Calculate the pOH, pH, and percent protonation of solute in the following aqueous solution.

0.043 M quinine, given that the pKa of its conjugate acid is 8.52

nevermind I figured out what I did

To calculate the pOH, pH, and percent protonation of the solute in the given aqueous solution, we need to consider the dissociation of quinine and the equilibrium between its conjugate acid and base forms.

1. First, let's find the concentration of the conjugate acid of quinine. Since quinine is a base, it will partially dissociate into its conjugate acid (quinineH+). The concentration of the conjugate acid ([quinineH+]) is given by the equation:

[quinineH+] = [quinine] * percent protonation

Given that the concentration of quinine ([quinine]) is 0.043 M, we can find the concentration of the conjugate acid.

2. To find the percent protonation, we use the equation:

percent protonation = (10^(pH - pKa)) / (1 + 10^(pH - pKa))

Given that the pKa of quinine's conjugate acid is 8.52, we can now substitute the known values into the equation and solve for the percent protonation.

3. Once we have the percent protonation, we can find the concentration of the conjugate acid using the equation from step 1.

4. With the concentration of the conjugate acid, we can find the concentration of the hydroxide ion ([OH-]). The concentration of hydroxide ions is equal to the concentration of the conjugate acid since they are in a 1:1 ratio.

5. Finally, we can calculate the pOH and pH using the equations:

pOH = -log10([OH-])
pH = 14 - pOH

Now, let's apply these steps to find the values for this specific solution.

Step 1: Calculate the concentration of the conjugate acid
[quinineH+] = [quinine] * percent protonation

Step 2: Calculate the percent protonation
percent protonation = (10^(pH - pKa)) / (1 + 10^(pH - pKa))

Step 3: Calculate the concentration of the conjugate acid
[quinineH+] = [quinine] * percent protonation

Step 4: Calculate the concentration of [OH-]
[OH-] = [quinineH+]

Step 5: Calculate pOH and pH
pOH = -log10([OH-])
pH = 14 - pOH