Science And Technology Solutions Solutions for Class 9 Science Chapter 5 - Acids, Bases And Salts

Answer:

ans
a.Ammonium = NH₄⁺
Nitrate = NO₃^-
Hydride = H^-
Chloride = Cl^-
Ammonium is odd because Ammonium is cation and rest are anions.

b.Hydrogen chloride is odd because Hydrogen chloride is acid and rest are base.

c.Acetic acid = CH₃COOH
   Carbonic acid = H₂CO₃
   Hydrochloric acid = HCl
   Nitric acid = - HNO₃
   HCl is the only Diatomic Hetro-Nuclear compound and remaining are Poly Atomic compound.

d.Ammonium chloride is odd  because it is acidic salt and rest all are neutral salts.

e.Sodium carbonate is odd because the solutions of sodium nitrate, sodium sulphate and sodium chloride are neutral. But the solution of sodium carbonate is BASIC.

f.Calcium oxide = CaO
   Magnesium oxide = MgO
   Zinc oxide = ZnO
   sodium oxide = Na₂O
   ZnO is odd because it is amphoteric in nature and other ions are basic in nature

g.Common salt is odd because on heating, there is no change in color of compound. But in rest of the compounds, there is change is colour.

h. Potassium hydroxide is odd because in a reaction, when Sodium chloride reacts with acetic acid, then Sodium acetate is formed. There is no role of Potassium hydroxide in the reaction given below.
$\mathrm{NaCl} +{\mathrm{CH}}_3\mathrm{COOH}\to {\mathrm{CH}}_3\mathrm{COONa} +\mathrm{HCl}$

Answer:

ans
a. When 5o mL water is added to 50 mL solution of copper sulphate, then reversible reaction occurs and the colour change from pale blue to white and then change back to blue when water is added again.

$CuS{O}_4 +H_{2}O \leftrightarrow CuS{O}_4.5H_{2}O ( BLUE VITRIOL).$

b. Phenolphthalein is an indicator used for determining the quantity of base. When two drops of Phenolphthalein indicator are added to 10 mL solution of Sodium hydroxide, then the solution turns pink in colour because the acidic part of phenophthalein reacts with the base, it forms sodium salt of phenolphthalein which has pink color.
For example : In acid base titration phenolphthalein is used to detect end point of base.

c. Copper is an unreactive metal and doesn’t react in normal circumstances with dilute acids.

There are actually two equations for the reaction of copper with nitric acid take place. It depends on whether the nitric acid is concentrated or not. If it is concentrated, then the ratio is 1:4  for copper to nitric acid. If it is dilute then the ratio is 3:8.

 

$$\begin{gathered} Cu + 4HN{O}_3 (dil.)\to Cu(N{O}_3{)}_2 + 2N{O}_{2 }+ 2H_{2}O \\ 3Cu + 8HN{O}_{3 }(conc.)\to 3Cu(N{O}_3{)}_2 + 2NO + 4H_{2}O \end{gathered}$$
Concentrated Nitric acid  is act as strong oxidising agent so it makes sense that a higher oxidation state of nitrogen (IV) oxide is formed .

d. When litmus paper is dipped into 2 mL of dilute HCl solution , then blue litmus paper is turned into red colour and there is no effect on red litmus paper. Again, if the same litmus paper is dipped into 2 mL of concentrated NaOH solution, then red litmus paper turns into blue colour but there is no effect on blue litmus paper.This is due to the respective properties of blue and red litmus paper with acid and base.
 

e. Magnesium oxide is a base. when base reacts with an acid, it forms salt and water.This reaction is known as neutralisation reaction.

$MgO\left(s\right) + 2HCl\left(l\right) = MgC{l}_{2 }\left(aq\right) + H_{2}O\left(l\right)$

MgO doesnot react with NaOH.As NaOH is a base and bases react only with oxides of non-metal to form salt and water because oxides of non-metals are said to be acidic in nature so neutralization reaction take place. But MgO is a oxide of metal so, reaction is not possible.

f. Zinc oxide is added to dilute HCl, then neutralization reaction takes place to form salt and water.
The reaction is as follows:

$ZnO\left(aq\right)+2HCl\left(aq\right)\to ZnC{l}_2\left(aq\right)+H_{2}O\left(l\right)$

Zinc oxide reacts  with sodium hydroxide to produce zincate sodium and water. This reaction takes place at a temperature of 500-600°C.it is exothermic reaction.
$\mathrm{ZnO} + 2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{ZnO}}_2 + {\mathrm{H}}_2\mathrm{O}$

g. Limestone is calcium carbonate. When limestone is added to a 10% solution of dilute HCl, then brisk effervescence of CO₂ is released due to the reaction of acid with carbonate of metals.

