Question

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?

Answer 1

Answer: The pH after 0.02 mol of $Ba(OH)_{2}$ 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.

      $B(aq) + H_{2}O(l) \rightleftharpoons BH^{+}(aq) + OH^{-}(aq)$

According to Henderson-Hasselbach equation,

       pH = $pK_{a} + log \frac{[B]}{[BH^{+}]}$

      9.31 = $pK_{a} + log \frac{0.17}{0.39}$

    $pK_{a}$ = 9.31 + 0.361

                = 9.671

Initial moles of B is as follows.

    $0.72 L \times \frac{0.17 M}{L}$

        = 0.1224 mol B

Now, the initial concentration of $BH^{+}$ is as follows.

      $0.72 L \times \frac{0.39 mol}{L}$

         = 0.281 mol $BH^{+}$

We will calculate the equilibrium moles after the addition of 0.02 moles $Ba(OH)_{2}$ which is 0.04 moles $OH^{-}$ as follows.

        $BH^{+}(aq) + OH^{-}(aq) \rightleftharpoons B(aq) + H_{2}O(aq)$

Initial:     0.281       0.04         0.1224

Change:  -0.04      -0.04        +0.04

Equilbm: 0.241        0             0.1624

Hence,

           pH = $pK_{a} + log \frac{[B]}{[BH^{+}]}$

                 = $9.671 + log \frac{0.1624/0.72}{0.241/0.72}$

                 = $9.671 + log (0.225/0.334)$

                 = 9.671 - 0.171

                 = 9.5

Thus, we can conclude that the pH after 0.02 mol of $Ba(OH)_{2}$ are added to 0.72 L of the solution is 9.5.