Topic 4 Acids Bases Unit 14 Acids alkalis Unit 15 Molarity, ph scale strengths of acids alkalis Unit 16 Salts neutralization Unit 17 Concentration of s volumetric analysis
Key C o ncepts Molarity, ph scale strengths of acids alkalis Concept of molarity ph scale Strengths of acids alkalis Acids Bases Acids alkalis Characteristics of acids alkalis Role of water for acids alkalis Concentration of s volumetric analysis Concentration of s (in g dm 3 ) Preparing stard s of acids alkalis Calculations involving volumetric analysis Salts neutralization Acid-base neutralization Preparation of salts Uses of neutralization
Topic 4 Acids Bases Unit 14 Acids alkalis Unit 14 Acids alkalis 14.1 Acids in our daily lives 14.2 Acids in the laboratory 14.3 Characteristics of dilute acids 14.4 The role of water for acids 14.5 Basicity of an acid 14.6 Bases alkalis 14.7 Alkalis in the home 14.8 Alkalis in the laboratory 14.9 Characteristics of s of alkalis 14.10 The role of water for alkalis 14.11 An introduction to analytical chemistry 14.12 Concentrated acids 14.13 Corrosive nature of concentrated acids alkalis 14.14 Hygroscopic deliquescent substances 14.1 14.5 Summary 1 An acid is a hydrogen-containing substance that gives hydrogen ions (H + (aq)) as the only type of positive ions when dissolved in water. 2 Concentrated acids contain acids dissolved in a small amount of water. Dilute acids contain acids dissolved in a large amount of water. 3 Characteristics of dilute acids include the following: a) Most dilute acids have a sour taste. b) Colours of indicators in acids: Indicator Litmus Methyl orange Phenolphthalein Colour in hydrochloric acid sulphuric acid nitric acid red red colourless c) Dilute acids react with reactive metals to give hydrogen gas. metal + dilute acid salt + hydrogen Mg(s) + 2HCl(aq) MgCl 2 (aq) + H 2 (g) Mg(s) + H 2 SO 4 (aq) MgSO 4 (aq) + H 2 (g) d) Dilute acids react with carbonates to give carbon dioxide gas. carbonate + dilute acid salt + water + carbon dioxide CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) CaCO 3 (s) + 2HNO 3 (aq) Ca(NO 3 ) 2 (aq) + H 2 O(l) + CO 2 (g) e) Dilute acids react with hydrogencarbonates to give carbon dioxide gas. hydrogencarbonate + dilute acid salt + water + carbon dioxide NaHCO 3 (s or aq) + HCl(aq) NaCl(aq) + H 2 O(l) + CO 2 (g) 2NaHCO 3 (s or aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O(l) + 2CO 2 (g) f) Oxides hydroxides of metals are called bases. When an acid reacts with a base, a salt water are formed. The reaction is called neutralization. acid + metal oxide salt + water acid + metal hydroxide salt + water H 2 SO 4 (aq) + CuO(s) CuSO 4 (aq) + H 2 O(l) 2HCl(aq) + Mg(OH) 2 (s) MgCl 2 (aq) + 2H 2 O(l)
Topic 4 Acids Bases Unit 14 Acids alkalis g) All dilute acids conduct electricity due to the presence of mobile ions. 4 Water must be present for an acid to show its acidic properties. 5 The maximum number of hydrogen ions produced by an acid molecule is called the basicity of the acid. Exam tips Example Solid acids do NOT contain hydrogen ions. Hydrogen ions are formed only when acid molecules dissociate in water. Solid citric acid does NOT contain hydrogen ions. So, it CANNOT turn dry blue litmus paper red. Questions often ask about citric acid. It is a weak acid. It is an electrolyte. It exists as a solid at room conditions. It contains ionizable hydrogen atoms. When citric acid is dissolved in water, citric acid molecules become mobile. Aqueous citric acid reacts with magnesium to give hydrogen while solid citric acid does NOT. Questions may ask students to give the observable change(s) of a chemical reaction. Dilute nitric acid is added to copper(ii) oxide. The black copper(ii) oxide dissolves in dilute nitric acid to give a blue. CuO(s) + 2HNO 3 (aq) Cu(NO 3 ) 2 (aq) + H 2 O(l) When asked about observable changes, do NOT just give the names of the products. Both oxides hydroxides do NOT give a gas when reacted with acids. Consider the following information on two reactions involving magnesium ribbons of the same shape. Reaction 1 Reacting mixture 1.5 g of Mg + 100 cm 3 of 1.0 mol dm 3 HCl(aq) b) Would you expect the initial rate of Reaction 2 to be higher, lower or the same as that of Reaction 1? Explain your answer. (3 marks) Answer a) Mg(s) + 2HCl(aq) MgCl 2 (aq) + H 2 (g) 1.5 g Number of moles of Mg present = 24.3 g mol 1 = 0.062 mol (1) Number of moles of HCl used = 1.0 mol dm 3 100 x 1 000 dm3 = 0.10 mol (1) According to the equation, 1 mole of Mg reacts with 2 moles of HCl. During Reaction 1, 0.10 mole of HCl reacted with 0.050 mole of magnesium. Therefore magnesium was in excess. (1) b) The initial rate of Reaction 2 was higher than that of Reaction 1. (1) During the reaction between magnesium the acids, magnesium would react with hydrogen ions in the acids. Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid. (1) Both are strong acids dissociate completely in water. Therefore 1.0 mol dm 3 sulphuric acid has a higher concentration of hydrogen ions. (1) Thus, the initial rate of Reaction 2 is higher than that of Reaction 1. Remarks* Questions often ask students to compare strong acids with weak acids. Comparing 20 cm 3 of 1 mol dm 3 CH 3 COOH(aq) with 20 cm 3 of 1 mol dm 3 HCl(aq) they contain different number of hydrogen ions; they have different ph values; they have different electrical conductivity; they react with magnesium at different rates; they give the same amount of hydrogen gas with equal mass of magnesium; they require the same number of moles of NaOH for complete neutralization; they give different colour change with the same quantity of universal indicator; when completely neutralized by 1 mol dm 3 NaOH(aq), HCl(aq) gives a larger temperature rise than CH 3 COOH(aq). Reaction 2 1.5 g of Mg + 100 cm 3 of 1.0 mol dm 3 H 2SO 4(aq) a) Calculate whether magnesium or hydrochloric acid is in excess in Reaction 1. (3 marks)
Topic 4 Acids Bases Unit 14 Acids alkalis 14.6 14.10 Summary Comparing 20 cm 3 of 1 mol dm 3 CH 3 COOH(aq) with 10 cm 3 of 1 mol dm 3 H 2 SO 4 (aq) they have different ph values; they have different electrical conductivity; they react with magnesium at different rates; they give the same amount of hydrogen gas with equal mass of magnesium; they require the same number of moles of NaOH for complete neutralization. 1 A base is a compound which reacts with an acid to give a salt water only. Bases are usually oxides or hydroxides of metals. acid + base salt + water 2 Alkalis are bases that are soluble in water. 3 The properties of s of alkalis depend on the presence of mobile hydroxide ions (OH (aq)). 4 Characteristics of s of alkalis include the following: a) Solutions of alkalis have a bitter taste. b) Dilute s of alkalis have a slippery feel. c) Colours of indicators in s of alkalis: Indicator Litmus Methyl orange Phenolphthalein potassium hydroxide Colour in sodium hydroxide blue yellow aqueous ammonia d) Adding dilute sodium hydroxide to s containing some metal ions gives precipitates as shown in the following table: Adding NaOH(aq) to containing Colour of precipitate formed Precipitate dissolves in excess NaOH(aq)? red Ionic equation Ca 2+ (aq) white Ca 2+ (aq) + 2OH (aq) Ca(OH) 2(s) Mg 2+ (aq) white Mg 2+ (aq) + 2OH (aq) Mg(OH) 2(s) Adding NaOH(aq) to containing Al 3+ (aq) Pb 2+ (aq) Zn 2+ (aq) Colour of precipitate formed white white white Precipitate dissolves in excess NaOH(aq)? (a colourless forms) (a colourless forms) (a colourless forms) Al 3+ (aq) + 3OH (aq) Al(OH) 3(s) + OH (aq) Pb 2+ (aq) + 2OH (aq) Ionic equation Pb(OH) 2(s) + 2OH (aq) Zn 2+ (aq) + 2OH (aq) Zn(OH) 2(s) + 2OH (aq) Al(OH) 3(s) [Al(OH) 4] (aq) Pb(OH) 2(s) [Pb(OH) 4] 2 (aq) Zn(OH) 2(s) Fe 2+ (aq) green Fe 2+ (aq) + 2OH (aq) Fe(OH) 2(s) Fe 3+ (aq) reddish brown Fe 3+ (aq) + 3OH (aq) Fe(OH) 3(s) [Zn(OH) 4] 2 (aq) Cu 2+ (aq) pale blue Cu 2+ (aq) + 2OH (aq) Cu(OH) 2(s) e) Adding dilute aqueous ammonia to s containing some metal ions gives precipitates as shown in the following table: Adding NH 3(aq) to containing Colour of precipitate formed Precipitate dissolves in excess NH 3(aq)? Ionic equation Mg 2+ (aq) white Mg 2+ (aq) + 2OH (aq) Mg(OH) 2(s) Al 3+ (aq) white Al 3+ (aq) + 3OH (aq) Al(OH) 3(s) Pb 2+ (aq) white Pb 2+ (aq) + 2OH (aq) Pb(OH) 2(s) Zn 2+ (aq) white (a colourless forms) Zn 2+ (aq) + 2OH (aq) Zn(OH) 2(s) Zn(OH) 2(s) + 4NH 3(aq) [Zn(NH 3) 4] 2+ (aq) + 2OH (aq) Fe 2+ (aq) green Fe 2+ (aq) + 2OH (aq) Fe(OH) 2(s) Fe 3+ (aq) reddish brown Fe 3+ (aq) + 3OH (aq) Fe(OH) 3(s) Cu 2+ (aq) pale blue (a deep blue forms) Cu 2+ (aq) + 2OH (aq) Cu(OH) 2(s) Cu(OH) 2(s) + 4NH 3(aq) [Cu(NH 3) 4] 2+ (aq) + 2OH (aq) f) Heating solids or s of ammonium compounds with s of alkalis liberates ammonia gas. 2NH 4 Cl(s or aq) + Ca(OH) 2 (aq) NH 4 Cl(s or aq) + NaOH(aq) 2NH 3 (g) + CaCl 2 (aq) + 2H 2 O(l) NH 3 (g) + NaCl(aq) + H 2 O(l) g) A of alkali reacts with an acid to give a salt water only. h) All s of alkalis conduct electricity due to the presence of mobile ions.
