Chap 12 Preparation of Salts (Explained) 2018 PDF

Acid, Bases & Salts – Preparation of Salts (Explained!) Part A : Preparation Of Soluble Salts Excess (Insoluble) Reactive Metal + Acid 1. Reaction of a metal and an acid Acid : HCl, HNO3, H2SO4 Metal : Any metal above Hydrogen in Reactivity Series. eg. Mg, Zn, Fe...

Acid, Bases & Salts – Preparation of Salts (Explained!) Part A : Preparation Of Soluble Salts Excess (Insoluble) Reactive Metal + Acid 1. Reaction of a metal and an acid Acid : HCl, HNO3, H2SO4 Metal : Any metal above Hydrogen in Reactivity Series. eg. Mg, Zn, Fe Example: Zinc and dilute sulfuric acid Equation: Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Procedure: 1. Excess zinc is added to dilute sulfuric acid until no more zinc will react. (Reason : To ensure all the acid has been completely used up.) 2. When effervescence has stopped, the solution is filtered to remove the excess zinc. (Reason : effervescence stopped means all acid is used up.) (Reason : To remove the excess, unreacted zinc by filtration, salt solution not contaminated with any reactants.) 3. Heat to boil off about 1/3 the solution. (Reason : To get a hot SATURATED solution so crystallization can take place.) 4. Leave it to cool. (Reason : so crystallization can take place.) 5. The crystals are then filtered and rinsed with cold distilled water. (Reason : to remove any surface impurities) 6. Crystals are then dried between filter papers. Note : 1. Alternatively, the solution can be evaporated to dryness by leaving in open air to obtain the zinc sulfate crystals. 2. Do NOT HEAT IT strongly because the crystals will lose its water of crystallization upon heating to dryness.) 3. It is not suitable for Group 1 metals the reaction is too explosive and hence dangerous. (never react group I metals with acids!!!) 4. Gentle heating must be applied when the solution is being evaporated. Many salts, e.g. nitrate salts, decompose when heated strongly. Such salts can be obtained by concentrating, crystallization, filtration and finally drying between filter papers. © Mr Victor Lee @mychemclass.com Page 1 Acid, Bases & Salts – Preparation of Salts (Explained!) Excess Insoluble base/carbonate + Acid 2. Reaction of an insoluble base or carbonate and an acid Acid : HCl, HNO3, H2SO4 Base/Carbonate : Any insoluble base (oxide, hydroxide) or carbonate eg. MgO, CuO, MgCO3, CuCO3 Example: Copper (II) oxide (or carbonate) and dilute sulfuric acid Equation: CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) or CuCO3(s) + H2SO4(aq) → CuSO4(aq) + CO2(g) + H2O(l) Procedure: 1. Excess copper (II) oxide or carbonate is added to a gently heated dilute sulfuric acid. (Reason : Heating to speed up the rate of reaction.) 2. When no more of the copper (II) oxide or copper(II) carbonate will dissolve, the solution is filtered. (Reason : Excess copper(II) oxide or carbonate is to ensure all the acid has been completely used up. Since the acid has been used up, when you remove the excess, unreacted oxide or carbonate by filtration, you will get copper(II) sulfate solution that is not contaminated with any reactants.) 3. The solution is then heated to saturation and then cooled. 4. The copper (II) sulfate crystals formed are filtered and rinsed with cold distilled water. 5. They are then dried between filter papers. *explanation for making saturated solution and then crystallisation are explained in previous page. © Mr Victor Lee @mychemclass.com Page 2 Acid, Bases & Salts – Preparation of Salts (Explained!) Titration for SPA Salts ONLY 3. Reaction of a soluble base (alkali) or carbonate with an acid Acid : HCl, HNO3, H2SO4 Base/Carbonate : Any soluble base (oxide, hydroxide) or carbonate Example: Sodium hydroxide (or carbonate) and dilute sulfuric acid Equation: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Procedure: Name of Procedure : Titration 1. Use a pipette to transfer 25.0 cm3 of sodium hydroxide into a conical flask. 2. Add a few drops of pH indicator (eg. screened methyl orange) to the conical flask. Screened methyl orange will turn green in alkaline solution. (Reason : The colour of the indicator will tell us if the solution has been neutralized, when we add the dilute sulfuric acid to the fixed volume of sodium hydroxide solution in the conical flask.) 