Skip to main content

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry

Activity 2.1

Indicator

Acid

Base

Red litmus

No Change

Blue

Blue Litmus

Red

No change

Phenolphthalein

Colourless

Pink

Methyl Orange

Pink  

Yellow

Indictors are substance which change colour in acidic or basic media.

Activity 2.2
  • There are some substances whose odour changes in in acidic or basic media.
  • These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution.
Activity 2.3
  • Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it.
  • => You will observe bubbles of hydrogen gas on the surface of zinc granules.
  • Zn + H2SO4 --> ZnSO4 + H2
  • => Pass the Hydrogen gas through the soap solution.
  • Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it.
  • Take a burning candle near a gas filled bubble. Candle burn with pop sound.
  • The metal combines with the acid and forms a compound called a salt.

Activity 2.4

  • Place a few pieces of granulated zinc metal in a test tube.
  • And add 2 mL of sodium hydroxide solution and warm the contents of the test tube.
  • There is a release of Hydrogen gas.
  • NaOH + Zn → Na2ZnO2 (Sodium zincate) +  H2

 Activity 2.5

  • Take about 0.5 g of sodium carbonate (Na2CO3) in test tube A
  • Take about 0.5 g of sodium hydrogen carbonate(NaHCO3) in test tube B. 
  • Add about 2 mL of dilute HCl to both the test tubes.
You will observe gas evolved

Test tube A: Na2CO3 (s)   +  2HCl(aq) → 2NaCl(aq) +  H2O(l) + CO2 (g)

Test tube B: NaHCO3 (s)  + HCl(aq) → NaCl(aq) + H2O(l) + CO2 (g)

  • Pass the gas produced in each case through lime water (calcium hydroxide solution) .
  • => You will observe lime water turn milky due to formation of calcium carbonate.

On passing the carbon dioxide gas evolved through lime water,

Ca(OH) 2 (aq) + CO2 (g) → CaCO3 ( s)+ H2O(l)

On passing excess carbon dioxide through lime water milkiness disappear due to formation of

CaCO3 + H2O(l)+ CO2 (g) → Ca(HCO)3  

  • Limestone, chalk and marble are different forms of calcium carbonate.
  • All metal carbonates and hydrogen carbonates react with acids to give a corresponding salt, carbon dioxide and water.

Activity 2.6  Base + Acid → Salt + water

  • Take about 2 mL of dilute NaOH solution in a test tube and add two drops of phenolphthalein solution.
  • The colour of the solution changes to pink
  • Add dilute HCl solution to the NaOH solution drop by drop
  • The pink colour of solution disappears due to formation of Salt.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Acid and base when react cancel their properties and form neutral Salt. This reaction is called neutralization reaction.

  • Now add a few drops of NaOH to the above mixture.
  • The pink colour of phenolphthalein reappear as salt became basic.

Activity 2.7

Take a small amount of copper oxide in a beaker and add dilute hydrochloric acid slowly while stirring.

  • You will notice that the colour of the solution becomes blue-green and the copper oxide dissolves.
  • The blue-green colour of the solution is due to the formation of copper(II) chloride in the reaction.

Cu O + H Cl  → Cu Cl2 + HO

Activity 2.8

Take solutions of glucose, alcohol, hydrochloric acid, sulphuric acid as electrolyte in voltaic cell one by one.

  • The bulb will start glowing in the case of acids as acid contain H+(aq)
  • In glucose and alcohol solutions bulb does not glow as they cannot ionize to release ions.
  • Glowing of the bulb indicates that there is a flow of electric current through the solution. The electric current is carried through the solution by ions.

Activity 2.9

Take about 1g solid NaCl in a clean and dry test tube and Add some concentrated sulphuric acid to the test tube.

  • We observe hydrogen chloride gas coming out of the delivery tube
  • Test the gas evolved successively with dry and wet blue litmus paper.
  • Wet the blue litmus paper change colour to red.
  • This happen because acid release H+ Ion in aqueous solution.

Explanation :  2NaCl+H2SO4(conc.)→Na2SO4+2HCl

  • Explanation: Dry HCl does not contain any hydrogen ions (H+) in it , so it does not show acidic behaviour so HCl gas does not change the colour of the dry of the dry blue litmus paper.
  • When HCl gas it tested with wet blue litmus paper then HCl gas first dissolves in water to form HCl acid solution.
  • HCl(g)   − dissolve in water→ HCl(aq)
  • HCl(aq) → H+(aq) + Cl(aq)
  • HCl solution forms hydrogen ions which can turn blue litmus paper to red.
  • Or, HCl+H2O→H3O+  +  Cl
  • Bases generate hydroxide (OH) ions in water. Bases which are soluble in water are called alkalis.
  • NaOH(s) +  H2O→ Na+ (aq) + OH-
Activity 2.10
Add few drops of concentrated H2SO4 to test tube containing water and swirl the beaker slowly.
  • Touch the base of the beaker to feel a change in temperature.
  • =>It is an exothermic process
Repeat the above Activity with sodium hydroxide pellets you will get same observation.
  • The acid must always be added slowly to water with constant stirring as it is exothermic process.
  • Precaution :  If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns.
  •  Increasing volume of water during mixing an acid or base with water results in decrease in the concentration of ions (H3O+  / OH) per unit volume of solution. Such a process is called dilution and the acid or the base is said to be diluted.
 Activity 2.11

Test the pH values of acidic and basic solution using pH Paper

  • Acid- pH 1 to 6 (acidic nature decreases),
  • Neutral water pH 7
  • Base – pH 8 to 14 (Basic nature increases)

Activity 2.12 Ph in daily life

  • The ideal soil pH for the growth of plants is 6.0 to 7.0 (slightly acidic to neutral) range.
  • Acidity of soil neutralise by adding calcium hydroxide

Activity 2.13. Naming of salt

  • Salts having the same positive or negative radicals are said to belong to a family.
  • NaCl and Na2SO4 belong to the family of sodium salts.
  • NaCl and KCl belong to the family of chloride salts.
  • Potassium sulphate, sodium sulphate, calcium sulphate, magnesium sulphate, copper sulphate belong to the family of Sulphate salts
  • Sodium nitrate and Potassium nitrate belong to the family of Nitrte salts
  • Sodium carbonate, Zinc carbonate, Iron carbonate belong to the family of carbonate salts        

Activity 2.14

  • Among the following salts, which salts are acidic, basic or neutral?
  • Acidic salts: Aluminium chloride, zinc sulphate, copper sulphate, ammonium chloride.
  • Basic salts: Sodium acetate, sodium carbonate.
  • Neutral salts: Sodium chloride, potassium nitrate, sodium sulphate.
  • Copper sulphate crystals CuSO4. 5H2O  is blue in colour which seem to be dry but contain water of crystallisation.
Test: When we heat the Copper sulphate crystals , water is removed and the salt turns white.

If you moisten the crystals again with water, Copper sulphate crystals reappear blue in colour

Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt. E.g. Na2CO3.10H2O is not really wet.

CuSO4. 5H2O (Blue) ---Heat--> CuSO4 (white) + 5H2O

For more study material visit JSUNILTUTORIAL 

Key terms class10chapteracidbasessaltsncertactivity,acids

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca