Form 4 Chemistry Textbook PDF

CHAPTER 6 Acid, Base and Salt Keywords Basicity of acids pH and pOH Strength of acids and alkalis Molarity Standard solutions Neutralisation Titration Insoluble salts Recrystallisation Double decomposit...

CHAPTER 6 Acid, Base and Salt Keywords Basicity of acids pH and pOH Strength of acids and alkalis Molarity Standard solutions Neutralisation Titration Insoluble salts Recrystallisation Double decomposition reactions Limestone cave, Taman Negara Mulu What will you learn? 6.1 The Role of Water in Showing Acidic and Alkaline Properties 6.2 pH Value 6.3 Strength of Acids and Alkalis 6.4 Chemical Properties of Acids and Alkalis 6.5 Concentration of Aqueous Solution 6.6 Standard Solution 6.7 Neutralisation 6.8 Salts, Crystals and Their Uses in Daily Life 6.9 Preparation of Salts 6.10 Effect of Heat on Salts 6.11 Qualitative Analysis 134  Bulletin How are stalactites and stalagmites formed in limestone caves? Limestone caves consist of calcium Stalactites carbonate, CaCO3. When rainwater falls on the caves seep through the limestone, the following reaction takes place to produce calcium bicarbonate salt, Ca(HCO3)2. H2O(l) + CO2(g) + CaCO3(s) → Ca(HCO3)2(aq) Stalagmites The flowing water will carry the soluble calcium bicarbonate, Ca(HCO3)2 through the crevices at the roof of the caves. When the water comes in contact with air in the caves, a small portion of calcium bicarbonate, Ca(HCO3)2 reverts back to calcium carbonate, CaCO3, due to water and carbon dioxide losses. Calcium carbonate, CaCO3 starts to precipitate on these crevices. Hence, the formation of stalactites begins gradually. Water that drips from the ends of the stalactites will fall on the floor of the cave. Over the time, stalagmites will form in the same way as stalactites. This is why stalactites and stalagmites are found together in the caves. Formation of stalactites and stalagmites http://bit.ly/2ISEfPQ What is the relationship between pH value and concentration of hydrogen ions, H+? Why are all alkalis bases but not all bases are alkalis? How does a laboratory assistant prepare a standard solution? 135 THEME 3 Interaction between Matter 6.1 The Role of Water in Showing Acidic and Alkaline Properties Situation in Figure 6.1 shows the uses of acidic and alkaline substances in daily life. Identify which subtances are acidic and which substances are alkaline. Mom, why do we In order to neutralise Dad, how can I make need to brush our the acid on our teeth. Try scrubbing teeth every morning? this coin shiny again? it with lime. Figure 6.1 Acidic and alkaline substances in daily life Acids g When acid is dissolved in water, the hydrogen atoms in acid Learnin tandard molecules are released in the form of hydrogen ions, H+. Therefore, S based on the Arrhenius theory, acid is defined as follows: At the end of the lesson, pupils are able to: 6.1.1 Define acid and alkali Chemical substances ionise in water to produce hydrogen 6.1.2 State the meaning of ions, H+. basicity of an acid 6.1.3 Investigate the role of water in showing acidic HCl(aq) → H+(aq) + Cl–(aq) HNO3(aq) → H+(aq) + NO3–(aq) and alkaline properties through experiment When hydrogen chloride gas is dissolved in water, hydrogen chloride molecules will ionise in water to produce hydrogen ions, H+ and chloride ions, Cl–. However, do the hydrogen ions, H+ remain in the aqueous solution? Literally, hydrogen ions, H+ produced will combine with the water molecules, H2O to form Chemistry hydroxonium ions, H3O+. Although hydroxonium ions, H H H H H3O+ are the actual ions existing in the aqueous H C1 + O O+ + C1– solution that gives the acidic properties, to simplify H explanation, we often use hydrogen ion, H+ to represent Figure 6.2 Formation of hydroxonium ion, H3O+ hydroxonium ions, H3O+. 136  Acid, Base and Salt CHAPTER 6 Basicity of Acids Basicity of acids refers to the number of hydrogen ions, H+ that can be produced by an acid molecule that ionises in water. Hydrochloric acid, HCl is monoprotic acid because it can produce one hydrogen ion, H+ per acid molecule. How about diprotic acid and triprotic acid? Acids Monoprotic acid Diprotic acid Triprotic acid Hydrochloric acid, HCl Sulphuric acid, H2SO4 Phosphoric acid, H3PO4 produces one H+ ion produces two H+ ions produces three H+ ions per acid molecule per acid molecule per acid molecule Figure 6.3 The classification of acids based on the basicity of the acids Formic acid, HCOOH is used in the coagulation of latex. Is formic acid, HCOOH a diprotic acid? Why? Alkalis Brain Teaser Base is a substance that reacts with acids to produce salt and water only. Metal oxides and metal hydroxides are bases. For example, magnesium oxide, MgO and calcium hydroxide, Ca(OH)2 are bases because they react with acids to produce salt and water only. MgO(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) Brain Teaser Salt Observe the chemical equation below. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(l) Is magnesium a base? Why? Salt A base that is soluble in water is called an alkali. Potassium hydroxide, KOH and sodium hydroxide, NaOH are alkalis because they are soluble in water. When sodium hydroxide pellets, NaOH is dissolved in water, sodium ions, Na+ and hydroxide ions, OH– that can move freely in water are produced. Water Na+ Na+ OH– Na+ OH– Na+ OH– OH– Figure 6.4 Dissociation of sodium hydroxide, NaOH into ions that move freely in water 137 THEME 3 Interaction between Matter An alkali is defined as follows: Chemical substances that ionise in water to produce hydroxide ions, OH–. What will happen to ammonia molecule when ammonia gas is dissolved in water? Why is aqueous ammonia produced an alkali? H H H H N H H + H N+ + O– O H H H Ammonia gas, NH3 Water, H2O Ammonium ion, NH4+ Hydroxide ion, OH– NH3(aq) + H2O(l)   NH4+(aq) + OH–(aq) Figure 6.5 Formation of hydroxide ion, OH– from ammonia molecule By dissolving ammonia gas in water, aqueous ammonia is produced. Aqueous ammonia is an alkali because the ammonia molecules ionise partially to produce hydroxide ions, OH–. Uses of Acids, Bases and Alkalis Acids, bases and alkalis are not just chemical substances in the laboratory but they are also widely found in daily life. Toothpaste which is alkaline, functions to neutralise acid on the teeth, while vinegar is an acidic substance used to make pickled chillies. Photograph 6.1 Uses of acid and alkali in daily life Activity 6.1 Century Discussing the uses of acids and alkalis in daily life using 21st Skills CT examples of acidic and alkaline substances 1. Carry out the activity in groups. 