SJChO Notes · Chapter 5
5

Acids, Bases and Salts

Singapore Junior Chemistry Olympiad

Lemon juice, vinegar and stomach acid are acids. Oven cleaner and soap are alkalis. This chapter explains what makes a substance an acid or a base, how acids and bases react, how to measure acidity, how to make salts, and how a titration measures exactly how much acid or alkali a solution holds.

Learning objectives

By the end of this chapter, you can:

Foundation O Level
  • describe acids and alkalis in terms of H+ and OH− ions, and their reactions
  • explain the difference between a strong and a concentrated acid
  • use the pH scale and indicators
  • classify oxides as acidic, basic, amphoteric or neutral
  • use the solubility rules to choose a method of making a salt
  • calculate concentrations from titration results
Core A Level
  • apply the Arrhenius, Brønsted–Lowry and Lewis theories
  • identify conjugate acid–base pairs
  • choose an indicator for a titration
  • use back-titration to analyse an insoluble solid
Prerequisites Chapter 1 (filtration and crystallisation), Chapter 3 (ions, formulae, lone pairs and dative bonds) and Chapter 4 (ionic equations, moles and concentration). Fill in the blanks as the lesson goes.

5.1Acids

DefinitionAn acid is a substance that produces hydrogen ions, H+, when dissolved in water.

When an acid dissolves, its molecules ionise: HCl(aq) → H+(aq) + Cl−(aq). Common acids are hydrochloric acid, HCl, nitric acid, HNO3, sulfuric acid, H2SO4, and ethanoic acid, CH3COOH, the acid in vinegar. The H+ ions give all acids the same properties: they taste sour, turn blue litmus red, and their solutions conduct electricity because they contain ions.

5.1.1The three reactions of acids

Learn these three reactions. Each forms a salt, named after the acid: sulfates from sulfuric acid, nitrates from nitric acid and chlorides from hydrochloric acid.

1acid + metal → salt + hydrogen

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

  • Test: hydrogen puts out a lighted splint with a ‘pop’.
  • Only reactive metals react. Copper and silver do not react with dilute acids.
  • Lead seems not to react with hydrochloric or sulfuric acid, because a layer of insoluble lead(II) chloride or lead(II) sulfate coats it.
2acid + base → salt + water

CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)

  • This is neutralisation (Section 5.3).
3acid + carbonate → salt + water + carbon dioxide

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

  • Test: carbon dioxide turns limewater milky.

Worked exampleWrite the ionic equation for CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g).

1. CaCO3(s) + 2H+(aq) + 2Cl−(aq) → Ca2+(aq) + 2Cl−(aq) + H2O(l) + CO2(g)
Split the aqueous ionic compounds. The solid stays whole.
2. CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
Cl− is a spectator ion.
Check your understanding
  1. Name the salt and the gas formed when zinc reacts with dilute hydrochloric acid.
  2. Why does solid citric acid not conduct electricity, while its solution in water does?

5.2Strong and weak acids

The strength of an acid is how far it ionises in water (Figure 5.1).

DefinitionA strong acid is completely ionised in aqueous solution. A weak acid is only partially ionised.

Hydrochloric, nitric and sulfuric acids are strong: HCl(aq) → H+(aq) + Cl−(aq). Ethanoic acid and citric acid are weak. Only some ethanoic acid molecules ionise, and the ions can join again, so the equation has a reversible arrow: CH3COOH(aq) ⇌ CH3COO−(aq) + H+(aq).

hydrochloric acid: all ionsethanoic acid: mostly molecules−−−−−−−−++++++++−−++
Figure 5.1 A strong acid is fully ionised. A weak acid is mostly un-ionised molecules (numbers for illustration).
Exam tipStrong is not the same as concentrated. Concentration is how much acid is dissolved in a given volume. A strong acid can be dilute, and a weak acid can be concentrated.
Check your understanding
  1. Solutions of hydrochloric acid and ethanoic acid have the same concentration. Which has the higher concentration of H+ ions, and why?

5.3Bases and alkalis

DefinitionA base is a metal oxide or metal hydroxide. An alkali is a base that is soluble in water, producing hydroxide ions, OH−.

5.3.1The hydroxide ion

A hydroxide ion, OH−, is an oxygen atom covalently bonded to a hydrogen atom, with one extra electron. The extra electron gives it a charge of 1−, and its oxygen keeps three lone pairs. Being negative and having lone pairs to spare, OH− strongly attracts H+ ions and joins with them to form water:

OH−(aq) + H+(aq) → H2O(l)

This is what makes a hydroxide a base: it removes the H+ ions that make a solution acidic. The oxide ion, O2−, in a metal oxide does the same, taking two H+ ions to form one water molecule.

