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![3.9 g of potassium is burnt completely in the air as shown in the following equation;
4K(s) + O2(g) → 2K2O(s)
What is the mass of potassium oxide produced?
[Relative atomic mass: K, 39; O, 16]
Solutions
Tip: Solve the question step by step
Step 1: Write Chemical Equation
4K(s) + O2(g) → 2K2O(s)
4 mol of K react with 1 mol of O2 produce 2 mol K2O
Step 2: Calculate the number of mole
[Get the information from the question]
Step 3: Find the coefficient From Balance Chemical Equation
FBCE;
4 mol of K produce 2 mol K2O
Thus;
0.1 mol of K produce 2/4 mol K2O = 0.2 mol K2O
FBCE;
[Sebelah kiri] [Sebelah kanan]
Bil. mol yang telah dikira Bil. Mol yang hendak ditentukan
4 mol K = 2 mol K2O
0.1mol K = 2/4 x 0.1mol K2O = 0.05 mol K2O
No. of mol of K2O = 0.05 mol
Step 4: Solve the questions
Thus;
Mass of K2O = 0.05 mol × Molar mass
= 0.05 mol× 55 g mol-1
= 2.75 g
Example 3:
30
No. of mol K =
mass
Molar mass
=
3.9 g
39 gmol-1
0.1 mol=](https://image.slidesharecdn.com/notesupdatessalts-131006000433-phpapp01/85/Notes-updates-salts-30-320.jpg)
![Acids reacts with calcium carbonate, CaCO3 in limestone to form a salt and carbon dioxide, CO2.
A piece of limestone reacted completely with 100 cm3
of 31.5 g dm-3
nitric acid, HNO3.
[Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Ca, 40. Molar volume: 24 dm3
mol-1
at room
conditions]
a. Calculate the mass of salt produced.
b. What is the volume of carbon dioxide, CO2 liberated at room conditions?
Step 1: Write Chemical Equation
Chemical Equation: 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O
Step 2: Calculate the number of mole
Get the information from the question;
FBCE; 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O
2 mol HNO3 = 1 mol Ca(NO3)2
0.05 mol HNO3 = ½ x 0.05 mol Ca(NO3)2 = 0.025 mol Ca(NO3)2
No. of mol of Ca(NO3)2 = 0.025 mol
Mass of Ca(NO3)2 = 0.025 mol × 40 + 2[14 + 3(16)] g mol-1
= 4.1 g
31
No. of mole of HNO3 =
Molarity × Volume
1000
=
0.5 mol dm-3
× 100 cm3
1000
= 0.05 mol
Concentration of HNO3 = 31.5 g dm3
=
Molar mass of HNO3
31.5 g dm3
= 0.5 mol dm-3
=
31.5 g dm3
1 + 14 + 48 g mol-1
Change the
concentration
given in g dm-3
to mol dm-3
first](https://image.slidesharecdn.com/notesupdatessalts-131006000433-phpapp01/85/Notes-updates-salts-31-320.jpg)
![FBCE; 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O
2 mol HNO3 = 1 mol CO2
0.05 mol HNO3 = ½ x 0.05 mol CO2 = 0.025 mol CO2
No. of mol of CO2 = 0.025 mol
Volume of CO2 = 0.025 mol × 12 + 2(16) dm3
mol-1
= 1.1 dm3
Example 4:
Pb(NO3)2 compound decomposes when heated as shown in the following equation.
If 6.62 g of Pb(NO3)2 compound is heated, calculate;
[Relative atomic mass: N, 14; O, 16; Pb, 207; 1 mol of gas occupies 22.4 dm3
at s.t.p.]
(i) mass of PbO that is produced
(ii) volume of nitrogen dioxide produced at s.t.p
(ii) volume of oxygen produced at s.t.p
Solution:
FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2
2 mol Pb(NO3)2 = 2 mol PbO
0.02 mol Pb(NO3)2 = 0.02 mol PbO
No of mol PbO = 0.02 mol
Mass of PbO = 0.02 x 223 = 4.46 g
32
No of mol Pb(NO3
)2
=
mass
Molar mass
=
6.62 g
331 gmol-1
0.02 mol=
2Pb(NO3)2 → 2PbO + 4NO2 + O2](https://image.slidesharecdn.com/notesupdatessalts-131006000433-phpapp01/85/Notes-updates-salts-32-320.jpg)
![FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2
2 mol Pb(NO3)2 = 4 mol NO2
0.02 mol Pb(NO3)2 = 4/2 x 0.02 mol O2 = 0.04 mol O2
No of mol O2 = 0.04 mol
Volume of O2 = 0.04 x 22.4 dm3
= 0.896 dm3
// 896 cm3
FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2
2 mol Pb(NO3)2 = 1 mol O2
0.02 mol Pb(NO3)2 = ½ x 0.02 mol O2 = 0.01 mol O2
No of mol O2 = 0.01 mol
Volume of O2 = 0.01 x 22.4 dm3
= 0.224 dm3
// 224 cm3
Numerical Problems involving stoichiometric reactions in the precipitation of salts
Question 1:
A student prepare copper(II) nitrate, Cu(NO3)2 by reacting copper(II) oxide, CuO with 200 cm3
of 2.0
moldm-3
nitric acid, HNO3. Calculate the mass of copper(II) oxide, CuO needed to react completely
with the acid.
[Relative atomic mass: Cu, 64 ; O, 16]
Question 2:
X cm3
of 0.5 moldm-3
sulphuric acid, H2SO4 is added to 100 cm3
of 1.0 moldm-3
lead(II) nitrate
solution to produce lead(II) sulphate, PbSO4.