$2HCl + CaC{O}_3 \to CaC{l}_{2 }+ C{O}_2 +H_{2}0$

h. When pieces of blue vitriol are heated in a test tube, then crystal structure of blue vitriol broke down to forn a colourless powder and water come out. This water is called water of crystallization. when water is added to the same test tube, then white powder turned into blue colour again.This is due to reversible reaction take place from anhydrous salt to hydrous salt and vice-versa.
$CUS{O}_4.5H_{2}O\leftrightarrow CUS{O}_4 +5H_{2}O$
 

i. When a dilute solution of sulfuric acid is electrolysed, gases are produced at both the anode and the cathode electrode.

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The gas produced at the cathode burns with a 'pop' sound, when a sample is lit with a lighted splint. This shows that the gas is hydrogen.

The gas produced at the anode relights a glowing splint dipped into a sample of the gas. This shows that the gas is oxygen.

The gases are produced when ions move towards the electrodes.

At the cathode:

2H⁺ +2e^- → H₂

At the anode:

4OH^- - 4e^- → 2H₂O + O₂

 

Answer:

Oxides are of three types :

1) Acidic Oxides: CO₂ (Carbon dioxide), SO₃ (Sulfur trioxide)

2) Basic Oxides: CaO (Calcium oxide), MgO (Magnesium oxide), Na₂O (Sodium oxide)

3) Amphoteric Oxides: ZnO (Zinc oxide), Al₂O₃ (Aluminium oxide), Fe₂O₃ (Ferric oxide)

Answer:

a. Atomic number of Sodium(Na) atom is 11.
Electronic configuration is :
Na = 2, 8, 1
So it contains 1 valence electron. In order to achieve the nearest noble gas configuration, it loses one electron to form Sodium ion.
Na⁺ = 2,8

Atomic number of Chlorine(Cl) atom is 17.
Electronic configuration is :
Cl = 2, 8, 7
So it contains 7 valence electron. In order to achieve the nearest noble gas configuration, it gains one electron to form Chloride ion.
Cl^- = 2,8
An Ionic bond is formed between sodium ion and chloride ion by complete transfer of electron from sodium to chlorine.



b. Atomic number of Magnesium(12) atom is 12.
Electronic configuration is :
Mg = 2, 8, 2
So it contains 2 valence electron. In order to achieve the nearest noble gas configuration, it loses two electrons to form Magnesium ion.
Mg²⁺ = 2,8

Atomic number of Chlorine(Cl) atom is 17.
Electronic configuration is :
Cl = 2, 8, 7
So it contains 7 valence electron. In order to achieve the nearest noble gas configuration, it gains one electron to form Chloride ion.
Cl^- = 2,8
An Ionic bond is formed between Magnesium ion and two Chloride ion by complete transfer of one electron to each Chlorine ion.

Answer:

a.HCl(aq) + H2O(l)→H3O⁺(aq)+Cl^-(aq)
Explanation:
Hydrochloric acid(HCl) is a strong acid, so HCl is ionize completely in aqueous solution.In other words, every molecule of hydrochloric acid that is added to water will donate its proton H⁺ to water molecule to form a hydronium cation, and H₃O⁺and chloride ions Cl^-  is formed.

b.Reaction:
NaCl(s) + H₂O(aq) -> Na⁺(aqueous) + Cl^-(aqueous) + H₂O(l)
Explanation:
When sodium chloride reacts with water the Na⁺ part of NaCl is attracted to the oxygen side of the water molecules, while the Cl^- side is attracted toward the hydrogen's side of the water molecule.

   

Answer:

a)7.3 g HCl in 100 mL solution:


$\mathrm{Concentration} \mathrm{of} \mathrm{HCl} \mathrm{in} \mathrm{g} {\mathrm{L}}^{-1 }=\frac{Mass of solute \mathrm{in} \mathrm{gram}}{\mathrm{Volume} \mathrm{of} \mathrm{HCl} \mathrm{in} litre}$
                                     
                                      $=\frac{7.3 \times 1000}{100}=73$ g L^-1

In terms of gram per litre:
7.3 g of HCl in 100 mL has concentration = 73 g L^{-1
 
Molecular mass of HCl = $1+35.5=36.5$}

$\mathrm{Molarity} =\frac{\mathrm{Mass} \mathrm{of} \mathrm{solute} \mathrm{in} moles}{\mathrm{Volume} \mathrm{of} \mathrm{HCl} \mathrm{in} \mathrm{L}}$

            $=\frac{7.3 \times 1000}{36.5 \times 100}=2$ mol L^-1

In terms of moles per litre:
7.3 g of HCl in 100 mL has concentration = 2 mol L^-1

 
b)2g NaOH in 50 mL solution


$Concentration of NaOH in g L^{-1 }=\frac{Mass of NaOH in g}{Volume of NaOH in L}$
                                     
