Doc Brown's Advanced A Level Chemistry Revision Notes

Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5.7

5.7 Acids, Bases and Salts – Revision and A level update

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.

Chemical Equilibrium Notes Parts 5 & 6 Index


5.7 Acids, Bases and Salts – An A level update

5.7.1 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 proton (H+) donation–acceptance theory of acids and bases (Bronsted–Lowry) is covered in more theoretical detail in section 5.1

    • Acids are substances that form hydrogen ions, H+(aq), when dissolved in water e.g.

      • hydrochloric acid HCl gives H+(aq) and Cl(aq) ions

      • sulphuric acid H2SO4 gives 2H+(aq) and SO42–(aq) ions

      • nitric acid HNO3 gives H+(aq) and NO3(aq) ions.

    • Alkalis are substances that form hydroxide ions (OH(aq)) in water e.g.

      • sodium hydroxide NaOH gives Na+(aq) and OH(aq) ions

      • calcium hydroxide Ca(OH)2 gives Ca2+(aq) and 2OH(aq)  ions.

      • Note: an alkali is a base soluble in water.

    • 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 ...

      • ACID + ALKALI ==> SALT + WATER... e.g.

      • hydrochloric acid + sodium hydroxide ==> sodium chloride + water

      • HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l)

      • 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

          •  HCl(g) + H2O(l) ==> H3O+(aq) + Cl(aq)

          • This is how hydrochloric acid is formed which you write simply as HCl.

        • e.g. water acting as an acid – proton donor with a weak BUT stronger base like the alkaline gas ammonia

          • NH3(aq) + H2O(l) (c) doc b NH4+(aq) + OH(aq)

          • This is why ammonium solution is alkaline – sometimes wrongly called 'ammonium hydroxide' instead of aqueous ammonia.


 

5.7.2 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

      • Test for hydrogen gas – squeaky pop with lit splint (can often see condensed water on side of test tube)

      • 2H2(g) + O2(g) ==> 2H2O(l) + energy!

  • 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

    • NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)

  • 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

    • Ca(OH)2(s) + 2HCl(aq) ==> CaCl2(aq) + 2H2O(l

  • 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

      • Test for carbon dioxide gas – it gives a white precipitate of calcium carbonate (cloudiness) when bubbled into limewater (calcium hydroxide solution).

      • Ca(OH)2(aq) + CO2(g) ==> CaCO3(s) + H2O(l)

    • 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. Sometimes the equations are written with the 'fictitious' 'ammonium hydroxide'

e.g. NH4OH(aq) + HCl(aq) ==> NH4Cl(aq) + H2O(l)


 

5.7.3 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.

    • sodium hydroxide + carbon dioxide ==> sodium carbonate + water

      • 2NaOH(aq) + CO2(g) ==> Na2CO3(aq) + H2O(l)

      • This reaction can be used to remove carbon dioxide gas from a mixture of gases.

    • calcium hydroxide + carbon dioxide ==> calcium carbonate + water

      • Ca(OH)2(aq) + CO2(g) ==> CaCO3(s) + H2O(l)

      • The formation of the white precipitate of calcium carbonate is used as a test for the gas carbon dioxide.

TOP and LINKS


 

5.7.4 Methods of making Salts which are water soluble

soluble salt preparation from soluble base-acid neutralisationMETHOD (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.

soluble salt preparation from insoluble base-acid neutralisationMETHOD (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

  • The apparatus for the preparation of aluminium chlorise (c) doc b

  • e.g. if dry chlorine gas Cl2 is passed over heated iron or aluminium, the chloride is produced. The experiment (shown above) should be done very carefully by the teacher in a fume cupboard.

    • 2Al(s) + 3Cl2(g) ==> 2AlCl3(s)

    • The aluminium can burn intensely with a violet flame, white fumes of aluminium chloride sublime from the hot reacted aluminium and the white solid forms on the cold surface of the flask (its often discoloured yellow from the trace chlorides of copper or iron that may be formed).

    • 2Fe(s) + 3Cl2(g) ==> 2FeCl3(s)

    • The iron (e.g. as steel wool) glows red and brown fumes of iron(III) chloride stream off, the brown solid collects on the cold flask surface.

    • Note (i): Both these chlorides react exothermically and hydrolyse with water to give the metal hydroxide and fumes of hydrogen chloride, and so dry conditions are needed.

    • Note (ii): Both these chlorides cannot be made in an anhydrous form from aqueous solution neutralisation. This is because on evaporation the compounds contain 'water of crystallisation'. On heating the hydrated salt  hydrolyses and decomposes into water, the oxide or hydroxide and fumes of hydrogen chloride, and maybe some impure anhydrous chloride, basically it a mess in terms of trying to make pure AlCl3 and FeCl3 in this way


 

5.7.5 What pH changes go on in a neutralisation reaction?

Simple introduction using a strong acid–strong base reaction

pH graphs of neutralisationpH scale diagram

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!)


 

5.7.6 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


 

5.7.7 Dilution calculations and simple titration calculations

volumetric apparatus

  • 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 AS A2 volumetric analysis acid–base titration calculations


 

5.7.8 Doing an acid base titration

volumetric apparatus for a 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

 

5.7.9 Water of crystallisation

Calculation of % water of crystallisation given the hydrated salt formula

  • Calculate the % of water in hydrated magnesium sulphate MgSO4.7H2O

    • Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1

    • relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246

    • 7 x 18 = 126 is the mass of water

    • so % water = 126 x 100 / 246 = 51.2%

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


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