[Relative atomic mass: Pb, 20; O, 16; S, 32]
a. Calculate the value of X.
b. Calculate the mass of lead(II) sulphate obtained.
Start to do exercises from any book.
I will help and guide you to master this topic.
Prepared by;
Kamal Ariffin Bin Saaim
SMKDBL
33](https://image.slidesharecdn.com/notesupdatessalts-131006000433-phpapp01/85/Notes-updates-salts-33-320.jpg)
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Similar to Notes updates salts
Notes updates salts
- 1. SALTS What is salt? Inthe preparation of salts, we must identify the type of salt. This can be done by analysing the cations and the anions that are present in salts. Two types of salts • Soluble salt – salts that can be dissolve in water at room temperature • Insoluble salt – salts cannot be dissolve in water at room temperature Type of salt Solubility in water Sodium salts Potassium salts Ammonium salts All dissolves in water Nitrate salts All dissolves in water Chloride salts All dissolves in water, Except: Lead(II) chloride, PbCl2 Silver chloride, AgCl Mercury chloride, HgCl Sulphate salts All dissolves in water Except: Lead(II) sulphate, PbSO4 Barium sulfat, BaSO4 Calcium sulfat, CaSO4 Carbonate salts All did not dissolves in water, Except: Sodium carbonate, Na2CO3 Potassium carbonate, K2CO3 Ammonium carbonate, (NH4)2CO3 Special Properties of lead(II) chloride and lead(II) iodide NOTES: Lead halide such as lead(II) chloride (PbCl2), lead(II) bromide (PbBr2), and lead(II) iodide (PbI2) did not dissolve in cold water but dissolve in hot water. 1 White precipitate of PbCl2 White precipitate dissolves in hot water White precipitate formed when the water is cooled down. Salt is an ionic compound formed when the hydrogen ion, H+ from acid is replaced by a metal ion or ammonium ion, NH4 + PbCl2 are soluble in hot water.
- 2. Use of salts; ItemUse Example Food preparation Flavor Monosodium glutamate (MSG) Sodium chloride Preservatives Sodium chloride - salted fish Sodium benzoate - sauce Sodium nitrite - processed meat, burger Baking powder Sodium hydrogen carbonate Agriculture Nitrogen fertilizers Potassium nitrate Sodium nitrate Pesticide Copper(II) sulphate Iron(II) sulphate Medicine Reduce stomach acidic (gastric) Calcium carbonate Calcium hydrogen carbonate Sniff salt (fainted) Ammonium carbonate Plaster of Paris (cement to support broken bone) Calcium sulphate A. Preparation of Salt The procedure of preparation salt depends to the type of salt. a. Insoluble salt is prepared through precipitation reaction. b. Soluble salt is prepared by one of these reactions; i. Acid and alkali ii. Acid and metal oxide iii. Acid and metal carbonate iv. Acid and reactive metal 2 Yellow precipitate of PbI2 Yellow crystals formed when the water is cooled down. Yellow precipitate dissolves in hot water PbI2 are also soluble in hot water.
- 3. a. Insoluble Salts i.Preparing Insoluble Salts 1. Insoluble salts can be prepared through precipitation reactions or double decomposition reactions. 2. Precipitation or double decomposition reaction involves; - two aquoues solutions/soluble salts were mix together - one of the solutions contains the cations of the insoluble salt. - one of the solutions contains the anions of the insoluble salt. - the ions of the two aqueous solutions above interchange to produce two new compound which is insoluble salt or precipitate, and aqueous solution. - the precipitate produced is obtained by filtration. The residue left in the filter paper is the insoluble salt. The filtrate is soluble salt. - the residue/precipitate (insoluble salt) then rinsed with distilled water to remove any other ions as impurities. 3 Anion (Non-metal ion) Cation (Metal ion) nn mm
- 4. Chemical and ionicequations Chemical equation : MX(aq) + NY(aq) → MY(s) + NX(aq) solution solution precipitate solution 4 Na+ Na+ NO3 - NO3 - PbCl2 Pb2+ ions combined with Cl- ions to form white precipitate Na+ ions and NO3 - ions do not take part in the reaction and are free to move in the solution Ionic equation: Pb2+ + 2Cl- PbCl2 Glass rod Residue/precipitate (Insoluble salt) Filtrate (Soluble salt) Mixture of solutions Filter funnel Filter paper Retort stand
- 5. Ionic equation :M+ (aq) + Y- (aq) → MY(s) Study this reaction carefully In the formation of the precipitate of barium sulphate, BaSO4, the chemical equation can be written: BaCl2(aq) + Na2SO4 (aq) BaSO4(s) + 2NaCl (aq) Ions Ba2+ + Cl- + Na+ + SO4 2- BaSO4 + Na+ + Cl- Ionic equation : Ba2+ + SO4 2- BaSO4 (shows the ions that take part in the reaction to form precipitate/insoluble salts) More examples; Insoluble Salt Ions Ionic equation ZnCO3 Zn2+ , CO3 2- Zn2+ + CO3 2- ZnCO3 AgCl Ag+ , Cl- Ag+ + Cl- AgCl BaSO4 Ba2+ , SO4 2- Ba2+ + SO4 2 BaSO4 PbCl2 Pb2+ , Cl- Pb2+ + Cl- PbCl2 PbSO4 Pb2+ , SO4 2- Pb2+ + SO4 2- PbSO4 CaCO3 Ca2+ , CO3 2- Ca2+ + CO3 2- CaCO3 ii. Preparation and purification of insoluble salts Preparation of Plumbum(II) iodide Chemical equation : Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq) Ionic equation : Pb2+ (aq) + 2I- (aq) → PbI2 (s) Step 1: Preparation 5
- 6. 1. 20 cm3 lead(II)nitrate 0.1 mol dm-3 solution is measured with measuring cylinder 50 ml, and poured into a beaker. 2. 20 cm3 potassium iodide 0.1 mol dm-3 solution is measured with measuring cylinder 50 ml and poured into a beaker contains lead(II) nitrate solution. 3. The mixture is stirred with a glass rod. A yellow precipitate is formed. 4. The mixture is filtered to obtain the yellow solids of lead(II) iodide as the residue. Step 2: Purification 6 + → 20 cm3 Lead(II) nitrat 0.1 mol dm-3 20 cm3 potassium iodide 0.1 mol dm-3 Glass rod Precipitate of lead(II) iodide (yellow) Sodium nitrate solution Mixture of solutions Filter funnel Filter paper Retort stand Beaker
