Co, Ni,
Cu
-
Note: Using sulphuric acid and calcium carbonate you don't
get much of a fizz! because the calcium sulphate salt
formed, is not very soluble, and coats the remaining
calcium carbonate inhibiting the reaction! This
will happen with any reaction between an acid and a water
insoluble reactant which forms an insoluble solid product!
-
magnesium
carbonate + sulphuric acid ==> magnesium sulphate + water +
carbon dioxide
-
MgCO3(s) +
H2SO4(aq) ==> MgSO4(aq) + H2O(l)
+ CO2 (g) -
or sodium hydrogencarbonate + nitric acid
==> sodium nitrate + water
+ carbon dioxide
-
NaHCO3(s) +
HNO3(aq) ==> NaNO3(aq) + H2O(l) + CO2(g)
-
ammonia + acid ==> ammonium salt
-
Note that no water is
formed. -
e.g.
ammonia + hydrochloric acid
==>
ammonium chloride -
NH3(aq) +
HCl(aq) ==>
NH4Cl(aq) -
or ammonia +
sulphuric acid ==> ammonium sulphate
-
2NH3(aq) +
H2SO4(aq) ==> (NH4)2SO4(aq)
The name of the
particular salt formed depends on
(i) the metal
name, which becomes the first part of salt name,
and (ii) the acid e.g. H2SO4
sulphuric acid on neutralisation makes a ... sulphate; HCl hydrochloric
acid makes a ... chloride;
HNO3
nitric
acid makes a nitrate etc.
See
list of
compound formulae and their solubility
at the bottom of the page. The first
part of the salt name is
ammonium derived from ammonia (with metals or their compounds the
metal retains its original name), but the second part of the salt name
is always derived from the acid as in NOTE (a) above.
Ammonia
is an alkaline gas that is very soluble in water. It is a weak alkali
or soluble base and is readily neutralised by acids in solution to form ammonium
salts which can be crystallised on evaporating the resulting solution.
5.0.3 Some important
reactions of Bases (alkali = soluble base)
-
Neutralisation with acids is
dealt with above. -
Ammonium salts are decomposed
when mixed with a base e.g. the alkali sodium hydroxide.
-
e.g.
sodium hydroxide +
ammonium chloride ==>
sodium chloride + water + ammonia -
NaOH(aq) + NH4Cl(aq)
==>
NaCl(aq) + H2O(l) + NH3(g)
-
The ammonia is readily detected
by its pungent odour (strong smell) and by turning damp red
litmus blue.
-
The ionic equation is:
NH4+(aq)
+ OH–(aq)
==>
H2O(l) + NH3(l)
-
This reaction can be used to
prepare ammonia gas and as a simple chemical test for an
ammonium salt (see also the
"Chemical
Tests" and "Gas
Preparation–Collection" pages).
-
Alkali's (soluble bases) are
used to produce the insoluble hydroxide precipitates of many metal
ions from their soluble salt solutions.
-
e.g.
sodium hydroxide + copper(II) sulphate ==>
sodium sulphate + copper(II) hydroxide
-
2NaOH(aq) + CuSO4(aq)
==>
Na2SO4(aq) + Cu(OH)2(s)
a blue precipitate
-
ionically:
Cu2+(aq)
+ 2OH–(aq)
==> Cu(OH)2(s)
-
This reaction can be used as a
simple test to help identify certain metal ions.
-
Aqueous
solutions of alkalis like sodium hydroxide ('caustic soda') and
calcium hydroxide ('limewater') react with the acidic gas carbon
dioxide to form carbonate compounds if the gas is bubbled into
their solutions.
5.0.4 Methods
of making Salts which are water soluble
 METHOD
(a)
Neutralising an acid with a soluble base
e.g. the hydroxide of an alkali metal like sodium hydroxide or ammonia
solution. Steps (1) to (3)
below is
called a titration.
Typical common soluble bases
(alkalis) used for preparing soluble salts:
NaOH sodium hydroxide, KOH potassium
hydroxide, NH3 ammonia
(1)
A known volume of
acid is pipetted into a conical flask and universal indicator added.
The acid is titrated with the alkali from the burette.
(2)
The acid is added until the indicator turns
green, pH 7 neutral.
This means all the acid has been neutralised to form the salt
(3)
The volume of alkali needed for neutralisation is then noted, this
is called the endpoint volume. (1)–(3) are repeated with both known volumes
mixed together BUT without the contaminating universal indicator.
