A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in Na. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4.

Question

BlackTom AI · Accepted Answer

Henderson-Hasselbalch relates pH to pKa and the conjugate base/acid ratio: pH = pKa + log([A-]/[HA]).

1) Relevant concepts
- A buffer of HF/ F- resists pH change; adding HCl converts F- to HF by: H+ + F- → HF.
- Use stoichiometry to find new moles, then concentrations (accounting for volume change), then Henderson-Hasselbalch with pKa from Ka.

2) Step-by-step reasoning and stoichiometry
- Initial moles (1.00 L): n(HF) = 0.250 M × 1.00 L = 0.250 mol; n(F-) = 0.250 mol.
- Moles HCl added: 1.00 M × 0.1000 L = 0.1000 mol H+.
- Reaction: H+ consumes F- to form HF. New moles:
  n(F-)final = 0.250 − 0.1000 = 0.150 mol
  n(HF)final = 0.250 + 0.1000 = 0.350 mol
- Total volume = 1.00 L + 0.1000 L = 1.100 L.
- Final concentrations:
  [HF] = 0.350 mol / 1.100 L = 0.31818... M
  [F-] = 0.150 mol / 1.100 L = 0.13636... M
  Note: the ratio [A-]/[HA] = 0.150/0.350 = 3/7 ≈ 0.428571 (volume cancels).

3) pKa and calculation
- Ka(HF) = 3.5 × 10^-4 → pKa = −log(3.5×10^-4)
  log(3.5×10^-4) = log(3.5) + log(10^-4) = 0.5441 − 4 = −3.4559 so pKa = 3.4559 ≈ 3.456.
- log([A-]/[HA]) = log(0.428571) ≈ −0.3680.
- pH = 3.4559 + (−0.3680) = 3.0879 ≈ 3.09.

4) Why the correct answer is correct
The exact stoichiometry of strong acid neutralizing the conjugate base shifts the ratio of base/acid to 3/7; that ratio with the HF pKa yields pH ≈ 3.09 via Henderson-Hasselbalch. Volume change was included, but cancels in the ratio.

5) Why other choices are incorrect
- 4.11 and 3.82 are too high; they correspond to much larger [A-]/[HA] ratios than the actual 0.4286.
- 3.46 equals pKa (would occur if [A-]=[HA]), but here [A-]<[HA], so pH must be lower than 3.456.
- 2.78 is too low; it would imply a much smaller [A-]/[HA] than 0.4286.
Thus none of the other options match the calculated ratio and pKa.

6) Conclusion
Calculated pH = 3.09, which matches the provided correct answer 3.09.

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in Na. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4.单项选择题

类似问题

What amounts of a weak acid and a strong base, when combined, will lead to the formation of a buffer?

The gradient of the straight line, which passes through (3, 5) and (5, 3) is:

Calculate the De Broglie wavelength of an electron moving with a velocity of [math: 7.26×106ms−1] .Give your answer in standard form to 2 decimal places without units.Example format: [math: 1.23×10−45N] should be written as [ 1.23e-45 ]

A 0.18 m section of wire is placed in a uniform magnetic field of 0.40 T. The wire carries a current of 1.70 A. The angle between the magnetic field and the wire is [math: 67.0∘]. What is the magnitude of the force experienced by the wire?

Two point charges, [math: q1=+7μC] and [math: q2=−7μC] , are placed 0.21 m apart in a vacuum. What is the magnitude of the net electric field, at the midpoint between the two charges?*Note: [math: 1.23×104NC−1] is written as 1.23e4 N/C

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单