- 7. 5. The residueis rinsed with distilled water to remove other ions in it. 6. The yellow solid is dried by pressing between two pieces of filter paper. EASY LAH ! b. Soluble Salt 7 Glass rod Distilled water Precipitate of lead(II) iodide Precipitate of lead(II) iodide Filter paper
- 8. i. Preparaing SolubleSalt - Sodium salts - Potassium salts Acid + alkali salts + water - Ammonium salts Soluble Salts Acid + metal oxide salts + water - Others salts Acid + reactive metal salts + hydrogen gas Acid + metal carbonate salt + water + carbon dioxide Notes: Reactive metal is magnesium, aluminium, and zinc Unreactive metal is iron, lead, silver a. Sodium, potassium or ammonium salts prepared from acid and alkali reaction. Salt Alkali Acid Chemical equation NaCl NaOH HCl NaOH + HCl → NaCl + H2O K2SO4 KOH H2SO4 2KOH + H2SO4 → K2SO4 + 2H2O NH4NO3 NH3/NH4OH HNO3 NH3 + HNO3 → NH4NO3 + H2O CH3COONa NaOH CH3COOH NaOH + CH3COOH → CH3COONa + H2O Note: To prepare the above salts, titration technique is use. b. Soluble salt (except sodium, potassium and ammonium salt) is prepared using these methods - Acid and metal - Acid and metal oxide - Acid and metal carbonate Name of Salt Acid that must be used Substance that can be use to react with acid Metal Metal oxide Metal carbonate ZnCl2 HCl Zn ZnO ZnCO3 Mg(NO)3 HNO3 Mg MgO MgCO3 CuSO4 H2SO4 × CuO CuCO3 Pb(NO3)2 HNO3 × PbO PbCO3 Write a chemical equation for each experiment above. 8
- 9. Remember this notesok 1. Metal that is less reactive from hydrogen such as copper, lead and silver/argentum did not react with dilute acid. 2. Metal, metal oxide and metal carbonate above is a solid that cannot dissolves in water, hence during reaction that solid must be added excessively to make sure all hydrogen ions in acid is completely reacted. Excess solid can be expelling through filtration. 3. Impure soluble salt can be purified through crystallization process. ii. Preparation and purification of soluble salts A. Preparing soluble salt through reaction between acid and alkali. Preparation of Soluble Sodium, Potassium and ammonium Salts Soluble salts of sodium, potassium and ammonium can be prepared by the reaction between an acid and alkali. Acid (aq) + alkali (aq) Salt (aq) + Water (l) Procedure : • Using pipette, 25 cm3 of alkali solution is measured and transferred into a conical flask. • Two drops of phenolphthalein are added to the alkali solution. • Dilute acid is place in a burette. The initial reading is recorded. • Acid is added slowly into the alkali solution while shaking the conical flaks, until the pink solution turn colourless. • The final reading of the burette is recorded. • The volume of acid added, V cm3 is calculated. • The experiment is repeated by adding V cm3 of acid to 25 cm3 of alkali solution in a beaker without using phenolphthalein as an indicator. • The mixture is transferred into a evaporating dish. • The mixture is heated until saturated and the saturated solution is allowed to cool at room temperature. • Salt crystals formed are filtered and rinsed with a little of cold distilled water. • Salt crystals are dried by pressing it between filter papers (or in oven) The reaction between acid and alkali is known as what process? Refer to acids and bases notes ok. Don’t worry I’ll help you.! Kita bukan along kita cuma nak tolong. 9
- 10. Example: Preparing sodiumchloride Step 1: Preparation (Titration) 1. 25.0 cm3 sodium hydroxide solutions is pipette into conical flask. 2. Two drops of phenolphthalein indicator are added into conical flask. The colour of solution is recorded. 3. A 50 cm3 burette is filled with hydrochloric acid. The initial burette reading is recorded. 4. Hydrochloric acid is added gradually from a burette into conical flask and swirling the conical flask. 5. Titration is stopped when phenolphthalein changes from pink to colourless. The final burette reading is recorded. 6. The volume of hydrochloric acid used is calculated. 7. The experiment is repeated by adding hydrochloric acid (known volume) to 25.0 cm3 sodium hydroxide in a beaker without using phenolphthalein. Step 2: Preparation (Crystallization) 8. The mixture is transferred into a evaporating dish. 10 Retort stand Burette Hydrochloric acid Conical flask 25 cm3 NaOH + phenolphthalein indicator Bunsen burner Evaporating dishSalt solution
- 11. 9. The colourlesssolution is slowly heated/evaporated until its saturated or to about one-third (1/3) of the original volume. 10. The saturated solution is then cooled to allow crystallization to occur. Step 3: Purification 10. The white crystals formed are then filtered, rinsed with a little distilled water and dried by pressing between filter paper. Note: Phenolphthalein indicator is used at the beginning of the experiment to determine the volume of hydrochloric acid that is required to react completely with 25 cm3 of sodium hydroxide. However experiment is repeated without using phenolphthalein so that the salt prepared will not contaminated by the indicator. 11 Glass rod Distilled water Copper(II) sulphate Filter funnel