2. Gather information from reading materials or websites on examples of acidic and alkaline substances as well as their uses in various fields. TABLET ANTACID Agriculture Industries Medicine Food industry 138  Acid, Base and Salt CHAPTER 6 3. Based on the information gathered, discuss the followings: (a) Identify the acid, base or alkali in each substance that you have found (b) State the use of acid, base or alkali found in the substance 4. Pin up your group work on the bulletin board to share the information and references with the other groups. The Role of Water to Show Acidity and Alkalinity Mom, why is the new Try to add some water soap taken out of its and rub the soap. What box not slippery? do you feel now? It's slippery, mom! Figure 6.6 Role of water to show alkalinity Based on the conversation in Figure 6.6, why is the water added to the soap? Is water needed to allow acids or alkalis to show acidic or alkaline properties? Experiment 6.1 Aim: To study the role of water in showing acidic properties. Problem statement: Is water needed to allow an acid to show its acidic properties? Hypothesis: Water is needed for an acid to show its acidic properties. Variables: (a) Manipulated : Presence of water (b) Responding : Colour change on blue litmus paper (c) Fixed : Type of acid Materials: Solid oxalic acid, C2H2O4, distilled water and blue litmus paper Apparatus: Test tubes and test tube rack Procedure: 1. Add a spatula of solid oxalic acid, C2H2O4 in a test tube. 2. Insert a piece of dry blue litmus paper into the test tube. 137 THEME 3 Interaction between Matter 3. Observe any changes to the colour of the blue litmus paper. Record your observations. Acid is corrosive. Be careful 4. After that, add 2 cm3 distilled water and shake well. when handling acid. If in 5. Observe any changes to the colour of the blue litmus paper. contact with acid, run Record your observations. continuous flow of water on the affected area. Results: Table 6.1 Contents Observations Solid oxalic acid, C2H2O4 Solid oxalic acid, C2H2O4 + water Interpreting data: 1. State the change in colour of the blue litmus paper that is used to detect acidic properties. 2. Based on the observations, state a suitable inference. 3. What are the conditions needed for an acid to show its acidic properties? Conclusion: Is the hypothesis acceptable? What is the conclusion of this experiment? Discussion: 1. Name the ion that is responsible for showing the acidic properties. 2. Solid oxalic acid, C2H2O4 had differences in observation compared to the solid oxalic acid, C2H2O4 that has been dissolved in water. Give a reason. 3. What is the operational definition for acid in this experiment? Prepare a complete report after carrying out this experiment. Acids only show acidic properties in the presence of water. Literacy Tips When an acid is dissolved in water, acid molecules will ionise to produce hydrogen ions, H+. The presence of hydrogen ions H+ Reflect on the properties of acid: allows the acid to show its acidic properties. Therefore, blue litmus Sour taste paper changes to red. Without water, solid oxalic acid, C2H2O4 Corrosive only exist as molecules. Hydrogen ions, H+ are not present. Thus, Has pH value less than 7 the colour of blue litmus paper remains unchanged. Changes moist blue litmus paper to red 140  Acid, Base and Salt CHAPTER 6 Experiment 6.2 Aim: To study the role of water in showing alkaline properties. Problem statement: Is water needed to allow an alkali to show its alkaline properties? Hypothesis: Make a suitable hypothesis for this experiment. Variables: State all the variables involved in this experiment. Sodium hydroxide, NaOH is corrosive. A pellet of sodium hydroxide, NaOH is sufficient to carry out this experiment. Sodium hydroxide If in contact with the alkaline Sodium solution, NaOH hydroxide solution, run water over the produced area continuously until it no pellet, NaOH Red litmus paper longer feels slippery. Figure 6.7 Method to test the alkaline properties of sodium hydroxide, NaOH Procedure: 1. Based on Figure 6.7, list out the apparatus and materials required for this experiment. 2. Plan the procedure for this experiment with your group members. 3. Determine the method used to collect data and prepare a suitable table. 4. Carry out the experiment with your teacher̕s permission. 5. Record your observations in the table provided. Results: Record your observation in a table. Interpreting data: 1. Based on the observations, state a suitable inference. 2. What is the condition of the litmus paper needed to detect alkaline properties? Conclusion: Is the hypothesis acceptable? What is the conclusion of this experiment? Discussion: 1. Name the ion responsible to show alkaline properties. 2. Explain the difference in observation between using a pellet of sodium hydroxide, NaOH and sodium hydroxide solution, NaOH. 3. Give the operational definition for alkali in this experiment. Prepare a complete report after carrying out this experiment. 137 THEME 3 Interaction between Matter Akalis only shows alkaline properties when they are dissolved Literacy Tips in water. Without water, hydroxide ions, OH– in the sodium hydroxide pellet, NaOH cannot move freely and are still tied in its Reflect on the properties of lattice structure. So, the pellet of sodium hydroxide, NaOH does an alkali: Tastes bitter and feels not show alkaline properties. The red litmus paper cannot change slippery colour. When a pellet of sodium hydroxide, NaOH is dissolved in Corrosive water, hydroxide ions, OH– are produced and able to move freely Has pH value more than 7 in water. Thus, sodium hydroxide solution, NaOH shows alkaline Changes moist red properties. Hence, the moist red litmus paper turns blue. litmus paper to blue The presence of water also enables ammonia gas, NH3 to ionise and produce hydroxide ions, OH– that are responsible for its alkaline properties. Therefore, the moist red litmus paper turns blue. Without water, ammonia gas, NH3 only exists as molecules. Hydroxide ions, OH– are not present. So, red litmus paper remains unchanged. TestYourself 6.1 1. State the meaning of the following terms: (a) Acid (b) Alkali 2. Carbonic acid is a mineral acid with the formula, H2CO3. What is the basicity of carbonic acid? Explain why. 3. Figure 6.8 shows a conversation between Khairul and his teacher. Khairul, what is Teacher, the cleaning powder your problem? is alkaline. Why doesn't the red litmus paper turn blue? Figure 6.8 (a) What possible mistake was committed by Khairul in his test? (b) How can you help Khairul in his test? Explain why. 142  Acid, Base and Salt CHAPTER 6 6.2 pH Value The pH Values of Acids and Alkalis g Learnin tandard Zahir, what are you doing? S At the end of the lesson, I am taking a sample of pupils are able to: water to test the pH value 6.2.1 State the meaning of pH of water in this aquarium. and its uses 6.2.2 Calculate pH values of The clownfish that acids and alkalis I am rearing need 6.2.3 Investigate the water with pH relationship between values between pH value and the 8.0 to 8.3 in concentration of hydrogen order for them ions and hydroxide ions to be healthy. through experiment Figure 6.9 Clownfish need water with specific pH values Based on the pH value mentioned by Zahir, do clownfish require acidic or alkaline water? Why do you say so? The pH scale which ranges from 0 to 14 is used to show the acidity and alkalinity of an aqueous solution. Solutions with pH value less than 7 is acidic while solutions with pH value more than 7 is alkaline. Universal indicator, pH meter or pH paper is commonly used to determine the pH value. Referring to Figure 6.10, what is the relationship between pH value and degree of acidity or alkalinity? More acidic More alkaline Neutral 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Figure 6.10 The pH scale What is actually ̒pH̕? In chemistry, pH is a logarithmic measure of the concentration of hydrogen ions in an aqueous solution. pH = –log [H+] where log is logarithm of base 10 and [H+] is the concentration of hydrogen ions in mol dm–3 of the solution. Using that formula, we can determine the pH value of an acid by calculation. 