Most bases, such as copper(II) oxide and magnesium hydroxide, are insoluble. Sodium hydroxide, potassium hydroxide and calcium hydroxide dissolve, so they are alkalis. Aqueous ammonia is also an alkali, although it contains no metal: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq). Like acids, alkalis are strong or weak. Sodium hydroxide is fully ionised, so it is a strong alkali. Ammonia is only partly ionised, so it is a weak alkali. Alkalis feel soapy and turn red litmus blue.

5.3.2The two reactions of bases

1base + acid → salt + waterneutralisation

For example, hydrochloric acid is neutralised by aqueous sodium hydroxide:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Whatever the acid and alkali, the only ions that react are the H+ of the acid and the OH− of the alkali (Figure 5.2):

H+(aq) + OH−(aq) → H2O(l)

H⁺+OH⁻H₂ONa⁺ and Cl⁻:spectator ions
Figure 5.2 Hydrochloric acid neutralised by sodium hydroxide: only H+ and OH− react.
  • Na+ and Cl− are spectator ions. They stay in solution as sodium chloride, the salt.
  • The salt of a strong acid and a strong alkali, such as sodium chloride or sodium nitrate, gives a neutral solution.
2alkali + ammonium salt → salt + water + ammonia

KOH(aq) + NH4Cl(s) → KCl(aq) + H2O(l) + NH3(g)

  • The mixture is warmed.
  • The OH− takes an H+ from the NH4+ ion, leaving NH3.
  • Test: ammonia has a sharp smell and turns damp red litmus paper blue.
Check your understanding
  1. Write the balanced equation for magnesium hydroxide reacting with hydrochloric acid.
  2. Is every base an alkali? Explain.

5.4The pH scale and indicators

The pH scale, usually from 0 to 14, shows how acidic or alkaline a solution is (Figure 5.3).

  • pH < 7: more H+ than OH−, so the solution is acidic. The lower the pH, the more acidic.
  • pH = 7: equal H+ and OH−, so the solution is neutral, like pure water.
  • pH > 7: more OH− than H+, so the solution is alkaline.

Each step of one pH unit is a tenfold change in the concentration of H+. A solution of pH 4 has 1000 times more H+ than one of pH 7.

01234567891011121314acidicneutralalkalinemore H⁺ ions: more acidicmore OH⁻ ions: more alkalinestrong acidweak acidneutralweak alkalistrong alkali
Figure 5.3 The pH scale and the colours of Universal Indicator.

pH can be measured with a pH meter or with an indicator, a dye whose colour depends on pH. Universal Indicator is a mixture of dyes that gives a different colour at each pH. Other indicators have one colour change over a narrow range:

indicatorcolour in acidchanges over pHcolour in alkali
methyl orangered3 to 5yellow
litmusred5 to 8blue
phenolphthaleincolourless8 to 10pink
thymolphthaleincolourless9 to 10.5blue

Most plants grow best in neutral or slightly acidic soil. Soil that is too acidic is treated with a base such as calcium hydroxide (slaked lime), which neutralises the excess acid.

Check your understanding
  1. Solution P has pH 2 and solution Q has pH 5. How many times greater is the concentration of H+ in P?
  2. What colour is methyl orange in a solution of pH 9?

5.5Oxides

How an oxide behaves depends on whether its element is a metal or a non-metal (Figure 5.4).

Does it react with acids?Does it react with alkalis?Does it react with alkalis?yesnoamphotericZnO, Al₂O₃, PbObasicCuO, CaO, MgOacidicCO₂, SO₂, P₄O₁₀neutralH₂O, CO, NOyesnoyesno
Figure 5.4 Classifying oxides.
1Basic oxidesoxides of most metals

They react with acids to form a salt and water.

MgO(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2O(l)

  • Most are insoluble in water.
  • Sodium oxide and potassium oxide dissolve to form alkalis: Na2O(s) + H2O(l) → 2NaOH(aq)
2Amphoteric oxidesZnO, Al2O3, PbO

They react with both acids and alkalis to form a salt and water.

with an acid: ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(l)

with an alkali: ZnO(s) + 2NaOH(aq) → Na2ZnO2(aq) + H2O(l)

  • With sodium hydroxide, zinc oxide forms the salt sodium zincate.
3Acidic oxidesoxides of most non-metals

They react with alkalis to form a salt and water.

CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l)

Most dissolve in water to form acids.

SO2(g) + H2O(l) → H2SO3(aq), sulfurous acid

  • SO3 dissolves to give sulfuric acid, and CO2 gives carbonic acid.
  • Silicon dioxide is insoluble, but it reacts with hot concentrated sodium hydroxide, so it is acidic too.
4Neutral oxidesH2O, CO, NO

They react with neither acids nor alkalis.

Check your understanding
  1. A white oxide is insoluble in water but dissolves in both dilute hydrochloric acid and aqueous sodium hydroxide. What type of oxide is it? Name one possible oxide.

5.6Theories of acids and bases Core · A Level

The definitions so far are the Arrhenius theory: an acid dissociates in water to give H+(aq), and a base gives OH−(aq). It works only in water, and it cannot explain why ammonia, with no OH group, is a base. Two wider theories fix this.

Brønsted–Lowry theory: an acid is a proton (H+) donor and a base is a proton acceptor.

A proton is far too small to exist on its own. When an acid dissolves, it gives its proton to a lone pair on a water molecule, forming the hydronium ion, H3O+ (Figure 5.5). H+(aq) and H3O+(aq) mean the same thing. A Brønsted–Lowry base must have a lone pair to accept the proton.

HCl+H₂OH⁺ movesH₃O⁺+Cl⁻base and its conjugate acidacid and its conjugate baseacid: HClbase: H₂O
Figure 5.5 Proton transfer from HCl to water, and the two conjugate pairs.

Two species that differ by one H+ are a conjugate acid–base pair. When an acid gives away its proton, what is left is its conjugate base. When a base accepts a proton, it becomes its conjugate acid. In HCl + H2O → H3O+ + Cl−, Cl− is the conjugate base of HCl, and H3O+ is the conjugate acid of H2O.

Worked exampleIdentify the conjugate pairs in NH3 + H2O ⇌ NH4+ + OH−.

1. NH3 gains H+ to become NH4+: base and conjugate acid.
The base accepts the proton.
2. H2O loses H+ to become OH−: acid and conjugate base.
Here water acts as an acid. With HCl it acted as a base.

Water can act as an acid or a base, depending on what it reacts with. An acid that can donate one proton per molecule, such as HCl or HNO3, is monobasic. H2SO4 is dibasic and H3PO4 is tribasic.

Lewis theory: an acid is an electron-pair acceptor and a base is an electron-pair donor.

Ammonia and boron trifluoride show why this happens (Figure 5.6). Boron is in Group 13, so in BF3 it shares its three valence electrons and ends up with only six electrons round it, two short of an octet. The nitrogen in NH3 has a lone pair that is not used in bonding. Nitrogen gives this pair to boron, forming a dative bond (Chapter 3), and boron reaches its octet: NH3 + BF3 → H3N→BF3.

  • NH3 donates the electron pair, so it is the Lewis base.
  • BF3 accepts the electron pair, so it is the Lewis acid.

No proton moves, so this is an acid–base reaction only in the Lewis sense. Every Brønsted–Lowry base is also a Lewis base, because it accepts a proton with a lone pair.

NHHHlone pair on N+BFFFB: only 6 electrons,space for 2 moreNHHHBFFFdative bond: B now has 8Lewis base:donates the pairLewis acid:accepts the pair
Figure 5.6 A Lewis acid–base reaction. Boron has six outer electrons until nitrogen donates its lone pair. The lone pairs on fluorine are not shown.
Check your understanding
  1. Give the conjugate base of H2SO4 and the conjugate acid of NH3.
  2. In HCl(g) + NH3(g) → NH4Cl(s), which reactant is the Brønsted–Lowry acid? Why is this not an acid–base reaction in the Arrhenius sense?

5.7Salts

A salt is an ionic compound made of a cation and an anion. It forms when the H+ of an acid is replaced by a metal ion or an ammonium ion.

So far, every salt has come from an acid. There is a second way. When two solutions are mixed and the cation from one and the anion from the other form an insoluble salt, the salt appears as a solid in the liquid, a precipitate (Figure 5.7). This is a precipitation reaction:

Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
ionic equation: Ba2+(aq) + SO42−(aq) → BaSO4(s)

Na+ and NO3− are spectator ions and stay in solution. The precipitate is separated by filtration (Chapter 1).

mix: a white precipitate formsfilter: the precipitate is the residue
Figure 5.7 Precipitation of barium sulfate.