                                      $=\frac{2 \times 1000}{50}= 40$ g L^-1
In terms of gram per litre:
2 g of NaOH in 50 mL has concentration = 40 g L^{-1
 
Molecular mass of NaOH = $23 +16 + 1 = 40$}

$\mathrm{Molarity} =\frac{\mathrm{Mass} \mathrm{of} \mathrm{solute} \mathrm{in} moles}{\mathrm{Volume} \mathrm{of} NaOH \mathrm{in} \mathrm{L}}$

            $=\frac{2 \times 1000}{40 \times 50}=1$ mol L^-1

In terms of moles per litre:
2 g of NaOH in 50 mL has concentration = 1 mol L^-1


c)3 g CH₃COOH in 50 mL solution


$Concentration of C{H}_{3}COOH in g L^{-1 }=\frac{Mass ofC{H}_{3}COOH in g}{Volume of C{H}_{3}COOH in L}$
                                     
                                      $=\frac{3 \times 1000}{100}= 30$ g L^-1
In terms of gram per litre:
3 g of CH₃COOH in 100 mL has concentration = 30 g L^{-1
 
Molecular mass of}  CH₃COOH ^{= $2 \times 12 + 2 \times 16 + 4 \times 1 = 60$}

$Molarity =\frac{Mass of solute in moles}{Volume of solution in L}$

            $=\frac{3 \times 1000}{60 \times 100}=0.5$ mol L^-1

In terms of moles per litre:
3 g of  CH₃COOH in 100 mL has concentration = 0.5 mol L^-1

d)4.9 g H₂SO₄  in 200 mL solution


$Concentration of H_{2}S{O}_{4 }in g L^{-1 }=\frac{Mass of H_{2}S{O}_4 in g}{Volume of H_{2}S{O}_{4 }in L}$
                                     
                                                $=\frac{4.9 \times 1000}{200}= 24.5$ g L^-1
In terms of gram per litre:
4.9 g of  H₂SO₄ in 200 mL has concentration = 24.5 g L^{-1
 
Molecular mass of}   H₂SO₄ ^{= $2 \times 12 + 2 \times 16 + 4 \times 1 = 60$}

$Molarity =\frac{Mass of solute in moles}{Volume of solution in L}$

            $=\frac{4.9\times 1000}{98 \times 200}=1$ mol L^-1

In terms of moles per litre:
4.9 g of   H₂SO₄ in 200 mL has concentration = 1 mol L^-1

Answer:

Ans7.This is an activity based question in which you are supposed to collect rainwater from different places and compare there results on the basis of following parameters :
a) pH of water
b)Action of water on blue litmus paper
c)Action of water on red litmus paper
d)Effect of indicator like phenolphthalein and methyl orange
Conclusion :
If we take samples of rain water from different places, we observe the following results :

• pH of water is in between 1-6
• Blue litmus paper turns red
• No change in colour of red litmus paper
• No change in colour of rain water on adding phenolphthalein
• rain water turns red in colour on adding methyl orange

Answer:

Ans 8.

a. The number of ionizable hydrogen (H+) ions present in one molecule af an acid is called its basicity.

 For example :
HCl   ---------> H⁺ + Cl^-
Basicity of HCl is 1.

H₂SO₄ ------> 2H⁺ + SO₄^2-
Basicity of  H₂SO₄ is 2.

Based on Basicity acids were classified into different types :
1. Mono-basic acids
2. Di-basic acids
3. Tri-basic acids

1. Mono-basic acids :

Acids, which on ionisation produce one hydronium ion on reaction with water.
Acids, which on ionisation produce one hydrogen ion.
For example : HCl, HNO₃ etc.

2. Di-basic acids :

Acids, which on ionisation produce two hydronium ion on reaction with water
Acids, which on ionisation produce two hydrogen ion

For example : H₂SO_4, H₂CO₃ etc.

3. Tri-basic acids :

Acids, which on ionisation produce three hydronium ion on reaction with water
Acids, which on ionisation produce three hydrogen ion

For example : H₃PO₄, H₃PO₃ etc.
 
b. Neutralization reaction :

A neutralization reaction is a reaction in which an acid and a base reacts to form water and a salt. It involves the combination of H⁺ ions and OH^- ions to generate water.
The neutralization of a strong acid and strong base has a pH equal to 7.
The neutralization of a strong acid and weak base will have a pH of less than 7.
The neutralization of a strong base neutralizes a weak acid will be greater than 7.

When a solution is neutralized, it means that salts are formed from equal weights of acid and base.

