chapter 8 salts

Salts

What is a salt ?

• A salt is formed in a reaction between an acid & a base. Acid + base → salt + water

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

• A salt is an ionic compound consisting a cation (metal ion or ammonium ion) from a base and an anion from an acid.

• A salt is a compound formed when the hydrogen ion in an acid is replaced by a metal ion or ammonium ion.

Acid SaltHydrochloric acid, HCl Chloride

saltsSodium chloride, NaClAmmonium chloride, NH4Cl

Nitric acid, HNO3 Nitrate salts

Potassium nitrate, KNO3

Aluminium nitrate, Al(NO3)2

Sulphuric acid, H2SO4 Sulphate salts

Ammonium sulphate, (NH4)2SO4

Magnesium sulphate, MgSO4

Carbonic acid, H2CO3 Carbonate salts

Iron(II) carbonate, FeCO3

Calcium carbonate, CaCO3

Phosphorus acid, H3PO4 Phosphate salts

Iron(III) phosphate, FePO4

Ammonium phosphate, (NH4)3PO4

Ethanoic acid, CH3COOH

Ethanoate salts

Lead(II) ethanoate, (CH3COO)2Pb

Copper(I) ethanoate, CH3COOCu

Solubility of Salts in water

Type of salt Solubility in waterAmmonium salts All are solubleSodium / potassium salts

All are soluble

Ethanoate salts All are solubleNitrate salts All are solubleChloride salts All are soluble except AgCl, HgCl2 & PbCl2Sulphate salts All are soluble except BaSO4, CaSO4 &

PbSO4

Carbonate salts All are insoluble except Na2CO3, K2CO3 & (NH4)2CO3

Lead(II) salts All are insoluble except Pb(NO3)2 & (CH3COO)2Pb

Preparation of salts

Soluble salts• Acid + alkali (potassium, sodium & ammonium salts)• Acid + base• Acid + metal• Acid + metal carbonate

Insoluble salts• Double decomposition reaction

Sodium, potassium & ammonium salts :• Method : titration• Reaction : neutralisation

4 stages involved :• A titration is carried out to determine the exact volume of an acid

needed to neutralise a fixed volume of an alkali with the help of an indicator.

NH4OH(aq) + HCl(aq) → NH4Cl(aq) + H2O

• This volume of an acid is then added straight to the same volume of alkali without any indicator to obtain a pure salt solution.

• Crystallisation is carried out to obtain crystals of the salt.

• Recrystallisation is done to obtain pure crystals of the salt.

1

3

2

4

Preparing insoluble salts :• Method : precipitation• Reaction : double decomposition

• 2 aqueous solutions of 2 different soluble salts are mixed to form the insoluble salt.

Example : lead(II) sulphatePb(NO3)2(aq) + K2SO4(aq) → PbSO4(s) + 2KNO3(aq)

• The insoluble salt is obtained by filtration.

How to select suitable methods for preparation of salts ?

Salt

Is the salt soluble ?

Reaction : Double decomposition

Procedure : Choose 2 aqueous solutions containing cation & anion of the insoluble salt.

Mix the 2 solutions.

Filter, wash & dry the precipitate.

No

Is it a Na+, K+ or NH4

+ salt ?

Yes

Reaction : Acid + metal Acid + metal oxide Acid + metal hydroxide Acid + metal carbonate

Procedure : Add excess solid to hot dilute acid.

Filter off the unreacted solid.

Evaporate to saturate the salt solution.

Cool to crystallisation to occur.

Filter, wash & dry the crystals

No

Reaction : Neutralisation reaction (acid + alkali)

Procedure : Use titration method to neutralise a given volume of acid to obtain the salt solution. Evaporate to saturate the salt solution.Cool to crystallisation to occur.Filter, wash & dry the crystals

Yes

Constructing ionic equation using the continuous variation method :

• Carry out a reaction between a fixed volume of reactant A with varying volumes of a second reactant B.

• Determine the volume of reactant B required to react completely with the fixed volume of reactant A.

• Calculate the number of moles of reactants A & B respectively.

• Determine the simplest mole ratio of reactant A to reactant B to form one mole of the insoluble salt.

• Use the ratio to construct the ionic equation.

Example 110 cm3 of 0.25 mol dm−3 lead(II) nitrate solution reacts completely with 5 cm3 of 0.10 mol dm−3 potassium iodide soluton. A yellow precipitate of lead(II) iodide is formed. construct the ionic equation for the formation of lead(II) iodide.Solution :Number of moles of Pb2+ ion Number of moles of I− ion

Simplest mole ratio of Pb2+ ion to I− ion

mol0.00251000

100.251000MV

mol0.0051000

51.01000MV

2:10025.0005.0:

0025.00025.0

005.0:0025.0

Ionic equation :

Pb2+(aq) + 2I−(aq) → PbI2(s)

Question 1In the preparation of copper(II) sulphate, a student added 4.0 g of copper(II) oxide to 1.25 mol dm−3 sulphuric acid. Calculate the volume of the acid needed to react completely with the copper(II) oxide. [Relative atomic mass : O, 16 ; Cu, 64]

Question 2Excess sodium chloride is added to 50.0 cm3 of silver nitrate solution. 2.87 g of silver chloride is precipitated. Calculate the concentration of the silver nitrate solution in mol dm−3.[Relative atomic mass : Cl, 35.5 ; Ag, 108 ]

Question 3Excess aluminium powder is added to 300.0 cm3 of 2.0 mol dm−3 hydrochloric acid. The mixture is then warmed to speed up the reaction. Calculate the mass of salt formed.[Relative atomic mass : Al, 27 ;Cl, 35.5]

Question 4150.0 cm3 of 1.0 mol dm−3 ammonia solution is completely neutralised with phosphoric acid using a titration method. Calculate the mass of ammonium phosphate formed.[Relative atomic mass : H, 1 ; N, 14 ; O, 16 ; P,31]

Question 56.20 g of copper(II) carbonate is added to 100 cm3 of 1.46 g dm−3 hydrochloric acid. Calculate the mass of copper(II) chloride produced.[ Relative atomic mass : H, 1 ; C, 12 ; O, 16 ; Cl, 35.5 ; Cu, 64 ]

Colour → solubility in water → chemical reactions → confirmatory tests

Yellow precipitate is formed.

Some yellow precipitate is dissolved.

Golden yellow crystals are formed.

Light blue precipitate is formed.

Dark blue precipitate is formed.

Dark blue precipitate is formed.

Greenish brown solution is formed.