137 THEME 3 Interaction between Matter Example 1 Calculate the pH value of nitric acid, HNO3 with 0.5 mol dm–3 of hydrogen ion, H+. Solution Given the concentration of H+ = 0.5 mol dm–3 pH = –log [0.5] Use the formula pH = –log [H+] = –(– 0.301) = 0.301 pH value of nitric acid, HNO3 = 0.3 Example 2 Determine the molarity of hydrochloric acid, HCl with pH value of 2.0. Solution pH = –log [H+] 2.0 = –log [H+] log [H+] = –2.0 [H+] = 10–2 = 0.01 mol dm–3 Molarity of hydrochloric acid, HCl = 0.01 mol dm–3 The concentration of hydroxide ion, OH– is used to calculate the value of pOH of an alkali based on the following formula, where [OH–] represents the concentration of hydroxide ions in mol dm–3 of the alkali solution. pOH = –log [OH–] Given that the sum of pH value and pOH value is 14, the pH value of an alkali can be calculated by using the following relationship: pH + pOH = 14 pH = 14 – pOH Example 3 Calculate the pOH value for sodium hydroxide solution, NaOH with 0.1 mol dm–3 hydroxide ions, OH–. Solution Given that the concentration of hydroxide ion, OH– = 0.1 mol dm–3 pOH = –log [0.1] Use the formula pOH = –log [OH–] = –(–1) =1 pOH value of sodium hydroxide solution, NaOH = 1.0 144  Acid, Base and Salt CHAPTER 6 Example 4 Calculate the pH value for potassium hydroxide, KOH that has 0.01 mol dm–3 hydroxide ions, OH–. Solution Given concentration of hydroxide ion, OH– = 0.01 mol dm–3 pOH = –log [0.01] Use the formula pOH = –log [OH–] = –(–2) =2 pOH value of potassium hydroxide solution, KOH = 2.0 Consider the relationship: pH value of potassium hydroxide solution, KOH = 14.0 – pOH pH + pOH = 14 = 14.0 – 2.0 = 12.0 Example 5 Determine the molarity of lithium hydroxide solution, LiOH with pH value 12.0. Solution pH + pOH = 14.0 12.0 + pOH = 14.0 pOH = 14.0 – 12.0 = 2.0 pOH = –log [OH–] 2.0 = –log [OH–] log [OH–] = –2.0 [OH–] = 10–2 = 0.01 mol dm–3 Molarity of lithium hydroxide solution, LiOH = 0.01 mol dm–3 Did you know that the decimal place of pH is related to the significant numbers in the concentration of hydrogen ions given? If the value of given concentration has two significant numbers, the pH value should be rounded to two decimal places. The pH value can be calculated based on the concentration of hydrogen ions, H+ in an acid, or the concentration of hydroxide ions, OH– in an alkali. So, the pH scale allows us to compare the concentration of hydrogen ions, H+ or the hydroxide ions, OH– in an aqueous solution. The relationship between hydrogen ions, H+ or hydroxide ions, OH– with the pH value can be studied in Experiment 6.3. 137 THEME 3 Interaction between Matter Experiment 6.3 Aim: To study the relationship between the concentration of hydrogen ions, H+ and pH value of acid. Problem statement: Does the concentration of hydrogen ions, H+ of an acid affect its pH value? Hypothesis: The higher the concentration of hydrogen ion, H+, the lower the pH value of the acid. Variables: (a) Manipulated : Concentration of hydrogen ions, H+ (b) Responding : pH value (c) Fixed : Type of acid Materials: 0.1 mol dm–3, 0.01 mol dm–3 and 0.001 mol dm–3 hydrochloric acid, HCl Apparatus: 100 cm3 beaker and pH meter Procedure: 1. Pour 20 cm3 of hydrochloric acid, HCl of different concentrations into three beakers. 2. Measure the pH value of each hydrochloric acid, HCl with the pH meter. 3. Record the pH values in Table 6.2. Results: Table 6.2 Concentration of hydrochloric acid, HCl (mol dm–3) 0.1 0.01 0.001 Concentration of hydrogen ions, H+ (mol dm–3) pH value Interpreting data: 1. Based on the results obtained, how does the pH value change when the concentration of hydrochloric acid, HCl decreases? 2. State the changes in the concentration of hydrogen ions, H+ when the concentration of hydrochloric acid, HCl decreases. 3. What is the relationship between the concentration of hydrogen ions, H+ and pH value? Conclusion: Is the hypothesis acceptable? What is the conclusion of this experiment? Discussion: 1. When an acidic solution is diluted, what are the changes in the: (a) Concentration of hydrogen ions, H+? (b) pH value? (c) Degree of acidity of the aqueous solution? 2. State the relationship between the concentration of hydrogen ions, H+, pH values and degree of acidity of an acidic aqueous solution. Prepare a complete report after carrying out this experiment. 146  Acid, Base and Salt CHAPTER 6 Experiment 6.4 Aim: To study the relationship between the concentration of hydroxide ions, OH– and pH value of an alkali. Problem statement: Does the concentration of hydroxide ions, OH– of an alkali affect its pH value? Hypothesis: Make a suitable hypothesis for this experiment. Variables: State all variables involved in this experiment. Materials: 0.1 mol dm–3, 0.01 mol dm–3 and 0.001 mol dm–3 sodium hydroxide solution, NaOH Apparatus: 100 cm3 beaker and pH meter Procedure: 1. Plan the procedure to measure the pH value of sodium hydroxide solution, NaOH. 2. Your plan should include the pH meter. 3. Carry out the experiment with your teacher̕s permission. 4. Record the pH values obtained in your report book. Results: Record the pH values in a table. Interpreting data: 1. Based on the data obtained, how does the pH value change when the concentration of sodium hydroxide solution, NaOH decreases? 2. When the concentration of sodium hydroxide solution, NaOH decreases, what are the changes that occur to the: (a) Concentration of hydroxide ions, OH–? (b) pH value? (c) Degree of alkalinity of sodium hydroxide solution, NaOH? 3. State the relationship between the concentration of hydroxide ions, OH–, pH value and degree of alkalinity of sodium hydroxide, solution NaOH. Conclusion: Is the hypothesis acceptable? What is the conclusion of this experiment? Prepare a complete report after carrying out this experiment. When the concentration of acid increases, more acid molecules ionise to produce hydrogen ions, H+. The higher the concentration of hydrogen ions, H+, the lower the pH value. Acidity increases when the pH value of the acid solution decreases. Concentration of hydrogen ions, H+ , pH value On the other hand, the higher the concentration of hydroxide ions, OH–, the higher the pH value. Alkalinity increases when the pH value of the alkaline solution increases. Concentration of hydroxide ions, OH– , pH value 137 THEME 3 Interaction between Matter Most substances found in our daily lives contain acids or Chemistry & Us alkalis. The determination of pH values for these substances can be done in Activity 6.2. Purple cabbages change colour at different pH values. Activity 6.2 Determining the pH values of various items in daily life 1. You are supplied with the following items: Soap water Milk tea Carbonated drink Lime juice Coffee Tap water 2. In pairs, measure the pH value of each item using the universal indicator. 3. Record the items with similar pH values. 4. Prepare a pH indicator using a purple cabbage. Visit websites to know how to prepare this pH indicator. Use the pH indicator that you have prepared to measure the pH value of each of the above items. 5. Using a suitable graphic management tools, present your findings. 