- 12. B. Preparing solublesalt through reaction between acid i. Metal oxide. ii. Metal iii. Metal carbonate Procedure To Prepare a Soluble Salt (not Na, K or NH4 + ) • 50 cm3 of acid is measured using a measuring cylinder and poured into a beaker. The acid is heated slowly. • Using a spatula, metal / metal oxide / metal carbonate powder is added a little at a time while stirring the mixture with a glass rod. • The addition of the solid powder is stopped when some solids no longer dissolve anymore. (the solid is excess and all the acid is completely neutralised by the solid) • The mixture is filtered to remove the excess solid powder. • The filtrate is transferred to an evaporating dish. • The filtrate is heated until saturated. (The filtrate is evaporated to about one-third (1/3) of the original volume) • The saturated solution is then allowed to cool to room temperature and the salt crystals are formed. • The crystals are filtered and rinsed with a little cold distilled water. • Salt crystals are then dried by pressing it between filter papers. 12 Heating Acid Powder of : Metal oxide Metal carbonate Metal Excess of solid powder Filtrate (Salt solution) Heating Saturated solution Crystals Filtrate
- 13. Example: Preparing copper(II)sulphate (Sulphuric acid and copper(II) oxide powder) Step 1: Preparation 1. 50 cm3 sulphuric acid 0.1 mol dm-3 is put in a beaker and is heated. 2. Using spatula copper(II) oxide powder is added a little at a time to the hot sulphuric acid while stirring continuously with glass rod. 3. The addition of copper(II) oxide is stopped when solids powder remain undissolved. 4. The mixture is filtered to remove the excess copper(II) oxide. 5. The filtrate is transferred to an evaporating dish. 13 xxxxxxxxxxxxxxxx Copper(II) oxide Glass rod Beaker Wire gauze 50 cm3 sulphuric acid 0.1 mol dm-3 Tripod Bunsen burner Spatula Stir Glass rod Reactant mixture Excess copper(II) oxide Copper(II) sulphate solution
- 14. 6. The filtrateis slowly heated/evaporated until its saturated, or to about one-third (1/3) of the original volume. 7. The saturated solution is then allowed to cool to room temperature. Step 3: Purification 8. The crystals are filtered and rinsed with a little cold distilled water. 9. Salt crystals are then dried by pressing it between filter papers. REMEMBER…. THIS NOTES OK Unreactive metal such as lead (Pb), copper (Cu), and silver (Ag) cannot react with dilute asid. So to prepare salt contains lead ions (Pb2+ ), copper ions (Cu2+ ) or silver ions (Ag+ ), we must use either oxide powder or carbonate powder only. Example: CuO + H2SO4 CuSO4 + H2O (ok) CuCO3 + H2SO4 CuSO4 + H2O + CO2 (ok) Cu + H2SO4 no reaction (not ok) 14 Bunsen burner Evaporating dish ×××××××××××× Copper(II) sulphate solution Glass rod Distilled water Copper(II) sulphate
- 15. B. Physical Characteristicsof Crystals. A salt is made up of positive and negative ions. When these ions are packed closely with a regular and repeated arrangement in an orderly manner, a solid with definite geometry known as crystal lattice is formed. All crystals have these physical characteristics: a) Reqular geometry shapes, such as cubic or hexagonal. b) Flat faces, straight edges and sharp angles. c) Same angle between adjacent faces. d) All crystals of the same salt have the same shape although the sizes may be different. Start to memorize the solubility of a salt in water OK. It will help you a lot to better understand this chapter. KNOWLEDGE IS POWER The solubility of a salt in water depends on the types of cations and anions present. C. Qualitative Analysis of Salts 15 Salt Solubility in water Sodium, potassium and ammonium salts (Na+ , K+ , NH4 + ) • All are soluble Nitrate salt (NO3 - ) • All are soluble Chloride salt (Cl - ) • All chloride salts are soluble in water except PbCl2, AgCl and HgCl2 Sulphate salt (SO4 2- ) • All sulphate salts are soluble in water except PbSO4, BaSO4 and CaSO4 Carbonate salt (CO3 2- ) • All carbonate salts are insoluble except Na2CO3, K2CO3 and (NH4)2CO3
- 16. What is Qualitativeanalysis? In the qualitative analysis of salts, we need to identify the ions that are present in salts. This can be done by analysing their physical and chemical properties. Observations on the physical properties of salts 1. Colour and solubility in water Certain physical properties of salts such colour and solubitity in water are observed to help us infer certain cations and anions that are present in salts. The table shows the colour of salts in solid , in aqueous solution and the solubility of salts in water Salt Colour in solid Solubility in water Colour in Aqueous solution 1. Ammonium chloride NH4Cl white soluble colourless 2. Ammonium nitrateNH4(NO3)3 white soluble colourless 3. Calcium carbonate CaCO3 white insoluble - 4. Calcium nitrate Ca(NO3)2 white soluble colourless 5. Magnesium sulphate MgSO4 white soluble colourless 6. Magnesium carbonate MgCO3 white insoluble - 7. Zinc sulphate Zn SO4 white soluble colourless 8. Zinc nitrate Zn(NO3)2 white soluble colourless 9. Lead(II) chloride , PbCl2 white insoluble - 10. Lead(II) sulphate , PbSO4 white insoluble - 11. Lead(II) carbonate , PbCO3 white insoluble - 12. Copper(II) chloride , CuCl2 Blue soluble Blue 13 Copper(II) sulphate , PbSO4 Blue soluble Blue 14. Copper(II) carbonate , PbCO3 Green insoluble - 15. Iron(II) sulphate , FeSO4 Green soluble Pale green 16. Iron(III) chloride , FeCl3 Brown / Yellow soluble Brown/Yellow/ Yellowish brown 17. Sodium nitrate , NaNO3 white soluble colourless 18, Sodium carbonate , Na2CO3 white soluble colourless 19. Potassium nitrate , KNO3 white soluble colourless 20. Potassium carbonate , K2CO3 white soluble colourless The table shows the colour of different cations in the solid form or in aqueous solution 16 Qualitative analysis is a chemical technique used to determine what substances are present in a mixture but not their quantities.