(4)
The solution
is transferred to an evaporating dish and heated to partially evaporate
the water causing crystallisation or can be left to slowly
evaporate – which tends to give bigger and better crystals.
(5)
The residual liquid can be decanted away and the
crystals can be carefully collected and dried by 'dabbing' with a filter
paper OR the crystals can be collected by filtration (below) and dried (as
above).
Note
(i)
You can put the acid in the
burette and the alkali in the flask.
(ii)
Parts (1) to (3) are
known specifically as an acid–base (alkali) titration, and the
general method is known as a volumetric titration by which it possible to find out
exactly what volume ratios are needed for neutralisation. So
knowing one concentration, you can calculate the other.
(iii)
Apparatus used: (1) pipette and conical flask; (2)–(3) burette
and conical flask; (4) evaporating (crystallising) dish, bunsen
burner, tripod and gauze; (5) filter paper.
(iv)
Other indicators e.g. phenolphthalein can be used instead (pink alkaline,
colourless acid).
(v) The burette and
pipette are both used for the accurate
measurement of volume.
 METHOD
(b)
Reacting
an acid with a metal or with an insoluble base
e.g. an insoluble metal oxide, hydroxide
or carbonate, often of a Group 2 metal like calcium, magnesium or a
Transition Metal like nickel,
copper or zinc. Copper metal won't dissolve in acids, but its oxide
and carbonate will.
Typical common insoluble bases used for
preparing soluble salts:
MgO magnesium oxide, MgCO3
magnesium carbonate
CaO Calcium oxide, CaCO3 calcium carbonate,
Ca(OH)2 calcium hydroxide,
NiO nickel(II) oxide, ZnO zinc oxide, Zn(OH)2,
zinc hydroxide, ZnCO3 zinc carbonate
(1)
The required
volume of acid is measured out into the beaker with a measuring cylinder. The
insoluble metal, oxide, hydroxide
or carbonate is weighed out and the solid added in small portions to the acid in the
beaker with stirring.
(2) The mixture may be heated to speed up the
reaction. When no more of the solid dissolves it means ALL the acid is
neutralised and there should be a little excess solid.
(3) The hot solution (with care!) is
filtered
to remove the excess solid metal/oxide/carbonate, into an evaporating dish.
(4) The hot solution is left to cool and crystallise. Then
collect and dry the crystals with a filter paper.
Note
(i)
Apparatus used: (1) balance, measuring cylinder, beaker and glass
stirring rod. (2) beaker/rod, bunsen burner,
tripod and gauze; (3)–(4) filter funnel and filter paper, evaporating
(crystallising) dish.
(ii)
A measuring cylinder is adequate for measuring the acid volume, you do not
need the accuracy of a pipette or burette required in method (a).
(iii) How to calculate amounts
required and % yield is dealt with in
Chemical Calculations Part 14.
METHOD (c) An insoluble salt
can be made by
mixing two solutions of soluble salts
in a process is called precipitation.
One solution contains the 1st required ion, and the other solution
contains the 2nd required ion. The precipitated salt can then be filtered off with a filter funnel and paper. The
collected solid is washed with distilled water to
remove any remaining soluble salt impurities and removed from the
filter paper to be dried. Examples ...
-
(i) Silver chloride is made by
mixing solutions of solutions of silver nitrate and sodium chloride.
-
silver nitrate + sodium chloride ==>
silver chloride + sodium nitrate
-
AgNO3(aq) + NaCl(aq)
==> AgCl(s) + NaNO3(aq)
-
in terms of
ions it could be written as
-
Ag+NO3–(aq)
+ Na+Cl–(aq) ==> AgCl(s) +
Na+NO3–(aq)
-
or: Ag+(aq)
+ NO3–(aq)
+ Na+(aq)
+ Cl–(aq) ==> AgCl(s) +
Na+(aq)
+ NO3–(aq)
-
but the
spectator
ions are
nitrate NO3– and
sodium Na+
which do not change at all,
-
so the ionic
equation is simply: Ag+(aq)
+ Cl–(aq) ==> AgCl(s)
-
Note that ionic
equations omit ions that do not change there chemical or physical
state.
-
In this case the
nitrate, NO3–(aq) and sodium
Na+(aq)
ions do not change physically or chemically and are called
spectator ions,
-
BUT the aqueous
silver ion, Ag+(aq), combines with the aqueous
chloride ion, Cl–(aq), to form the insoluble
salt silver chloride, AgCl(s), thereby changing their
states both chemically and physically.