10 Topic 4 Acids Bases Unit 14 Acids alkalis 11 Exam tips Example Remarks* Questions often ask about nickel(ii) compounds. Ni 2+ (aq) is green in colour. Nickel(II) hydroxide can be precipitated by adding dilute sodium hydroxide to a of nickel(ii) salt. Ni 2+ (aq) + 2OH (aq) Ni(OH) 2 (s) green precipitate Nickel(II) carbonate can be precipitated by mixing Ni 2+ (aq) CO 3 2 (aq). Ni 2+ (aq) + CO 3 2 (aq) NiCO 3 (s) Be careful with the spelling of the chemicals ammonium chloride, ammonium sulphate, etc. Cu 2+ (aq) ions can be separated from a mixture containing Cu 2+ (aq) K + (aq) ions by adding NaOH(aq). Cu 2+ (aq) ions give a precipitate with NaOH(aq) while K + (aq) ions do not. When reading questions, do NOT mix up the two words separate distinguish. Do NOT confuse the colour of Fe 3+ (aq) ions with that of Fe(OH) 3 (s). For each of the following pairs of species, suggest a chemical test to distinguish between them write the chemical equation(s) of the reaction(s) involved. a) Mg 2+ (aq) Pb 2+ (aq) (4 marks) b) NH 4 NO 3 (s) KNO 3 (s) (3 marks) Answer a) Add dilute sodium hydroxide to each species. (1) Only Pb 2+ (aq) gives a white precipitate which is soluble in excess NaOH(aq). (1) Pb 2+ (aq) + 2OH (aq) Pb(OH) 2 (s) (1) Pb(OH) 2 (s) + 2OH (aq) [Pb(OH) 4 ] 2 (aq) (1) b) Warm each solid with dilute sodium hydroxide. (1) Only NH 4 NO 3 (s) gives a gas that turns moist red litmus paper blue. (1) NH 4 NO 3 (s) + NaOH(aq) NH 3 (g) + NaNO 3 (aq) + H 2 O(l) (1) Questions often ask about the chemical test for identifying an ammonium compound. Remember it is necessary to warm the sample under test with sodium hydroxide / calcium hydroxide. The ammonium compound gives an alkaline gas, NH 3. 14.11 14.14 Summary 1 The chemical reactions learnt in this unit can be used to identify ions present in a sample. 2 Concentrated acids alkalis are corrosive. 3 Hygroscopic deliquescent substances ( anhydrous calcium chloride, silica gel concentrated sulphuric acid) are often used as drying agents. Exam tips Questions often ask about the reagents (or methods) that can be used to distinguish between two species. Examples: aluminium sulphate lead(ii) ethanoate solid sodium carbonate solid calcium carbonate iron(ii) sulphate iron(iii) sulphate magnesium nitrate silver nitrate solid ammonium chloride solid potassium chloride dilute sulphuric acid dilute nitric acid dilute sulphuric acid dilute nitric acid dilute hydrochloric acid dilute nitric acid dilute hydrochloric acid dilute nitric acid distinguished by hydrochloric acid (only lead(ii) ethanoate produces a white precipitate, PbCl 2 ) distinguished by testing water solubility distinguished by colour / dilute aqueous ammonia / dilute sodium hydroxide (they give precipitates of different colours with dilute aqueous ammonia / dilute sodium hydroxide ) distinguished by potassium chloride (only silver nitrate produces a white precipitate, AgCl) distinguished by warming with sodium hydroxide / calcium hydroxide (only ammonium chloride gives an alkaline gas, NH 3 ) distinguished by heating with copper (only dilute nitric acid gives gas bubbles) distinguished by barium chloride (only dilute sulphuric acid gives a white precipitate, BaSO 4 ) distinguished by heating with copper (only dilute nitric acid gives gas bubbles) distinguished by silver nitrate (only dilute hydrochloric acid gives a white precipitate, AgCl)
