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.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).