In-text Questions (Page 18)<\/span><\/p>\nQuestion 1.
\nYou have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
\nAnswer:
\nWe dip the red litmus paper in any one of the three test tubes. If there is no change in the colour of the litmus paper, it means this test tube contains water or acidic solution.<\/p>\n
If the colour of the red litmus paper turns blue, it means this test tube contains basic solution.<\/p>\n
Now, again we dip this litmus paper in any one of the remaining two test tubes. If we do not observe any change in the colour of the litmus paper it is surely water and if a colour change is noticed, then it will be an acidic solution.<\/p>\n
In-text Questions (Page 22)<\/span><\/p>\nQuestion 1.
\nWhy should curd and sour substances not be kept in brass and copper vessels?
\nAnswer:
\nBrass is an alloy of copper and zinc, Brass and copper vessels reacts with atmospheric oxygen and to form metal oxides. The nature of metal oxides is basic. Curd and sour substances contains acids, e.g., curd contains lactic acid and orange contains citric acid. If we keep curd and sour substances in brass and copper vessels the layer of oxides present on the surface of brass and copper reacts with the acids present in curd and sour substances to form salt and water and taste of sour substances chages from sour to bitter. So curd and sour substance are not be kept in brass and copper vessels.<\/p>\n
Question 2.
\nWhich gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
\nAnswer:
\nHydrogen has is usually liberated when an acid reacts with a metal.
\ne.g.
\n
\nThis gas buns with a pop sound when a burning candle is placed near the evolved gas.<\/p>\n
Question 3.
\nA Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
\nAnswer:
\nMetal compound ‘A’ is CaCO3<\/sub>
\n<\/p>\n<\/p>\n
In-text Questions (Page 25)<\/span><\/p>\nQuestion 1.
\nWhy do HCl, HNO3<\/sub>, etc show acidic character in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
\nAnswer:
\nA substance is said to be an acid, when it produces H+<\/sup> ions in its aqueous solution and turns blue litmus paper into red. HCl, HNO3<\/sub>, etc. ionise in their aqueous solutions and produce H+<\/sup> ions. They turn the colour of blue litmus paper into red. When electric current is passed through the aqueous solutions of acids, the H+<\/sup> ions reach the cathode and form hydrogen. But in their aqueous solutions alcohol and glucose do not produce H+<\/sup> ions and they do turn blue litmus into red. So they do not show acidic character in their aqueous solution.<\/p>\nQuestion 2.
\nWhy does an aqueous solution of an add conduct electricity?
\nAnswer:
\nAcids are ionised in their aqueous solution and produce H+<\/sup> (aq) ions. These ions are responsible for the flow of electricity in the aqueous solutions of acids. H+<\/sup> ions migrates towards cathode and gain electron to produce hydrogen gas. So when electricity is passed through the aqueous solution of an acid, hydrogen gas is liberated at cathode.<\/p>\nSo an aqueous solution of an acid conducts electricity because it contains H+<\/sup> (aq) ions or hydronium ions.<\/p>\nQuestion 3.
\nWhy does dry HCl gas not change the colour of the dry litmus paper?
\nAnswer:
\nThe separation of H+<\/sup> ion from HCl molecules cannot occur in the absence of water. In fact H+<\/sup> ion is an atom that has lost an eletron, it is simply isolated proton and is smallest positive ion. It is attracted very strongly towards electrons of anything. So the separation of H+<\/sup> ion from HCl molecules cannot occur and the reaction.
\nH – Cl(g) H+<\/sup> \u2192 + Cl–<\/sup><\/p>\ncannot occur in the absence of water, because highly concentrated positive charge of TP would join back with the Cl–<\/sup> ion to get back electron and reform covalent \u2018H – Cl\u2018 molecules. It means dry HCl does not contain H+<\/sup> ions so it does not change the colour of the dry litmus paper.<\/p>\nQuestion 4.