- 17. Observation Inference Blue solutionIon copper (Cu2+ ) present Pale green solution Ion Iron(II) Fe2+ present Yellow/Yellowish- brown/brown solution Ion Iron (III) Fe3+ present Green solid Hydrated Fe 2+ , CuCO3 Brown solid Hydrated Fe 3+ salt White solid Salts of Na+ , K+ ,NH4 + , Mg 2+ , Ca 2+ Al 3+ , Zn 2+ , Pb 2+ (If the anions are colourless Colourless solution Na+ , K+ ,NH4 + , Mg 2+ , Ca 2+ , Al 3+ , Zn 2+ , Pb 2+ The table shows the solubility of different types of salts in water Compounds Solubility in water Sodium salts Potassium salts Ammonium salts All are soluble Nitrate salts All are soluble Chloride salts All are soluble except AgCl, HgCl and PbCl2 (soluble in hot water) Sulphate salts All are soluble except BaSO4, PbSO4 and CaSO4 Carbonate salts All are insoluble except sodium carbonate, potassium carbonate and ammonium carbonate 2. Tests for gases 17
- 18. Gases are oftenproduced from reactions carried out during laboratory tests on salts. By identifying the gases evolved,it is possible to infer the types of cations and anions that are present in a salt. The table shows the test and the result of different gases Gas Test Result Oxygen gas, O2 Test with a glowing wooden splinter Wooden splinter is rekindled /lighted Hydrogen gas , H2 Test with a lighted wooden splinter Gas explodes with a pop sound Carbon dioxide gas , CO2 Bubble the gas produced into lime water Lime water turns milky Ammonia gas, NH3 Test with moist red litmus paper Moist red litmus paper turns blue Chlorine gas, Cl2 Test with moist blue litmus paper Moist blue litmus paper turns red and then turns white Hydrogen chlorine gas , HCl Test with a drop of concentrated ammonia NH3 solution Dense white fumes Sulphur dioxide gas , SO2 Bubble the gas produced into purple acidified potassium manganate (VII), KMnO4 solution Purple acidified potassium manganate (VII),KMnO4 solution decolourises Nitrogen dioxide gas , NO2 Test with moist blue litmus paper moist blue litmus paper turns red 3. Action of heat on salts 18
- 19. Effect of heaton carbonate salts Carbonaate salt Colour of salt before heating Colour of residue Effect on lime water Hot cold Copper (II) carbonate, CuCO3 Green powder Black powder Black powder The gas liberated turns lime water milky/chalky Zinc carbonate , ZnCO3 White solid Yelow solid White solid The gas liberated turns lime water milky/chalky Lead(II) carbonate, PbCO3 White solid Brown sold Yelow solid The gas liberated turns lime water milky/chalky Sodium carbonate, Na2CO3 White solid White solid White solid No change Calcium carbonate, CaCO3 White solid White solid White solid The gas liberated turns lime water milky/chalky Potassium carbonate, K2CO3 White solid White solid White solid No change Magnesium carbonate, MgCO3 White solid White solid White solid The gas liberated turns lime water milky/chalky Effect of heat on nitrate salts Nitrate Salt Colour of salt before heating Colour of residue Test on gases liberated Hot cold Copper (II) nitrate, Cu(NO3)2 Blue solid Black powder Black powder A brown gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter Zinc nitrate, Zn(NO3)2 White solid Yellow solid White solid A browan gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter Lead(II) nitrate, Pb(NO3)2 White solid Brown solid Yellow solid A browan gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter Sodium nitrate, NaNO3 White solid White solid White solid A colourless gas that rekindles a glowing splinter is liberated Calcium nitrate, Ca(NO3)2 White solid White solid White solid A browan gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter 19
- 20. Potassium nitrate, KNO3 White solidWhite solid White solid A colourless gas that rekindles a glowing splinter is liberated Magnesium nitrate, Mg(NO3)2 White solid White solid White solid A browan gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter Iron(II) nitrate, Fe(NO3)2 Pale Green solid Pale Green solid Pale Green solid A browan gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter Iron(III) nitrate, Fe(NO3)3 Brown solid Reddish- Brown solid Reddish- Brown solid A browan gas that turns blue litmus paper red is liberated. The gas liberated also ignites a glowing splinter The table shows the comparison of the effect of heat on carbonate and nitrate salts Metal Effect of heat on carbonate salt Effect of heat on nitrate salt Potassium Sodium Are not decomposed by heat Decompose to nitrite salt and oxygen gas. Calcium Magnesium Aluminium Zinc Iron Tin Lead Copper Decompose to metal oxide and carbon dioxide gas. Decompose to metal oxide, nitrogen dioxide gas and oxygen gas. Mercury Silver Gold Decompose to metal, carbon dioxide gas and oxygen gas. Decompose to metal , nitrogen dioxide gas and oxygen gas. Most sulphate salts are not decomposed by heat. Only a few sulphate such as iron(II) sulphate,zinc sulphate and copper sulphate decompose to sulphur dioxide or sulphur trioxide gas when heated. All chloride salts are stable when heated except ammonium chloride. Ammonium chloride sublimes and decomposes to produce ammonia gas and hydrogen chloride gas. The table shows the deduction of the types of ion present based on the gas produced Type of gas produced Type of ion present(anion) CO2 Carbonate ion (CO3 2- ) present except Na2CO3 and K2CO3 O2 Nitrate ion (NO3 - ) present NO2 Nitrate ion (NO3 - ) present except NaNO3 and KNO3 SO2 Sulphate ion (SO4 2- ) present NH3 Ammonim ion (NH4 + ) present 20
