Question

What is the percent dissociation of 0.187 M weak acid (HA) with a Ka of 4.3x10-7 at 18.9°C? Enter to 2 decimal places.

Answer 1

Explanation:

Let the reaction equation for the dissociation of weak acid is as follows.

                      $HA \rightarrow H^{+} + A^{-}$

Initial:         0.87            0         0

Change:      -x               +x         +x

Equilibrium:  0.87 - x    +x         +x

Hence, expression for the dissociation constant will be as follows.

           $k_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$

Now, putting the given values into the above formula as follows.

        $k_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$

       $4.3 \times 10^{-7} = \frac{x \times x}{(0.187 - x)}$

               x = 0.000283

Hence, at equilibrium the concentration of hydrogen ions is 0.000283.

or,     $[H^{+}] = 2.83 \times 10^{-4}$

            $[H^{+}] = C \times \alpha$

Also,   $\alpha = \frac{[H^{+}]}{C}$

                      = $\frac{0.000283}{0.187}$

                      = 0.00151

And, the percentage of dissociation is $0.00151 \times 100$ = 0.151%

Thus, we can conclude that percent dissociation of given weak acid is 0.151%.