\nWhile diluting an acid, why is it recommended that the acid should be added to water and not water to the acid ?
\nAnswer:
\nThe process of dissolving an add or a base in water is a highly exothermic process. Be careful while mixing concentrated sulphuric acid with water During dilution of an acid, the acid must always be added slowly to water with constant stirring. It water is added to a concentrated acid the heat generated may cause the mixture lo splash out from the container and cause bums. The glass container may also break due to heal release during dilution. So it is recommended that the acid should be added to water and not water to the acid.<\/p>\n
Question 5.
\nHow is the concentration of hydronium ions (H3<\/sub>O)+<\/sup> affected when a solution of an acid is diluted?
\nAnswer:
\nAn aqueous solution of an acid contains the specific number of H3<\/sub>O+<\/sup> ions per unit volume. These H3<\/sub>O+<\/sup> ions are responsible for the acidic character of an acid. When water is added in an acid it results in decrease in inconcentration of H3<\/sub>O+<\/sup> ions per unit volume of the acid because the specified volume of an acid contains the definite no. of H3<\/sub>O+<\/sup> ions, when its volume increases by addition of water, the no. of H3<\/sub>O+<\/sup> ions decreases per unit volume of the acid. So we can say that on dilution the concentration of H3<\/sub>O+<\/sup> ions decreases per unit volume of the acid.<\/p>\nQuestion 6.
\nHow is the concentration of hydroxide ions (OH–<\/sup>) affected when excess base is dissolved in a solution of sodium hydroxide?
\nAnswer:
\nA solution of sodium hydroxide contain Na+<\/sup> and OH–<\/sup> ions. When a base is dissolved in a solution of sodium hydroxide, it results in increase of concentration of OH–<\/sup> ions per unit volume because added base is ionised and produce a large no. of hydroxide ions (OH–<\/sup>) in the solution of sodium hydroxide.<\/p>\n<\/p>\n
In-text Questions (Page 28)<\/span><\/p>\nQuestion 1.
\nYou have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
\nAnswer:
\nThe hydrogen ion concentration of a solution is calculated by the formula :
\n[H+<\/sup>] = 10-pH<\/sup>
\n\u2234 The conc. of H+<\/sup> ions in solution ‘A’ = 10-6<\/sup>
\nand the conc, of H+<\/sup> ions in solution ‘B’ = 10-8<\/sup>
\nSo solution ‘A’ has more hydrogen ion concentration than ‘B’.
\nSolution A’ is acidic and solution ‘B’ is basic.<\/p>\nQuestion 2.
\nWhat effect does the concentration of H+<\/sup>(aq) ions have on the nature of the solution?
\nAnswer:
\nOn the basis of concentration of H+<\/sup> (aq) ions in a solution we can justify the nature of the solution i.e., whether it is acidic, basic or neutral. A solution which has the concentration of H+<\/sup> (aq) ion equal to 10-7<\/sup> mole\/litre will be neutral and if the concentration is slightly less than 10-7<\/sup> then it will be basic and if the cone, of H+<\/sup> (aq) ions in a solution is slightly higher than 10-7<\/sup> then it will be acidic.<\/p>\nQuestion 3.
\nDo basic solutions also have H+<\/sup>(aq) ions? If yes, then why are these basic?
\nAnswer:
\nYes, basic solutions also have H+<\/sup> (aq) ions and similarly acidic solutions have OH–<\/sup> (aq). because acids or bases shows acidic or basic character in water. Distilled water also has 10-7<\/sup> mol\/L, H+<\/sup> and 10-7<\/sup> mole\/L, OH–<\/sup> ions at 298K.<\/p>\nIf an aqueous solution has an equal concentration of hydrogen ions and hydroxide ions in it, it is said to be neutral. Now, if an aqueous solution has more of hydrogen ions and less of hydroxide ions, it will be acidic solution. On the other hand, if an aqueous solution has more of hydroxide ions and less of hydrogen ions it will be basic in nature.<\/p>\n
Question 4.
