Question

To indicate the pH of a substance, one can use a dye that is both an acid and that appears as different colors in its protonated and deprotonated forms. Suppose that you have a 0.001 M solution of a dye with a pKa of 7.2. From the color, the concentration of the protonated form, HA, is 0.0002 M. Assume that the remainder of the dye is in the deprotonated form, A−. What is the pH of the solution? Give your answer to one decimal place. g

Answer 1

Answer:

The pH of the solution is 7.8

Explanation:

The concentration of the solution is 0.001M and  the dye could be in its protonated and deprotonated forms. If the concentration of the protonated form [HA] is 0.0002 M the concentration of the deprotonated form will be the subtraction between the concentration of the bye and the concentration of the protonated form:

[A-] = 0.001M - 0.0002M = 0.0008M

Also, the Henderson-Hasselbalch equation is

$pH = pKa + Log \frac{[A^{-}]}{[HA]}$

this equation shows the dependency between the pH of the solution, the pKa and the concentration of the protonated and deprotonated forms. Thus, replacing in the equation

$pH = 7.2 + Log \frac{0.0008M}{0.0002M} = 7.8$