22 mL of 0.37 mol per litre acetic acid is titrated by way of a standardized 0.29 mol/L KOH solution. Calculate the pH of the solution after roughly 18 mL of the solution of KOH is added. The Ka of acetic acid proves to be 1.8 x 10^-5

The equation is

CH3COOH + NaOH ==> CH3COONa + HOH

mol CH3COOH initially = M x L = ??
mol NaOH initially = M x L
Look to see how much has reacted, how much sodium acetate (the salt) has been formed, and how much CH3COOH remains unreacted. You will find that there is salt and an excess of CH3COOH and this constitutes a buffer solution (a weak acid and its conjugate base). Then use the Henderson-Hasselbalch equation and calculate pH. Post your work if you get stuck.

dear DR Bod

I had the same problem and worked it out as follows:

the equation: CH3COOH + NaOH ==> CH3COONa + H20
MOL CH3COOH = M x L
MOL NaON = M x L
now, to find how much sodium acetate formed, do i have to add the moles of NaOH and CH3COOH together? and how do i find the amount of CH3COOH that has not reacted?

To calculate the pH of the solution after adding approximately 18 mL of the KOH solution, we need to determine the amount of acetic acid and KOH remaining, and use that information to calculate the concentration of the resulting species.

1. Calculate the moles of acetic acid initially present:
Moles of acetic acid = volume(L) x concentration(mol/L)
Moles of acetic acid = 0.022 L x 0.37 mol/L
Moles of acetic acid = 0.00814 mol

2. Determine the limiting reagent between acetic acid and KOH:
The reaction between acetic acid (CH3COOH) and KOH (KOH) is as follows:
CH3COOH + KOH -> CH3COOK + H2O

The stoichiometry of the balanced chemical equation is 1:1. Since the moles of acetic acid (0.00814 mol) and KOH (0.018 L x 0.29 mol/L = 0.00522 mol) are not in a 1:1 ratio, KOH is the limiting reagent.

3. Calculate the moles of KOH used in the reaction:
Moles of KOH = initial moles - remaining moles
Moles of KOH = 0.00522 mol

4. Calculate the moles of acetic acid remaining:
Moles of acetic acid = initial moles - moles of KOH used
Moles of acetic acid = 0.00814 mol - 0.00522 mol
Moles of acetic acid = 0.00292 mol

5. Calculate the concentration of acetic acid remaining:
Concentration of acetic acid = moles of acetic acid remaining / volume(L)
Concentration of acetic acid = 0.00292 mol / 0.022 L
Concentration of acetic acid = 0.133 mol/L

6. Calculate the concentration of acetate ions (from the reaction of KOH with acetic acid):
Since the reaction between acetic acid and KOH is 1:1, the concentration of acetate ions is equal to the concentration of KOH used:
Concentration of acetate ions = concentration of KOH used
Concentration of acetate ions = 0.00522 mol / 0.018 L
Concentration of acetate ions = 0.29 mol/L

7. Calculate the pOH of the solution using the concentration of acetate ions:
pOH = -log[OH-]
pOH = -log(0.29 mol/L)
pOH ≈ 0.54

8. Calculate the pH of the solution using the pOH:
pH + pOH = 14
pH = 14 - pOH
pH ≈ 14 - 0.54
pH ≈ 13.46

Therefore, the pH of the solution after adding approximately 18 mL of the KOH solution is approximately 13.46.

To calculate the pH of the solution after the addition of KOH, we first need to determine the moles of acetic acid and the moles of KOH that reacted.

1. Calculate the moles of acetic acid:
The volume of acetic acid solution is given as 22 mL, which we can convert to liters:
22 mL ÷ 1000 mL/L = 0.022 L

Now, using the molarity of acetic acid (0.37 mol/L), we can calculate the moles of acetic acid:
Moles of acetic acid = Molarity × Volume
Moles of acetic acid = 0.37 mol/L × 0.022 L

2. Calculate the moles of KOH:
The volume of KOH solution added is given as roughly 18 mL, which we can also convert to liters:
18 mL ÷ 1000 mL/L = 0.018 L

Using the molarity of KOH (0.29 mol/L), we can calculate the moles of KOH:
Moles of KOH = Molarity × Volume
Moles of KOH = 0.29 mol/L × 0.018 L

3. Determine the limiting reactant:
To determine which reactant is limiting, compare the moles of KOH and acetic acid. The reactant with fewer moles is the limiting reactant.
In this case, it appears that KOH is the limiting reactant, as it will be completely consumed before the acetic acid.

4. Calculate the moles of acetic acid remaining:
Since KOH is the limiting reactant, all of the KOH will react with the acetic acid. Therefore, the remaining moles of acetic acid will be:
Moles of acetic acid remaining = Initial moles of acetic acid - Moles of KOH reacted

5. Calculate the concentration of acetic acid after the reaction:
The volume of the solution does not change, so the new concentration can be calculated using the moles of acetic acid remaining and the volume of the solution:
Concentration of acetic acid = Moles of acetic acid remaining / Volume of solution

6. Calculate the pKa and then the pH:
The Ka of acetic acid is given as 1.8 × 10^(-5), so the pKa can be calculated as:
pKa = -log(Ka)

Finally, the pH can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([concentration of acetic acid] / [concentration of acetate ion])

In this case, since acetic acid is a weak acid, most of it does not dissociate into acetate ions, and the [concentration of acetate ion] can be assumed to be negligible compared to the [concentration of acetic acid].

By plugging in the values, you should be able to calculate the pH of the solution after the addition of roughly 18 mL of KOH.