\nUnder what soil condition do you think a fanner would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
\nAnswer:
\nIf the soil has acidic nature, then famer would treat the soil of his fields with quick lime or slaked lime or chalk to maintain its pH near about neutral.<\/p>\n
In-text Questions (Page 33)<\/span><\/p>\nQuestion 1.
\nWhat is the common name of the compound CaOCl2<\/sub>?
\nAnswer:
\nBleaching powder.<\/p>\nQuestion 2.
\nName the substance which cm treatment with chlorine yields bleaching powder,
\nAnswer:
\nDry slaked lime : Ca(OH)2<\/sub>
\nChlorine : Cl2<\/sub><\/p>\nQuestion 3.
\nName the sodium compound which is used for softening hard water.
\nAnswer:
\nWashing soda : Sodium carbonate (Na2<\/sub>CO3<\/sub>)<\/p>\n<\/p>\n
Question 4.
\nWhat will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved.
\nAnswer:
\nOn heating sodium hydrocarbonate, it produces sodium carbonate, carbondioxide and water.
\n<\/p>\n
Question 5.
\nWrite an equation to show the reaction between Plaster of Paris and water,
\nAnswer:
\n<\/p>\n
Class 10 Science Chapter 2 Acids, Bases and Salts Textbook Questions and Answers<\/h3>\n
Page No. 34<\/span><\/p>\nQuestion 1.
\nA solution turns red litmus blue, its pH is likely to be:
\n(a) 1
\n(b) 4
\n(c) 5
\n(d) 10
\nAnswer:
\n(d) 10<\/p>\n
Question 2.
\nA solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains.
\n(a) NaCl
\n(b) HCl
\n(c) LiCl
\n(d) KCl
\nAnswer:
\n(b) HCl<\/p>\n
Question 3.
\n10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount of HCl solution (the same solution as before) required to neutralise it will be:
\n(a) 4 mL
\n(b) 8 mL
\n(c) 12 mL
\n(d) 16 mL
\nAnswer:
\n(d) 16 mL<\/p>\n
Question 4.
\nWhich one of the following types of medicines is used for treating indigestion ?
\n(a) Antibiotic
\n(b) Analgesic
\n(c) Antarid
\n(d) Antiseptic
\nAnswer:
\n(c) Antacid.<\/p>\n
Question 5.
\nWrite word equations and then balanced equations for the reactions taking place when:
\n(a) dilute sulphuric add reacts with zinc granules.
\n(b) dilute hydrochloric acid reacts with magnesium ribbon.
\n(c) dilute sulphuric acid reacts with aluminium powder.
\n(d) dilute hydrochloric acid reacts with iron fillings.
\nAnswer:
\n(a) Dilute sulphuric acid reacts with zinc granules to form zinc sulphate and hydrogen.