12 Topic 4 Acids Bases Unit 14 Acids alkalis 13 dilute sodium hydroxide dilute aqueous ammonia of cane sugar of sodium chloride distinguished by a containing aluminium ions / lead(ii) ions (only dilute sodium hydroxide gives a white precipitate which dissolves in excess alkali) distinguished by electrical conductivity (only of sodium chloride can conduct electricity) Questions often ask about the hazard warning labels that should be displayed on the bottle of a certain chemical. Example You are provided with three unlabelled bottles, each containing one of the white powders listed below: KBr(s), SiO 2 (s) glucose a) Outline the physical tests you would use to distinguish the three substances from one another. (4 marks) b) Describe a chemical test you would use to distinguish between KBr(s) from glucose. (2 marks) Answer a) Add water to each solid. (1) Only SiO 2 (s) is insoluble. (1) Test the electrical conductivity of the obtained. (1) KBr(aq) conducts, but glucose does not. (1) b) Heat each solid strongly. (1) Only glucose chars. (1) Questions often ask about the set-ups drying agents that can be used to dry a certain gas. Ammonia CANNOT be dried by concentrated sulphuric acid as there is a reaction between them. Ammonia can be dried by calcium oxide. Hydrogen chloride gas CANNOT be dried by calcium oxide as there is a reaction between them. Hydrogen chloride gas can be dried by anhydrous calcium chloride / concentrated sulphuric acid. The set-up shown below CANNOT be used to dry a gas because the end of the delivery tube for the incoming gas should be put under the drying agent to ensure that all the incoming gas can pass through the drying agent. Remarks* Part (a) of the question asks for physical tests, so do NOT give chemical tests. The solids CANNOT be distinguished by determining their boiling points because the boiling point of SiO 2 (s) is so high (2 230 C) that it CANNOT be determined using apparatus commonly available in a school laboratory. moist gas dry gas concentrated sulphuric acid
14 Topic 4 Acids Bases Unit 15 Molarity, ph scale strengths of acids alkalis 15 Unit 15 Molarity, ph scale strengths of acids alkalis 15.1 Concentration of a 15.2 The ph scale 15.3 Determining ph values of s 15.4 Strong weak acids 15.5 Comparing the strength of acids 15.6 ph the extent of dissociation of acids 15.7 Strong weak alkalis 15.8 Comparing the strength of alkalis 15.9 Strength versus concentration 15.1 15.9 Summary 1 The molarity of a is the number of moles of solute dissolved in 1 dm 3 of the. Molarity of a (mol dm 3 ) = number of moles of solute (mol) volume of (dm 3 ) 2 The ph of a is log 10 of the molar concentration of hydrogen ions in that. ph = log 10 [H + ] 3 The ph scale is used to measure the degree of acidity or alkalinity of a. Its value ranges from 0 to 14. increasing acidity increasing alkalinity 4 Methods used to determine the ph values of s include: a) using universal indicator; b) using a ph meter; c) using a data-logger with a ph sensor. 5 A strong acid is an acid that almost completely dissociates in water. A weak acid is an acid that only partially dissociates in water. Compared to a weak acid of the same concentration, a strong acid has a lower ph, a higher electrical conductivity a higher rate of reaction with metals. 6 A strong alkali is an alkali that almost completely dissociates to give hydroxide ions (OH (aq)) in water. A weak alkali is an alkali that only partially dissociates to give hydroxide ions in water. Compared to a weak alkali of the same concentration, a strong alkali has a higher ph a higher electrical conductivity. 7 Strength of an acid or alkali concerns the extent of dissociation of the acid or alkali in water. Concentration of an acid or alkali concerns the amount of the acid or alkali in a unit volume of. Exam tips Adding water to KOH(aq) decreases its degree of alkalinity. Thus, the ph of KOH(aq) decreases. Adding water to HCl(aq) decreases its degree of acidity. Thus, the ph of HCl(aq) increases. Concentrated ammonia is a weak alkali.
