Okay, so I need to solve this chemistry problem. Let me read it again carefully. The question is about titrating a diluted acetic acid solution with NaOH. They're asking for the pH at 25% titration and at the equivalence point. The options are A, B, C, D. Let's break this down step by step.

First, the initial setup: We start with 20.00 cm³ (which is 20.00 mL) of 0.05 M acetic acid. Then we dilute it by adding 20.00 cm³ of water. So the total volume after dilution is 40.00 mL. Then we titrate this diluted solution with 0.1 M NaOH. The question is about the pH at two points: 25% of the titration and at the equivalence point. The Ka of acetic acid is given as 1.85x10^-5.

Let me start by understanding what happens during the titration. Acetic acid is a weak acid, and NaOH is a strong base. When they react, they form sodium acetate and water. The reaction is:

CH3COOH + NaOH → CH3COO^-Na^+ + H2O

So, the moles of acetic acid will react with moles of NaOH. Since NaOH is a strong base, it will completely dissociate, so the amount of NaOH added will determine the moles of acetate ions in solution.

First, let's calculate the initial moles of acetic acid. The initial volume is 20.00 mL (0.02000 L) and concentration is 0.05 M. Moles = Molarity × Volume = 0.05 mol/L × 0.02000 L = 0.001 mol. Then, when we add 20.00 mL of water, the total volume becomes 40.00 mL. But the number of moles of acetic acid remains the same (0.001 mol) because dilution doesn't change the amount of solute, just the concentration.

Wait, but when we dilute, the concentration decreases. The new concentration is (0.001 mol) / (0.04000 L) = 0.025 M. That's the diluted acetic acid concentration. But for the titration, the important thing is the moles of acid and base reacting.

Now, the titration is done with 0.1 M NaOH. Let's figure out the volume of NaOH needed to reach the equivalence point. At equivalence, all the acetic acid has been neutralized to sodium acetate. So the moles of NaOH needed will be equal to the moles of acetic acid initially present (0.001 mol). Since the concentration of NaOH is 0.1 M, the volume required is moles / concentration = 0.001 / 0.1 = 0.01 L = 10.00 mL. So the equivalence point is at 10.00 mL of NaOH added.

But the question asks for the pH at 25% titration and at equivalence. Let's handle each part separately.

First, at 25% titration. That means 25% of the moles of NaOH required to reach equivalence have been added. Since the equivalence point requires 10.00 mL, 25% would be 2.5 mL of NaOH added. Wait, but wait, is that correct? Wait, 25% of the way to the equivalence point. So 25% of 10.00 mL is 2.50 mL. So at 2.50 mL of NaOH added, we have a solution that is a mixture of acetic acid and its conjugate base (sodium acetate). This is a buffer solution. So we can use the Henderson-Hasselbalch equation here.

But wait, let me confirm. The total volume after adding 2.50 mL of NaOH is the initial 40.00 mL (from the diluted acid) plus 2.50 mL, so 42.50 mL. The moles of NaOH added would be 0.1 M × 0.0025 L = 0.00025 mol. Since each mole of NaOH neutralizes one mole of acetic acid to form one mole of acetate ion, the moles of acetic acid remaining would be initial moles (0.001) minus moles of NaOH added (0.00025) = 0.00075 mol. The moles of acetate formed are 0.00025 mol.

So the concentrations in the solution would be:

[CH3COOH] = 0.00075 mol / 0.0425 L ≈ 0.01765 M

[CH3COO^-] = 0.00025 mol / 0.0425 L ≈ 0.005882 M

Then, using the Henderson-Hasselbalch equation:

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

Ka is given as 1.85x10^-5, so pKa = -log(1.85x10^-5) ≈ 4.735

So pH ≈ 4.735 + log(0.005882 / 0.01765) = 4.735 + log(0.3333) = 4.735 - 0.477 ≈ 4.258

Wait, that's approximately 4.26. Let me check the calculation again. The ratio [A^-]/[HA] is 0.005882 / 0.01765 ≈ 0.3333. Log of 0.3333 is log(1/3) ≈ -0.477. So 4.735 - 0.477 ≈ 4.258, which rounds to 4.26. So that's the pH at 25% titration.

Now, at the equivalence point. At equivalence, all the acetic acid has been converted to acetate ions. The solution at equivalence point contains the sodium acetate and excess water. The acetate ion is the conjugate base of a weak acid, so it will hydrolyze in water to produce OH- ions, making the solution basic.