\nSulphuric acid + Zinc \u2192 Zinc Sulphate + Hydrogen
\nH2<\/sub>SO4<\/sub> + Zn \u2192 ZnSO4<\/sub> + H2<\/sub><\/p>\nStep I.<\/p>\n
\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n |
\nH<\/td>\n | 2<\/td>\n | 2<\/td>\n<\/tr>\n |
\nS<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n |
\nO<\/td>\n | 4<\/td>\n | 4<\/td>\n<\/tr>\n |
\nZn<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n The above equation is a balanced equation because it contains the equal no. of atoms of different elements in reactants to the no. of atoms of different elements in products.<\/p>\n (b) Hydrochloric add + Magnesium \u2192 Magnesium chloride + Hydrogen \nHCl + Mg \u2192 MgCl2<\/sub> + H2<\/sub> \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nH<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n | \nCl<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n | \nMg<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II. Balance hydrogen and chlorine by multiplying ‘HCl’ with \u20182′ we get \n2 HCl + Mg \u2192 MgCl2<\/sub> = H2<\/sub><\/p>\n(c) Sulphuric acid + Aluminium \u2192 Aluminium sulphate + Hydrogen \nH2<\/sub>SO4<\/sub> + Al \u2192 Al2<\/sub>(SO4<\/sub>)3<\/sub> + H2<\/sub> \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nH<\/td>\n | 2<\/td>\n | 2<\/td>\n<\/tr>\n | \nS<\/td>\n | 1<\/td>\n | 3<\/td>\n<\/tr>\n | \nO<\/td>\n | 4<\/td>\n | 12<\/td>\n<\/tr>\n | \nAl<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II. Balance oxygen by multiplying ‘H2<\/sub>SO4<\/sub>‘ with ‘3’ we get: \n3H2<\/sub>SO4<\/sub> + Al \u2192 Al2<\/sub>(SO4<\/sub>)3<\/sub> + H2<\/sub> \nStep III. Balance aluminium by multiplying ‘Al’ in reactants side with ‘2’ we get; \n3H2<\/sub>SO4<\/sub> + 2Al \u2192 Al2<\/sub>(SO4<\/sub>)3<\/sub> + H2<\/sub> \nStep IV. Balance hydrogen by multiplying ‘H2<\/sub>‘ in product side with ‘3’ we get: \n3H2<\/sub>SO4<\/sub> + 2Al \u2192 Al2<\/sub>(SO4<\/sub>)3<\/sub> + 3H2<\/sub> \nThis is a balance equation.<\/p>\n(d) Hydrochloric acid + Iron \u2192 Iron (III) chloride + Hydrogen \nHCl + Fe \u2192 FeCl3<\/sub> + H2<\/sub> \nStep I.<\/p>\n\n\n\nElement<\/td>\n | No. of atoms in Reactants<\/td>\n | No. of atoms in Products<\/td>\n<\/tr>\n | \nH<\/td>\n | 1<\/td>\n | 2<\/td>\n<\/tr>\n | \nCl<\/td>\n | 1<\/td>\n | 3<\/td>\n<\/tr>\n | \nFe<\/td>\n | 1<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Step II. Balance chlorine by multiplying ‘FeCl3<\/sub>‘ with ‘2’ and ‘HCl’ with ‘6’ we get: \n6HCl + Fe \u2192 2FeCl3<\/sub> + H2 \nStep III. Balance hydrogen and iron by multiplying ‘Fe’ with ‘2’ and ‘H2<\/sub>‘ with ‘3’ we get: \n6HCl + 2Fe \u2192 2FeCl3<\/sub> + 3H2<\/sub> \nThis is a balance equation.<\/p>\n<\/p>\n Question 6. \nCompounds such as alcohols and glucose also contain hydrogen but are not categorised as adds. Describe an activity to prove it. \nAnswer: \nA substance is said to be an acid when it can produce H+<\/sup> ions in its aqueous solution and turns blue litmus paper to red.<\/p>\nHCl, HNO3<\/sub> etc. are ionized in their aqueous solution and furnished H+<\/sup> ions. When electricity is passed through aqueous solutions of acids, hydrogen liberates at cathode.<\/p>\nCompounds such as alcohols and glucose contain hydrogen but they do not ionise in their aqueous solution and not produce H+<\/sup> ions, and they do not turn blue litmus to red so they are not categorised as acids.<\/p>\nActivity: Take four test tubes namely, A, B, C and D. Add ethyl alcohol (alcohol) aq. solution of glucose ; hydrochloric acid and nitric acid, 2 mL of each in test tube A, B, C and D respectively. Now add 1-2 drops of blue litmus solution in each test tube. We observe that the colour of test tube ‘A’ and ‘B’ remain unchanged however the colour of test tube and ‘C’ and ‘D’ turns red. If proves that alcohol and glucose are not categorised as acid.