Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. What is the pH after the following titrant additions: 40.0 mL?

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To start, establish the amounts involved: the initial moles of acetic acid (HA) are 50.0 mL × 0.100 M = 0.0050 mol. The NaOH titrant adds 40.0 mL × 0.100 M = 0.0040 mol of OH−. After the neutralization reaction HA + OH− → A− + H2O, the remaining HA is 0.0050 − 0.0040 = 0.0010 mol and the produced A− is 0.0040 mol.

Calculate the total volume after mixing: 50.0 mL + 40.0 mL = 90.0 mL = 0.0900 L. This gives concentrations [HA] = 0.0010 mol / 0.0900 L ≈ 1.11×10^−2 M and [A−] = 0.0040 mol / 0.0900 L ≈ 4.44×10^−2 M.

Since the solution now contains a weak acid and its conjugate base, it behaves as a buffer. Use the Henderson–Hasselbalch equation: pH = pKa + log([A−]/[HA]). The pKa of acetic acid is about 4.76.

Compute the ratio [A−]/[HA]: (0.0040/0.0900) ÷ (0.0010/0.0900) = 0.0040/0.0010 = 4.0. The log of 4.0 is approximately 0.602.

Plug into the buffer formula: pH ≈ 4.76 + 0.602 ≈ 5.36. This aligns with rounding to two decimals, yielding about pH ≈ 5.36, which is consistent with the provided correct answer around 5.35.

Now assess the other options:
- 8.64 and 8.65 would require a solution rich in base relative to acid (a much larger ratio of A− to HA), which is not the case here since only 40 mL of 0.100 M NaOH has been added to 50 mL of 0.100 M acetic acid, yielding a modest A− formation.
- 7.00 would indicate a near-neutral solution, but with a sizable acetate concentration relative to remaining acetic acid, the pH shifts into the acidic buffer region around 5–5.5, not 7.
- 5.35 is consistent with the calculated buffer pH using the Henderson–Hasselbalch relationship and the stoichiometry of the neutralization.

In summary, the buffer calculation gives pH ≈ 5.36, so the option closest to that value is 5.35, while the other options are not supported by the molar ratios and the buffer formula.

CHM-1020-LD01.2025SU Practice Exam II

Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. What is the pH after the following titrant additions: 40.0 mL?

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