Download Copy of Acids, bases, salts answer key

STD : X
Date : 26-06-16
SCI & MATHS
Hirav Trivedi’s Group TuiTions
Time : 1.5 hrs.
Total Marks : 40
Q – 1. Answer the following questions in brief: (Each of one mark)
[05]
1. Mention names of four methods of expressing concentration of a solution.
Ans. : The concentration of a solution is expressed in terms of normality, molality,
formality, ppm, percentage proportion, etc.
2. Mention names of two strong acids and two strong bases.
Ans. : The two strong acids are hydrochloric acid and Sulphuric acid. Two strong bases
are Sodium Hydroxide and Potassium Hydroxide.
3. Give definitions:
(a) Bronsted – Lowry acid : The substance which donates a proton [H+ ion] to other
substance is called Bronsted-Lowry acid.
(b) pOH of solution: The negative logarithm to the base 10 of concentration of OH- in
aqueous solution is called pOH.
(c) 1 M concentration: If 1 mole solute is dissolved in 1 litre solution, then the
concentration of solution is 1 molar or molarity (1 M).
Q – 2. Mention the formula, name and physical state of the products of the following
reactions:
[06]
1. H2SO4(aq)
+ 2KOH(aq)

K2SO4(aq)
+
2H2O(l)
Sulphuric Potassium
Potassium
Water
acid
Hydroxide
Sulphate
2. 2HCl(aq)
+ Na2CO3(aq)  2NaCl(aq) + H2O(l)
+
CO2(g)
Hydrochloric
Acid
3. NaOH(aq)
+
Sodium
Hydroxide
4. HNO3(aq)
+
Nitric acid
5. Na2SO4(aq)
Sodium
Sulphate
6. AgNO3(aq)
Silver
Nitrate
+
+
Sodium
Carbonate
HCl(aq)

Hydrochloric
Acid
NH4OH(aq)
Ammonium
Hydroxide
Ba(OH)2(aq) 
Sodium
Water
Carbon Dioxide
Chloride
NaCl(aq)
+ H2O(l)
Sodium
Water
Chloride
 NH4NO3(aq)
+ H2O(l)
Ammonium
Water
Nitrate
BaSO4(s)
+
2NaOH(aq)
Barium
hydroxide
Barium
Sulphate
NaCl(aq) 
Sodium
Chloride
AgCl(s) +
Silver
Chloride
Sodium
Hydroxide
NaNO3(aq)
Sodium
Nitrate
2nd Floor, Rajcastle, Near Kalpataru, Ellora Park, Baroda – 390007 Ph: +91-265-3053166
1
STD : X
Date : 26-06-16
SCI & MATHS
Hirav Trivedi’s Group TuiTions
Time : 1.5 hrs.
Total Marks : 40
Q – 3. Answer the following: (Each of Two marks)
[08]
1. Write two chemical properties of salt.
Ans : Reaction of salt with acid :
Salt reacts with certain acids and forms other salt and acid.
Salt
+
Acid

Other salt
+
a)
BaCl2(aq)
+
H2SO4(aq)

BaSO4(s)
+
b)
AgNO3(aq)
+
HCl(aq)

AgCl(s)
+
 Reaction of salt with base :
Salt reacts with certain bases and forms other salt and base :
Salt
+
Base

Other salt
+
a)
Na2SO4(aq)
+
Ba(OH)2(aq) 
BaSO4(s)
+
b)
K2SO4(aq)
+
Ca(OH)2(aq) 
CaSO4(s)
+
 Reaction of salt with other salt :
Salt reacts with other salt to form new salts :
Salt
+
Other salt

New salt
+
a)
AgNO3(aq)
+
NaCl(aq)

AgCl(s)
b)
CaCl2(aq)
+
Na2SO4(aq)