<\/p>\n Question 7. \nWhy does distilled water not conduct electricity, whereas rainwater does ? \nAnswer: \nDistilled water does not contain ions (H+<\/sup> or OH–<\/sup>) or it contains a very poor concentration of H+<\/sup> or OH–<\/sup> ions \u2248 1 \u00d7 10-7<\/sup> moles per litre. So it does not conduct electricity, whereas rainwater has some acidic character due to the presence of acids such as H2<\/sub>SO4<\/sub>, HNO3<\/sub> etc. So rainwater has a sufficient amount or concentration of H+<\/sup> ions, so it conducts electricity.<\/p>\nQuestion 8. \nWhy do acids not show acidic behaviour in the absence of water ? \nAnswer: \nA substance is said to be an acid when it can furnish H+<\/sup> ions in its aqueous solution. The separation of H+<\/sup> ions from the molecules of an acid cannot occur in the absence of water. The molecule of water are polar in nature, so the negative pole of ‘H2<\/sub>O’ molecule pull the H+<\/sup> ions from acids towards itself and separate them from acids e.g, \nHCl + H2<\/sub>O \u2192 H3<\/sub>O+<\/sup> + Cl–<\/sup>. Hydrogen ions cannot exist alone, but they exist after combining with water molecules. Thus hydrogen ions must always be shown as H+<\/sup> (aq) of hydronium ion (H3<\/sub>O+<\/sup>).<\/p>\nIn absence of water ‘H+<\/sup>‘ are does not furnish by an acid so it does not show acidic behaviour.<\/p>\nQuestion 9. \nFive solutions A, B, C, D and E when tested with universal indicater showed pH as 4, 1, 11, 7 and 9, respectively. Which solution is: \n(a) neutral ? \n(b) strongly alkaline ? \n(c) strongly acidic \n(d) weakly acidic ? \n(e) weakly alkaline ? \nArrange the pH in increasing order of hydrogen ion concentration. \nAnswer: \n(a) Solution ‘D’ is neutral because it has pH 7. \n(b) Solution ‘C’ is strongly alkaline because it has pH 11. \n(c) Solution ‘B’ is strongly acidic because it has pH 1. \n(d) Solution ‘A’ is weakly acidic because it has pH 4. \n(e) Solution ‘E’ is weakly alkalin because it has pH 9.<\/p>\n Hydrogen ion concentration is inversely proportional to the pH of the solution. It means, higher the ‘pH’ lower will be the hydrogen ion concentration.<\/p>\n The ‘pH’ of the given solution can be arranged m increasing order of hydrogen-ion concentration as follows : \npH : 11 > 9 > 7 > 4 > 1 \nHydrogen ion concentration : C < E < D < A < B<\/p>\n Question 10. \nEqual lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A which acetic acid (CH3<\/sub>COOH) is added to test tube B. In which test tube will the fizzing occurs more vigorously and why ? \nAnswer: \nFizzing occurs more vigourosly in test tube A’ because hydrochlonic acid is a strong acid while acetic acid is a weak acid.<\/p>\nMagnesium metal reacts with acids and produce repective salt and hydrogen gas. \n<\/p>\n Question 11. \nFresh milk has a pH of 6. How do you think the pH will change as it turns into curd ? Explain your answer. \nAnswer: \nThe fresh milk has a pH of 6. Milk has slightly sweet taste But when it turns into curd its taste has changed from lightly sweet taste to intense sour taste. It means its pH has changed. The intense sour taste of curd is the presence of the strong acidic character. So when milk changes into curd its pH decreases because acidic character increases.