6. Pin up your work in class to share with others. Sewage pipe Battery Lemon Tomato Milk Blood Antacid Soap cleaner Milk Baking soda Ammonia Bleach Gastric juice Vinegar Coffee Water Baking soda Ammonia Bleach solution Figure 6.11 The pH value of few items in daily life tested with the universal indicator TestYourself 6.2 1. Write the formula to calculate the pH value of acid. 2. Calculate the pH value for hydrochloric acid, HCl that contains 0.001 mol dm–3 hydrogen ions, H+. 3. Determine the pH value for calcium hydroxide, Ca(OH)2 with concentration of 0.05 mol dm–3. [pH + pOH = 14]. 148  Acid, Base and Salt CHAPTER 6 6.3 Strength of Acids and Alkalis Observe Figure 6.12. What is meant by a strong acid and a weak acid? Can you tell the similarities or the differences between Hydrogen ions, H+ are HCl is a strong these two acids? Teacher, both acids produced when acid is acid, CH3COOH is are monoprotic. dissolved in water. a weak acid. Figure 6.12 Similarities and differences between two acids Strong Acids and Weak Acids g The strength of an acid depends on the degree of dissociation or Learnin tandard ionisation of the acid in water. S At the end of the lesson, Strong Acids pupils are able to: 6.3.1 Define strong acid, A strong acid is an acid that ionises completely in water to weak acid, strong alkali and weak alkali produce a high concentration of hydrogen ions, H+. Hydrochloric 6.3.2 Explain the strength acid, HCl is a strong acid because all molecules of hydrogen of acid and alkali chloride, HCl that dissolve in water are ionised completely based on its degree of to hydrogen ions, H+ and chloride ions, Cl–. No molecule of dissociation in water hydrogen chloride, HCl exists in this solution. HCl(aq) → H+(aq) + Cl–(aq) Chemistry Hydrogen ions, H+ produced from acid molecules Hydrogen will combine with water Hydrogen chloride chloride gas, HCl molecules to form molecule, HCl hydroxonium ions, H3O+. The hydroxonium ions, H3O+ is the product of a dative bond formed between Water Hydrogen ion, H+ hydrogen ion, H+ with water molecule. Chloride ion, Cl– Figure 6.13 Complete ionisation of hydrogen chloride, HCl 137 THEME 3 Interaction between Matter Weak Acids A weak acid is an acid that ionises partially in water to produce low concentration of hydrogen ions, H+. Ethanoic acid, CH3COOH is a weak acid because the molecules of ethanoic acid, CH3COOH ionise partially in water. The degree of dissociation of ethanoic acid molecules is 1.54%. In other words, from 100 molecules of ethanoic acid, CH3COOH, only one molecule of ethanoic acid, CH3COOH ionises to hydrogen ions, H+ and ethanoate ions, CH3COO–. The rest still exist as molecules of ethanoic acid, CH3COOH. CH3COOH(aq)   H+(aq) + CH3COO–(aq) The reversible arrow shows that ethanoic acid, CH3COOH molecule can form hydrogen ions, H+ and ethanoate ions, CH3COO–. These ions can also combine again to Ethanoic acid, form the acid molecules. CH3COOH molecule Water Glacial ethanoic acid, CH3COOH Ethanoic acid, CH3COOH molecule Ethanoic acid, Hydrogen ion, H+ CH3COOH Ethanoate ion, CH3COO– Figure 6.14 Partial ionisation of ethanoic acid, CH3COOH Strong Alkalis and Weak Alkalis Alkalis also consist of strong alkalis and weak alkalis depending on their degree of ionisation in water. Strong Alkalis A strong alkali is an alkali that ionises completely in water to produce a high concentration of hydroxide ions, OH–. Sodium hydroxide, NaOH is a strong alkali that undergoes complete dissociation when dissolved in water. Only sodium ions, Na+ and hydroxide ions, OH– are present in the solution. NaOH(aq) → Na+(aq) + OH–(aq) Sodium ion, Na+ Literacy Tips Dissociation is also known Sodium hydroxide Hydroxide ion, OH– as ionisation. solution, NaOH Figure 6.15 Complete ionisation of sodium hydroxide solution, NaOH 150  Acid, Base and Salt CHAPTER 6 Weak Alkalis A weak alkali is an alkali that ionises partially in water to produce a low concentration of hydroxide ions, OH–. Ammonia solution, NH3 is a weak alkali because ammonia molecules, NH3 ionise partially in water. The degree of dissociation of ammonia, NH3 is 1.3%. In other words, from 100 molecules of ammonia, NH3 only one molecule of ammonia, NH3 will receive hydrogen ion, H+ from water molecule. So, only a small number of hydroxide ions, OH– is present in the solution. NH3(aq) + H2O(l)   NH4+(aq) + OH–(aq) Ammonia molecule, NH3 Ammonia gas, NH3 Ammonia molecule, NH3 Hydroxide ion, OH– Water Ammonium ion, NH4+ Figure 6.16 Partial ionisation of ammonia solution, NH3 Activity 6.3 Century Carry out a simulation to explain the strength of acids and alkalis 21st Skills CT 1. Visit the website given. 2. Vary the regulator for acid strength and observe the degree Simulation on acid of dissociation and number of hydrogen ions, H present. + and alkali 3. Repeat step 2 for alkali and observe the degree of http://bit.ly/31cGoMQ dissociation and number of hydroxide ions, OH present. – 4. Interpret the information on strength of acids and alkalis based on the degree of dissociation. 5. Relate the concentration of hydrogen ions, H+ and hydroxide ions, OH– with the degree of dissociation of the acid and alkali. 6. Display your findings in an interesting presentation. TestYourself 6.3 1. Give the meaning of the following terms: (a) Strong acid (c) Strong alkali (b) Weak acid (d) Weak alkali 2. Why does ammonia solution, NH3 that has the same concentration as potassium hydroxide, KOH have a lower pH value? 3. The pH value 0.1 mol dm-3 nitric acid, HNO3 is different from the pH value 0.1 mol dm–3 oxalic acid, H2C2O4? Explain. 137 THEME 3 Interaction between Matter 6.4 Chemical Properties of Acids and Alkalis g Learnin tandard After 30 Carbon S Balloon filled with seconds dioxide gas At the end of the lesson, baking soda pupils are able to: 6.4.1 Summarise the chemical properties of acids by carrying out the Vinegar reactions between: Acid and base Acid and reactive Figure 6.17 To inflate the balloon using vinegar and baking soda metal Acid and metal Have you ever inflated a balloon using vinegar and baking soda? carbonate Is this process related to the chemical properties of acid? 6.4.2 Summarise the chemical properties of alkalis Chemical Properties of Acid by carrying out the reactions between: The properties of acid is divided into physical and chemical Alkali and acid properties. The properties of acid such as having a sour taste, Alkali and metal ion changing moist blue litmus paper to red and having pH values less Alkali and ammonium salt than 7 are the physical properties of acid. The chemical properties of acid on the other hand refer to the reactions between acid and other substances. Carry out activity 6.4 to study the chemical properties of acid. Balloon inflation http://bit.ly/2Mh3M7y Activity 6.4 Aim: To study the chemical properties of acids. Materials: Copper(II) oxide powder, CuO, zinc powder, Zn, marble chips, CaCO3, 1.0 mol dm–3 sulphuric acid, H2SO4, 2.0 mol dm–3 nitric acid, HNO3, 2.0 mol dm–3 hydrochloric acid, HCl, limewater, wooden splinter and filter papers Apparatus: 100 cm3 beaker, glass rod, filter funnel, retort stand with clamp, evaporating dish, Bunsen burner, pipeclay triangle, delivery tube and rubber stopper, tripod stand, spatula, test tubes and test tube holder Reaction between acid and base Procedure: 1. Pour 20 cm3 of 1.0 mol dm–3 sulphuric acid, H2SO4 into a beaker. Heat the acid by using a small flame. 