- 21. Tests for anions Reagent/ Condition Observation Anion Ionic Equation (if any) 2 cm3 the unknown solution + dilute hydrochloric acid / nitric acid / sulphuric acid pour into a test tube gas liberated is immediately bubbled through lime water. Effervescence. Colourless gas turns lime water milky. CO3 2- ion CO3 2- + 2H+ CO2 + H2O 2 cm3 of nitric acid + 2 cm3 of the unknown solution pour into a test tube + 2 cm3 silver nitrate solution White precipitate is formed. Cl- ion Ag+ + Cl- AgCl 2 cm3 of dilute hydrochloric acid / nitric acid + 2 cm3 of the unknown solution pour into a test tube + 2 cm3 of barium chloride / barium nitrate solution shake well White precipitate is formed. SO4 2- ion Ba2+ + SO4 2 - BaSO4 2 cm3 of the unknown solution pour into a test tube 2 cm3 of dilute sulphuric acid + 2 cm3 of iron(II) sulphate solution shake well. Then drop carefully and slowly a few drops of concentrated sulphuric acid along the side of a slanting test tube into the mixture without shaking it. Brown ring is formed at the boundary between the concentrated H2SO4 (top layer) and aqueous solution of the mixture (bottom layer) NO3 - ion - 21
- 22. Tests for cations ConfirmatoryTest for Fe2+ , Fe3+ , Pb2+ , NH4 + Ions Confirmatory Test for Fe2+ Reagent Observation Conclusion Potassium hexacyanoferrate(II) solution Pale blue precipitate Fe2+ ion is present Dark blue precipitate Fe3+ ion is present Potassium hexacyanoferrate(III) solution Dark blue precipitate Fe2+ ion is present Greenish-brown solution Fe3+ ion is present Potassium thiocyanate solution Pale red colouration Fe2+ ion is present Blood red colouration Fe3+ ion is present Confirmatory Test for Pb2+ Method Observation Ionic Equation • Using aqueous solution of chloride - 2 cm3 of any solution of Cl- + 2 cm3 of any solution of Pb2+ dilute with 5 cm3 of distilled water heat until no further change occurs allow the content to cool to room temperature using running water from the tap - A white precipitate is formed When heated – dissolve in water to form colourless solution When cooled – white precipitate reappear Pb2+ + 2Cl- PbCl2 • Using aqueous solution of iodide - 2 cm3 of any solution of I- + 2 cm3 of any solution of Pb2+ dilute with 5 cm3 of distilled water heat until no further change occurs allow the content to cool to room temperature using running water from the tap - A yellow precipitate is formed When heated – dissolve in water to form colourless solution When cooled – yellow precipitate reappear Pb2+ + 2I- PbI2 Confirmatory Test for NH4 + Method Observation • 2 cm3 of any solution of NH4 + + 2 cm3 of NaOH / KOH / Ca(OH)2 heat put a piece of moist red litmus paper at the mouth of the test tube - Moist red litmus paper turns blue Reaction with Nessler’s Reagent • 2 cm3 of any solution of NH4 + + 2 cm3 of Nessler’s Reagent shake well - A brown precipitate is formed 22
- 23. Reaction of CationsWith NaOH Pb2+ Zn2+ Al3+ : White precipitate dissolves/larut in excess NaOH Ca2+ Mg2+ : White precipitate insoluble/tidak larut in excess NaOH 23 Cations + NaOH (aq) Precipitate producedNo precipitate White precipitate Coloured precipitate Green Blue Brown Fe2+ Cu2+ Fe2+ NH4 + K+ Na+ Dissolve in excess NaOH (aq) to form colourless solution Sodium hydroxide solution is poured slowly into 2 cm3 of the solution to be tested in a test tube, until in excess. Insoluble in excess NaOH (aq) NH3 gas produced warm Zn2+ Al3+ Pb2+ Ca2+ Mg2+ Easylah
- 24. Reaction of CationsWith NH3 Zn2+ : White precipitate dissolves/larut in excess NH3 Pb2+ Al3+ Mg2+ : White precipitate insoluble/tidak larut in excess NH3 24 Cations Precipitate producedNo precipitate White precipitate Coloured precipitate Green Blue Brown Fe2+ Cu2+ Fe2+ NH4 + K+ Na+ Aqueous ammonia solution is poured slowly into 2 cm3 of the solution to be tested in a test tube until in excess. NH3 gas produced warm Zn2+ Al3+ Pb2+ Ca2+ Mg2+ Easylah + NH3 (aq) + excess NH3 (aq) Dark blue solution Dissolve in excess NH3 (aq) to form colourless solution Insoluble in excess NH3 (aq)
- 25. Zn2+ ion is theonly cation that form white precipitate and dissolves in both excess NaOH and NH3 solutions. Mg2+ ion is the only cation that form white precipitate and insoluble in both excess NaOH and NH3 solutions. Ca2+ ion in the only cation that form white precipitate in NaOH solutions, but no precipitate in NH3 solution. Fe2+ , Fe3+ and Cu2+ ions is easy to spot because the ions shows coloured precipitate. Pb2+ ion and Al3+ ion form white precipitate and dissolves in excess NaOH solution, but insoluble in excess NH3 solutions. Example: lead(II) nitrate solution and aluminium nitrate solution Sodium sulphate solution is added slowly into 2 cm3 of the solution to be tested in a test tube. If a white precipitate is formed, then then the solution tested is lead(II) nitrate. If no change occurs, then the solution tested is aluminium nitrate. 25 Based from the observation, I can conclude that How to differentiate between Pb2+ and Al3+ ? A chemical tests can be carried out in the laboratory to differentiate between Pb2+ and Al3+ . (Please refer to Confirmatory Test for Pb2+ , in ealier notes). Now let see some questions about salt. Try to solve it by yourself first and then compare with the answers provided.
