24 mL of 0.39 mol/L acetic acid is titrated with a standardized 0.33 mol/L KOH solution. Calculate the pH of the solution after 17 mL of the KOH solution has been added. Assume the Ka of acetic acid is 1.8 x 10-5.

This combinations forms a buffered solution. Determine how much potassium acetate is formed and use the Henderson-Hasselbalch equation.

To calculate the pH of the solution after adding 17 mL of KOH, we need to consider the neutralization reaction between acetic acid (CH3COOH) and potassium hydroxide (KOH). The balanced equation for this reaction is:

CH3COOH + KOH -> CH3COOK + H2O

Based on the equation, we can determine the number of moles of acetic acid and KOH that react. Let's start by calculating the number of moles of acetic acid in 24 mL of a 0.39 mol/L solution:

moles of CH3COOH = volume (in L) x concentration (in mol/L)
= 24 mL x (1 L / 1000 mL) x 0.39 mol/L
= 0.00936 mol

Since the reaction between acetic acid and KOH occurs in a 1:1 ratio, the number of moles of KOH will also be 0.00936 mol.

Now, let's calculate the remaining moles of KOH after adding 17 mL of the KOH solution:

remaining moles of KOH = initial moles of KOH - moles of KOH used in titration
= 0.00936 mol - (17 mL x (1 L / 1000 mL) x 0.33 mol/L)
= 0.00936 mol - 0.00561 mol
= 0.00375 mol

We can now determine the concentration of KOH left in the mixture:

concentration of KOH = remaining moles of KOH / final volume of the solution (in L)
= 0.00375 mol / ((24 mL + 17 mL) x (1 L / 1000 mL))
= 0.00375 mol / 0.041 L
= 0.0914 mol/L

Since the reaction is a neutralization reaction, the concentration of CH3COOH and CH3COOK will be the same as they react in a 1:1 ratio.

Now, we can calculate the concentration of acetic acid (CH3COOH):

concentration of CH3COOH = 0.0914 mol/L

Applying the Henderson-Hasselbalch equation for weak acids:

pH = pKa + log (concentration of CH3COOK / concentration of CH3COOH)

Given that the pKa of acetic acid (CH3COOH) is 1.8 x 10-5, we can substitute the values into the equation:

pH = 1.8 x 10-5 + log (0.0914 / 0.0914)
= 1.8 x 10-5 + log (1)
= 1.8 x 10-5 + 0
= 1.8 x 10-5

Therefore, the pH of the solution after adding 17 mL of KOH is approximately 1.8 x 10-5.