Question

At the equivalence point of a KHP/NaOH titration, you have added enough OH- to react with all of the HP- such that the only species left in solution are P2- (the conjugate base of HP-) and the spectator ions Na+ and K+. If a titration of KHP with NaOH resulted in [P2-] = 0.042 M at the equivalence point, what is the pH of the solution? Ka for HP- = 3.9 x 10-6

Answer 1

Explanation:

The given data is as follows.

  $[P^{2-}]$ = 0.042 M,      $K_{a}$ for $HP^{-} = 3.9 \times 10^{-6}$

According to the given situation $P^{2-}$ acts as a base.The reaction equation will be as follows.

            $P^{2-} + H_{2}O \rightleftharpoons HP^{-} + OH^{-}$

Relation between $K_{b}$ and $K_{a}$ are as follows.

                   $K_{a} \times K_{b} = K_{w}$

                     $K_{b} = \frac{1 \times 10^{-14}}{K_{a}}$

                                      =  $\frac{1 \times 10^{-14}}{3.9 \times 10^{-6}}}$

                                      = $2.6 \times 10^{-9}$

Also,      $K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}$

Let us take $[OH^{-}] = [HP^{-}]$ = x

So,                       $K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}$

                           $2.6 \times 10^{-9} = \frac{x \times x}{0.042}$

                      x = $1.04 \times 10^{-5}$

$[OH^{-}] = [HP^{-}]$ = $1.04 \times 10^{-5}$

                          pOH = - log$[OH^{-}]$

                                   = - log ($1.04 \times 10^{-5}$)

                                   = 4.99

As it is known that pH + pOH = 14

so,                  pH + 4.99 = 14

                         pH = 9.01

Thus, we can conclude that pH of the solution is 9.01.