IONIC EQUILIBRIUM

                                        Question Answer 

1. Find the pH of 1x10^-5  H₂SO₄.
> Given, normality of H₂SO₄, = 1x10.5 N
Basicity of H2SO = 2
We know that,
Molarity (M)=Normality(N)/ Basicity of acid
                       =1×10^-5/2

                       =5×10^-6 M

H₂SO₄ ionizes  as
H₂SO_{4       →}                            2H⁺  + SO₄
1 M                              2    M
5x 10^-6M        1 × 10^-5 M
So, [H⁺] = 1× 10^-5 M
We know that
pH = -log[H­⁺] =- log|1 × 10 1=5
Hence, pH of 1×10- N H₂SO₄ is 5

 

2.Write the conjugate acid and base of NH₃.
>The conjugate of acid of NH is NH₃  ion which is shown by the reaction.
NH₃      +     H₂O            →                           NH₄⁺       +          OH^-

Base           acid                      conjugate acid         conjugate base
Similarly, the conjugate base of NH, is NH₃  which is shown by the reaction.
NH₃      +     NH₃              →                         NH₄⁺         +        NH₂^-  

Base            acid                       conjugate acid       conjugate base 

3. is the solution acidic basic or neutral in which concentration of hydroxyl ion is
3.33 x ^10-6 mol L^-1?
>The concentration of OH-ion - 3.33× 10^-10 mol L^-1
pH=?
We know that
POH =-log OH = - log   3.33x 10-11 - 9.47
pH = 14 - 9.47 =4.53
The pH of aqueous solution is less than 7. So, the aqueous solution is acidic in nature.

4. Is the solution acidic, basic or neutral in which the hydrogen ion concentration in
3 x 10^-5 molL^-1?
>Given
The concentration of H⁺= 3×10^-5 mol l^-1
pH=?
We know that,
pH=-log   H⁺    =-log   3×10‑^-5   =4.52
The pH of aqueous solution is less than 7. So, the aqueous solution is acidic in nature

5. Calculate the pH of the solution by dissolving  1g of NaOH in 1 litre of its solution
> Given
The concentration of NaOH= 1 g/L
Molecular weight of NaOH = (23 + 16+1) = 40
We know that
Molarity = g/l//molecular weight

                 =1/40=0.025M
Now,
NaOH        ⇀⇁            Na ⁺     +     OH^-

0.025 M                  0.025M     0.025M

poH=-log    OH^- =-log  0.025   =1.60
We know that,
pH = 14-poH- (14 - 1.6) = 12.4
Hence, the pH of 1g/L NaOH solution is 12.4.

7.Is an aqueous solution containing hydroxyl ion concentration 3.33×10^-1 mol/l  acidic, basic or neutral?
>Given,
The concentration of [OH^-] =3.33×10^-1  mol /L
pH = ?
We know that,
pOH = -log[OH-] =-log(3.33 × 10^-1) =0.477
pH = 14 -0.477 = 13.52
The pH of aqueous solution is more than 7. So, the aqueous solution is basic in nature.

8.Define

i) lonic product of water

ii)pH of solution

>i. lonic product of water: The product of the concentration of hydronium ion and
hydroxyl ion in pure water is constant at particular temperature which is known as ionic
product of water. It is denoted by K_w.
Kw = 1.008 × 10^-14 = [H₃O⁺]  OH at 25°C.
ii. pH of solution: The magnitude of negative power to which 10 must be raised to express
the hydronium ion concentration is called pH of solution
pH =-log[H⁺]

9. State Ostwald's dilution law and mention It's limitation.
>According to the theory of electrolytic dissociation of an electrolyte, when dissolved in
water undergoes spontaneous dissociation into +vely and -vely charged ions and in the
case of a weak electrolyte like CH₃, COOH, NH₄OH, etc. there exists a definite.
equilibrium between the unionized molecules of electrolyte and the Ions present in
solution at a given temperature. It is therefore, to be expected that the law of mass
action can be applied to this equilibrium. This application was first carried out by
Ostwald and the result is known as Ostwald dilution law.
α=√KV where, V=dilution
Limitations:- Weak electrolyte obeys Ostwald's dilution law fairly well, but in the case a
strong electrolyte it fails completely. In the case of strong electrolytes, the value of K far
from remaining constant rapidly falls with dilution. This is due to the fact strong
electrolyte are completely ionized at all dilutions i.e. for strong electrolyte is equal to 1 and therefore, it is expected from the equation α=  λν/λα
that λν = λα , but in practice it is not so. It means that the equation α=  λν/λα  does not hold good for strong electrolyte. It is observed that the value of λν is lower than λα even though α=1

10.Define:
i)Degree of ionization
ii) Ostwald's dilution law

> i) Degree of ionization: It is the fraction of the total number of molecules of electrolyte
present as free ions in the solution. It is represented by

α = No. of molecules split into ions/Total no. of molecules of the electrolyte
ii. Ostwald dilution law: According to the theory of electrolytic dissociation of an
electrolyte, when dissolved in water undergoes spontaneous dissociation into +vely and
-vely charged ions and in the case of a weak electrolyte like CH₃COOH, NH₄OH, etc
there exists a definite equilibrium between the unionized molecules of electrolyte and
the ions present in solution at a given temperature It is therefore to be expected that the
law of mass action can be applied to this equilibrium. This application was first carried
out by Ostwald and the result is known as Ostwald dilution law.

11.What are the limitation of Lewis acid and base?
>Following are the limitation of Lewis acid and base:-
1. This concept cannot tell anything about the strength of acid and base.
2. This concept is unable to explain the catalytic activity and amphoteric nature of several
compounds
3. This concept does not consider about ionization process of acid and base.
4. Relative strength of acid and base are not quantitatively estimated by this concept

i) Degree of ionization: It is the fraction of the total number of molecules of electrolyte
present as free ions in the solution. It is represented by

α = No. of molecules split into ions/Total no. of molecules of the electrolyte
ii. Ostwald dilution law: According to the theory of electrolytic dissociation of an
electrolyte, when dissolved in water undergoes spontaneous dissociation into +vely and
-vely charged ions and in the case of a weak electrolyte like CH₃COOH, NH₄OH, etc
there exists a definite equilibrium between the unionized molecules of electrolyte and
the ions present in solution at a given temperature It is therefore to be expected that the
law of mass action can be applied to this equilibrium. This application was first carried
out by Ostwald and the result is known as Ostwald dilution law.
11.What are the limitation of Lewis acid and base?
>Following are the limitation of Lewis acid and base:-
1. This concept cannot tell anything about the strength of acid and base.
2. This concept is unable to explain the catalytic activity and amphoteric nature of several
compounds
3. This concept does not consider about ionization process of acid and base.
4. Relative strength of acid and base are not quantitatively estimated by this concept

12.Calculate the pH of 0.1 N H₂SO_­4.
> Given, normality of H₂SO₄ =0.1N
     Basicity of H₂SO₄ = 2
We know that,
Normality (N) = Molarity (M) × Basicity of acid
or, Molarity (M) = Normality (N)/Basicity of acid =0.1/2=0.05M
H₂SO₄  ionizes as

H₂SO₄                       →                        2H⁺   +    SO₄^--

1M  2M                             

0.05M                            0.1M

 1 mole of H₂SO₄  produces 2 mole of H⁺

0.05 mole of H₂SO₄ produces 0.1 mole of H⁺

so  H⁺    =0.1M

We know that 

pH=-log H⁺

   =-log 0.1

   =1

Hence, pH of 0.1 N  H₂SO_{4  is 1}