The concentration of acetate ions at equivalence can be calculated. The total moles of acetate is 0.001 mol (same as initial moles of acetic acid). The total volume is 40.00 mL + 10.00 mL = 50.00 mL = 0.05 L. So concentration of acetate is 0.001 mol / 0.05 L = 0.02 M.

Acetate ion (CH3COO^-) reacts with water:

CH3COO^- + H2O ↔ CH3COOH + OH^-

The Kb for acetate ion can be calculated from Ka. Since Kw = Ka * Kb, so Kb = Kw / Ka = (1.0x10^-14) / (1.85x10^-5) ≈ 5.405x10^-10.

Now, using the formula for weak base hydrolysis:

Kb = [CH3COOH][OH^-] / [CH3COO^-]

Assuming x is the concentration of OH^- produced:

Kb = x² / (C - x) ≈ x² / C (since C is much larger than x)

So x² ≈ Kb * C = 5.405x10^-10 * 0.02 = 1.081x10^-11

x ≈ sqrt(1.081x10^-11) ≈ 1.04x10^-6 M

So [OH^-] ≈ 1.04x10^-6 M

pOH = -log(1.04x10^-6) ≈ 5.98

pH = 14 - 5.98 ≈ 8.02

Wait, but none of the options have 8.02 as the equivalence point. Let me check the options again. The choices are:

A. 4.73; 7.00

B. 3.17; 6.73

C. 4.26; 8.52

D. 4.57; 6.92

Hmm, 8.02 is closest to 8.52 (option C) but not exact. Maybe I made a mistake in the calculation.

Wait, let's recalculate the Kb. Wait, the initial concentration of acetate is 0.02 M. So Kb = 5.405e-10. Then, the expression is:

Kb = x² / (0.02 - x) ≈ x² / 0.02

So x² = Kb * 0.02 = 5.405e-10 * 0.02 = 1.081e-11

x = sqrt(1.081e-11) ≈ 1.04e-6 M. So [OH^-] is 1.04e-6, pOH is about 5.98, pH is 8.02. But the options don't have 8.02. The closest is option C at 8.52. Maybe I made a mistake in the Kb calculation.

Wait, let's check the calculation of Kb again. The given Ka is 1.85e-5. So Kw is 1e-14. So Kb = 1e-14 / 1.85e-5 ≈ 5.405e-10. That's correct. Then, 0.02 M acetate. The calculation seems right. But perhaps the assumption that x is negligible is too rough. Let me check if x is indeed negligible compared to 0.02.

x is 1.04e-6, which is 0.0001% of 0.02. So the approximation holds. So the pH at equivalence should be approximately 8.02. But the options don't have that. Hmm. Let me check the options again. Option C says 8.52. Wait, maybe I made a mistake in the calculation of Kb. Let me check.

Wait, maybe I miscalculated the concentration of acetate at equivalence. Let's recalculate that. The moles of acetate is 0.001 mol, and the total volume is 50.00 mL, so 0.05 L. So 0.001 / 0.05 = 0.02 M. That's correct. So the concentration is 0.02 M. Then Kb is 5.405e-10. So x² = 5.405e-10 * 0.02 = 1.081e-11. x = 1.04e-6. So [OH^-] = 1.04e-6. pOH = 5.98. pH = 8.02. So the correct pH at equivalence should be approximately 8.02. But the closest option is C: 8.52. Hmm, maybe there's a mistake in the problem statement. Alternatively, perhaps I made an error in the calculation of the initial concentration.

Wait, let's re-examine the initial steps. The initial solution is 20.00 mL of 0.05 M acetic acid. Then diluted with 20.00 mL water. So total volume is 40.00 mL. The moles of acetic acid are 0.001 mol. Then, when titrated with 0.1 M NaOH, the equivalence point requires 0.001 mol NaOH, which is 0.01 L, so 10.00 mL. That's correct. So the equivalence point is at 10.00 mL NaOH. Then, after adding 2.5 mL NaOH, we have 25% of the way to equivalence.

Wait, but the options for equivalence point are 8.52 (option C) and 6.92 (option D). Let me check if I miscalculated the Kb. Wait, perhaps the problem is that when you have a solution of sodium acetate, the pH can be calculated using the formula for a weak base. Alternatively, maybe the solution is treated as a buffer, but at equivalence, it's not a buffer anymore. So the pH comes from the hydrolysis of the acetate ion.

Alternatively, maybe the problem is that the solution is diluted, but no, the equivalence point solution is 0.02 M acetate. Wait, perhaps I should use the exact formula instead of the approximation. Let's try solving the quadratic equation without the approximation.

