Question at position 5 A buffer solution is prepared by dissolving 0.23 mol of HC7H5O2 and 0.27 mol of NaC7H5O2 in water to yield 1.00 L of solution. The addition is 0.05 mol of NaOH to this solution causes the pH to increase only slightly. The pH does not increase drastically because the NaOH reacts with the _______ present in the buffer. The Ka of HC7H5O2 is 6.3 x 10–5. HC7H5O2H3O+OH–NaC7H5O2This is a buffer solution. The pH does not change upon addition of acid or base.单项选择题
A
HC7H5O2
B
H3O+
C
OH–
D
NaC7H5O2
E
This is a buffer solution. The pH does not change upon addition of acid or base.

Question

Question at position 5 A buffer solution is prepared by dissolving 0.23 mol of HC7H5O2 and 0.27 mol of NaC7H5O2 in water to yield 1.00 L of solution. The addition is 0.05 mol of NaOH to this solution causes the pH to increase only slightly. The pH does not increase drastically because the NaOH reacts with the _______ present in the buffer. The Ka of HC7H5O2 is 6.3 x 10–5. HC7H5O2H3O+OH–NaC7H5O2This is a buffer solution. The pH does not change upon addition of acid or base.单项选择题

A

HC7H5O2

B

H3O+

C

OH–

D

NaC7H5O2

E

This is a buffer solution. The pH does not change upon addition of acid or base.

BlackTom AI · Accepted Answer

A buffer consists of a weak acid (HC7H5O2) and its conjugate base (C7H5O2−); added OH– is consumed by the weak acid, minimizing pH change.

1) Relevant knowledge points and concepts
- A buffer resists pH change because a weak acid and its conjugate base can neutralize added base or acid.
- Henderson–Hasselbalch equation: pH = pKa + log([A−]/[HA]).
- Ka and pKa relation: pKa = −log10(Ka).

2) Step-by-step reasoning
- Initial moles in 1.00 L: HA = 0.23 mol, A− = 0.27 mol (so concentrations equal to moles).
- Added base: 0.05 mol NaOH supplies 0.05 mol OH−. In a buffer, OH− is neutralized by the weak acid (HA): HA + OH− → A− + H2O.
- After reaction: HA moles = 0.23 − 0.05 = 0.18 mol; A− moles = 0.27 + 0.05 = 0.32 mol.

3) Calculations with intermediate steps
- Ka = 6.3 × 10−5 → pKa = −log10(6.3 × 10−5) = 5 − log10(6.3) ≈ 5 − 0.7993 = 4.2007.
- Initial pH = pKa + log(0.27/0.23) = 4.2007 + log(1.1739) = 4.2007 + 0.0699 ≈ 4.2706.
- Final pH = pKa + log(0.32/0.18) = 4.2007 + log(1.7778) = 4.2007 + 0.2499 ≈ 4.4506.
- pH change ≈ 0.18 (a small increase), demonstrating buffer action.

4) Why the correct answer is correct
OH− is neutralized primarily by the weak acid HC7H5O2, which donates a proton to form its conjugate base. That conversion consumes the added base and keeps pH from rising drastically.

5) Why other options are incorrect
- H3O+: While H3O+ reacts with OH−, the buffer’s hydronium concentration is small and the dominant neutralization pathway in a buffer is OH− consuming HA, not free H3O+.
- OH−: This is the reagent added, not the species that reacts with OH−.
- NaC7H5O2 (A−): A− is the conjugate base; it will not react with added OH− (both are basic).
- “This is a buffer solution. The pH does not change upon addition of acid or base.”: Incorrect—buffer resists but does not prevent pH change entirely; we calculated a small change (~0.18 pH units).

6) Conclusion
Therefore the OH− reacts with the weak acid HC7H5O2, so the correct answer is HC7H5O2, matching the provided answer.

类似问题

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