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Correct Answer: D
Correct Answer: D
Method / Topic: General Chemistry
The correct option matches the required rule and the given information.
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Correct Answer: D
Method / Topic: General Chemistry
Apply Henderson–Hasselbalch directly: pH = pKa + log([A−]/[HA]) = 4.76 + log(10). Since log(10) = 1, the pH is 5.76. A tenfold excess of conjugate base over weak acid places the pH exactly one unit above the pKa. This result follows from the balanced chemical relationship and the stated conditions; no unstated biological assumption is needed. On MCAT chemistry problems, first identify the governing principle, write the relevant equation with units, and only then substitute values. Check whether the
prevents dosage, dilution, and preparation errors.
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PASSAGE 3 — ACETATE BUFFER A researcher prepares 1.00 L of a buffer containing 0.100 mol CH3COOH and 0.100 mol CH3COO−. The acid dissociation constant of acetic acid is Ka = 1.8 × 10−5, so pKa ≈ 4.76. Volume changes after small additions may be neglected. The Henderson–Hasselbalch equation is pH = pKa + log([A−]/[HA]). QUESTION 8: If [CH3COO−]/[CH3COOH] = 10, what is the pH?
Explanation
Correct Answer: D
Method / Topic: General Chemistry
The correct option matches the required rule and the given information.
Short Explanation
Detailed Explanation
Correct Answer: D
Method / Topic: General Chemistry
Apply Henderson–Hasselbalch directly: pH = pKa + log([A−]/[HA]) = 4.76 + log(10). Since log(10) = 1, the pH is 5.76. A tenfold excess of conjugate base over weak acid places the pH exactly one unit above the pKa. This result follows from the balanced chemical relationship and the stated conditions; no unstated biological assumption is needed. On MCAT chemistry problems, first identify the governing principle, write the relevant equation with units, and only then substitute values. Check whether the
prevents dosage, dilution, and preparation errors.
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