a. If an acid with a pKa of 5.3 is in an aqueous solution

Solution:

Step 1

(a)

Here we have to calculate the percentage of the acid present in its acidic form.

Acid dissociation constant(pKa):-The term "pKa" is a measure of the strength of an acid in solution. It is defined as the negative base 10 log of the acid dissociation constant.

pKa = -log Ka

pH:-pH of a solution is defined as the -ve logarithm of hydrogen ion concentration [H+]

pH= -log [H+]

Given:

pKa of an unknown acid is = 5.3

pH of the aqueous solution = 5.7

Here we have to calculate the percentage of the acid present in its acidic form.

Explanation:

percentage of the acid present in its acidic form can be calculated by using the following relationship

  =

The concentration of [H3O+] can be expressed as [H3O+] = 10-pH

Thus the concentration of H3O+ will be ,  [H3O+] = 10-5.7

                                           [H3O+] =2.0 10-6

The relationship between acid dissociation constant and pKa is given by

pKa = -logKa

Ka = 10-pKa

 Ka = 10-5.3

 Ka = 5.0 10-6

Thus the percentage of acid is equal to            2.0 10-6

                                         =     --------------------------    100  

                                              2.0 10-6 + 5.0 10-6

   2.0 10-6

                                         =     --------------------------    100  

                                                    (2.0  + 5.0) 10-6

     2.0 10-6

                                         =     --------------------------    100  

                                                      7.0 10-6

                                         = 28.571%  29%

Thus 29% of the acid is present in its acidic form.