- 26. Example 1: Describe chemicaltests that can be carried out in the laboratory to differentiate between (a) lead(II) nitrate solution and aluminium nitrate solution Sodium sulphate solution is added slowly into 2 cm3 of the solution to be tested in a test tube. If a white precipitate is formed, then then the solution tested is lead(II) nitrate. If no change occurs, then the solution tested is aluminium nitrate. (b) aluminium nitrate solution and zinc nitrate solution Aqueous ammonia solution is poured slowly into 2 cm3 of the solution to be tested in a test tube until in excess. If a white precipitate that dissolves in excess aqueous ammonia solution is formed, than the solution tested is zinc nitrate. If a white precipitate that is insoluble in excess aqueous ammonia solution is formed, than the solution tested is aluminium nitrate. (c) ammonium chloride solution and potassium chloride solution Nessler’s Reagent is added to 2 cm3 of the solution to be tested in a test tube. If a brown precipitate is formed, then the solution tested is ammonium chloride. If no change occurs, then the solution tested is potassium chloride (d) iron(II) sulphate solution and iron(III) sulphate solution Potassium hexacyanoferrate(II) solution is poured into 2 cm3 of the solution to be tested in a test tube. If a dark blue precipitate is formed, then the solution tested is iron(III) chloride. If no change occurs, then the solution tested is iron(II) chloride. Or Potassium hexacyanoferrate(III) solution is poured into 2 cm3 of the solution to be tested in a test tube. If a greenish-brown solution is formed, then the solution tested is iron(III) chloride. If no change occurs, then the solution tested is iron(II) chloride. Or Potassium thiocyanate solution is poured into 2 cm3 of the solution to be tested in a test tube. 26
- 27. If a bloodred colouration is formed, then the solution tested is iron(III) chloride. If no change occurs, then the solution tested is iron(II) chloride. (e) sodium chloride and sodium sulphate Silver nitrate solution is poured into 2 cm3 of the solution to be tested in a test tube. If a white precipitate is formed, then the solution tested is sodium chloride. If no change occurs, then the solution tested is sodium sulphate. Or Barium chloride solution is poured into 2 cm3 of the solution to be tested in a test tube. If a white precipitate is formed, then the solution tested is sodium sulphate. If no change occurs, then the solution tested is sodium chloride. Example 2: 1. State three examples of a) soluble salts b) insoluble salts Potassium carbonate Magnesium carbonate Lead(II) nitrate Lead(II) sulphate Ammonium chloride Argentum chloride 2. Which of the following salts is soluble 3. Identify the gas that turns moist red litmus paper blue Ammonia gas 4. Gas X has the following properties Gas X is carbon dioxide gas 27 Lead(II) chloride Sodium carbonate Calcium sulphate Barium sulphate • Colourless • Acidic gas • Turns lime water milky Salt P Metal oxide X Gas Y
- 28. 5. Heat + Colourof metal oxide X is yellow when hot and white when cold. Gas Y turns lime water milky. a) Name gas Y : carbon dioxide gas b) Name metal oxide X : zinc oxide c) Name salt P : zinc carbonate d) Write an equation to represent the action of heat on salt P ZnCO3 (s) ZnO (s) + CO2 (g) 6. A sample of copper(II) nitrate, Cu(NO3)2 was heated strongly. Write down the expected observation. Copper(II) nitrate decompose to produce black colour of residue when hot and cold. A brown gas that changed moist blue litmus paper to red and colourless gas that lighted up a glowing wooden splinter are produced. 28
- 29. D. Numerical probleminvolving stoichiometric reaction in the preparation of salt Example 1; Ammonium phosphate, (NH4)3PO4 is use as a fertilizer. 29.8g of this salt is prepared by neutralizing phosphoric acid, H3PO4 with ammonium gas, NH3. Calculate the volume of ammonium gas, NH3 reacted at room conditions. ( Relative atomic mass; H, 1: N, 14: P, 31; O, 16; Molar volume; 24 dm3 mol-1 at room conditions) Solutions; a. Calculate the number of moles 2.88 g [3(14) + 12(1) + 31 + 4(16) = 0.2 mol b. Write a balanced chemical equation Compare the mole ratio of NH3 and (NH4)3PO4 H3PO4(aq) + 3NH3(aq) (NH4)3PO4(aq) c. Calculate the number of moles of NH3 base on the mole ratio = 3 X 0.2 mol = 0.6 mol d. Calculate the volume of NH3 Volume = number of mole X volume = 0.6 mol X 24 dm3 mol -1 = 14.4 dm3 Example 2: 29 A balanced chemical equation for a reaction in preparation of a salt can be used to calculate the stoichiometric quantities of the following Masses of reactants Volumes and concentrations of reactants Masses of products Volumes of products = 3 mol 1 mol