<\/p>\n <\/p>\n Question 12. \nA milkman adds a very small amount of baking soda to fresh milk. \n(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ? \n(b) Why does this milk take a long time a set as curd ? \nAnswer: \n(a) Milk contains lactose and pH of milk is approximately 6. The micro-organism which converts milk into curd work effectively in slightly acidic medium. Baking soda has slightly alkaline nature. So milkman adds small quantity of baking soda to protect it from curding because in alkaline medium micro-organism do not function properly and thus milk is prevented from rancidity or curding. So milkman makes milk alkaline.<\/p>\n (b) The process of converting milk into curd takes place in slightly acidic medium, Micro-organism converts lactose of milk into lactic acid. But in alkaline medium curding takes place very slowly, so this milk takes long to set as curd.<\/p>\n Question 13. \nPlaster of Paris should be stored in moisture-proof container. Explain why ? \nAnswer: \nPlaster of Paris (CaSO4<\/sub>. \\(\\frac {1}{2}\\)H2<\/sub>O) has a very remarkable Property of setting into a hard mass on wetting with water or moisture. The setting of Plaster of Paris is due to its hydration to form crystals of gypsum which set to form a hard, solid mass as follows : \n \nSo, to protect the hydration of Plaster of Paris it is stored in moisture proof container.<\/p>\nQuestion 14. \nWhat is a neutralisation reaction ? Given two examples. \nAnswer: \nNeutralisation reaction : A reaction in which an acid reacts with a base and to form a salt and water is called neutralisation reaction. \nExamples: \n(i) \n \n(ii) \n<\/p>\n Question 15. \nGiven two important uses of washing soda and baking soda. \nAnswer: \nUses of washing soda :<\/p>\n \n- Washing soda is used in laundary. In other words, washing soda is used as a cleansing agent for domestic purposes. It is a main component of many dry soap powders.<\/li>\n
- It is used in the manufacture of many useful sodium compounds like caustic soda, borax, glass and soap.<\/li>\n<\/ul>\n
Uses of baking soda : \n(i) Baking soda (sodium hydrogen carbonate) is used as an antacid in medicine to remove acidity of the stomach.<\/p>\n (ii) Baking soda is used in making powder. Baking powder contains baking soda in acid like tartaric acid. When baking powder is mixed in the dough for preparing cake or bread is heated, sodium hydrogen carbonate present it decomposes it give carbon dioxide and sodium carbonate. This carbon dioxide gas bubbles out causing the cake or bread to rise and becomes light and spongy. The tartaric acid present in baking powder reacts with sodium carbonate and neutralises it. If tartaric acid is not present in baking power, then the cake will taste bitter due to the presence of sodium carbonate in it.<\/p>\n Class 10 Science Chapter 2 Acids, Bases and Salts Textbook Activities<\/h3>\nActivity 2.1 (Page 18)<\/span><\/p>\n\n- Collect the following samples from the science laboratory-hydrochloric acid (HCl), sulphuric acid (H2<\/sub>SO4<\/sub>), nitric acid (HNO3<\/sub>), acetic add (CH3<\/sub>COOH), sodium hydroxide (NaOH), calcium hydroxide [Ca(OH)2<\/sub>], potassium hydroxide (KOH), magnesium hydroxide [Mg(OH)2<\/sub>], and ammonium hydroxide (NH4<\/sub>OH).<\/li>\n
- Put a drop of each of the above solutions on a watch-glass and test with a drop of the following indicators as shown in Table 2.1.<\/li>\n
- What change in colour did you observe with red litmus, blue litmus, phenolphthalein and methyl orange solutions for each of the solutions taken?<\/li>\n
- Tabulate your observations in Table 2.1.<\/li>\n<\/ul>\n
Table 2.1 \n<\/p>\n Activity 2.2 (Page 18)<\/span><\/p>\n\n- Take some finely chopped onions in a plastic bag along with some strips of clean cloth. Tie up the bag tightly and leave overnight in the fridge. The cloth strips can now be used to test for acids and bases.<\/li>\n
