Question

assume that for the pH 9.2 phenolphthalein sample, you used half as much phenolphthalein solution (i.e your sample is 9 ml buffer, 0.5 ml phenophthalein solution and 0.5 ml water). what should happen to the value of Kc you determine?

Answer 1

Explanation:

It is given that pH is equal to 9.2. So, we will calculate the value of $pK_{a}$ as follows.

        pH = $pK_{a} + log \frac{[H_{3}O^{+}]}{[\text{solution}]}$

Putting the given values into the above formula as follows.

    pH = $pK_{a} + log \frac{[H_{3}O^{+}]}{[\text{solution}]}$

      $pK_{a} = pH - log \frac{\frac{1}{0.5}}{\frac{1}{10}}$

                  = 7.9

             $K_{a} = 10^{7.9}$

                       = $1.26 \times 10^{-8}$

Since, the value of $K_{a}$ comes out to be positive. This means that there will occur an increase in the value of $K_{a}$ or $K_{c}$.