Question

Shown below is the equilibrium between phenol and phenolate. The pKa of phenol is 10.00. At what pH is phenol 85% ionized?

a) Present the equilibrium representation between phenol and phenolate.
b) Instruct the respondent to analyze and determine the pH at which phenol is 85% ionized.
c) Encourage a step-by-step explanation or calculation process, demonstrating an understanding of acid-base equilibrium and ionization.

Ensure clarity in the question to prompt a thoughtful calculation and explanation of the pH at which phenol is 85% ionized in the given equilibrium.

Answer 1

The pH at which phenol is 85% ionized is found to be 9.75, determined by utilizing the equilibrium expression for the acid dissociation constant (Ka) and the percent ionization.

Step-by-step explanation:

To determine the pH at which phenol is 85% ionized, one must first understand the equilibrium between phenol (PhOH) and the corresponding phenolate ion (PhO−). The equilibrium can be written as:

PhOH <--> PhO− + H+

The dissociation constant (Ka) of phenol can be calculated using its pKa:

Ka = 10-pKa = 10-10.00

For phenol to be 85% ionized, the ratio of the concentration of phenolate ion to the total concentration of phenol must be 0.85/0.15. Using the expression for Ka and incorporating the percent ionization:

Ka = [PhO−][H+] / [PhOH]

We can substitute [PhO−] = 0.85 [PhOH Total] and [PhOH] = 0.15 [PhOH Total], leading to:

Ka = (0.85 [PhOH Total])[H+] / (0.15 [PhOH Total])

By canceling out [PhOH Total] and rearranging, we get:

[H+] = Ka * 0.15 / 0.85

Substitute the known value of Ka:

[H+] = 10-10 * 0.15 / 0.85

Calculating [H+] gives us 1.76 x 10-10 M. To find the pH:

pH = -log[H+] = -log(1.76 x 10-10) = 9.75

This calculation of pH indicates that, at a pH of 9.75, phenol is 85% ionized.