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 is a buffered solution. Calculate mols acetic acid and mols KOH; you will have some salt produced as well as some acetic acid remaining. Then pH = pKa + log[(base)/(acid)]. Post your work if you get stuck.

mols acetic acid = 0.024 Lx 0.39 mol/L

= 0.00936 mols

mols KOH = 0.33 mol/L x 0.017 L
= 0.00561 mols

pH = +log(0.00936)(0.00561)
= -4.279
= 14 + -4.279
=9.7

I am stuck

KOH + CH3COOH ==> CH3COOK + HOH

mols KOH = 0.00561. correct.
mols CH3COOH = 0.00936. correct.
But from here on is new territory.
You must recognize that KOH and CH3COOH are strong base and weak acid (which you probably know already) so what happens is these react to produce how much of the salt? It will produce, of course, the lesser of the two, which in this case is 0.00561 mols of the CH3COOK. So how much of the CH3COOH is used? Obviously 0.00561 mol has been used. How much of the CH3COOH is left unreacted? The difference which is 0.00936 - 0.00561 = 0.00375.
So you see there is CH3COOH unreacted PLUS the salt of the weak acid which makes this a buffered solution. That is solved by the Henderson-Hasselbalch equation.
Concn CH3COOH = mols/L (L here will be 17 mL + 24 mL = 41 mL or 0.041 L).
concn CH3COOL (the salt) is mols/L.

HH equation is
pH = pKa + log [(base)/(acid)/]
The base in this case is the conjugate base of acetic acid which is the salt.

To calculate the pH of the solution, we need to determine the amount of acetic acid remaining and the amount of acetate ion formed after 17 mL of the KOH solution has been added. This will allow us to solve for the concentration of the acid and its conjugate base, which we can then use to calculate the pH.

Step 1: Calculate the moles of acetic acid initially present:
Moles of acetic acid = concentration x volume
Moles of acetic acid = 0.39 mol/L x 0.024 L = 0.00936 mol

Step 2: Calculate the moles of KOH used:
Moles of KOH = concentration x volume
Moles of KOH = 0.33 mol/L x 0.017 L = 0.00561 mol

Step 3: Determine the limiting reagent:
Since acetic acid and KOH react in a 1:1 molar ratio, the limiting reagent is the one with fewer moles, which is KOH in this case.

Step 4: Calculate the moles of acetic acid and acetate ion after reaction:
Moles of acetic acid remaining = initial moles - moles of KOH used
Moles of acetic acid remaining = 0.00936 mol - 0.00561 mol = 0.00375 mol

Moles of acetate ion formed = moles of KOH used
Moles of acetate ion formed = 0.00561 mol

Step 5: Calculate the concentration of acetic acid and acetate ion:
Concentration of acetic acid = moles / volume
Concentration of acetic acid = 0.00375 mol / 0.024 L = 0.15625 mol/L

Concentration of acetate ion = moles / volume
Concentration of acetate ion = 0.00561 mol / 0.024 L = 0.2325 mol/L

Step 6: Calculate the pH of the solution:
pH = -log [H+]

[H+] is the concentration of H+, which can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-] / [HA])

From the equation, we can see that [A-] represents the concentration of the acetate ion and [HA] represents the concentration of acetic acid.

pH = -log (1.8 x 10^-5) + log (0.2325 / 0.15625)
pH = 4.74 + log (1.487)

Using a calculator, we can find that log (1.487) is approximately 0.173.

pH = 4.74 + 0.173
pH ≈ 4.913

Therefore, the pH of the solution after 17 mL of the KOH solution has been added is approximately 4.913.