When a 0.089m aqueous solution of a certain acid is prepared, the acid is 12.% dissociated. calculate the ph of the solution. round your answer to 2 decimal places.

pH = 1.98

Explanation:

Given a general acid dissociation:

HA + H₂O  ⇆  H₃O⁺   + A⁻    

The pH is -log[ H₃O⁺ ]

Therefore we need to determine [ H₃O⁺ ] to answer this question, and we should use the data of % dissociation of the acid.

Percent dissociation is

% dissociation = [ H₃O⁺ ] / [ HA ]₀ x 100

where [ HA ]₀  is the original acid concentration, so we can calculate [ H₃O⁺ ] , and then the pH.

12 =  [ H₃O⁺ ] /0.089 M ⇒  [ H₃O⁺ ]  = (12 x 0.089 /100) M

                                                          = 1.07 x 10⁻² M

and pH = - log ( 1.07 x 10⁻² ) = 1.98