-
If you use
barium chloride the word and symbol equations are ...
-
barium
chloride + silver nitrate ==> silver chloride + barium
nitrate
-
BaCl2(aq)
+ 2AgNO3(aq) ==> 2AgCl(s) +
Ba(NO3)2(aq)
-
which can be
written as
-
Ba2+(aq)
+ 2Cl–(aq) + 2Ag+(aq) +
2NO3–(aq) ==> 2AgCl(s)
+ Ba2+(aq) + 2NO3–(aq)
-
the spectator
ions are Ba2+ and
NO3–
-
so the ionic
equation is: Ag+(aq)
+ Cl–(aq) ==> AgCl(s)
-
(ii) Lead(II) iodide,
a yellow precipitate (insoluble in water!) can be made
by mixing lead(II) nitrate solution with e.g. potassium iodide solution.
-
lead(II) nitrate + potassium iodide
==> lead(II) iodide + potassium nitrate
-
Pb(NO3)2(aq) +
2KI(aq) ==> PbI2(s) + 2KNO3(aq)
-
which can be
written as
-
Pb2+(aq)
+ 2NO3–(aq)
+ 2K+(aq)
+ 2I–(aq) ==> PbI2(s) +
2K+(aq) + 2NO3–(aq)
-
the ionic equation is: Pb2+(aq)
+ 2I–(aq) ==> PbI2(s)
-
because the
spectator ions are nitrate
NO3– and
potassium K+.
-
In a similar
way you can make lead(II) chloride by e.g. using dilute
hydrochloric acid
-
lead(II) nitrate +
hydrochloric acid
==> lead(II) chloride + nitric acid
-
Pb(NO3)2(aq)
+ 2HCl(aq) ==> PbCl2(s) + 2HNO3(aq)
-
Pb2+(aq)
+ 2NO3–(aq)
+ 2H+(aq)
+ 2Cl–(aq) ==> PbCl2(s)
+ 2H+(aq) + 2NO3–(aq)
-
the ionic equation is:
Pb2+(aq)
+ 2Cl–(aq) ==> PbCl2(s)
-
because the
spectator ions are nitrate
NO3– and
hydrogen H+.
-
and you can
make lead(II) bromide by e.g. using sodium bromide
-
lead(II) nitrate +
sodium bromide
==> lead(II) bromide + sodium nitrate
-
Pb(NO3)2(aq)
+ 2NaBr(aq)
==> PbBr2(s) + 2NaNO3(aq)
-
Pb2+(aq)
+ 2NO3–(aq)
+ 2Na+(aq)
+ 2Br–(aq) ==> PbBr2(s)
+ 2Na+(aq) + 2NO3–(aq)
-
the ionic equation is:
Pb2+(aq)
+ 2Br–(aq) ==> PbBr2(s)
-
because the
spectator ions are nitrate
NO3– and
sodium Na+.
-
(iii) Calcium carbonate,
a white precipitate, forms on
e.g. mixing calcium chloride and sodium carbonate solutions ...
-
calcium chloride + sodium carbonate
==> calcium carbonate + sodium chloride
-
CaCl2(aq) +
Na2CO3(aq)
==> CaCO3(s) + 2NaCl(aq)
-
Ca2+(aq)
+ 2Cl–(aq) + 2Na+(aq) +
CO32–(aq) ==> CaCO3(s) + 2Na+(aq)
+ 2Cl–(aq)
-
ionically: Ca2+(aq)
+ CO32–(aq) ==> CaCO3(s)
-
because the
spectator ions are chloride
Cl– and sodium Na+.
-
(iv) Barium sulphate,
a white precipitate, forms on
mixing e.g. barium chloride and dilute sulphuric acid ...
-
barium chloride + sulphuric acid
==> barium sulphate + hydrochloric acid
-
BaCl2(aq) +
H2SO4(aq)
==> BaSO4(s) + 2HCl(aq)
-
Ba2+(aq)
+ 2Cl–(aq) + 2H+(aq) +
SO42–(aq) ==> BaSO4(s) + 2H+(aq)
+ 2Cl–(aq)
-
ionic
equation: Ba2+(aq)
+ SO42–(aq) ==> BaSO4(s)
-
because the
spectator ions are chloride
Cl– and
hydrogen H+.
