Upload
kavitha-thayagarajan
View
157
Download
1
Tags:
Embed Size (px)
Citation preview
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.