the acid dissociation constant for acetic acid is 1.85x10^-5. Calculate the pH at the equivalent point and the pH at one-half the equivalence point for the titration of 25.00 mL of 0.100 M acetic acid with 0.100 M NaOH? Its been a little while since ive done one of these problems. Can someone give step by step with explanation. Just need to see one worked out to get me started. Thanks

Sure, I'll give it a shot! But beware, my explanations may contain a pinch of humor. Let's dive in!

Step 1: Find the moles of acetic acid.
To find the moles, we need to multiply the volume (converted to liters) by the molarity. So, 25.00 mL (which is 0.02500 L) multiplied by 0.100 M gives us 0.00250 moles of acetic acid.

Step 2: Determine the number of moles of NaOH needed for complete neutralization.
Since acetic acid and NaOH react in a 1:1 ratio, the number of moles of NaOH needed is also 0.00250 moles.

Step 3: Find the volume of NaOH needed for complete neutralization.
To do this, we'll divide the number of moles of NaOH by its molarity. So, 0.00250 moles divided by 0.100 M gives us 0.02500 L, which is equal to 25.00 mL.

Step 4: Determine the volume of NaOH at the half-equivalence point.
The half-equivalence point occurs when half of the acetic acid has reacted. Since we started with 25.00 mL of acetic acid, the half-equivalence point will be at 12.50 mL.

Step 5: Calculate the moles of acetic acid at the half-equivalence point.
Using the same ratio as before, we'll multiply the volume (converted to liters) by the molarity. So, 12.50 mL (which is 0.01250 L) multiplied by 0.100 M gives us 0.00125 moles of acetic acid.

Step 6: Find the moles of NaOH reacted at the half-equivalence point.
Since acetic acid and NaOH react in a 1:1 ratio, the number of moles of NaOH reacted will also be 0.00125 moles.

Step 7: Calculate the volume of NaOH reacted at the half-equivalence point.
To find the volume, we'll divide the number of moles of NaOH reacted by its molarity. So, 0.00125 moles divided by 0.100 M gives us 0.01250 L, which is equal to 12.50 mL.

Now, let's calculate the pH at the equivalent point and the pH at one-half the equivalence point.

Step 8: For the equivalent point, the acetic acid has completely reacted with the NaOH. So, we have a solution of sodium acetate. The salt, sodium acetate, is the conjugate base of acetic acid. To find the pH, we'll use the equation:

pOH = -log10 (concentration of sodium acetate)
pOH = -log10 (0.00250 mol / 0.02500 L) (concentration of sodium acetate is equal to the moles divided by the volume in liters)
pOH = -log10 (0.100) (dividing 0.00250 mol by 0.02500 L gives us 0.100 M)
pOH = 1

To find the pH, we'll subtract the pOH value from 14:
pH = 14 - pOH
pH = 14 - 1
pH = 13

So, at the equivalent point, the pH is 13.

Step 9: For one-half the equivalence point, we have a mixture of acetic acid and its conjugate base, sodium acetate. Calculate the moles of acetic acid remaining by subtracting the moles of NaOH reacted from the initial moles of acetic acid:
0.00250 mol - 0.00125 mol = 0.00125 mol

Now, divide the moles of acetic acid remaining by the total volume (in liters) of the mixture at one-half the equivalence point:
0.00125 mol / 0.01250 L = 0.100 M

Since the acetic acid is a weak acid, we'll use the dissociation constant (Ka) to calculate the pH. The Ka value for acetic acid is 1.85 x 10^-5.

pH = -log10 (square root of Ka x concentration of acetic acid)
pH = -log10 (√(1.85 x 10^-5 x 0.100))
pH = -log10 (√(1.85 x 10^-6))
pH = -log10 (4.303 x 10^-4)
pH = -(-3.365)
pH ≈ 3.365

So, at one-half the equivalence point, the pH is approximately 3.365.

I hope this step-by-step explanation helped to refresh your memory! If you have any more questions or need further assistance, feel free to ask.

Sure! I can help you with that.

Step 1: Write the balanced equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):

CH3COOH + NaOH → CH3COONa + H2O

Step 2: Calculate the number of moles of acetic acid used in the titration:

Moles of acetic acid = volume (in L) × concentration (in mol/L)
= 25.00 mL × (1 L / 1000 mL) × 0.100 mol/L
= 0.00250 mol

Step 3: Since acetic acid and sodium hydroxide react in a 1:1 ratio, the number of moles of sodium hydroxide used in the titration will also be 0.00250 mol.

Step 4: Calculate the volume of sodium hydroxide required to reach the equivalence point:

Volume of sodium hydroxide (in L) = moles of sodium hydroxide / concentration of sodium hydroxide
= 0.00250 mol / 0.100 mol/L
= 0.025 L
= 25.00 mL

Step 5: Determine the pH at the equivalence point:
At the equivalence point, all the acetic acid will react with the sodium hydroxide, resulting in the formation of sodium acetate and water. The solution will be a sodium acetate salt dissolved in water.