-
Or you can use
sulphate salts like sodium sulphate, so the word and symbol
equations are ..
-
barium chloride +
sodium sulfate
==> barium sulfate + sodium chloride
-
BaCl2(aq)
+ Na2SO4(aq)
==> BaSO4(s) + 2NaCl(aq)
-
The ionic
equation is the same: Ba2+(aq)
+ SO42–(aq) ==> BaSO4(s)
-
because the
spectator ions are sodium Na+ and
chloride Cl–
-
(v) Lead(II)
sulphate, a white precipitate, forms in a similar way e.g.
-
lead(II) nitrate +
sodium sulphate ==> lead(II) sulphate + sodium nitrate
-
Pb(NO3)2
(aq) + Na2SO4 (aq)
==> PbSO4
(s) + 2NaNO3 (aq)
-
ionically: Pb2+(aq)
+ SO42–(aq) ==> PbSO4(s)
-
because the
spectator ions are sodium
Na+ and
nitrate
NO3–
-
NOTE:
A precipitation reaction is generally defined as 'the formation of
an insoluble solid on mixing two solutions or bubbling a gas into a
solution'.
METHOD (d) By direct
combination of the elements to form anhydrous salts
5.0.5 What
pH changes go on in a neutralisation reaction?
Simple
introduction using a strong acid–strong base reaction
 
The
graphs show how the pH changes when an alkali (soluble base) and an acid
neutralise each other and what you see visually using universal indicator
(univ. ind.).
This what is happening in the salt preparation
method (a) above. Note: you can prepare a
salt by doing the acid–alkali addition either way round but in either case the
volume of acid or alkali needed for neutralisation = the volume reading X at pH
7 (univ. ind.
green).
Red graph line: If you add acid
to an alkali (univ. ind. = blue), the pH starts at about 13 and only falls little at first
as the colour changes from purple ==> blue. Then the pH falls much more steeply
as the indicator colour changes from 'bluey' green ==> dark green ==> pale
green. The solution is then neutralised at pH 7. This is the point where
the salt is 100% formed. With further addition of excess acid, the pH falls and then
levels out to about pH 1 as the colour changes further from green ==>
yellow ==> orange.
Blue graph line: If you add
alkali to an acid (univ. ind. = red), the pH starts at about 1 and only rises
a little
at first with the colour still quite red. Then on further addition of
alkali the pH rises more sharply
as the colour changes from red ==> orange ==> yellow and eventually at the
neutralisation point at pH 7 the univ. ind. is
green. This is the point where the salt is 100% formed. With excess alkali the pH
continues to rise
and then levels out to about 13 as the indicator colour changes through
dark green ==> blue ==> purple. Universal indicator, and most
other acid–base indicators, work for strong acid and alkali titrations, but
universal indicator is a somewhat crude indicator for other acid–alkali
titrations because it gives such a range of colours for different pH's. Examples of more accurate and 'specialised' indicators are:
- titrating a strong alkali with a
strong acid (or vice versa):
- e.g. for sodium hydroxide (NaOH)
– hydrochloric/sulphuric acid (HCl/H2SO4)
titrations, use ...
- phenolphthalein indicator (pink
in alkali, colourless in acid–neutral solutions), the end–point is the
pink <==> colourless change.
- Litmus works too, the end point
is the red <==> purple/blue colour change.
- titrating a weak alkali with a
strong acid:
- e.g. for titrating ammonia (NH3)
with hydrochloric/sulfuric acid (HCl/H2SO4), use
...
- methyl orange indicator (red in
acid, yellowish–orange in neutral–acid), the end–point is an 'orange'
colour, not easy to see accurately.
- screened methyl orange indicator
is a slightly different dye–indicator mixture that is reckoned to be
easier to see than methyl orange, the end–point is a sort of 'greyish
orange', but still not easy to do accurately.
- titrating a weak acid with a strong alkali:
- e.g. for titrating ethanoic acid
(CH3COOH) with sodium hydroxide (NaOH), use ...
- phenolphthalein indicator (pink
in alkali, colourless in acid–neutral solutions, pink in alkali), the
end–point is the first permanent pink.
- methyl red indicator (red in
acid, yellow in neutral–alkaline), the end–point is 'orange'.
- titrating a weak acid with a weak
alkali (or vice versa):
- These are NOT practical
titrations because the pH changes at the end–point are not great
enough to give a sharp colour change with any indicator.