3. Use the burette to add dilute sulfuric acid to the alkali in the conical flask drop by drop. (Reason : We have to add drop by drop because we may add too much and miss the end-point when the pH indicator changes colour. The end-point is the neutral point when the alkali is neutralized by the acid. Screened methyl orange will turn grey. ) 4. At the end point, the indicator colour changes from green (alkaline) to grey (neutral). This is point of neutralisation. The volume of acid needed to neutralise the alkali is then recorded. (Screened methyl orange will turn red when excess acid is added, turning solution acidic) 5. The experiment is then repeated, this time without the indicator. The same volumes of acid and alkali needed for the neutralisation are mixed. 6. The salt solution obtained is then heated to get a saturated solution, then cooled to crystallise. (Reason : We are interested to know how much acid is needed to neutralized the alkali, but we do NOT want the indicator as impurities in our salt, so we must repeat the experiment without any indicator. However, make sure that the amount of acid and alkali is exactly the same as when indicator was used, because that is the volume of both to produce the neutral salt.) Experimental Techniques : 1. The pipette should be rinsed with the pipette solution before being filled with it for the actual titration. 2. The burette should be rinsed with the burette solution before being filled with it for the actual titration. (you may rinse the pipette and burette with distilled water before the respective solutions if extensive cleaning of pipette/burette is needed, eg. when you filled them with the wrong solutions initially) 3. The conical flask must be rinsed only with distilled water. © Mr Victor Lee @mychemclass.com Page 3 Acid, Bases & Salts – Preparation of Salts (Explained!) Precipitation of silver chloride, lead(II) chloride, lead(II) sulfate, barium sulfate, calcium sulfate, and ALL carbonates (except Sodium, Potassium, Ammonium) Part B : Preparation Of Insoluble Salts 1. Precipitation using two solutions Reactants : Any TWO soluble compounds that contain the required cation and anion of the insoluble salt you want. Example: Preparation of lead(II) chloride Reactants : Lead (II) nitrate and sodium chloride solutions Equation: Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) Ionic equation: Pb2+(aq) + 2Cl–(aq) → PbCl2(s) Procedure: 1. Mix the 2 solutions, with one containing the cation and the other one containing the anion. (Reason : We want the cation and the anion to combine and form the insoluble salt.) 2. The precipitate formed is the salt we want. Mixture is filtered to obtain the residue which is then rinsed with distilled water to remove impurities in the form of unreacted lead(II) nitrate, sodium chloride or sodium nitrate. (Note : The precipitate is the insoluble particles of lead(II) chloride formed) 3. The insoluble salt is then dried in between filter papers. 2. From an insoluble compound Important : If we use an insoluble reactant to produce an insoluble product, the insoluble product formed on the surface of reactant will COVER UP the insoluble reactant, preventing it from reacting completely. Method : Step 1 : convert the insoluble reactant to a soluble salt using nitric acid Step 2 : convert the soluble salt into the insoluble salt that you want. Example: Preparation of lead(II) sulfate from lead(II) carbonate Equations: Step 1 : PbCO3(s) + 2HNO3(aq) → Pb(NO3)2(aq) + CO2(g) + H2O(l) Step 2 : Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq) Procedure: 1. Excess lead(II) carbonate is first dissolved in dilute nitric acid to form soluble lead(II) nitrate. 2. The salt solution is then filtered to remove excess lead(II) carbonate. (Reason :Insoluble lead(II)carbonate will interfere with the Step 2 reaction.) 3. The lead(II) nitrate solution is then mixed with dilute sulfuric acid. The insoluble salt, lead(II) sulfate, precipitates out and can be obtained by filtration. 4. The salt is then rinsed and dried between filter papers. © Mr Victor Lee @mychemclass.com Page 4