2. Acids, bases and salts

Acids, Bases, and Salts

1. You 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?

Answer

We make three pieces of the litmus paper. Now, dip each piece one by one into into each test tube.

The solution of the tube who changed the colour of litmus paper is basic.

Now, pour the identified basic solution into othe two test tube. The solution of the test tube who changed colour is identified as acid.

The solution of the test tube did not change colour is identified as distilled water.

Understanding the chemical properties of acids and base

1. Why should curd and sour substances not be kept in brass and copper vessels?

Answer
We know that curd and sour substances are acids whereas and brass and copper vessels are metals.
Metals react with acids and displaces hydrogen atoms as hydrogen gas.
Therefore, curd and sour substances should not be kept in brass and copper vessels.
2. Which 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?
Answer
When an acid reacts with a metal hydrogen gas is usually liberated.

Take about 5 mL of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
Pass the gas being evolved through the soap solution.
Bubbles are formed in the soap solution.
Take a burning candle near a gas filled bubble.
The gas is started to burn by a pop sound and tested as hydrogen gas.
3. 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.
Answer
The suitable balanced chemical equation for the reaction is as follow.
CaCO₃ + 2HCl → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

1. Why do HCl, HNO3, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?

Answer

The dissociation of HCL or HNO₃ to form hydrogen ions always occurs in the presence of water. Hydrogen ions (H⁺) combine with H₂O to form hydronium ions (H₃O⁺)

The reaction is as follows

HCL + water - H⁺ + Cl⁻

H⁺ + H₂O - H₃O⁺

Although aqueous solutions of glucose and alcohol contain hydrogen ions. Hence, they do not show acidic charactor.

2. Why does an aqueous solution of an acid conduct electricity?

Answer

Acids dissociate in aqueous solution to form ions. These ions are responsible for conduction of electricity.

3. Why does dry HCl gas not change the colour of the dry litmus paper?

Answer

Colour of the litmus paper is changed by the hydrogen ions. Dry HCL gas does not contain H⁺ ions. It is only in the aqueous solution that an acid dissociates to give ions. Since in this case, neither HCL is in the aqueous nor the litmus paper is wet.

4. While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?

Answer

The process of dissolving an acid or a base in water is a high exothermic one. Care must be taken while mixing concentrated nitric acid or sulpheric acid with water. The acid must always be added slowly to water with constant stirring. If water is added to concentrated acid, the heat generated may cause the mixture to splash out and cause burns. The glass container may also break due to axcessive local heating.

5. How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted?

Answer

When an acid is diluted. The concentration of hydronium ions (H₃O⁺) per unit volume decreases. This means that the strength of the acid decreases.

6. How is the concentration of hydroxide ions (OH–) affected when excess base is dissolved in a solution of sodium hydroxide?

Answer

The concentration of hydroxide ions (OH⁻) would increase when excess base is dissolved in a solution of sodium hydroxide.

12. Electricity

Learning points

Electric current and circuit

Electric Potential and Potential Difference

Circuit Diagram

Ohm's Law

Factors on which the resistance of a conductor depends

Resistance of a system of resistors

Resistors in series

Resistors in parallel

Heating effect of electric current

Practical Applications of heating effect of electric current

Electric Power

Questions Answers

NCERT Textbook.

1. What does an electric circuit mean?

Answer

An electric circuit is pathway in which current can follow. A circuit has electric devices, switching devices, source of energy etcetera. These are connected by conducting wire.

2. Define the unit of current.

Answer

When 1C of charge flows through a conductor in 1 s, it is called 1 ampere(A) current.

I = Q/t

The unit of electric current is ampere (A).

3. Calculate the number of electrons constituting one coulomb of charge.

Answer

We know that one electron possesses 1.6×10–¹⁹ C.

Therefore, 

Number of electron  = Total charge/Charge on 1 electron

or, No. of electron = 1/1.6×10–¹⁹

or, No. of electron = 6.25×10¹⁸ electrons. Ans.

1. Name a device that helps to maintain a potential difference across a conductor.

Answer

A cell, battery, power supply, etc helps to maintain potential difference across a conductor. 

2. What is meant by saying that the potential difference between two points is 1 V?

Answer 

When 1 J of work is required to move a charge of 1 C from one point to another, then it is said that the potential difference between two points is 1 V.

V = W/Q

or, 1 V = 1J/1C

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer 

From question it is clear that 

Work done = ?

Potential difference = 6 V

Charge = 1C.

We know that 

Potential difference = Work done/Charge

Therefore, 

Work done = Potential difference × Charge

or, Work done = 6 V×1C = 6 Joules Ans.

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be

(a) 1:2 

(b) 2:1

(c) 1:4

(d) 4:1

Answer

According to combination of resistance we know that

In the case of series

R(ES) = R1 + R2

From question it is clear that

R1 = R2 = R

R(S) = 2R

Similarly,

In the case of parallel

1/R(P) = 1/R1 + 1/R2

1/R(P) = 1/R + 1/R

R(P) = R/2

For same potential difference V,

We know that

V = IR

I = V/R

I(S) = V/R(S) = V/2R

Similarly,

I(P) = V/R(P) = 2V/R

H = I^2RT

Here,

I = Current follow through the conductor

R = Resistance

T = Time

Therefore,

H(S) = (V/2R)^2.R.T = V^2T/4R  ------- (i)

Similarly,

H(P) = (2V/R)^2 . R.T = 4V^2T/R ------- (ii)

Now, from (i) and (ii)

Therefore, ratio of heat produced in series and parallel

H(S)/H(P) = 1:4 Ans.

Continues 

Answers of the rest questions.....