What is pKa and what is the relation of pKa with acidity and basicity.

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This pKa is something related to equilibrium constant, how let see.
 Let us consider the following reaction
            A   +   B     ⇋   C   +   D                                          
in the above reaction initially some of A and B reacts to form C and D as products (In this reaction remember all the A and B are not converted to C and D).When the amount of C and D reaches a particular conentration the reaction goes to backward direction with the same rate as the forward reaction goes on. In these types of reaction the composition of reaction mixture remains constant in a particular physical conditions.
For the above reaction:-
             equlibrium constant  , Keq = [A][B]/[C][D]
    A],[B],[C],[D] are the concentrations of A,B,C and D respectively.

If there is a reaction is in equilibrium involving acid in water then:
               AH  +  H2O    ⇋    A-   +  H3O+    ...................(eq 1)
 For this the equilibrium constant is :
                        Keq =[A-][H3O+]/ [AH][H2O]
Here notice that concentration of H2O, [H2O] is always constant with dilute solutions of acids(55.56 mol/dm3). So there is an another term that define equilibrium is acidity constant Ka.
    Ka =  [A-][H3O+]/ [AH]       (Ka = Keq[H2O] )

Now we can compare the acidity with pKa.

  • pH scale tells about the acidity.
  • Similarly pKa also tells about how much acidic the compound is .
  • More the value of ka, more it tend to loose H+, means more acidic the compound is.

Relation between pH and pKa

We know that-
                          Ka =  [A-][H3O+]/ [AH]
taking -log both sides : 
            -log(Ka) = -log( [A-][H3O+]/ [AH] )
            -log(Ka) = -log[H3O+] -log( [A-]/[AH] )
                   pKa = pH -log( [A-]/[AH] )
This is the relation between acidity(pH) and pKa.
  • The value of pH becomes equal to pKa when the concentration of [AH] become equal to [A-].    pKa = pH -log( [A-]/[AH] )
            when [A-] = [AH]
            -log( [A-]/[AH] ) = -log([AH]/[AH]) = -log(1) = 0
            -log(Ka) = -log[H3O+]
              ⇨ pKa =  pH.

Similarly for basicity this trend is opposite .

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