CaSO4(s)
Acid
2HCl(aq)
HNO3(aq)
Base
2NaOH(aq)
2KOH(aq)
New salt
+
NaNO3(aq)
+
2NaCl(aq)
2.
Deduce pH + pOH = 14.
Ans. :
+
 We express the concentration of hydrogen [H ] ions in aqueous solution on a pH scale.
 Similarly, the concentration of hydroxide [OH] in aqueous solution can also be expressed.
 For representing hydroxide, the abbreviation ‘pOH’ is used.
Mathematical form of ‘pOH’ :
 The negative logarithm to the base 10 of concentration of hydroxide [OH] ions in aqueous
solution is called pOH.
 Thus, pOH = log10[OH].
Derivation of pH + pOH = 14 :
[H3O+] = [OH] = 1 × 107M
 ([H3O+] × [OH]) = 107 × 107 = 1014 M
 (log10[H3O+] + log10[OH] = 14 log1010 ( Taking log on both the sides)
 (log10[H3O+] log10[OH]) = 14 log1010 (Multiplying 1 on both the sides)]
But, log10[H3O+] = pH and log10[OH] = pOH and log1010 = 1
 pH + pOH = 14
+
 Note that in acidic aqueous solution, as the concentration [H3O ] increases, the solution
becomes more and more acidic.
 Even if the solution becomes highly acidic, it will still have some ions of hydroxide [OH].
 Similarly, for a basic aqueous solution, even if the solution is highly basic, it will still have
some negative ions of hydronium [H3O+] ions in it.
+
 But under any of these circumstances, the product of concentrations i.e. [H3O ] × [OH] =
4
10 will always remain constant in a solution.
+
 Thus, [H3O ] × [OH] = 14 = constant
2nd Floor, Rajcastle, Near Kalpataru, Ellora Park, Baroda – 390007 Ph: +91-265-3053166
2
STD : X
Date : 26-06-16
SCI & MATHS
Hirav Trivedi’s Group TuiTions
Time : 1.5 hrs.
Total Marks : 40
3. Write chemical equations of four neutralization reactions.
Ans. : Reaction of acid and base gives salt and water. This reaction is called neutralization
reaction.
Acid
+
Base

Salt
+
Water
1) HCl(aq)
+
NaOH(aq)

NaCl(aq)
+
H2O(l)
Hydrochloric acid
Sodium hydroxide
Sodium chloride
Water
2) H2SO4(aq)
+
2KOH(aq)

K2SO4(aq)
+
2H2O(l)
Sulphuric acid
Potassium hydroxide Potassium sulphate
Water
3) 2HNO3(aq)
+
Ca(OH)2(aq) 
Ca(NO3)2(aq) +
2H2O(l)
Nitric acid
Cacium hydroxide
Calcium nitrate(salt)
4) H2CO3(aq)
+
2NaOH(aq)

Na2CO3(aq)
+
2H2O(l)
Hydrogen carbonate Sodium hydroxide
Sodium carbonate
5) HCl(aq)
+
NH4OH(aq)

NH4Cl(aq)
+
H2O(l)
Hydrochloric acid Ammonium hydroxide Ammonium Chloride (Salt)
 When base reacts with acid, salt and water are produced.
 Since, base neutralizes the effect of acid, this reaction is called a neutralization reaction.
Base
+
Acid

Salt
+
Water
1) NaOH(aq)
+
HCl(aq)

NaCl(aq)
+
H2O(l)
Sodium hydroxide
Hydrochloric acid
Sodium chloride
2) 2KOH(aq)
+
H2SO4(aq)

K2SO4(aq)
+
2H2O(l)
Potassium hydroxide
Sulphuric acid
Potassium sulphate
3) Ca(OH)2(aq)
+
2HNO3(aq)

Ca(NO3)2(aq) +
2H2O(l)
Calcium hydroxide
Nitric acid
Cacium Nitrate
4) 2NaOH(aq)
+
H2CO3(aq)

Na2CO3(aq)
+
2H2O(l)
Soidum hydroxide
Hydrogen carbonate
Sodium carbonate
5) NH4OH(aq)
+
HCl(aq)

NH4Cl(aq)
+
H2O(l)
Ammonium hydroxide
Hydrochloric acid
Ammonium chloride



4. Explain giving example, Arrhenius acid-base theory. Mention the limitation of this
theory.
Ans. :
A Swedish scientist Svante Arrhenius clarified the concept of acid and base. According to him,
acid is a substance, which contains hydrogen and produces hydrogen [H+] ion in its aqueous
solution. Whereas base is a substance containing hydroxide [OH] ion in its aqueous solution.
Thus, as per Arrhenius acidbase theory, acid ionizes in water and produces [H+] and base ionizes
in water and produces [OH] ions.
The concept of ionization is involved on the basis of this theory.

