A 50.00 mL sample of 0.0950 M acetic acid (Ka = 1.8 × 10−5) is being titrated with 0.0848 M NaOH. What is the pH after 28.00 mL of NaOH has been added?

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We begin by identifying the reaction and calculating the quantities involved. The acetic acid  (CH3COOH) has initial moles = 0.0500 L × 0.0950 M = 0.00475 mol. The NaOH added has moles = 0.02800 L × 0.0848 M = 0.0023744 mol. Since the base amount is less than the acid, we form a buffer consisting of the weak acid and its conjugate base (acetate). The remaining acid after neutralization is 0.00475 − 0.0023744 = 0.0023756 mol, and the produced conjugate base (acetate) is 0.0023744 mol.

The total volume after addition is 50.0 mL + 28.0 mL = 78.0 mL = 0.078 L. Therefore, the concentrations in the buffer are: [HA] = 0.0023756 / 0.078 ≈ 0.03044 M and [A−] = 0.0023744 / 0.078 ≈ 0.03044 M. These values are essentially equal, indicating a nearly equimolar buffer.

Using the Henderson–Hasselbalch equation for a buffer: pH = pKa + log([A−]/[HA]). The Ka for acetic acid is 1.8 × 10−5, so pKa ≈ −log10(1.8×10−5) ≈ 4.7447. With [A−] ≈ [HA], the ratio [A−]/[HA] ≈ 1, and log(1) = 0. Thus pH ≈ pKa ≈ 4.745, which rounds to about 4.74.

Now, evaluate each answer option:
- 3.18: This is far below the pKa region and would occur if the solution were dominated by the weak acid with little conjugate base present. The calculated [A−]/[HA] ratio near 1 rules this out, so this is incorrect.
- 3.06: Similar to 3.18, this value lies well into a more acidic pH region than our buffer calculation supports; the buffer composition does not support such a low pH here, making this option incorrect.
- 4.44: This value is lower than the expected pH for a near-equimolar acetic acid/acetate buffer at this Ka; while in the correct range, it does not align with the calculated pH and would imply a significantly larger imbalance between [A−] and [HA], so this is misleading.
- 5.04: This would correspond to a much more basic solution or a buffer with substantially more conjugate base than acid, which is not the case since moles of acid and base consumed were nearly equal; thus this option is incorrect.
- 4.74: This option matches the calculated pH from the Henderson–Hasselbalch treatment for a near-equimolar acetic acid/acetate buffer formed after partial neutralization, making it the correct choice.

In summary, the calculation shows the system behaves as a weak acid–conjugate base buffer with nearly equal concentrations of CH3COOH and CH3COO− after adding 28.0 mL of NaOH, yielding a pH approximately equal to pKa, which is about 4.74. The other values arise from scenarios inconsistent with the buffer composition at this equivalence stage.

CHM-1020-LD01.2025SU Practice Exam II

A 50.00 mL sample of 0.0950 M acetic acid (Ka = 1.8 × 10−5) is being titrated with 0.0848 M NaOH. What is the pH after 28.00 mL of NaOH has been added?

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