The equation is:

Kb = x² / (0.02 - x) = 5.405e-10

Multiply both sides by (0.02 - x):

x² = 5.405e-10 * (0.02 - x)

x² + 5.405e-10 x - 1.081e-11 = 0

This is a quadratic equation: ax² + bx + c = 0, where a = 1, b = 5.405e-10, c = -1.081e-11

The solution is x = [-b ± sqrt(b² + 4ac)] / 2a

Plugging in the numbers:

b² = (5.405e-10)^2 ≈ 2.921e-19

4ac = 4 * 1 * (-1.081e-11) = -4.324e-11

So discriminant = 2.921e-19 + 4.324e-11 ≈ 4.324e-11 (since 2.921e-19 is negligible compared to 4.324e-11)

sqrt(4.324e-11) ≈ 6.576e-6

So x = [-5.405e-10 ± 6.576e-6]/2

We take the positive root:

x = (6.576e-6 - 5.405e-10)/2 ≈ 6.576e-6 / 2 ≈ 3.288e-6

Wait, but that's inconsistent with the previous approximation. Wait, this can't be right. Wait, let me check the discriminant again.

Wait, discriminant is b² + 4ac. Wait, b is 5.405e-10, so b² is (5.405e-10)^2 = (5.405)^2 * 1e-20 = 29.218 * 1e-20 = 2.9218e-19

4ac is 4 * 1 * (-1.081e-11) = -4.324e-11

So discriminant = 2.9218e-19 + (-4.324e-11) = negative number? Wait, that can't be. Wait, no, 4ac is negative, so 4ac = -4.324e-11. So discriminant = b² + 4ac = 2.9218e-19 + (-4.324e-11) = -4.324e-11 + 2.9218e-19 ≈ -4.324e-11 (since 2.9e-19 is negligible). But that would make the discriminant negative, which is impossible. Wait, that can't be right. So perhaps I made a mistake in the sign of the equation.

Wait, the original equation after multiplying both sides is:

x² = 5.405e-10 * (0.02 - x)

Which rearranges to:

x² + 5.405e-10 x - 1.081e-11 = 0

So a = 1, b = 5.405e-10, c = -1.081e-11

Then discriminant is b² - 4ac = (5.405e-10)^2 - 4*1*(-1.081e-11) = 2.9218e-19 + 4.324e-11 = 4.324e-11 + 2.9218e-19 ≈ 4.324e-11

So sqrt(4.324e-11) ≈ 6.576e-6

So x = [-5.405e-10 + 6.576e-6]/2 ≈ (6.576e-6 - 5.405e-10)/2 ≈ 6.576e-6 / 2 ≈ 3.288e-6 M

Wait, but this is different from the previous approximation. So x ≈ 3.288e-6 M. Then [OH^-] = 3.288e-6 M, pOH = -log(3.288e-6) ≈ 5.477, pH = 14 - 5.477 ≈ 8.523

Ah! So that's where 8.52 comes from. So the correct pH at equivalence point is approximately 8.52, which matches option C. So earlier, when I used the approximation, I got pH ≈ 8.02, but when solving the quadratic properly, it's 8.52.

So that explains why option C is correct. So the pH at 25% titration is 4.26 (option C), and at equivalence point is 8.52 (option C). So the correct answer is C.

Wait, but let me confirm the 25% titration again. Earlier, I calculated pH ≈ 4.26. Let me check the Henderson-Hasselbalch calculation again.

At 25% titration, 2.5 mL of NaOH added. Moles of NaOH = 0.1 M * 0.0025 L = 0.00025 mol. Moles of CH3COOH initially = 0.001 mol. So moles of CH3COOH remaining = 0.001 - 0.00025 = 0.00075 mol. Moles of CH3COO^- formed = 0.00025 mol. Total volume = 40 mL + 2.5 mL = 42.5 mL = 0.0425 L.

Concentrations:

[CH3COOH] = 0.00075 / 0.0425 ≈ 0.017647 M

[CH3COO^-] = 0.00025 / 0.0425 ≈ 0.005882 M

Henderson-Hasselbalch:

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

pKa = 4.735 (since Ka = 1.85e-5)

log(0.005882 / 0.017647) = log(0.3333) ≈ -0.477

So pH ≈ 4.735 - 0.477 ≈ 4.258 ≈ 4.26. That's correct. So the pH at 25% is 4.26, and at equivalence point is 8.52. Both match option C: 4.26; 8.52.

So the correct answer is C.
</think>