Neutralization reaction has application in daily life:

1)Self defence by animals and plants through chemical warfare :

Bee stings are acidic in nature, household remedy for a bee sting is baking soda or sodium bicarbonate, which is a basic substance.
A wasp stings are mildly basic, household remedy for this will be vinegar, also known as acetic acid.
These simple treatments ease these painful stings by a process called neutralization.

2)pH in our digestive system :
An acidic stomach due to eating too much spicy food, can be relieved by taking an antacid. The antacid is alkaline/basic in nature and helps to neutralize the stomach's acidity or you may take magnesium hydroxide(Milk of magnesia) and sodium hydrogen carbonate(Baking soda).

3) pH change as the cause of tooth decay :
When we eat food containing sugar, then bacteria present in our mouth break down the sugar to form acids(such as lactic acid). Thus acid is formed in the mouth after digestion. This will lead to the cause of tooth decay. The best way to prevent tooth decay is to clean the mouth after eation food with toothpaste, which is basic in nature. This will result in neutralization of acid by base.
 
4)soil pH and plant growth :
Most of the plants grow best when the pH of the soil is close to 7 that's neural. If the soil is too acidic or too basic(alkaline), the plants grow badly.
The acidic soil is neautralize by treatment with materials like quicklime(calcium oxide) or slaked lime(calcium hydroxide) or chalk(calcium carbonate).
If the soil is too basic, then alkalinity can be reduced by adding decaying organic matter(manure or composite)which contains acidic materials.

c.Electrolysis of water :

Electrolysis of water is the decompositon into oxygen and hydrogen gas due to an electric current passed through the water.

The following equation represents the electrolysis of water :H2O(l)

In pure water, at the negatively charged cathode, a reduction reaction takes place, with electrons (e⁻) from the cathode being given to hydrogen cations to form hydrogen gas.

Reduction at cathode: 2 H⁺ + 2e⁻ → H₂

On positively charged anode, an oxidation reaction occurs, generating oxygen gas by giving electrons to the anode :

Oxidation at anode: 2 H₂O → O₂(g) + 4 H⁺(aq) + 4e⁻

The same half reactions can also be balanced with base as listed below. To add half reactions they must be balanced with either acid or base.

Combining either half reaction pair yields the same overall decomposition of water into oxygen and hydrogen:

Overall reaction: 2 H₂O(l) → 2 H₂(g) + O₂(g)

The number of hydrogen molecules produced is thus twice the number of oxygen molecules.  The produced hydrogen gas has therefore twice the volume of the produced oxygen gas. The number of electrons pushed through the water is twice the number of generated hydrogen molecules and four times the number of generated oxygen molecules.

2H2O(l)⟶2H2(g)+O2(g)

Answer:

a)When an acids dissolve in water. The H⁺ ion from acid always goes to the nearest water molecule to form hydronium ion.
HCl(aq)+H₂O(aq)→H3O⁺(aq)+Cl-(aq)
For example : When Hydrochloric acid(HCl) is a strong acid. When it is dissolved in water, HCl is ionized completely in aqueous solution.  Hydrochloric acid will donates its proton H⁺ to water molecule to form a hydronium cation H₃O⁺and chloride anions Cl^- .

b)Buttermilk spoils if kept in a copper or brass container because buttermilk is actually lactic acid. Lactic acid reacts with the container material and produces poisonous complex. It is actually the reaction between acid and metal . This reaction is called as electro chemical reaction.

Answer:

$$\begin{gathered} a. \mathrm{HCl}\left(\mathrm{l}\right) + \mathrm{NaOH}\left(\mathrm{l}\right) \to \mathrm{NaCl}\left(\mathrm{s}\right) + {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) + \mathrm{Energy} \\ \mathrm{b}. \mathrm{Zn}\left(\mathrm{s}\right)+ {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{ZnSO}}_4\left(\mathrm{s}\right) +{\mathrm{H}}_{2 }\left(\mathrm{g}\right) \\ \mathrm{c}. \mathrm{CaO}\left(\mathrm{s}\right) + 2{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Ca}({\mathrm{NO}}_3{)}_2 (\mathrm{s})+ {\mathrm{H}}_2\mathrm{O}(\mathrm{l}) \\ \mathrm{d}. \mathrm{KOH}(\mathrm{l}) + {\mathrm{CO}}_2(\mathrm{g})\to {\mathrm{KHCO}}_{3 } \\ \mathrm{e}. {\mathrm{NaHCO}}_3(\mathrm{s})+ 2\mathrm{HCl}(\mathrm{l})\to \mathrm{NaCl}(\mathrm{s}) + {\mathrm{H}}_2\mathrm{O} (\mathrm{l})+ {\mathrm{CO}}_2 \end{gathered}$$

Answer:

Answer:

Ans7.