 +
H 2O
For Acids :
HA(aq) 
H (aq
A (aq
)

)
When disolved in water
(Where A represents a negative ion)
Example :

HNO3(l)

H (aq
)
H O
2
+
NO 3( aq )
Nitric acid
Hydrogen ion
Nitrate ion
Here when nitric acid dissolved in water, it produces hydrogen [H+] ions and nitrate [NO3] ion
H O


HCl(l)
H+(aq)
+
Cl (aq
)
Hydrochloric acid
Hydrogen ion
Chlorine ion

2
2nd Floor, Rajcastle, Near Kalpataru, Ellora Park, Baroda – 390007 Ph: +91-265-3053166
3
STD : X
For bases :
Date : 26-06-16
MOH(s)
Hirav Trivedi’s Group TuiTions
Time : 1.5 hrs.
Total Marks : 40
SCI & MATHS
H O
2
M+(aq)
+

OH (aq
)
(Where M+ represents a metal ion or ammonium ion NH4+)
H O

NaOH(s)
Na+(aq)
+
OH (aq

)
Sodium metal ion
Hydroxide ion
 Here, sodium hydroxide, a base when dissolved in water, gives out positive sodium metal ion and
hydroxide ion.
H O

KOH(s)
K
+
OH (aq

)
(aq )
Potassium hydroxide
Potassium metal ion
Hydroxide ion
Limitations of Arrhenius theory :
Arhhenius’ theory became quite popular and was widely accepted yet it had the following limitations:
 This theory was applicable only to aqueous solutions.
 Substances like Ammonia (NH3) do not contain hydroxide (OH) ion, even then its aqueous
solution acts as a base. The theory could not explain the reason for this.
 Although convenient to use in equations, the symbol [H+] does not really represent the structure of
ion present in the aqueous solution.
o H+ is highly unstable because it does not contain electron and so the individual independent
existence of H+ does not exist. Rather H+ attaches to water molecule i.e. solvent to give out a
more stable ion called the hydronium ion [H36O+]
o Thus,
H+
+
H2O 
H3O+
Hydrogen ion
Hydronium ion
2
2
Q – 4. Answer the Following: (Each of three marks)
[09]
1. Discuss methods to measure pH of aqueous solution.
 Ans. : There are three ways to measure the pH of an aqueous solution. They are :
1) To obtain approximate range of pH :
Through litmus paper :
o In this method, approximate range in which the pH value may lie is obtained.
o Red and blue litmus papers are used to measure the approximate pH range of aqueous solution.
Blue litmus paper :
o If a blue litmus paper is dipped in a solution and it turns red, then it can be said that the solution is
acidic.
o In this case the approximate pH range of the solution will be 0  6.
o If the blue litmus paper does not change colour, it means the solution is neutral and so its pH value
will be 7.
Red litmus paper :
o If a red litmus paper is dipped in a solution and it turns blue, then it can be said that the solution is
basic.
o In this case, the approximate pH range of the solution will be 8  14.
o If the red litmus paper does not change colour, then it means the solution is neutral and so its pH
value will be 7.
2) To obtain approximate value of pH :
o In this method, approximate value of pH is obtained. This method can be considered one step
ahead of measuring the range of pH since this method gives approximate value.
A.Through pH paper :
o pH paper is a very light yellow coloured paper, which changes colour when dipped in a solution.
o If on dipping, the pH paper turns pink, it means the approximate pH value of the solution is 2 and
the solution is acidic.
o If the pH paper turns light green, it means the approximate pH value of the solution is 7 and it is
neutral.
2nd Floor, Rajcastle, Near Kalpataru, Ellora Park, Baroda – 390007 Ph: +91-265-3053166
4
STD : X
Date : 26-06-16
SCI & MATHS
Hirav Trivedi’s Group TuiTions
Time : 1.5 hrs.
Total Marks : 40
o If the pH paper turns blue, it means the approximate pH value of the solution is 10 and it is basic.
B. Through universal indicator :
o Universal indicator is a mixture of many different
indicators or say dyes which when added to a solution,
change in the colour of the solution and thus indicate
its pH value.
o The colour that the solution attains on adding
universal indicator in it is compared with the
colours shown on the label sticked on the bottle of
the indicator and thus pH of that solution is determined.
o Instead of liquid universal indicator, indicator paper is also
used in chemical reactions and biochemical reactions in the industry.
o A drop of solution is put on the universal indicator paper and the
change in colour that occurs on the paper is compared with a special
chart.
o For example, if the universal indicator paper turns
dark red, then it means pH value of the solution
will be 0 and the solution will be a strong acid.
3) To obtain exact value of pH :
A.Through pH meter :
o For calculating exact value of pH, an instrument called pH meter is used.
o Two electrodes of the pH meter are dipped in the solution whose exact pH is to be determined.
o The pH meter contains a needle which will show the exact pH value of the solution.
o To avoid variation in results, the pH meter is standardized by a solution of known pH before it is
used. This results in the exact measurement by the pH meter.
o pH meter is an essential instrument in chemical industries.
2. With reference to pH scale.
(i) Write formula of pH and pOH
(ii)Mention pH and concentration of H3O+ or OH- in acidic, basic and neutral
aqueous solutions.
(iii) Mention limitations of pH scale.
Ans. : pH = log10[H3O+] and pOH =  log10[OH]
(i)
Type of solution
Acidic aqueous solution
pH
<7
Concentration
[H3O+] > 107 M
Neutral aqueous solution
=7
[H3O+] = [OH] = 107 M
Basic aqueous solution
>7
[OH] > 107 M
(ii) Limitations :
 pH scale is applicable to aqueous solutions only.
+
 pH scale is applicable to aqueous solutions having concentration of hydronium [H3O ] ion less than
0
10 i.e. 1 M. As a result, pH values lie between 0 and 14.
2nd Floor, Rajcastle, Near Kalpataru, Ellora Park, Baroda – 390007 Ph: +91-265-3053166
5
STD : X
Date : 26-06-16
SCI & MATHS
Hirav Trivedi’s Group TuiTions
Time : 1.5 hrs.
Total Marks : 40
3. “The aqueous solution of the salt produced by neutralization of weak acid and
strong base possesses basic nature, while aqueous solution of salt produced by
neutralization of weak base and strong acid posses acidic nature” Explain.
Ans. :
(1) H2CO3(aq) +
2NaOH(aq)  Na2CO3(aq) + 2H2O(l)
Weak acid
Strong Base
Salt
water
(2) CH3COOH(aq) + NaOH(aq)
 CH3COONa(aq) + H2O(l)
Weak acid
Strong Base
salt
Water
Thus, the pH of aqueous solutions of salts Na2CO3 and CH3COOH produced by the
reactions (1) and (2) is more than 7, because the salts hydrolyse in water and produces OH- ion.
Thus, the aqueous solution of the salts produced by this reaction is basic in nature. E.g.