To choose how to make a salt, first decide whether it dissolves in water.

saltssolubleinsoluble
sodium, potassium and ammonium saltsallnone
nitratesallnone
chloridesmostAgCl, PbCl2
sulfatesmostBaSO4, PbSO4, CaSO4 (sparingly soluble)
carbonatesNa2CO3, K2CO3, (NH4)2CO3all others

There are three methods (Figure 5.8).

  • Acid + excess insoluble solid, for a soluble salt: add a metal, insoluble base or insoluble carbonate to the acid until no more reacts. The excess solid makes sure all the acid is used, and it is then filtered off. Heat the filtrate until it is saturated, cool it to crystallise the salt, filter, wash the crystals with a little cold water and dry them. Very reactive metals are too dangerous, and unreactive metals do not react.
  • Titration, for a soluble salt from soluble starting materials: sodium, potassium and ammonium salts. Neither reactant can be filtered off, so the exact volumes are found by titration with an indicator (Section 5.8). The reaction is repeated with the same volumes and no indicator, and the salt is crystallised.
  • Precipitation, for an insoluble salt: mix a solution containing its cation with a solution containing its anion. Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq). Filter, wash the precipitate with distilled water and dry it.
Is the salt soluble in water?Is the metal, base or carbonate soluble?Method 3: precipitationmix a solution of the cationwith a solution of the anionMethod 1: excess solidacid + excess metal,insoluble base orinsoluble carbonateMethod 2: titrationacid + alkali orsoluble carbonateyesnonoyes
Figure 5.8 Choosing a method to make a salt.

Worked exampleDescribe how to make lead(II) chloride from lead(II) carbonate, dilute nitric acid and aqueous potassium chloride.

1. Add excess lead(II) carbonate to dilute nitric acid and filter: PbCO3(s) + 2HNO3(aq) → Pb(NO3)2(aq) + CO2(g) + H2O(l)
PbCl2 is insoluble, so precipitate it. First make a soluble lead salt.
2. Add excess KCl(aq) to the filtrate: Pb(NO3)2(aq) + 2KCl(aq) → PbCl2(s) + 2KNO3(aq). Filter, wash and dry.
Every nitrate is soluble.
Check your understanding
  1. Suggest starting materials and a method to make (a) copper(II) sulfate, (b) potassium nitrate and (c) silver chloride.

5.8Acid–base titrations

A titration measures the volume of one solution that reacts exactly with a fixed volume of another (Figure 5.9). A pipette delivers a fixed volume of one solution into a conical flask with a few drops of indicator. The other solution is added from a burette until the indicator just changes colour.

burettetitrant, known concentrationconical flaskanalyte and indicatorwhite tile
Figure 5.9 A titration.
  • A standard solution is one whose concentration is accurately known.
  • The equivalence point is reached when the reactants have reacted exactly in the ratio of the equation. The end-point is when the indicator changes colour. A good indicator changes colour at the equivalence point.

Worked example5.00 g of impure sodium hydroxide is dissolved and made up to 1.00 dm3. 25.0 cm3 of this solution is neutralised by 15.15 cm3 of 0.102 mol dm−3 sulfuric acid. Find the percentage purity of the sodium hydroxide.

1. 2NaOH + H2SO4 → Na2SO4 + 2H2O
Ratio NaOH : H2SO4 = 2 : 1.
2. nH₂SO₄ = 0.102 × 0.01515 = 1.545 × 10−3 mol, so nNaOH in 25.0 cm3 = 3.091 × 10−3 mol
Volumes in dm3.
3. In 1.00 dm3: 3.091 × 10−3 × 1000/25.0 = 0.1236 mol, mass = 0.1236 × 40.0 = 4.945 g
Scale up from 25.0 cm3 to the whole solution.
4. Purity = 4.945 ÷ 5.00 × 100% = 98.9%
Pure mass ÷ sample mass.

Some salts crystallise with water molecules held in their lattice, called water of crystallisation. Copper(II) sulfate crystals are CuSO4·5H2O. A titration can find the number of water molecules.

Worked example8.65 g of Na2CO3·nH2O is made up to 250 cm3. 25.0 cm3 of this solution reacts with 24.00 cm3 of 0.125 mol dm−3 sulfuric acid: Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O. Find n.