Since sodium acetate is a strong electrolyte, it will completely dissociate in water, producing sodium ions (Na+) and acetate ions (CH3COO-). Sodium acetate does not undergo any further reaction with water. Therefore, the solution will be neutral, and the pH at the equivalence point will be 7.

Step 6: Determine the pH at one-half the equivalence point:
At one-half the equivalence point, half of the initial amount of acetic acid will react with half of the initial amount of sodium hydroxide. This means that half of the acetic acid has been neutralized.

The balanced equation for half-neutralization is:

CH3COOH + ½ NaOH → ½ CH3COONa + ½ H2O

Since the number of moles of sodium hydroxide used is half of the initial number of moles of acetic acid, let's recalculate the volume of sodium hydroxide used at half-neutralization:

Volume of sodium hydroxide at half-neutralization (in L) = 0.00250 mol / 0.050 mol/L (half the concentration)
= 0.05 L
= 50.00 mL

Step 7: Calculate the remaining number of moles of acetic acid after neutralization:

Moles of acetic acid remaining = initial moles of acetic acid - moles of sodium hydroxide used
= 0.00250 mol - 0.00250 mol
= 0 mol

Since all of the acetic acid has been neutralized, the remaining pH will depend on the hydrolysis of the sodium acetate formed. Acetate ions react with water to form acetic acid and hydroxide ions.

CH3COO- + H2O ⇌ CH3COOH + OH-

This reaction is a weak base reaction. To calculate the OH- concentration and convert it to pH, you will need to use the hydrolysis constant (Kb) for acetate ions.

To continue the calculation, the Kb for acetate ions (CH3COO-) is needed. Do you have the value for Kb?

To find the pH at the equivalent point and the pH at half the equivalence point for the titration of acetic acid with NaOH, we need to follow a step-by-step process.

First, let's calculate the number of moles of acetic acid (CH3COOH) present initially in the 25.00 mL of 0.100 M acetic acid solution.

Step 1: Calculate the number of moles of acetic acid.
Number of moles (n) = volume (V) × molarity (M)
n = 0.025 L × 0.100 mol/L
n = 0.0025 mol

Now, we need to determine the concentration of acetic acid after the addition of NaOH at the equivalent point.

Step 2: Calculate the volume of NaOH at the equivalent point.
In a neutralization reaction, the moles of acid will be equivalent to the moles of base at the equivalence point. Since the stoichiometry of the reaction between acetic acid and NaOH is 1:1, the number of moles of NaOH needed to reach the equivalence point will be the same as the number of moles of acetic acid.

Volume of NaOH at the equivalent point = number of moles of acetic acid = 0.0025 mol
Using the formula, concentration (C) = moles (n) / volume (V), we can calculate the concentration of acetic acid at the equivalent point.

Step 3: Calculate the concentration of acetic acid at the equivalent point.
Concentration (C) = moles (n) / volume (V)
C = 0.0025 mol / 0.025 L
C = 0.100 M

At the equivalent point, the acetic acid is exactly neutralized by NaOH, producing an equal concentration of acetate ions (CH3COO-) and hydroxide ions (OH-). In solution, the acetic acid dissociates, and the acetate ion acts as a weak base with water, forming hydroxide ions. Thus, the concentration of hydroxide ions (OH-) at the equivalent point is equal to the concentration of acetic acid (0.100 M).

Step 4: Calculate the pOH at the equivalent point.
pOH = -log(OH- concentration)
pOH = -log(0.100)
pOH = 1 (rounded to the nearest whole number)

The pH can now be found by subtracting the pOH from 14.

Step 5: Calculate the pH at the equivalent point.
pH = 14 - pOH
pH = 14 - 1
pH = 13

Next, we will calculate the concentration of acetic acid at half the equivalence point. At this point, half of the acetic acid has reacted with NaOH.

Step 6: Calculate the number of moles of acetic acid at half the equivalence point.
Number of moles (n) = initial moles of acetic acid - moles reacted
n = 0.0025 mol - 0.0025 mol/2 (half of the acetic acid has reacted)
n = 0.00125 mol

Step 7: Calculate the concentration of acetic acid at half the equivalence point.
Concentration (C) = moles (n) / volume (V)
C = 0.00125 mol / 0.025 L
C = 0.050 M

Now, we can find the concentration of hydroxide ions (OH-) using the same reasoning as before.

Step 8: Calculate the concentration of hydroxide ions (OH-) at half the equivalence point.
The concentration of OH- will be equal to the concentration of acetic acid at half the equivalence point (0.050 M).

Step 9: Calculate the pOH at half the equivalence point.
pOH = -log(OH- concentration)
pOH = -log(0.050)
pOH ≈ 1.30 (rounded to two decimal places)

Finally, we can calculate the pH at half the equivalence point.

Step 10: Calculate the pH at half the equivalence point.
pH = 14 - pOH
pH = 14 - 1.30
pH ≈ 12.70 (rounded to two decimal places)

Therefore, the pH at the equivalent point is 13, and the pH at half the equivalence point is approximately 12.70.