2. Add copper(II) oxide powder, CuO gradually into the acid by using a spatula. Stir the mixture with a glass rod. 3. Observe the change that takes place on copper(II) oxide, CuO that reacts with acid. Record your observation on the solution produced. 152  Acid, Base and Salt CHAPTER 6 4. Continue adding copper(II) oxide powder, CuO until it can no longer dissolve. 5. Filter out the excess copper(II) oxide, CuO from the mixture. 6. Pour the filtrate into an evaporating dish and heat the filtrate until one third of its initial volume remains. 7. Allow the saturated solution produced to cool until salt crystals are formed. 8. Filter the contents of the evaporating dish to obtain the salt crystals. Rinse the crystals with distilled water. 9. Dry the salt crystals between two pieces of filter papers. 10. Observe the physical properties of the salt crystals and record your findings. Glass rod Spatula Excess of Filter funnel copper(II) oxide powder, CuO Filter paper Residue (excess of copper(II) Hot 20 cm3 of 1.0 mol dm–3 oxide powder, CuO) sulphuric acid, H2SO4 Filtrate (copper(II) sulphate solution, CuSO4) Filtrate Evaporating dish Salt crystals Pipeclay triangle Heat Figure 6.18 Preparation of salt crystals from the reaction between acid and base Discussion: 1. What happens to the copper(II) oxide powder, CuO when added to sulphuric acid, H2SO4? 2. What is the colour of the solution produced from the reaction between sulphuric acid, H2SO4 and copper(II) oxide, CuO? 3. Write a chemical equation for the reaction between sulphuric acid, H2SO4 and copper(II) oxide, CuO. 4. From the chemical equation written above, complete the following equation in words: Acid + Base → + 137 THEME 3 Interaction between Matter Reaction between acid and reactive metal Procedure: 1. Plan a procedure to study the reaction between hydrochloric acid, HCl and zinc powder, Zn as shown in Figure 6.19. Glass rod Spatula Excess of zinc Filter funnel powder, Zn Filter paper Residue (excess of zinc powder, Zn) 20 cm3 of 2.0 mol dm–3 Filtrate (zinc chloride hydrochloric acid , HCl solution, ZnCl2) Filtrate Evaporating dish Salt crystals Pipeclay triangle Heat Figure 6.19 Preparation of salt crystals from the reaction between acid and reactive metal 2. Discuss with your teacher if you have encountered any problems when 5 cm3 of Burning planning the procedure. 2.0 mol dm–3 wooden 3. Make sure that you carry out the hydrochloric splinter chemical test on the gas released as acid, HCl shown in Figure 6.20. 4. Carry out this test with your Zinc powder, Zn teacher̕s permission. Figure 6.20 5. Record your observations. Discussion: 1. What is the observation that indicates acid has reacted with metal when zinc powder, Zn is added to hydrochloric acid, HCl? 2. Name the gas released in this activity. 3. Write a chemical equation for the reaction between hydrochloric acid, HCl and zinc, Zn. 4. From the chemical equation written above, complete the following equation in words: Acid + Reactive metal → + 154  Acid, Base and Salt CHAPTER 6 Reaction between acid and metal carbonate Procedure: 1. Plan a procedure to carry out this activity to study the reaction between nitric acid, HNO3 and marble chips, CaCO3. 2. Include safety measures taken in your procedure. 3. Discuss with your teacher if you have encountered any problems when planning the procedure. 4. Make sure that you carry out the chemical test on the gas released as shown in Figure 6.21. 5 cm3 of 2.0 mol dm–3 nitric acid, HNO3 Marble chips, CaCO3 Limewater Figure 6.21 5. Carry out this test with your teacher̕s permission. 6. Record your observations. Discussion: 1. What is the reason for using excess marble chips, CaCO3 to react with nitric acid, HNO3? 2. How do you remove the excess marble chips, CaCO3 from the salt solution produced? 3. For the reaction in this activity: (a) Name the salt produced (b) Name the gas released 4. Write a chemical equation for the reaction between nitric acid, HNO3 and marble chips, CaCO3. 5. From the chemical equation written above, complete the following equation in words: Acid + Metal carbonate → + + Prepare a complete report after carrying out this activity. From Activity 6.4 that was carried out, it can be summarised that acids have the following chemical properties: Acids react with bases to produce salt and water Acids react with reactive metals to produce salt and hydrogen gas, H2 Acids react with metal carbonates to produce salt, water and carbon dioxide gas, CO2 137 THEME 3 Interaction between Matter Chemical Properties of Alkalis Chemical properties of alkalis can be determined through Activity 6.5. Activity 6.5 Aim: To study the chemical properties of alkali. Materials: Benzoic acid powder, C6H5COOH, 1.0 mol dm–3 sodium hydroxide solution, NaOH, ammonium chloride powder, NH4Cl, copper(II) sulphate solution, CuSO4, distilled water, filter paper and red litmus paper Apparatus: 100 cm3 beaker, glass rod, filter funnel, retort stand with clamp, evaporating dish, Bunsen burner, pipeclay triangle, tripod stand, dropper, spatula, test tube, boiling tube and test tube holder Figure 6.22, Figure 6.23 and Figure 6.24 show three reactions involving alkalis. Glass rod Spatula Filter funnel Benzoic acid powder, C6H5COOH Filter paper Residue (excess acid powder) Filtrate 20 cm3 of 1.0 mol dm–3 sodium hydroxide solution, NaOH Evaporating dish Filtrate Salt crystals Pipeclay triangle Heat Figure 6.22 Preparation of salt crystals from the reaction between alkali and acid Moist red Ammonium chloride litmus paper 1.0 mol dm–3 powder, NH4Cl sodium hydroxide solution, NaOH 1.0 mol dm–3 sodium hydroxide Copper(II) sulphate Blue solution, NaOH Heat solution, CuSO4 precipitate Figure 6.23 Heating the mixture of alkali Figure 6.24 Addition of alkali to metal and ammonium salt to produce ions to produce insoluble metal ammonia gas hydroxide precipitate 156  Acid, Base and Salt CHAPTER 6 Procedure: 1. Based on Figure 6.22 to Figure 6.24, plan a laboratory activity to study the chemical properties of alkalis. 2. Plan and write out the procedure for the laboratory activity to be discussed with your teacher. 3. Record your observations in a report book. 4. Write an equation in words to summarise the chemical properties of alkalis. Prepare a complete report after carrying out this activity. From Activity 6.5 that was carried out, we can summarise that alkalis have the following chemical properties: Alkalis react with acids to produce salt and water When a mixture of alkali and ammonium salt is heated, ammonia gas, NH3 is released Addition of an alkali to most metal ions, will produce an insoluble metal hydroxide precipitate Table 6.3 summarises the chemical properties of acids and alkalis Table 6.3 Chemical properties of acids and alkalis Acid + Base → Salt + Water Example: 2HNO3(aq) + CuO(s) → Cu(NO3)2(aq) + H2O(l) Nitric acid Copper(II) oxide Copper(II) nitrate Water Chemical Acid + Reactive metal → Salt + Hydrogen gas properties of acids Example: H2SO4(aq) + Zn(s) → ZnSO4(aq) + H2(g) Sulphuric acid Zinc Zinc sulphate Hydrogen gas Acid + Metal carbonate → Salt + Water + Carbon dioxide gas Example: 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g) Hydrochloric acid Calcium carbonate Calcium