16 Topic 4 Acids Bases Unit 16 Salts neutralization 17 Questions often ask students to compare the properties of a strong acid / alkali with those of a weak acid / alkali of the same concentration. 0.5 mol dm 3 NaOH(aq) 0.5 mol dm 3 NH 3 (aq) higher ph lower ph higher electrical conductivity lower electrical conductivity larger temperature rise when smaller temperature rise when completely neutralized by completely neutralized by 1 mol dm 3 HCl(aq) 1 mol dm 3 HCl(aq) Suppose two identical zinc strips are added to s of monobasic acids X Y of the same concentration. Unit 16 Salts neutralization 16.1 Acid-base reactions 16.2 Heat change during neutralization 16.3 Formation of salts 16.4 Naming of salts 16.5 Soluble insoluble salts 16.6 Preparation of soluble salts 16.7 Preparation of insoluble salts 16.8 Uses of neutralization zinc strip of monobasic acid X of monobasic acid Y From the rate of the reaction between each acid the zinc strip, it can be deduced that acid X is weaker than acid Y. Example Explain whether you agree with the following statement. A is a stronger acid than B, so the ph of an aqueous of A must be lower than that of B. (3 marks) Answer Not agree A is a stronger acid than B only means that the degree of dissociation of A is larger than that of B. (1) However, the ph of an aqueous of an acid depends on both its degree of dissociation concentration. (1) As such, an aqueous of the stronger acid A may have a higher ph than that of the weaker acid B if the concentration of acid B is higher than that of acid A by an adequate amount. (1) Remarks* To compare the strength of a strong acid a weak acid, remember to use the same concentration in order to fairly compare their ph values.
18 Topic 4 Acids Bases Unit 16 Salts neutralization 19 16.1 16.8 Summary 1 Neutralization is the combination of hydrogen ions hydroxide ions (or oxide ions) to form water molecules. 2 In neutralization reactions, salt water are the only products. acid + alkali salt + water acid + insoluble metal hydroxide acid + insoluble metal oxide salt + water salt + water HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 3HNO 3 (aq) + Fe(OH) 3 (s) Fe(NO 3 ) 3 (aq) + 3H 2 O(l) H 2 SO 4 (aq) + CuO(s) CuSO 4 (aq) + H 2 O(l) 3 Neutralization reactions are exothermic reactions. Heat is released in these reactions. 4 Rules for naming salts are as follows: a) the name of the metal or ammonium ions always comes first; b) if the metal can form more than one type of positive ion, write a Roman numeral in brackets to show the charge of the ion; c) if the salt contains only two elements, the name ends in ide; d) if the salt contains three or more elements, one of which being oxygen, the name ends in ite or ate. 5 The following table summarizes methods for the preparation of salts. Salt Soluble salts Insoluble salts Method of preparation actions of acid on metal / insoluble base / insoluble carbonate action of acid on alkali / soluble carbonate (titration) precipitation Example(s) Preparing zinc sulphate: Zn(s) + H 2SO 4(aq) ZnSO 4(aq) + H 2(g) ZnO(s) + H 2SO 4(aq) ZnSO 4(aq) + H 2O(l) ZnCO 3(s) + H 2SO 4(aq) ZnSO 4(aq) + H 2O(l) + CO 2(g) Preparing potassium chloride: KOH(aq) + HCl(aq) KCl(aq) + H 2O(l) K 2CO 3(aq) + 2HCl(aq) 2KCl(aq) + H 2O(l) + CO 2(g) Preparing silver chloride: Ag + (aq) + Cl (aq) from from AgNO 3(aq) NaCl(aq) AgCl(s) 6 Uses of neutralization: a) Acidic soil can be neutralized by adding quicklime (calcium oxide) to it. b) Liquid waste from factories containing acids can be neutralized by adding slaked lime (calcium hydroxide) to it. c) Fertilizers such as ammonium nitrate ammonium sulphate are prepared by neutralization reactions. d) Antacids containing bases such as magnesium hydroxide can cure the pain caused by excess acid in the stomach. Exam tips Example 20 cm 3 of 1 mol dm 3 HCl(aq) mixed with 20 cm 3 of 1 mol dm 3 NaOH(aq) 40 cm 3 of 1 mol dm 3 HCl(aq) mixed with 40 cm 3 of 1 mol dm 3 NaOH(aq) 20 cm 3 of 1 mol dm 3 HCl(aq) mixed with 20 cm 3 of 1 mol dm 3 NaOH(aq) 20 cm 3 of 2 mol dm 3 HCl(aq) mixed with 20 cm 3 of 2 mol dm 3 NaOH(aq) 20 cm 3 of 1 mol dm 3 CH 3 COOH(aq) mixed w i t h 2 0 c m 3 o f 1 mol dm 3 NaOH(aq) 20 cm 3 of 1 mol dm 3 HCl(aq) mixed with 20 cm 3 of 1 mol dm 3 NaOH(aq) Examination questions often ask students to compare the temperature rises upon mixing acids alkalis. Temperature rise the same temperature rise of second mixture is twice that of the first mixture temperature rise of first mixture is less than that of the second mixture Number of moles of water formed 0.02 mole 0.04 mole 0.02 mole 0.04 mole 0.02 mole 0.02 mole Heat released* 1.14 kj 2.28 kj 1.14 kj 2.28 kj < 1.14 kj 1.14 kj Explanation The first mixture (total volume 40 cm 3 ) absorbs 1. 1 4 k J w h i l e t h e second mixture (total volume 80 cm 3 ) absorbs 2.28 kj. Hence the two mixtures give the same temperature rise. The total volumes of the two mixtures are the same. Hence the temperature rise of the second mixture is twice that of the first mixture. For neutralization in which either the acid or alkali or both are weak, the stard enthalpy change of neutralization w i l l b e l e s s t h a n 57.1 kj mol 1. CH 3 COOH(aq) is a weak acid. Hence the temperature rise of the first mixture is less than that of the second mixture. * For the reaction between a strong acid a strong alkali, the heat released is 57 kj for 1 mole of water formed. The heat of neutralization will be further discussed in Topic 9 Chemical Reactions Energy.
