A buffer that contains 0.17 M of a base, B and 0.39 M of its conjugate acid BH+, has a pH of 9.31. What is the pH after 0.02 mol of Ba(OH)2 are added to 0.72 L of the solution?

The pH after 0.02 mol of are added to 0.72 L of the solution is 9.5.

Explanation:

The chemical equation for the dissociation of base represented by B in water is depicted as follows.

According to Henderson-Hasselbach equation,

       pH =

      9.31 =

    = 9.31 + 0.361

                = 9.671

Initial moles of B is as follows.

        = 0.1224 mol B

Now, the initial concentration of is as follows.

         = 0.281 mol

We will calculate the equilibrium moles after the addition of 0.02 moles which is 0.04 moles as follows.

Initial:     0.281       0.04         0.1224

Change:  -0.04      -0.04        +0.04

Equilbm: 0.241        0             0.1624

Hence,

           pH =

                 =

                 =

                 = 9.671 - 0.171

                 = 9.5

Thus, we can conclude that the pH after 0.02 mol of are added to 0.72 L of the solution is 9.5.

thats a little if u want the answer just look it up on google thats what i do haha : )