-
Advanced level theory of indicators and
titrations
-
and
advanced acid–alkali
titration questions (GCE–AS–A2–IB students only!)
5.0.6 A
Summary of important formulae and solubility
The original acids
are hydrochloric acid HCl, sulfuric/sulphuric acid H2SO4,
nitric acid, HNO3
which give the salts when reacted
with a metal, oxide, hydroxide or carbonate.
|
Formulae of
bases: oxides, hydroxides and carbonates
'molecular' formula and the
'real' ionic formula |
Formulae of salts formed:
soluble chlorides, sulphates and nitrates
'molecular' formula and the
'real' ionic formula |
The metal (or other ion) involved |
|
M2O oxide (M+)2O2–, soluble, alkali
(O and S both in Group 6, so sulfides have similar formula e.g. Na2S)
MOH hydroxide M+OH–, soluble, alkali
M2CO3 carbonate (M+)2CO32–,
soluble mild alkali
MHCO3 hydrogencarbonate
M+HCO3–, soluble,
mild alkali
|
MCl chloride, M+Cl–
M2SO4 sulphate, (M+)2SO42–
MNO3 nitrate, M+NO3–
|
M = Li, Na, K,
usually Group 1
for the
M+ ion |
|
MO
oxide M2+O2–, often insoluble base
(O and S both in Group 6, so sulphides have the same formula e.g.
MgS, CuS)
M(OH)2 hydroxide M2+(OH–)2, often insoluble, alkali
if soluble
MCO3 carbonate M2+CO32–, often insoluble
|
MCl2 chloride M2+(Cl–)2
MSO4 sulphate* M2+SO42–
M(NO3)2 the nitrate M2+(NO3–)2
*CaSO4
is
not very soluble
|
M = Mg, Ca, Cu, Zn, Fe,
usually Group 2 or Transition metal
for the
M2+ ion
|
|
Al2O3,
Al(OH)3 (insoluble bases, amphoteric) |
AlCl3, Al2(SO4)3,
Al(NO3)3 |
Al3+ ion, aluminium in Group 3 |
|
the
alkaline soluble base ammonia, NH3, no
stable hydroxide i.e. NH4OH doesn't exist |
NH4Cl,
(NH4)2SO4, NH4NO3 |
the
ammonium ion, NH4+, in the salts from ammonia |
How
to work out formulae and balancing equations is explained on another web page
5.0.7
Dilution calculations and simple titration calculations
- It is very useful to be know exactly how much of a dissolved
substance is present in a solution of particular concentration or volume of a
solution. So we need a standard way of comparing the concentrations of
solutions.
- The concentration of an aqueous solution is usually
expressed in terms of moles of dissolved substance per cubic decimetre, mol
dm–3, this is called molarity, sometimes denoted in
shorthand as M.
- Note:
1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing
cm3/1000 gives dm3, which is handy to know since
most volumetric laboratory apparatus is calibrated in cm3 (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm3
(mol/litre).
- Equal volumes of solution of the
same molar concentration contain the same number of moles of solute i.e.
the same number of particles as given by the chemical formula.
- You need to be able to calculate
- the number of moles or mass of substance in an aqueous
solution of given volume and concentration
- the concentration of an aqueous solution given the
amount of substance and volume of water, for this you use the equation
....
- (1a) molarity (concentration) of Z
= moles of Z / volume in dm3
- you need to be able to rearrange
this equation ...
- therefore (1b) moles =
molarity (concentration) x volume in dm3
- and (1c) volume in dm3
= moles / molarity (concentration)
- You may also need to know that ...
- (2) molarity x formula mass of
solute = solute concentration in g/dm3,
- and dividing this by 1000 gives
the concentration in g/cm3, and
- (3) (concentration in g/dm3)
/ formula mass = molarity in mol/dm3,
- both equations (2) and (3)
result from equations (1) and (4), work it out for yourself.
- and don't forget by now you should
know:
- (4) moles Z
= mass Z / formula mass of Z
- and (5) 1 mole = formula mass in
grams
-
Example 1
-
What mass of
sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.5 dm3)
of a 0.5M solution? [Ar's: Na = 23, O = 16, H = 1]
-
1 mole of NaOH = 23 + 16 + 1 = 40g
-
for 1000 cm3 (1 dm3) of 0.5M you
would need 0.5 moles NaOH
-
which is 0.5 x 40 = 20g
-
however only 500 cm3 of solution is needed
compared to 1000 cm3
-
so scaling down:
mass NaOH
required = 20 x 500/1000 = 10g
-
Example 2
-
How
many moles of H2SO4 are there in 250cm3 of
a 0.8M sulphuric acid solution? What mass of acid is in this solution?