chloride Water Carbon dioxide gas Alkali + Acid → Salt + Water Example: 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l) Potassium hydroxide Sulphuric acid Potassium sulphate Water Chemical Alkali + Ammonium salt → Salt + Water + Ammonia gas properties of alkalis Example: KOH(aq) + NH4Cl(aq) → KCl(aq) + H2O(l) + NH3(g) Potassium hydroxide Ammonium chloride Potassium chloride Water Ammonia gas Alkali + Metal ion → Insoluble metal hydroxide + Cation from alkali Example: 2NaOH(aq) + Mg2+(aq) → Mg(OH)2(s) + 2Na+(aq) Sodium hydroxide Magnesium ion Magnesium hydroxide Sodium ion 137 THEME 3 Interaction between Matter TestYourself 6.4 1. Write a chemical equation for the reaction between hydrochloric acid, HCl and: (a) Barium hydroxide, Ba(OH)2 (b) Magnesium, Mg (c) Zinc carbonate, ZnCO3 2. Write an equation in words to summarise the reaction of an alkali solution and the following substances: (a) Dilute acids (b) Ammonium salts (c) Metal ions 6.5 Concentration of Aqueous Solution g Dad, why is the colour Because the concentration of Learnin tandard of my tea is different tea in our glasses are different. S from the one that you are drinking? At the end of the lesson, pupils are able to: 6.5.1 State the meaning of concentration of aqueous solution 6.5.2 Solve numerical problems involving concentration of solution Figure 6.25 Concentration of tea affects its colour Concentration of a solution is a measurement that shows the quantity of solute dissolved in a unit volume of solution, normally in 1 dm3 solution. The higher the quantity of solute, the higher the concentration of the solution. The quantity of solute dissolved can be measured in gram or mole, hence the concentration of a solution can be measured in unit g dm–3 or mol dm–3. Concentration in unit g dm–3, is the mass of solute found in 1 dm3 solution. Mass of solute (g) Concentration (g dm–3) = Volume of solution (dm3) Concentration in unit mol dm–3, is the number of moles of solute found in 1 dm3 solution. This concentration is called molarity. Number of moles of solute (mole) Molarity (mol dm–3) = Volume of solution (dm3) 158  Acid, Base and Salt CHAPTER 6 ÷ molar mass Literacy Tips Concentration Molarity The unit for molarity is mol dm–3 or (g dm–3) (mol dm–3) molar (M). You have to remember that mole is not the same as molar. × molar mass Mole is the unit for measuring matter Figure 6.26 Relationship between concentration and molarity while molar is the number of moles of solute in a given volume of solution. Example 6 Calculate the concentration in g dm–3, for each solution produced. (a) 40 g of solid copper(II) sulphate, CuSO4 is dissolved in water to produce 20 dm3 solution. (b) 18 g of sodium hydroxide pellets, NaOH is dissolved in water to produce 750 cm3 solution. Solution Mass of solute (g) (a) Concentration of copper(II) sulphate, CuSO4 = Volume of solution (dm3) 40 g = 20 dm3 = 2.0 g dm–3 (b) Concentration of sodium hydroxide, NaOH Mass of solute (g) = Volume of solution(dm3) 750 cm3 is converted 750 dm3 18 g to dm3 by dividing the 1000 = 0.75 dm3 volume with 1000. = 0.75 dm3 = 24.0 g dm–3 Example 7 Calculate the molarity of each solution prepared. (a) 10 mol of solid zinc chloride, ZnCl2 dissolved in water to produce 5 dm3 of solution. (b) 0.1 mol of solid calcium chloride, CaCl2 is dissolved in 500 cm3 of distilled water. Solution Number of moles of solute (mol) (a) Molarity of zinc chloride solution, ZnCl2 = Volume of solution (dm3) = 10 mol3 5 dm = 2.0 mol dm–3 (b) Molarity of calcium chloride solution, CaCl2 Number of moles of solute (mol) = Volume of solution (dm3) 500 cm3 is converted 500 dm3 = 0.1 mol to dm3 by dividing the 1000 0.5 dm3 volume with 1000. = 0.5 dm3 = 0.2 mol dm–3 159 THEME 3 Interaction between Matter Example 8 What is the concentration of nitric acid, HNO3 with a molarity of 0.5 mol dm–3 in unit g dm–3? [Relative atomic mass: H = 1, N = 14, O = 16] Solution Concentration = Molarity × Molar mass HNO3 RAM H RAM N RAM O = 0.5 mol dm–3 × [1 + 14 + 3(16)] g mol–1 = 0.5 mol dm–3 × 63 g mol–1 = 31.5 g dm–3 Example 9 Convert the concentration of 3.6 g dm–3 lithium hydroxide solution, LiOH to molarity, mol dm–3. [Relative atomic mass: H = 1, Li = 7, O = 16] Solution Concentration Molarity = Molar mass LiOH 3.6 g dm–3 = (7 + 16 + 1) g mol–1 RAM Li RAM O RAM H = 0.15 mol dm–3 We can calculate number of moles of solute dissolved in the solution if its molarity and the volume of the solution are known. If the volume of the solution is in Number of moles of solute (mol) cm3, thus, unit of volume needs to be Molarity = converted to dm3. Volume of solution (dm3) M = n n=M V ( 1000 ) V Therefore, n = MV Volume of 1000 solution is Therefore, n = MV Volume of solution in cm3. is in dm3. Example 10 Calculate the number of moles of potassium hydroxide, KOH found in 2 dm3 of 0.5 mol dm–3 potassium hydroxide solution, KOH. Solution Number of moles, n = MV This formula is applied because = 0.5 mol dm–3 × 2 dm3 the volume of solution is in dm3. = 1 mol KOH 160  Acid, Base and Salt CHAPTER 6 Example 11 A beaker contains 200 cm3 of 0.2 mol dm–3 lead(II) nitrate solution, Pb(NO3)2. How many moles of lead(II) nitrate, Pb(NO3)2 is in the beaker? Solution This formula is applied because Number of moles, n = MV 1000 the volume of solution is in cm3. = 0.2 mol dm × 200 cm –3 3 1000 = 0.04 mol Pb(NO3)2 Activity 6.6 Solving numerical problems related to concentration of solutions CT 1. 6 g of solid magnesium sulphate, MgSO4 is added into a beaker containing 200 cm3 of water. Calculate the concentration in g dm–3, for the solution produced. 2. 0.4 mol of zinc chloride, ZnCl2 is dissolved in water to produce 2 dm3 of solution. Calculate the molarity of the solution prepared. 3. What is the concentration of 0.5 mol dm–3 sulphuric acid, H2SO4 in g dm–3? [Relative atomic mass: H = 1, O = 16, S = 32] 4. The concentration of sodium chloride solution, NaCl is 1.989 g dm–3. Calculate the molarity of the solution in mol dm–3. [Relative atomic mass: Na = 23, Cl = 35.5] 5. Calculate the number of moles of sodium hydroxide, NaOH in 2.5 dm3 of 0.2 mol dm–3 sodium hydroxide solution, NaOH. 6. Given the molarity of 250 cm3 of barium hydroxide solution, Ba(OH)2 is 0.1 mol dm–3. How many moles of hydroxide ion, OH– is in the solution? TestYourself 6.5 1. What is meant by concentration in unit g mol–1? 2. State two units to measure concentration. 3. 0.03 mol of potassium nitrate, KNO3 is dissolved in 1.2 dm3 of distilled water. What is the molarity of the potassium nitrate solution, KNO3 produced? 4. Calculate the concentration of sulphuric acid, H2SO4 that has the molarity of 2.0 mol dm–3 in unit g dm–3. [Relative atomic mass: H = 1, O = 16, S = 32] 5. 