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 H2SO4?
- Tri-Protic acid: Which has three protons like phosphoric acid?
The method of calculating their Ka value is given below:
H2SO4 àH+ + SO42-, let’s say for first reaction K=Ka1
HSO4–àH+ + SO42-, let’s say for this K=Ka2
Please note Ka1 is always greater than Ka2 because as acid keeps on loosing proton ,its acidic strength keeps on decreasing .
Then for the complete reaction Ka =Ka1 x Ka2
Similarly, for bases we have: Kb = Kb1 x Kb2
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 Ka ,Kb, Kw .Similarly, we have Kh that is hydrolysis constant and degree of hydrolysis as ‘h’.
Different kind of salts
- Salt of weak acid and strong base
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:
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 :
K=[Ag+][Cl–]/[AgCl] (equation 1 )
Or, K[AgCl]= [Ag+][Cl–]
Or, Ksp=[Ag+][Cl–] (equation 2 )
If we compare equation 1 and equation 2 ,we get :
- If value of K>Ksp, then the precipitation occurs .
- If value of K<Ksp, then the precipitation doesn’t occurs.
What is the Kc 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 Kc =[SO3]2/[SO2]2[O2]
Substituting the values, we get :
At certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of iodine atoms :
I2 <–>2I .Calculate Kp 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
= (0.4 x 105)2/0.60 x 105=2.67 x 104
Find out the value of Kc for each of the following from the value of Kp:
2NOCl <–> 2NO + Cl2 ,Kp= 1.8 x 10-2 at 500k
CaCO3 <–> CaO + CO2 ,K p = 167 at 1073 K
Answer 3:- Kp and Kc are related to each other by the relation :
Kp=KcRTchange In gaseous moles
- Change in gaseous moles =3-2=1 ,Kp=kc(RT)1
Kc= 1.8 x 10-2/0.0831 x 500=4.33 x 10-4
- Change in gaseous moles = 1-0=1 , Kc=Kp/RT=167/0.0831 x 1073 =1.87
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 Kc= 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 Kc =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 :
Substituting the values, we get :
Kc= (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