- 30. 3.9 g ofpotassium is burnt completely in the air as shown in the following equation; 4K(s) + O2(g) → 2K2O(s) What is the mass of potassium oxide produced? [Relative atomic mass: K, 39; O, 16] Solutions Tip: Solve the question step by step Step 1: Write Chemical Equation 4K(s) + O2(g) → 2K2O(s) 4 mol of K react with 1 mol of O2 produce 2 mol K2O Step 2: Calculate the number of mole [Get the information from the question] Step 3: Find the coefficient From Balance Chemical Equation FBCE; 4 mol of K produce 2 mol K2O Thus; 0.1 mol of K produce 2/4 mol K2O = 0.2 mol K2O FBCE; [Sebelah kiri] [Sebelah kanan] Bil. mol yang telah dikira Bil. Mol yang hendak ditentukan 4 mol K = 2 mol K2O 0.1mol K = 2/4 x 0.1mol K2O = 0.05 mol K2O No. of mol of K2O = 0.05 mol Step 4: Solve the questions Thus; Mass of K2O = 0.05 mol × Molar mass = 0.05 mol× 55 g mol-1 = 2.75 g Example 3: 30 No. of mol K = mass Molar mass = 3.9 g 39 gmol-1 0.1 mol=
- 31. Acids reacts withcalcium carbonate, CaCO3 in limestone to form a salt and carbon dioxide, CO2. A piece of limestone reacted completely with 100 cm3 of 31.5 g dm-3 nitric acid, HNO3. [Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Ca, 40. Molar volume: 24 dm3 mol-1 at room conditions] a. Calculate the mass of salt produced. b. What is the volume of carbon dioxide, CO2 liberated at room conditions? Step 1: Write Chemical Equation Chemical Equation: 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O Step 2: Calculate the number of mole Get the information from the question; FBCE; 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O 2 mol HNO3 = 1 mol Ca(NO3)2 0.05 mol HNO3 = ½ x 0.05 mol Ca(NO3)2 = 0.025 mol Ca(NO3)2 No. of mol of Ca(NO3)2 = 0.025 mol Mass of Ca(NO3)2 = 0.025 mol × 40 + 2[14 + 3(16)] g mol-1 = 4.1 g 31 No. of mole of HNO3 = Molarity × Volume 1000 = 0.5 mol dm-3 × 100 cm3 1000 = 0.05 mol Concentration of HNO3 = 31.5 g dm3 = Molar mass of HNO3 31.5 g dm3 = 0.5 mol dm-3 = 31.5 g dm3 1 + 14 + 48 g mol-1 Change the concentration given in g dm-3 to mol dm-3 first
- 32. FBCE; 2HNO3 +CaCO3 → Ca(NO3)2 + CO2 + H2O 2 mol HNO3 = 1 mol CO2 0.05 mol HNO3 = ½ x 0.05 mol CO2 = 0.025 mol CO2 No. of mol of CO2 = 0.025 mol Volume of CO2 = 0.025 mol × 12 + 2(16) dm3 mol-1 = 1.1 dm3 Example 4: Pb(NO3)2 compound decomposes when heated as shown in the following equation. If 6.62 g of Pb(NO3)2 compound is heated, calculate; [Relative atomic mass: N, 14; O, 16; Pb, 207; 1 mol of gas occupies 22.4 dm3 at s.t.p.] (i) mass of PbO that is produced (ii) volume of nitrogen dioxide produced at s.t.p (ii) volume of oxygen produced at s.t.p Solution: FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2 2 mol Pb(NO3)2 = 2 mol PbO 0.02 mol Pb(NO3)2 = 0.02 mol PbO No of mol PbO = 0.02 mol Mass of PbO = 0.02 x 223 = 4.46 g 32 No of mol Pb(NO3 )2 = mass Molar mass = 6.62 g 331 gmol-1 0.02 mol= 2Pb(NO3)2 → 2PbO + 4NO2 + O2
- 33. FBCE; 2Pb(NO3)2 →2PbO + 4NO2 + O2 2 mol Pb(NO3)2 = 4 mol NO2 0.02 mol Pb(NO3)2 = 4/2 x 0.02 mol O2 = 0.04 mol O2 No of mol O2 = 0.04 mol Volume of O2 = 0.04 x 22.4 dm3 = 0.896 dm3 // 896 cm3 FBCE; 2Pb(NO3)2 → 2PbO + 4NO2 + O2 2 mol Pb(NO3)2 = 1 mol O2 0.02 mol Pb(NO3)2 = ½ x 0.02 mol O2 = 0.01 mol O2 No of mol O2 = 0.01 mol Volume of O2 = 0.01 x 22.4 dm3 = 0.224 dm3 // 224 cm3 Numerical Problems involving stoichiometric reactions in the precipitation of salts Question 1: A student prepare copper(II) nitrate, Cu(NO3)2 by reacting copper(II) oxide, CuO with 200 cm3 of 2.0 moldm-3 nitric acid, HNO3. Calculate the mass of copper(II) oxide, CuO needed to react completely with the acid. [Relative atomic mass: Cu, 64 ; O, 16] Question 2: X cm3 of 0.5 moldm-3 sulphuric acid, H2SO4 is added to 100 cm3 of 1.0 moldm-3 lead(II) nitrate solution to produce lead(II) sulphate, PbSO4. [Relative atomic mass: Pb, 20; O, 16; S, 32] a. Calculate the value of X. b. Calculate the mass of lead(II) sulphate obtained. Start to do exercises from any book. I will help and guide you to master this topic. Prepared by; Kamal Ariffin Bin Saaim SMKDBL 33