Protic and non- Protic acids || Hydrolysis of salt || Different kind of salts || Buffer solution || Solubility product ||

Class 11th Chemistry Online Class: Elevate Your CBSE Board Success

About Lesson

Protic and non- Protic acids

Protic acids: Are those which have Proton.

Example: HClàH⁺ + Cl^–

Non-Protic acid: A acid which has no Proton.

Example: carbon dioxide

Protic solvent: A solvent with proton like water.
Non- Protic solvent: A solvent with no proton like carbon tetrachloride etc.
Di-Protic acid: Which has two protons like H₂SO_4?
Tri-Protic acid: Which has three protons like phosphoric acid?

The method of calculating their K_a value is given below:

Let see,

H₂SO₄ àH⁺ + SO₄^2-, let’s say for first reaction K=K_a1

HSO₄^–àH⁺ + SO₄^2-, let’s say for this K=K_a2

Please note K_a1 is always greater than K_a2 because as acid keeps on loosing proton ,its acidic strength keeps on decreasing .

Then for the complete reaction K_a =K_a1 x K_a2

Similarly, for bases we have: K_b = K_b1 x K_b2

Hydrolysis of salt

It is the “interaction of anion and cation of salt with water to produce an acidic or basic solution”.

That is : Salt + water –>acid + base

BA + H₂OàHA + BOH

We can also say: “Hydrolysis as reverse of neutralization reaction”.

If salt has cation and anion, it reacts with water and result in acidic or basic solution (depending upon which is strong acid or base).

Like strong acid + weak base à acidic solution

Suppose we have salt of strong acid and weak base:

NH₄Cl +H₂OàNH₄OH + HCl

The resulting solution is acidic because NH₄OH is a weak base and HCl is strong acid.

On the other hand, if we have salt of strong base and weak acid à basic solution

For example: CH₃COONa +H₂Oà NaOH +CH₃COONa

The resulting solution is basic solution as NaOH is strong base. Like, we did K_{a ,}K_b,K_{w .}Similarly, we have K_h that is hydrolysis constant and degree of hydrolysis as ‘h’.

Different kind of salts

Salt of weak acid and strong base

Buffer solution

They are those solutions which has a constant value of pH , no matter what we add in to them .

Example: The pH of Blood is 7.4-7.6. Blood act as a buffer. As u all know, in spite of eating so many types of food items, the pH of blood remains same.

The buffer solution is defined as: A solution which resists the change in hydrogen ion concentration on addition of a small amount of acid or a base in to it.

Buffer action: It is the ability to resist the change in pH on addition of acid or base.

There are two types of buffers

Acidic buffer
Basic buffer

Acidic buffer: It contains equimolar concentration of weak acid and its salt with a strong base. For example:

We take any weak acid say, acetic acid. To it, if we add equimolar concentration of sodium acetate, then the concentration of acetate ion increases. This acetate ion will neutralize the effect of hydrogen ion and the pH remains same.

Basic buffer: It is equimolar concentration of solution of weak base and its salt with a strong acid.

Let’s say, ammonium hydroxide dissociates to ammonium ion and hydroxide ion. It has ammonium chloride which also gives ammonium ion. So, this ammonium ion combines with hydroxide ion and neutralizes its effect. Therefore, pH remains same.

A equation related to it was given by Henderson equation:

Solubility product

It is the amount of moles of solute dissolved in given amount of solvent.

Solubility of salt depends upon

Lattice enthalpy
Hydration enthalpy

If lattice enthalpy is low and hydration enthalpy is high, than the salt easily dissolves.

2.To predict the precipitate of salt

For example :

AgClàAg⁺ +Cl^–

K=[Ag⁺][Cl^–]/[AgCl] (equation 1 )

Or, K[AgCl]= [Ag⁺][Cl^–]

Or, K_sp=[Ag⁺][Cl^–] (equation 2 )

If we compare equation 1 and equation 2 ,we get :

If value of K>K_sp, then the precipitation occurs .
If value of K<K_sp, then the precipitation doesn’t occurs.

Numerical problems:

Problem 1:-

What is the K_c for the following equilibrium when the equilibrium concentration of each substance is :

2SO₂ + O₂ <–> 2SO₃

Given [SO₂]=0.60M,[O₂]=0.82M and [SO₃]=1.90M?

Answer 1: The K_c =[SO₃]²/[SO₂]²[O₂]

Substituting the values, we get :

K_c=(1.90)²/(0.60)²(0.82)=12.229 M^-1

Problem 2:-

At certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of iodine atoms :

I₂ <–>2I .Calculate K_p for equilibrium?

Answer 2: Total pressure of the mixture at equilibrium =10⁵ Pa

Partial pressure of iodine atoms = 40/100 x 10⁵ Pa

= 0.4 x 10⁵ Pa

Partial pressure of I₂ = 60/100 x 10⁵ Pa

= 0.60 x 10⁵ Pa

K_p=(pI)²/pI₂

= (0.4 x 10⁵)²/0.60 x 10⁵=2.67 x 10⁴

Problem 3:

Find out the value of K_c for each of the following from the value of K_p:

2NOCl <–> 2NO + Cl₂ ,K_p= 1.8 x 10^-2 at 500k

CaCO_{3 <–>}CaO + CO₂ ,K _p = 167 at 1073 K

Answer 3:- K_p and K_c are related to each other by the relation :

K_p=K_cRT^{change In gaseous moles}

Change in gaseous moles =3-2=1 ,K_p=k_c(RT)¹

K_c= 1.8 x 10^-2/0.0831 x 500=4.33 x 10^-4

Change in gaseous moles = 1-0=1 , K_c=K_p/RT=167/0.0831 x 1073 =1.87

Problem 4:-

Reaction between N₂ and O₂ takes place as follows :

2N₂ + O₂ <–> 2N₂O.

If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c= 2.0 x 10 ^-37,determine the composition of equilibrium mixture?

Answer 4:- let x moles of nitrogen gas take part in the reaction .Then according to the equation ,x/2moles of O₂ will react giving x moles of N₂O.

2N₂ + O₂ <–> 2 N₂O

Initial 0.482 mol 0.933mol 0

At equilibrium: 0.482-x 0.933-x/2 x/10

Molar concentration: 0.482-x/10 0.933-x/2/10 x/10

As K_c =2.0 x 10 ^-37 is very small ,this shows that the amount of nitrogen gas and oxygen gas reaction is to a very small extent and it can be neglected .Hence ,at equilibrium ,we have :

[N₂]=0.0482 mol/L

[O₂]=0.0933 mol/L

[N₂O]=0.1x

Substituting the values, we get :

K_c= (0.1x)²/(0.0482)²(0.0933)=2.0 x 10^-37

x= 6.6 x 10 ^-20

On solving :

N₂O=0.1x=6.6 x 10^-21 mol/L

[N₂]=0.0482 mol/L

[O₂]=0.0933 mol/L