20 Topic 4 Acids Bases Unit 16 Salts neutralization 21 NOT all salts are formed from neutralization. NOT all salt s are neutral. NaHSO 4 (aq) is acidic. The HSO 4 (aq) ions can dissociate to give hydrogen ions. HSO 4 (aq) H + (aq) + SO 4 2 (aq) Questions often ask about suitable method for preparing various salts. Lead(II) sulphate can be prepared by adding lead(ii) nitrate to dilute sulphuric acid. Pb 2+ (aq) + SO 4 2 (aq) PbSO 4 (s) Copper(II) sulphate can be prepared by adding excess copper(ii) oxide to dilute sulphuric acid. CuO(s) + H 2 SO 4 (aq) CuSO 4 (aq) + H 2 O(l) ii) Using the axes below, sketch the change of mass when a sample of CuSO 4 5H 2 O(s) is heated slowly. (2 marks) Mass (arbitrary unit) 100 150 200 250 300 Temperature ( C) Example Copper(II) sulphate crystals (CuSO 4 5H 2 O(s)) can be prepared in a laboratory by the following steps. Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Add excess copper(ii) oxide to dilute sulphuric acid warm the mixture. Remove the remaining copper(ii) oxide from the obtained. Evaporate the until it becomes saturated. Allow the saturated to cool down to obtain copper(ii) sulphate crystals. Separate the crystals from the saturated. Dry the crystals obtained. a) For Step 1, explain why copper(ii) oxide should be in excess. (1 mark) b) For Step 4, explain why crystals would be obtained when the saturated is cooled down. (1 mark) c) For Step 5, suggest how the crystals is separated from the saturated. (1 mark) d) For Step 6, suggest how the crystals can be dried. (1 mark) e) When CuSO 4 5H 2 O(s) is heated slowly such that the temperature rises steadily, it will lose four water molecules at about 110 C, then the last water molecule at about 250 C. i) Suggest a chemical test to show that water is being released. (2 marks) Answer a) To make sure that all the sulphuric acid has been reacted. / To make sure that the product is not contaminated with sulphuric acid. (1) b) The solubility of copper(ii) sulphate decreases when the temperature of the drops. (1) c) By filtration (1) d) Absorb the water by filter paper. / Place in a desiccator. (1) e) i) Treat the vapour evolved with dry cobalt(ii) chloride paper. (1) A colour change from blue to pink shows the presence of water. (1) ii) Mass (arbitrary unit) 100 150 200 250 300 Temperature ( C) (2)
22 Topic 4 Acids Bases Unit 17 Concentration of s volumetric analysis 23 Remarks* Questions often ask about the procedure for preparing CuSO 4 5H 2 O crystals. CuSO 4 5H 2 O crystals can be obtained by heating excess copper with concentrated sulphuric acid. Cu(s) + 2H 2 SO 4 (l) CuSO 4 (aq) + SO 2 (g) + 2H 2 O(l) Add water to the resulting mixture filter off the excess copper. Evaporate the filtrate to give a saturated copper(ii) sulphate. Allow the to cool to obtain CuSO 4 5H 2 O crystals. Dry the crystals in a desiccator. Unit 17 Concentration of s volumetric analysis 17.1 Concentration of a 17.2 Dilution 17.3 Volumetric analysis 17.4 Preparing a stard of an acid / alkali 17.5 Acid-alkali titration 17.6 ph change during a titration 17.7 Using an indicator in an acid-alkali titration 17.8 Equivalence point detection by temperature change 17.9 Applications of acid-alkali titrations 17.10 Back titration
24 Topic 4 Acids Bases Unit 17 Concentration of s volumetric analysis 25 17.1 17.4 Summary 1 Concentration of a (g dm 3 ) = mass of solute (g) volume of (dm 3 ) 2 The numbers of moles of solute in a before dilution after dilution are the same, i.e. (MV) before dilution = (MV) after dilution, where M = molarity, V = volume. Exam tips Questions often ask about the procedure of diluting a concentrated acid / alkali of known concentration. To prepare 250.0 cm 3 of 0.100 mol dm 3 Na 2 CO 3 (aq) from 1.00 mol dm 3 Na 2 CO 3 (aq), use a 25.0 cm 3 pipette a 250.0 cm 3 volumetric flask. When preparing a from a solid acid, a pipette is NOT required for transferring the acid to the volumetric flask. In the dilution process using a volumetric flask, distilled water is added to the flask until the bottom of the meniscus reaches the graduated mark on the flask. 17.5 17.10 Summary 1 Volumetric analysis is a method of finding out the concentration (or amount) of a substance in a sample. It relies on the precise measurement of volumes of s involved in chemical reactions. 