**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 H
_{2}SO_{4?} - Tri-Protic acid: Which has three protons like phosphoric acid?

**The method of calculating their K _{a} value is given below:**

Let see,

H_{2}SO_{4} àH^{+} + SO_{4}^{2-}, let’s say for first reaction K=K_{a1}

HSO_{4}^{–}àH^{+} + SO_{4}^{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

Or

BA + H_{2}OàHA + 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:

NH_{4}Cl +H_{2}OàNH_{4}OH + HCl

The resulting solution is acidic because NH_{4}OH 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: CH_{3}COONa +H_{2}Oà NaOH +CH_{3}COONa

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 :**

** 2SO _{2} + O_{2} <–> 2SO_{3}**

**Given [SO _{2}]=0.60M,[O_{2}]=0.82M and [SO_{3}]=1.90M?**

**Answer 1:** The K_{c} =[SO_{3}]^{2}/[SO_{2}]^{2}[O_{2}]

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 10 ^{5} Pa, iodine vapour contains 40% by volume of iodine atoms :**

I_{2 } **<–>**2I .Calculate K_{p} for equilibrium?

**Answer 2:** Total pressure of the mixture at equilibrium =10^{5} Pa

Partial pressure of iodine atoms = 40/100 x 10^{5} Pa

= 0.4 x 10^{5} Pa

Partial pressure of I_{2} = 60/100 x 10^{5} Pa

= 0.60 x 10^{5} Pa

K_{p}=(pI)^{2}/pI_{2}

= (0.4 x 10^{5})^{2}/0.60 x 10^{5}=2.67 x 10^{4}

**Problem 3:**

**Find out the value of K _{c} for each of the following from the value of K_{p}:**

** 2NOCl <–> 2NO + Cl _{2} ,K_{p}= 1.8 x 10^{-2} at 500k**

** CaCO _{3 <–> }CaO + CO_{2 } ,K _{p} = 167 at 1073 K**

**Answer 3:-** K_{p} and K_{c} are related to each other by the relation :

K_{p}=K_{c}RT^{change In gaseous moles}

- Change in gaseous moles =3-2=1 ,K
_{p}=k_{c}(RT)^{1}

or

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 N_{2} and O_{2} takes place as follows :

2N_{2} + O_{2} <–> 2N_{2}O.

If a mixture of 0.482 mol N_{2} and 0.933 mol of O_{2} is placed in a 10 L reaction vessel and allowed to form N_{2}O 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 O_{2} will react giving x moles of N_{2}O.

2N_{2} + O_{2} <–> 2 N_{2}O

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 :

[N_{2}]=0.0482 mol/L

[O_{2}]=0.0933 mol/L

[N_{2}O]=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 :

N_{2}O=0.1x=6.6 x 10^{-21} mol/L

[N_{2}]=0.0482 mol/L

[O_{2}]=0.0933 mol/L