0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid.

The value of for the weak acid is .

Explanation:

First, we will calculate the molarity of HA as follows.

     [HA] =

             =

             = 0.015

At equilibrium,

Initial:   0.015                  0          0

Change:  -x                     +x         +x

Equilibm: 0.015 - x          x           x

It is given that, at equilibrium

          [HA] = 0.015 - x = 0.013

             - x = 0.013 - 0.015

                  = 0.002

Now, expression for of this reaction is as follows.

                      =

                      =

                      =

Thus, we can conclude that the value of for the weak acid is .

just did the assignment and the answer was #1 +2 n

thats how its supposed to be