2 In volumetric analysis, just sufficient volume of a of known concentration is allowed to completely react with the substance being analyzed in a sample. The process of determining the just sufficient volume is called titration. From the measured volume known concentration of the used, we can calculate the concentration (or amount) of the substance in question. 3 A stard is a with accurately known concentration. 4 The point in an acid-alkali titration at which the acid alkali just react completely with each other is called the equivalence point. 5 An acid-alkali indicator is used to indicate the end point of an acid-alkali titration. Type of acid-alkali titration Strong acid-strong alkali Weak acid-strong alkali Strong acid-weak alkali Suitable indicator methyl orange or phenolphthalein phenolphthalein methyl orange Weak acid-weak alkali Example 20.0 cm 3 of 1.00 mol dm 3 NaCl(aq) are mixed with 10.0 cm 3 of 2.00 mol dm 3 Na 2 CO 3 (aq). What is the concentration of Na + (aq) ions in the resulting? (3 marks) Answer Number of moles of Na + (aq) ions in NaCl(aq) = 1.00 mol dm 3 20.0 x 1 000 dm3 = 0.0200 mol (1) Number of moles of Na + (aq) ions in Na 2 CO 3 (aq) = 2 x 2.00 mol dm 3 10.0 x 1 000 dm3 = 0.0400 mol (1) 6 Special apparatus used in volumetric analysis include electronic balance, volumetric flask, pipette, pipette filler burette. 7 The following diagram summarizes the steps for calculations in volumetric analysis. known concentration volume of A number of moles of A use coefficients in the balanced equation to find the mole ratio number of moles of substance in question mass, concentration or volume of substance in question Concentration of Na + (aq) ions in the resulting (0.0200 + 0.0400) mol = 20.0 + 10.0 dm 3 1 000 = 2.00 mol dm 3 (1) Remarks* In the above example, students need to use the correct volume of the resulting for calculations in order to arrive at the right answer.
26 Topic 4 Acids Bases Unit 17 Concentration of s volumetric analysis 27 8 The following diagram summarizes the steps to calculate the amount or concentration of the substance being analyzed in a sample. number of moles of acid / alkali originally added to the sample number of moles of acid / alkali left over after reaction with the substance being analyzed in the sample (information obtained from back titration) = number of moles of acid / alkali reacts with the substance being analyzed in the sample use coefficients in the balanced equation to find the mole ratio number of moles of the substance being analyzed in the sample The first titration is a trial the data should NOT be included in the calculation. Examination questions often ask students to sketch the ph change for the titration of a weak acid ( ethanoic acid) with dilute NaOH(aq). ph 14 12 10 8 6 4 2 equivalence point phenolphthalein changes colour within this ph range Exam tips amount or concentration of the substance being analyzed in the sample Students should be able to give the meaning of the term primary stard in titrimetric analysis. A primary stard of a substance can be prepared by dissolving a known mass of the substance in a solvent making up the to a known volume. As sodium hydroxide absorbs water / CO 2 readily in air, we CANNOT prepare a stard sodium hydroxide just by dissolving a known mass of the solid in distilled water making up to a known volume of. Questions often ask about the correct procedure for washing each glassware before titration. Wash a pipette a burette first with distilled water then with the it is going to deliver. Wash a conical flask a volumetric flask with distilled water only. Consider an aqueous of ethanoic acid with a ph value of 4. Adding solid calcium carbonate; dilute aqueous ammonia would increase the ph value of the acid. 0 0 10.0 20.0 30.0 40.0 Volume of NaOH(aq) added (cm 3 ) Students should be able to choose a suitable indicator based on the ph change during a titration. The equivalent point of a neutralization reaction can also be detected by measuring either temperature or electrical conductivity. The following graph shows the temperature change during the titration between NaOH(aq) HCl(aq). Temperature ( C) X equivalence point Y Volume of acid added (cm 3 ) equivalence point of titration; temperature is the highest Z