[Ar's:
H = 1, S = 32, O = 16]
-
formula mass of sulphuric
acid = 2 + 32 + (4x16) = 98, so 1 mole = 98g
-
if there was 1000 cm3
of the solution, there would be 0.8 moles H2SO4
-
but there is only 250cm3
of solution,
so scaling down ...
-
moles H2SO4
= 0.8 x (250/1000) = 0.2 mol
-
mass = moles x formula
mass, which is 0.2 x 98 = 19.6g
of H2SO4
-
Example 3
-
Using a 2.0 x
10–2 molar stock solution, what volume of it should be added to a
100cm3 volumetric flask to make 100 cm3 of a 5.0 x 10–3
mol dm–3 (0.005M) solution?
-
The ratio of the two
molarities is stock/diluted = 2.0 x 10–2/5.0 x 10–3 =
4.0 or a dilution factor of 1/4 (0.02/0.005).
-
Therefore 25 cm3
(1/4 of 100) of the 2.0 x 10–2 molar
solution is added to the 100 cm3 volumetric flask prior to
making it up to 100 cm3 with pure water to give the 5.0 x 10–3
mol dm–3 (0.005M) solution.
- Example 4
- : Given the equation
NaOH(aq)
+ HCl(aq) ==> NaCl(aq) + H2O(l)
- 25 cm3 of a sodium hydroxide solution was
pipetted into a conical flask and titrated with 0.2M hydrochloric acid.
Using a suitable indicator it was found that 15 cm3 of acid was
required to neutralise the alkali. Calculate the molarity of the sodium
hydroxide and concentration in g/dm3.
- moles HCl = (15/1000) x 0.2 = 0.003 mol
- moles HCl = moles NaOH (1 : 1 in equation)
- so there is 0.003 mol NaOH in 25 cm3
- scaling up to 1000 cm3 (1 dm3),
there are ...
- 0.003 x (1000/25) = 0.12 mol NaOH in 1 dm3
- molarity of NaOH is 0.12M or mol
dm–3
- since mass = moles x formula mass, and Mr(NaOH)
= 23 + 16 + 1 = 40
- concentration in g/dm3
is 0.12 x 40 = 4.41g/dm3
- Example 5
- Given the equation
2KOH(aq)
+ H2SO4(aq) ==> K2SO4 + 2H2O(l)
- 20 cm3 of a sulphuric acid solution was
titrated with 0.05M potassium hydroxide. If the acid required 36 cm3
of the alkali KOH for neutralisation what was the concentration of the acid?
- mol KOH = 0.05 x (36/1000) = 0.0018 mol
- mol H2SO4 = mol KOH / 2 (because
of 1 : 2 ratio in equation above)
- mol H2SO4 = 0.0018/2 = 0.0009 (in
20 cm3)
- scaling up to 1000 cm3 of solution = 0.0009
x (1000/20) = 0.045 mol
- mol H2SO4 in 1 dm3 =
0.045, so molarity of H2SO4
= 0.045M or mol dm–3
- since mass = moles x formula mass, and Mr(H2SO4)
= 2 + 32 + (4x16) = 98
- concentration in g/dm3
is 0.045 x 98 = 4.41g/dm3
-
More GCE advanced A level volumetric analysis acid–base titration calculations
5.0.8 Doing an acid-base titration
Doing a titration
An accurate volume of acid is
pipetted into the conical flasks using a suction bulb for health and safety
reasons. Universal indicator is then added, which turns red in the
acid.
The alkali, of known accurate
concentration, is put in the burette and you can conveniently
level off the reading to zero (the meniscus on the liquid surface should
rest on the zero –– graduation mark).
Note in stage 2. other
possibilities are:
A small amount of accurately
weighed solid acid is dissolved in water and titrated with alkali.
A small amount of accurately
weighed solid alkali is dissolved in water and titrated with acid.
After this, the method is
essentially the same as described below.
The alkali is then carefully added by
running it out of the burette in small quantities, controlling the flow
with the tap, until the indicator seems to be going yellow–pale green.
