KPK Board 10th class Chemistry Ch 2 Acid Base short questions answers

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When a clear liquid is placed in a beaker, how can you identify it as an acid, base or neutral (water)?

The nature of a substance can be checked by 1) Litmus paper 2) pH – paper.
1) Litmus paper
The liquid turn blue litmus red then it is acidic. If the liquid turn red litmus blue, it will be basic. If the liquid do not change the colour of red and blue litmus papers it will be neutral.
2) pH – paper
Find out the pH of the liquid with the help of pH meter. If its pH – value is greater then 7 it is basic.
If pH = 7 then it is neutral.
If pH < 7 then it is acidic.
pH paper: If pH paper turn _________ acidic
If green or blue __________________ 

Justify H^+ ion as Lewis acid?

Ans. According to Lewis concept a specie which can accept a lone pair of electrons is called an acid. It means that Lewis acid is an electron deficient specie. As H^+ ion has no electrons in its in its ist shell and completion of its ist shell by two electrons so it is an acid according to Lewis concept.
H3 N(°°) + H(+ ) H3 N H
(Base) (Acid)

Distinguish strong acids from weak acids. Give two examples of each.

Ans. Difference between strong acids and weak acids
Strong Acids: Strong acids are those which ionizer up to maximum extent in their aqueous solutions while weak acid are those which ionizes up to limited extents in their aqueous solutions.
Examples of strong acids:
Perchloric acid: HClO4
Hydro acid: HI
Weak Acids:
Acetic acid: CH3 COOH
Carbonic acid [H2 CO3]

Compare the physical properties of acids and bases.

Com parision between the physical properties of acids and bases is as under.

Acids Bases
Acids have sour taste . Bases have bitter taste.
Turn blue litmus paper red. Turn blue litmus paper blue.
It changes the colour of acid – base inclinator. It changes the colour of acid – base inclinator.
Aqueous solutions of acids conduct electricity Aqueous solutions of bases conduct electricity
Strong acids are corrosive to skin and fabrics. Strong acids are corrosive to skin.

A carbonated drink has (H+) = 3.2 ×10^(-3)^(-3). Classify the drink as neutral, acid or basic with reason.

For classification of a substance as acidic basic or neutral, we should know it pH or pOH value. If
pH = 7 The substance will be neutral.
pH < 7 The substance will be acidic.
pH > 7 The substance will be basic
We know that pH = – log [H^+]
Given [H^+] = 3.2 ×10^(-3)
Putting the value;
Ph = – log 3.2 ×10^(-3)
pH = – log 3.2 + ( – log 10^(-3))
pH = – 0.505 – ( log 10^(-3))
pH = 0.505 – ( – 3 log 10 )
pH = – 0.505 + × 1 As log 10 = 1
Ph = -0.505 + 3 = [2.495 = 2.5 ]
As ph value is less than 7 so a carbonated drink is acidic in nature.

Write the chemical name of an acid present in the following?

a) Apple : Maliac acid
b) Grape juice : Citric acid
c) Lemon juice : Citric acid
d) Sour milk : Lactic acid

What determine the strength of a base? Give one example of each solution that are strongly and weakly basic.

Hydroxide ion (OH^-) donation power of a substance in an aqueous solution determine the strength of a base. When a base provide maximum amount / number of hydroxide ions in aqueous solution it will be called a strong base, otherwise weak. Similarly the strength of a basic solution is directly related to the concentration of hydroxide ions in its aqueous solution. Higher the concentration of hydroxide ions in an aqueous solution stronger will be the base and Lower. The concentration of hydroxide ions weaker will be the base.
1) Na OH, aqueous solution is strongly basic.
2) NH4 OH, …………………………………………………………..weakly basic.
Calculate the pH and pOH of 0.5 solution of HCl. (pH = 0.301). (pOH = 13.69).

Solution: As HCl is a strong acid, so it ionize almost completely. The [ of acid.
H cl
0.5m 0.5m

As we know that;
pH + pOH = 14
0.301 + pOH = 14
pOH = 14 – 0.301 => [13.69]
Q.No.9. Calculate the pOH 0.005M [ pOh = 11.699]
Solution: As
0.005m 0.005m
Using the formula
pH = -log [H+]
pH = – log 5
pH = – log 5 + (- log )
pH = – log 5 – log
pH = – 0.698 – ( – 3 log 10 )
pH = -0.698 + 3 1 [ log 10 = 1 ]
pH = – 0.698 + 3
[pH = 2.302]
As we know that
pH + pOH = 14
pOH = 14 – pH
pOH = 14 – 2.302
[pOH = 11.698]

Calculate the pH of 0.2M solution of Na OH. ( pH = 13.30)

Solution: Na OH
0.2M 0.2M

pOH= log [OH-]
pOH = – log 2
pOH = – log 2 + ( – log )
pOH = – 0.301 – log

pOH = – 0.301 – ( – log 10)
pOH = – 0.301 + 1
[pOH = 0.698]

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