(1) Na2CO3(aq) ionisation
2Na+(aq) + CO3(2aq)

 HCO 3( aq) OH (aq)
CO3(2aq) + H2O(l) hydrolysis



 CH3COO (aq
(2) CH3COONa(aq) ionisation
) + Na (aq )
hydrolysis


 CH3COOH(aq) + OH (aq
CH3COO (aq
) + H2O(l)  
)
 (1) HCl(aq)
+ NH4OH(aq)  NH4Cl(aq) + H2O(l)
Strong acid
Weack Base
Salt
water
(2) HNO3(aq) + NH4OH(aq)  NH4NO3(aq) + H2O(l)
Strong Acid
Weak base
Salt
Water
Thus, the pH of aqueous solution of salts NH4Cl and NH4NO3 produced by the above
reactions (1) and (2) have pH less than 7, because the salt hydrolysis in water and produces
H3O+.e.g.,

 NH 4(aq) + Cl(aq)
(1) NH4Cl(aq) ionisation
NH 4(aq)
+

 NH3(g) + H3O+(aq)
H2O(l) hydrolysis

 NH 4(aq) + NO3( aq)
(2) NH4NO3(aq) ionisation
NH 4(aq)
+

 NH3(g) + H3O+(aq)
H2O(l) hydrolysis
Thus, the aqueous solutions of salts produced by neutralisation of weak base and strong
acid possess acidic nature.
MATHS
1. A man is standing on the top of a building 60 m high. He observes that the angle of
depression of the top and the bottom of a tower has measure 30 and 60 respectively.
Find the height of the tower.(refer text book example 7 Page No. 214)
[04]
2. A ladder rests against a wall at an angle having measure  with the ground. Its foot is
pulled away from the wall by a m keeping ladder on the ground. By doing this, its
upper end on the wall slides down by b m. Now the ladder makes an angle of measure
a cos   cos 
 with the ground. Then prove that 
.
[04]
b sin   sin 
(refer text book example 13 Page No. 218)
3. A flag-staff of height h stands on the top of a school building. If the angles of elevation
of the top and bottom of the flag-staff have measures  and  are respectively from a
h tan 
point on the ground, prove that the height of the building is
.
[04]
tan   tan 
(refer text book example 12 Page No. 217)
2nd Floor, Rajcastle, Near Kalpataru, Ellora Park, Baroda – 390007 Ph: +91-265-3053166
6