The conical flask should be carefully swirled after each addition of alkali
to ensure all the alkali reacts.
Near the end of the titration, the
alkali should added drop–wise until the universal indicator goes green. This is called the
end–point of the titration and the
green means that all the acid has been neutralised. The volume of alkali
needed to titrate–neutralise the acid is read off from burette scale, again
reading the volume value on the underside of the meniscus. The calculation
can then be done to work out the concentration of the alkali.
Universal indicator, and most
other acid–base indicators, work for strong acid and alkali titrations, but
universal indicator is a somewhat crude indicator for other acid–alkali
titrations because it gives such a range of colours for different pH's. Examples of more accurate and 'specialised' indicators are:
- titrating a strong alkali
with a strong acid (or vice versa):
- e.g. for sodium hydroxide (NaOH)
– hydrochloric/sulphuric acid (HCl/H2SO4)
titrations, use ...
- phenolphthalein indicator (pink
in alkali, colourless in acid–neutral solutions), the end–point is the
pink <==> colourless change.
- Litmus works too, the end point
is the red <==> purple/blue colour change.
- titrating a weak alkali with
a strong acid:
- e.g. for titrating ammonia (NH3)
with hydrochloric/sulfuric acid (HCl/H2SO4), use
...
- methyl orange indicator (red in
acid, yellowish–orange in neutral–acid), the end–point is an 'orange'
colour, not easy to see accurately.
- screened methyl orange indicator
is a slightly different dye–indicator mixture that is reckoned to be
easier to see than methyl orange, the end–point is a sort of 'greyish
orange', but still not easy to do accurately.
- titrating a weak acid with a strong alkali:
- e.g. for titrating ethanoic acid
(CH3COOH) with sodium hydroxide (NaOH), use ...
- phenolphthalein indicator (pink
in alkali, colourless in acid–neutral solutions, pink in alkali), the
end–point is the first permanent pink.
- methyl red indicator (red in
acid, yellow in neutral–alkaline), the end–point is 'orange'.
- titrating a weak acid with a
weak alkali (or vice
versa):
- These are NOT practical
titrations because the pH changes at the end–point are not great
enough to give a sharp colour change with any indicator.
-
Advanced A level
AS A2 volumetric analysis acid–base titration calculations
-
Advanced level theory of indicators and
titrations including titrating weak acids and weak bases
|
5.0.9 Water of crystallisation
Calculation of % water of crystallisation given the
hydrated salt formula
Deducing the formula of a hydrated salt from
experimental results
-
A known mass of the hydrated salt is gently
heated in a crucible until no further water is driven off and the weight
remains constant despite further heating. The mass of the anhydrous salt left
is measured.
The original mass of hydrated salt and the mass of the anhydrous salt
residue can be worked out from the various weighings.
-
The % water of
crystallisation and the formula of the salt are calculated as follows:
-
Suppose 6.25g of blue
hydrated copper(II) sulphate, CuSO4.xH2O, (x
unknown) was
gently heated in a crucible until the mass remaining was 4.00g. This
is the white anhydrous copper(II) sulphate.
-
The mass of anhydrous
salt = 4.00g, mass of water (of crystallisation) driven off =
6.25–4.00 = 2.25g
-
The % water of
crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%
-
[ Ar's
Cu=64, S=32, O=16, H=1 ]
-
The mass ratio of CuSO4
: H2O is 4.00 : 2.25
-
To convert from mass
ratio to mole ratio, you divide by the molecular mass of each
'species'
-
CuSO4 = 64
+ 32 + (4x18) = 160 and H2O = 1+1+16 = 18
-
The mole ratio of CuSO4
: H2O is 4.00/160 : 2.25/18
-
which is 0.025 : 0.125
or 1 : 5, so the formula of the hydrated salt is CuSO4.5H2O
WHAT NEXT?
INDEX of ALL my chemical equilibrium
context revision notes
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle–rules
* Part 2. Kc and Kp equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4
Partition between two
phases, solubility product Ksp, common ion effect,
ion–exchange systems *
Part 5. pH, weak–strong acid–base theory and
calculations * Part 6. Salt hydrolysis,
acid–base titrations–indicators, pH curves and buffers * Part 7.
Redox equilibria, half–cell electrode potentials,
electrolysis and electrochemical series
*
Part 8.
Phase equilibria–vapour
pressure, boiling point and intermolecular forces watch out for sub-indexes
to multiple sections or pages
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