- Take two of these cloth strips and check their odour.<\/li>\n
- Keep them on a clean surface and put a few drops of dilute HCl solution on one strip and a few drops of dilute NaOH solution on the other.<\/li>\n
- Rinse both cloth strips with water and again check their odour.<\/li>\n
- Note your observations.<\/li>\n
- Now take some dilute vanilla essence and clove oil and check their odour.<\/li>\n
- Take some dilute HCl solution in one test tube and dilute NaOH solution in another. Add a few drops of dilute vanilla essence to both test tubes and shake well. Check the odour once again and record changes in odour, if any.<\/li>\n
- Similarly, test the change in the odour of clove oil with dilute HCl and dilute NaOH solutions and record your observations.<\/li>\n<\/ul>\n
Observations: Vanilla indicator.<\/p>\n <\/p>\n Activity 2.3 (Page 19)<\/span><\/p>\n\n- Set the apparatus as shown in Fig.<\/li>\n
- Take about 5 mL of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.<\/li>\n
- What do you observe on the surface of zinc granules?<\/li>\n
- Pass the gas being evolved through the soap solution.<\/li>\n
- Why are bubbles formed in the soap solution?<\/li>\n
- Take a burning candle near a gas filled bubble.<\/li>\n
- What do you observe?<\/li>\n
- Repeat this Activity with some more acids like HCl, HNO3<\/sub>, and CH<\/span>3<\/sub>COOH,<\/span><\/li>\n
- Are the observations in all the cases the same or differed ?<\/li>\n<\/ul>\n
\nObservation : Bubbles of hydrogen gas are evolved from the surface of zinc granules. When hydrogen gas is passed through a soap solution it forms bubbles because hydrogen is lighter than soap solution and it filled in the bubbles of soap. Hydrogen gas burns with a pop sound.<\/p>\n
Hydrogen evolves vigorously with acids like HCl and HNO3<\/sub> but releases slowly with CH3<\/sub>COOH.<\/p>\nActivity 2.4 (Page 20)<\/span><\/p>\n\n- Place a few pieces of granulated zinc metal in a test tube.<\/li>\n
- Add 2 mL of sodium hydroxide solution and warm the contents of the test tube.<\/li>\n
- Repeat the rest of the steps as in Activity 2.3 and record your observations.<\/li>\n<\/ul>\n
Observations : Zinc reacts with \u2018NaOH\u2019 to forms sodium zincate and hydrogen. \n \nHydrogen bums with a pop near the candle.<\/p>\n Activity 2.5 (Page 20)<\/span><\/p>\n<\/p>\n \n- Take two test tubes, label them as A and B.<\/li>\n
- Take about 0.5 g of sodium carbonate (Na2<\/sub>CO3<\/sub>) in test tube A and about 0.5 g of sodium hydrogen carbonate (NaHCO3<\/sub>) in test tube B.<\/li>\n
- Add about 2 mL of dilute HCl to both the test tubes.<\/li>\n
- What do you observe?<\/li>\n
- Pass the gas produced in each case through lime water (calcium hydroxide solution) as shown in Fig. and record your observations.<\/li>\n<\/ul>\n
Observations : Sodium carbonate (Na2<\/sub>CO3<\/sub>) and sodium bicarbonate (NaHCO3<\/sub>) reacts with dilute HCl to form NaCl, H2<\/sub>O and CO2<\/sub>, Carbondioxide evolves with effervescence.<\/p>\nTest Tube A: \nNa2<\/sub>CO3<\/sub>(s) + 2 HCl (aq) \u2192 2NaCl (aq) H2<\/sub>O(l) + CO2<\/sub> \nTest Tube B: \nNaHCO3<\/sub> (s) + HCl (aq) \u2192 NaCl (aq) + H2<\/sub>O(l) + CO2<\/sub> (\u2191)<\/p>\nOn passing carbondioxide gas through lime water e.g. aq. Ca(OH)2<\/sub>, it turns milky. \n \nOn Passing carbondioxide for long time through lime water, the milky colour of water disapperas due to the formation of soluble Ca(HCO3<\/sub>)2<\/sub> (aq). \n<\/p>\nActivity 2.6 (Page 21)<\/span><\/p>\n | | | |