1.9 g MgY2 is dissolved in 100 cm3 of water to produce a solution with the molarity of 0.2 mol dm–3. What is the relative atomic mass of Y? [Relative atomic mass: Mg = 24] 161 THEME 3 Interaction between Matter 6.6 Standard Solution Have you seen the syrup dispenser as g shown in Photograph 6.2? Did you know Learnin tandard that the dispenser is filled with a standard S solution of sugar so that the machine can At the end of the lesson, dispense sugar at an amount requested pupils are able to: by customers? What do you know about 6.6.1 State the meaning of standard solution? standard solution 6.6.2 Describe the preparation Syrup dispenser Photograph 6.2 of a standard solution Syrup dispenser through activity: http://bit.ly/35Ajhiv From a solid substance Through dilution of an aqueous solution Meaning of Standard Solution 6.6.3 Solve numerical problems involving Most chemical reactions involve reactants in aqueous solution. preparation of standard In that case, the preparation of aqueous solution with specific solution and dilution concentrations is very important. Standard solution is a solution with known concentration. In the preparation of standard solutions, mass of solute and volume of distilled water are two parameters that have to be measured accurately. Preparation of a Standard Solution from a Solid Activity 6.7 Aim: To prepare 250 cm3 of standard solution of 1.0 mol dm–3 sodium carbonate, Na2CO3. Materials: Distilled water and solid sodium carbonate, Na2CO3 Apparatus: Electronic balance, filter funnel, 250 cm3 volumetric flask, dropper, wash bottle, 250 cm3 beaker and glass rod Procedure: 1. Determine the mass of sodium carbonate, Na2CO3 needed using the formula n = MV. 1000 2. Weigh the mass calculated using the electronic balance. 3. Add 100 cm3 of distilled water to the solid sodium carbonate, Na2CO3 in a beaker. 4. Stir the mixture with a glass rod until all the solid sodium carbonate, Na2CO3 is completely dissolved in the distilled water. 5. Transfer the sodium carbonate solution, Na2CO3 into a 250 cm3 volumetric flask via a filter funnel. 6. Rinse the beaker with distilled water. Make sure all the remaining solution is transferred into the volumetric flask. 7. Then, rinse the filter funnel with a little distilled water. All the remaining solution is transferred into the volumetric flask. 8. Remove the filter funnel. Add distilled water until it approaches the calibration mark on the volumetric flask. 162  Acid, Base and Salt CHAPTER 6 9. Using a dropper, add distilled water slowly until the meniscus level is aligned exactly on the calibration mark of the volumetric flask. 10. Close the volumetric flask with a stopper. Shake well by inverting the volumetric flask several times until the solution is homogenous. Note: Keep the standard solution of sodium carbonate, Na2CO3 that you have prepared for Activity 6.8. Distilled Solid sodium water carbonate, Na2CO3 Glass rod Filter funnel Stir Calibration Volumetric mark flask (a) Add distilled water (b) Dissolve the solid (c) Transfer the solution into a volumetric flask Dropper Stopper Distilled water Meniscus Calibration Meniscus Calibration level of mark level of mark solution solution (f) Close the volumetric flask with (e) Add distilled water until (d) Rinse the filter funnel with a stopper before shaking calibration mark distilled water Figure 6.27 Preparation of 1.0 mol dm–3 sodium carbonate solution, Na2CO3 from a solid Discussion: 1. Why must the beaker and filter funnel be rinsed with distilled water? 2. Why must all the remaining solution be transferred into the volumetric flask? 3. How can you ensure that the meniscus level aligns with the calibration mark of the volumetric flask? 4. Why does the volumetric flask need to be closed after the standard solution is prepared? Prepare a complete report after carrying out this activity. Sodium hydroxide, NaOH is not suitable to be used for the preparation of a standard solution because sodium hydroxide, NaOH is hygroscopic (absorbs water or moisture in the air). Sodium hydroxide, NaOH also absorbs carbon dioxide gas, CO2 in the air to form sodium carbonate, Na2CO3. 2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l) This causes difficulty to determine the exact mass of sodium hydroxide, NaOH. Therefore, the preparation of a standard solution of sodium hydroxide, NaOH with a known concentration could not be made. 163 THEME 3 Interaction between Matter Solid oxalic acid H2C2O4.2H2O can be used to prepare a standard solution in the laboratory. Chemistry Sodium carbonate, Na2CO3 that is used to prepare standard solutions is alkaline. When sodium carbonate, Na2CO3 is dissolved in distilled water, carbonate ions, CO32– react with water molecules to produce bicarbonate ions, HCO3– and hydroxide ions, OH–. The presence of hydroxide ions, OH– gives the alkaline properties to the solution. CO32–(aq) + H2O(l) → HCO3–(aq) + OH–(aq) Preparation of a Standard Solution by Diluting Aqueous Solution Another method of preparing solutions of known concentration is by dilution method. This method involves adding water to a concentrated standard solution, or known as stock solution, to produce a more diluted solution. During dilution, water that is added to the aqueous solution will alter the concentration of the solution but it would not alter the number of moles of solute contained in the solution. Add distilled water Solute Solute (a) Concentrated solution (b) Dilute solution Figure 6.28 Quantity of solute remains the same in both solutions of different concentrations Hence, Number of moles of solute before dilution = Number of moles of solute after dilution n1 = n2 M1V1 MV = 2 2 1000 1000 M1V1 = M2V2 where M1 is the molarity of aqueous solution (stock solution) before dilution. V1 is the volume of aqueous solution (stock solution) before dilution. M2 is the molarity of aqueous solution (prepared solution) after dilution. V2 is the volume of aqueous solution (prepared solution) after dilution. As an example, you wish to prepare 500 cm3 of 0.1 mol dm–3 copper(II) sulphate solution, CuSO4 from the stock solution of 2.0 mol dm–3 copper(II) sulphate, CuSO4. Use the following formula: M1V1 = M2V2 (2.0)(V1) = (0.1)(500) (0.1)(500) V1 = 2.0 = 25 cm3 Hence, 25 cm3 of stock solution of copper(II) sulphate, CuSO4 needs to be diluted using distilled water until 500 cm3 solution of copper(II) sulphate, CuSO4 is obtained. 164  Acid, Base and Salt CHAPTER 6 The preparation of a standard solution by dilution method can be carried out through Activity 6.8 using sodium carbonate solution, Na2CO3 prepared in Activity 6.7. Activity 6.8 Aim: To prepare 100 cm3 of standard solution of 0.2 mol dm–3 sodium carbonate, Na2CO3. Materials: Distilled water and 1.0 mol dm–3 sodium carbonate solution, Na2CO3 from Activity 6.7 Apparatus: 100 cm3 volumetric flask, dropper, filter funnel, pipette, wash bottle, pipette filler and 100 cm3 beaker Procedure: Pipette Stock Stock solution solution Calibration Beaker Beaker mark Volumetric flask (a) Pour stock solution from (b) Take out calculated volume of (c) Transfer V1 cm3 of solution Activity 6.7 into a beaker solution, V1 cm3 with a pipette into a volumetric flask Dropper Stopper Distilled water Meniscus Calibration Calibration level of mark mark solution (f) Close the volumetric flask and shake (e) Add distilled water slowly (d) Add distilled water until the well by inverting it several times with a dropper solution level approaches the until the solution is homogenous calibration mark Brain Teaser Figure 6.29 Preparation of 0.2 mol dm–3 sodium carbonate solution, Na2CO3 by dilution method 1. Based on Figure 6.29, plan a procedure to prepare a standard solution of 0.2 mol dm–3 sodium carbonate, Na2CO3 by Brain Teaser dilution method. 2. Include precautionary steps in the process of Why is the pipette not rinsed solution preparation. with distilled water but rinsed with 1.0 mol dm–3 sodium 3. Show your procedure to your teacher before carrying out carbonate solution, Na2CO3? this activity. 