1. nH₂SO₄ = 0.125 × 0.02400 = 3.000 × 10−3 mol = nNa₂CO₃ in 25.0 cm3
Ratio 1 : 1.
2. In 250 cm3: 0.03000 mol, so Mr = 8.65 ÷ 0.03000 = 288.3
One Na2CO3·nH2O per Na2CO3.
3. 106.0 + 18.0n = 288.3, so n = 10.1, which is 10
n is a whole number.

5.8.1Choosing an indicator Core · A Level

At the equivalence point the pH changes very quickly. The indicator must change colour within that sudden change, which depends on whether the acid and base are strong or weak.

titrationindicator
strong acid and strong basemethyl orange, phenolphthalein or thymolphthalein
strong acid and weak basemethyl orange
weak acid and strong basephenolphthalein or thymolphthalein
weak acid and weak baseno suitable indicator

5.8.2Back-titration Core · A Level

An insoluble solid, such as calcium carbonate or magnesium hydroxide, cannot be titrated directly. Instead, a known excess of acid is added so that all the solid reacts. The acid left over is then titrated with a standard alkali. The acid that reacted with the solid is found by difference (Figure 5.10).

amount of acid that reacted = amount added − amount left over
all HNO₃ added = 0.04000 mol (known)reacted withthe Mg(OH)₂left overfound by difference0.04000 − 0.01538 = 0.02462 moltitrated with KOH0.01538 molMg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O
Figure 5.10 Back-titration: the acid added splits into the part used by the solid and the part left over.

Worked exampleAn antacid tablet of mass 1.00 g contains insoluble Mg(OH)2. It is added to 100 cm3 of 0.400 mol dm−3 HNO3 (excess). The leftover acid needs 20.50 cm3 of 0.750 mol dm−3 KOH. Find the percentage of Mg(OH)2 in the tablet.

1. nHNO₃ added = 0.400 × 0.100 = 0.04000 mol. nHNO₃ left over = nKOH = 0.750 × 0.02050 = 0.01538 mol
KOH reacts 1 : 1 with HNO3.
2. nHNO₃ reacted = 0.04000 − 0.01538 = 0.02462 mol
By difference.
3. Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H2O, so nMg(OH)₂ = 0.01231 mol
Ratio 1 : 2.
4. Mass = 0.01231 × 58.3 = 0.7177 g, so 71.8%
Mr of Mg(OH)2 = 58.3.
Check your understanding
  1. 25.0 cm3 of aqueous sodium hydroxide is neutralised by 15.00 cm3 of 0.250 mol dm−3 hydrochloric acid. Find the concentration of the sodium hydroxide.
  2. Which indicator would you use to titrate aqueous ammonia with hydrochloric acid? Why?
  3. 2.50 g of impure CaCO3 is added to 50.0 cm3 of 1.00 mol dm−3 HCl (excess). The leftover acid needs 20.0 cm3 of 0.500 mol dm−3 NaOH. Find the percentage of CaCO3 (Mr 100.1).

Answers to Check your understanding

5.1

  1. Zinc chloride and hydrogen: Zn + 2HCl → ZnCl2 + H2
  2. In the solid there are no free ions. In water the acid ionises, and the ions carry the current.

5.2

  1. Hydrochloric acid. It is fully ionised, while most ethanoic acid molecules stay un-ionised.

5.3

  1. Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
  2. No. Only bases that dissolve in water are alkalis. Copper(II) oxide is a base but not an alkali.

5.4

  1. Three pH units, so 10 × 10 × 10 = 1000 times.
  2. Yellow

5.5

  1. Amphoteric, for example zinc oxide or aluminium oxide.

5.6

  1. HSO4− and NH4+
  2. HCl, because it gives a proton to NH3. There is no water and no H+(aq) or OH−(aq), so the Arrhenius theory does not apply.

5.7

  1. (a) Excess copper(II) oxide or carbonate with dilute sulfuric acid, filter, crystallise. (b) Potassium hydroxide or carbonate with dilute nitric acid by titration, then crystallise. (c) Precipitation: silver nitrate solution with a soluble chloride such as sodium chloride, filter, wash, dry.

5.8

  1. nHCl = 0.250 × 0.01500 = 3.750 × 10−3 mol = nNaOH. c = 3.750 × 10−3 ÷ 0.0250 = 0.150 mol dm−3
  2. Methyl orange. Hydrochloric acid is strong and ammonia is a weak base, so the sudden pH change at the equivalence point lies in the acidic range.
  3. HCl added 0.05000 mol, left over 0.01000 mol, so 0.04000 mol reacted. nCaCO₃ = 0.02000 mol, mass = 2.002 g, purity = 80.1%