28 Topic 4 Acids Bases Unit 17 Concentration of s volumetric analysis 29 The following graph shows the electrical conductivity change during the titration between H 2 SO 4 (aq) Ba(OH) 2 (aq). high Conductivity low Volume of H 2 SO 4 (aq) added to Ba(OH) 2 (aq) (cm 3 ) As sulphuric acid is added, it removes both the barium ions (by precipitation) hydroxide ions (by neutralization) Ba(OH) 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2H 2 O(l) At the equivalence point, all the barium ions hydroxide ions have been removed. Hence the electrical conductivity of the reaction mixture falls almost to zero. i) Draw a labelled diagram to show the set-up used in the titration. (3 marks) ii) Suggest a suitable indicator for the titration in Step 3, state the expected colour change at the end point. (2 marks) iii) Calculate the number of moles of excess hydrochloric acid in 25.0 cm 3 of the diluted from the data obtained in the titration. (2 marks) iv) Hence calculate the mass of aluminium hydroxide in one drug tablet. (3 marks) (Relative atomic masses: H = 1.0, C = 12.0, Al = 27.0) Answer a) Aluminium hydroxide in the tablet neutralizes the excess hydrochloric acid in the stomach. (1) Al(OH) 3 (s) + 3HCl(aq) AlCl 3 (aq) + 3H 2 O(l) (1) b) Chewing breaks down the tablets into smaller pieces. This increases the surface area of the tablets thus increases the reaction rate (brings faster relief of pain). (1) c) i) Example The drug tablets shown below can relieve stomach ache. It contains aluminium hydroxide, Al(OH) 3, as the only active ingredient. burette 0.190 mol dm 3 sodium hydroxide FORMULA: Each tablet contains Aluminium Hydroxide Dried Gel 200 mg, Magnesium Hydroxide 200mg, Dimethylpolysiloxane 20mg. DOSAGE: 2 tablets to be well chewed 30 minutes after meals, at bedtime, when symptoms occur or as directed by physician. Do not take more than 24 tablets in a 24 hour period. CAUTION: It is dangerous to exceed the stated does. Keep this drug out of the reach of children. Do not take this product if you are taking antibiotic drug containing Tetracycline. conical flask of drug tablet + indicator white tile (3) a) Explain how the drug tablets can relieve stomach ache with the help of an equation. (2 marks) b) Explain why the drug tablets should be chewed before swallowing. (1 mark) c) A student performed the following experiment to determine the amount of aluminium hydroxide contained in one drug tablet. Step 1 A drug tablet was dissolved in 50.0 cm 3 of 1.00 mol dm 3 hydrochloric acid to form a. Step 2 The was placed in a volumetric flask then diluted to 250.0 cm 3 with distilled water. Step 3 25.0 cm 3 of the diluted were titrated with 0.190 mol dm 3 sodium hydroxide using a suitable indicator. 18.1 cm 3 of sodium hydroxide were needed to reach the end point. (1 mark for correct set-up; 2 marks for correct labels; 0 mark if the set-up is not workable) ii) Methyl orange: from red to yellow / Phenolphthalein: from colourless to red or pink (2) iii) NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) 0.190 mol dm 3 18.1 cm 3 Number of moles of NaOH in 18.1 cm 3 = 0.190 mol dm 3 x 18.1 1 000 dm3 = 0.00344 mol (1) According to the equation, 1 mole of HCl requires 1 mole of NaOH for complete neutralization. i.e. number of moles of excess HCl in 25.0 cm 3 diluted = 0.00344 mol (1)
30 Topic 4 Acids Bases iv) Number of moles of HCl added in Step 1 = 1.00 mol dm 3 50.0 x 1 000 dm3 = 0.0500 mol (1) Number of moles of HCl left over after reaction with drug tablet in Step 1 250.0 cm 3 = 0.00344 mol x 25.0 cm 3 = 0.0344 mol Number of moles of HCl reacted with Al(OH) 3 in drug tablet = (0.0500 0.0344) mol = 0.0156 mol (1) Al(OH) 3 (s) + 3HCl(aq) AlCl 3 (aq) + 3H 2 O(l) According to the equation, 1 mole of Al(OH) 3 requires 3 moles of HCl for complete neutralization. 0.0156 i.e. number of moles of Al(OH) 3 in drug tablet = mol 3 = 0.00520 mol Mass of Al(OH) 3 in drug tablet = 0.00520 mol x 78.0 g mol 1 = 0.406 g (1) \ one drug tablet contains 0.406 g of aluminium hydroxide. Remarks* Questions often ask students to suggest a suitable indicator for a certain titration state the colour change. Do NOT omit the dilution factor in the calculations. Otherwise, a wrong answer would be obtained.