Calculate the ph of 0.005 methonoic acid solution given that the degree of dissociation is 0.057

pH = 2.77

Explanation:

Data Given:

degree of dissociation Ka = 0.057

Concentration of Methanoic acid = 0.005 M

pH = ?

Solution:

First we calculate Hydrogen ion concentration

Formula will be used

             [H⁺] = . . . . . . (1)

where

Ka = degree of dissociation

c = concentration

Put values in formula 1

                        [H⁺] = . . . . . . (1)

                        [H⁺] = 0.0017 M

Now Calculate the pH

As we know

          pH = -log [H⁺] . . . . . .(2)

Put values in equation

           pH = - log [ 0.0017 M]

           pH = - [-2.77]

           pH =    2.77

So,

pH of methanoic acid =    2.77