4. Carry out the procedure as planned. 5. Clean and keep the apparatus at their proper places after carrying out this activity. 165 THEME 3 Interaction between Matter Discussion: 1. What is the volume of the standard solution of 1.0 mol dm–3 sodium carbonate solution, Na2CO3 needed to prepare 100 cm3 of 0.2 mol dm–3 sodium carbonate solution, Na2CO3? 2. What is the size of pipette needed in this preparation process? 3. Why is the beaker not suitable to be used in preparing a standard solution by the dilution method? 4. Do you need to remove the last drop of the solution in the pipette? Why? Prepare a complete report after carrying out this activity. Examples 12 and 13 show samples of calculations involved in the preparation of a standard solution by dilution. Example 12 Figure 6.30 shows 75 cm3 of 2.0 mol dm–3 nitric acid, HNO3 that is diluted to x mol dm–3 when 25 cm3 distilled water is added. Calculate the value of x. 25 cm3 distilled water Nitric acid, HNO3 75 cm3 of 2.0 mol dm–3 x mol dm–3 nitric acid, HNO3 Before dilution After dilution Figure 6.30 Solution M1 = 2.0 mol dm–3 ; V1 = 75 cm3 Volume of solution M2 = x mol dm–3 ; V2 = (75 + 25) cm3 = Volume of HNO3 + Volume of distilled water = 100 cm3 2.0 mol dm–3 × 75 cm3 = x mol dm–3 × 100 cm3 Use the formula M1V1 = M2V2 x mol dm–3 = 2.0 mol dm ×3 75 cm –3 3 100 cm = 1.5 mol dm–3 Then, x = 1.5 Example 13 Determine the volume of 2.0 mol dm–3 hydrochloric acid, HCl needed to be pipetted into a volumetric flask 250 cm3 to produce 0.2 mol dm–3 hydrochloric acid, HCl. Solution M1 = 2.0 mol dm–3 ; V1 = ? M2 = 0.2 mol dm–3 ; V2 = 250 cm3 2.0 mol dm–3 × V1 = 0.2 mol dm–3 × 250 cm3 Use the formula M1V1 = M2V2 V1 = 0.2 mol dm × 250 cm3 –3 2.0 mol dm –3 = 25 cm3 166  Acid, Base and Salt CHAPTER 6 Activity 6.9 Solving calculation problems involved in the preparation of a standard CT solution by dilution 1. Calculate the volume of 2.0 mol dm–3 sodium carbonate solution, Na2CO3 needed to prepare 50 cm3 of 0.1 mol dm–3 sodium carbonate solution, Na2CO3. 2. What is the molarity of sodium hydroxide solution, NaOH when 30 cm3 distilled water is added to 50 cm3 of 0.5 mol dm–3 sodium hydroxide solution, NaOH? 3. Calculate the volume of solution produced when 50 cm3 of 1.2 mol dm–3 sodium nitrate solution, NaNO3 is diluted to 0.5 mol dm–3. 4. When 200 cm3 water is added to 50 cm3 concentrated sulphuric acid, H2SO4, sulphuric acid, H2SO4 with concentration 0.2 mol dm–3 is produced. Calculate the molarity of the initial concentration of sulphuric acid, H2SO4. TestYourself 6.6 1. What is meant by standard solution? 2. X cm3 of 0.15 mol dm–3 zinc nitrate solution, Zn(NO3)2 is pipetted into a 500 cm3 volumetric flask to produce 500 cm3 of 0.018 mol dm–3 zinc nitrate solution, Zn(NO3)2. Determine the value of X. 3. Calculate the new molarity of hydrochloric acid, HCl produced if 25 cm3 of 1.5 mol dm–3 hydrochloric acid, HCl is diluted to produce 150 cm3 of hydrochloric acid, HCl. 4. Determine the volume of distilled water needed to add to 50 cm3 of 0.2 mol dm–3 sodium thiosulphate solution, Na2S2O3 so that a 0.025 mol dm–3 sodium thiosulphate solution, Na2S2O3 is produced. 6.7 Neutralisation g Learnin tandard S When stung by a bee, the area that had been stung can be treated At the end of the lesson, with baking soda. Vinegar, on the other hand, is used to treat the pupils are able to: area that had been stung by a wasp. Why? 6.7.1 State the meaning of neutralisation 6.7.2 Determine the concentration of an unknown solution through titration method 6.7.3 Solve numerical problems involving neutralisation Photograph 6.3 Bee and wasp 167 THEME 3 Interaction between Matter Definition of Neutralisation Neutralisation is a reaction between an acid and an alkali (base) to produce salt and water only. In the reaction, the salt and water produced are neutral because the acid lose its acidity and the alkali lose its alkalinity. Acid + Alkali → Salt + Water For example, the neutralisation reaction between nitric acid, HNO3 with potassium hydroxide, KOH to produce potassium nitrate solution, KNO3 and water, H2O. HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l) In neutralisation, the actual reaction that occurs only involves the combination of hydrogen ions, H+, from the acid and the hydroxide ions, OH– from the alkali to produce water molecules, H2O. Hence, the ionic equation for the reaction is as follows: H+(aq) + OH–(aq) → H2O(l) The following shows how the ionic equations for neutralisation reaction can be obtained. Chemical equation: HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l) H+(aq) + NO3–(aq) + K+(aq) + OH–(aq) → K+(aq) + NO3–(aq) + H2O(l) Ions in nitric acid Ions in potassium Ions in potassium Water hydroxide solution nitrate solution molecules K+ and NO3– are considered as spectator ions that do not change in the reaction. Thus, these ions are cancelled out in the equation. Ionic equation: H+(aq) + OH–(aq) → H2O(l) Activity 6.10 Write chemical equations and ionic equations for neutralisation reactions CT 1. Complete and balance the following equations. After that, write the relevant ionic equation. (a) HCl(aq) + Ba(OH)2(aq) → (b) H2SO4(aq) + KOH(aq) → (c) HNO3(aq) + NaOH(aq) → 2. Play the role as a chemistry teacher by explaining your findings in front of your classmates. 168  Acid, Base and Salt CHAPTER 6 Applications of Neutralisation in Daily Life Figure 6.31 shows the application of neutralisation for a variety of uses in daily life. Medicine Hair health Dental health Agriculture TO OT Shampoo H PA ST E f Milk oe ia Magn s Me ny eg ark an Milk of magnesia Mg(OH)2 Weak alkali in the Toothpaste contains a Slaked lime, Ca(OH)2 relieves gastric pain by shampoo neutralises base that neutralises which is alkaline, is used neutralising the excessive acid on hair. lactic acid produced by to treat acidic soil. hydrochloric acid in bacteria in our mouth. the stomach. Figure 6.31 Applications of neutralisation in daily life Activity 6.11 Solving problems on soil fertility using suitable STEM 21st Century Skills CT fertilisers 1. Carry out this activity in groups. 2. Study the following problem statement: Apart from treating acidic soil, fertilisers need to be added to soil to replace nutrients such as nitrogen, potassium and phosphorus that have been absorbed by plants. There is a variety of fertilisers in the market. Which fertilisers are suitable for plants? 3. Gather information concerning the problem given above. (a) What type of crops were planted? (b) What are the type of elements required by the crops? (c) Identify the fertiliser that is suitable for the crops by considering the percentage of elements such as nitrogen, phosphorus, and other needs, fertiliser cost and the quantity needed for the area. 4. Present your group findings in a multimedia presentation. Neutralisation reaction is also applied in the production of fertilisers such as urea, potassium sulphate, K2SO4, ammonium nitrate, NH4NO3 and others. For example, urea can be produced from the neutralisation reaction between ammonia, NH3 and carbon dioxide, CO2. How about other fertilisers? Try to list out the acids and alkalis involved in the production of that fertilisers.