📚 Buffer Solutions | 缓冲溶液
Buffer solutions are a fundamental concept in acid-base equilibria, crucial for maintaining stable pH in chemical and biological systems. They resist changes in pH when small amounts of acid or alkali are added. Understanding their composition and mechanism is essential for A-Level Chemistry.
缓冲溶液是酸碱平衡中的一个基本概念，对于维持化学和生物体系中稳定的pH至关重要。当加入少量酸或碱时，它们能抵抗pH的变化。理解其组成和作用机制对A-Level化学至关重要。
1. Definition of Buffer Solutions | 缓冲溶液的定义
A buffer solution is one that resists a change in pH when small quantities of an acid or an alkali are added. It usually consists of a weak acid and its conjugate base, or a weak base and its conjugate acid, in roughly equal concentrations.
缓冲溶液是一种当加入少量酸或碱时能抵抗pH变化的溶液。它通常由弱酸及其共轭碱，或弱碱及其共轭酸以大致相等的浓度组成。
Such solutions are essential wherever a constant pH is required, for instance in enzyme-catalysed reactions and industrial processes.
这类溶液在需要恒定pH的任何地方都至关重要，例如在酶催化反应和工业流程中。
2. Types of Buffer Solutions | 缓冲溶液的种类
There are two main types: acidic buffers and basic buffers. An acidic buffer is made from a weak acid and its salt (providing the conjugate base), e.g., ethanoic acid and sodium ethanoate. A basic buffer is made from a weak base and its salt (providing the conjugate acid), e.g., ammonia and ammonium chloride.
主要有两种类型：酸性缓冲液和碱性缓冲液。酸性缓冲液由弱酸及其盐（提供共轭碱）组成，例如乙酸和乙酸钠。碱性缓冲液由弱碱及其盐（提供共轭酸）组成，例如氨和氯化铵。
Acidic buffers maintain pH in the acidic range (pH < 7), while basic buffers maintain pH in the alkaline range (pH > 7).
酸性缓冲液将pH维持在酸性范围(pH < 7)，而碱性缓冲液将pH维持在碱性范围(pH > 7)。
3. How an Acidic Buffer Works – Le Chatelier's Principle | 酸性缓冲的工作原理 – 勒夏特列原理
oic acid / sodium ethanoate buffer. The weak acid, CH₃COOH, partially dissociates: CH₃COOH ⇌ ssociates fully: CH₃COONa → CH₃COO⁻ + Na⁺, providing a high concentration of CH₃COO⁻ ions.
以乙酸/乙酸钠缓冲液为例。弱酸CH₃COOH部分电离：CH₃COOH ⇌ CH₃COO⁻ + H⁺。乙酸钠完全电离：CH₃COONa → CH₃COO⁻ + Na⁺，提供了高浓度的CH₃COO⁻离子。
When a small amount of acid (H⁺) is added, the equilibrium shifts to the left to remove the added H⁺, as CH₃COO⁻ ions combine with H⁺ to form CH₃COOH. Thus, the pH remains almost unchanged.
当加入少量酸(H⁺)时，平衡向左移动以消耗加入的H⁺，因为CH₃COO⁻离子与H⁺结合生成CH₃COOH。因此，pH几乎不变。
When a small amount of alkali (OH⁻) is added, the OH⁻ reacts with H⁺ to form water, disturbing the equilibrium. The equilibrium shifts to the right, the weak acid dissociates further to replace the H⁺ ions, maintaining pH.
当加入少量碱(OH⁻)时，OH⁻与H⁺反应生成水，破坏了平衡。平衡向右移动，弱酸进一步电离以补充H⁺离子，从而维持pH。
4. How a Basic Buffer Works | 碱性缓冲的工作原理
An ammonia/ammonium chloride buffer relies on the equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Ammonium chloride provides NH₄⁺ ions. Adding acid (H⁺) consumes OH⁻, shifting equilibrium to the right to produce more OH⁻. Adding alkali (OH⁻) shifts equilibrium to the left to remove OH⁻, as NH₄⁺ reacts with OH⁻ to form NH₃ and H₂O.
氨/氯化铵缓冲液依赖于平衡：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。氯化铵提供NH₄⁺离子。加入酸(H⁺)消耗OH⁻，平衡向右移动生成更多OH⁻。加入碱(OH⁻)则平衡向左移动以消除OH⁻，因为NH₄⁺与OH⁻反应生成NH₃和H₂O。
The key is the presence of both the weak base and its conjugate acid to neutralise added H⁺ or OH⁻ without large shifts in pH.
关键在于同时存在弱碱及其共轭酸，能够中和加入的H⁺或OH⁻而不引起pH的大幅变化。
5. The Henderson-Hasselbalch Equation for Acidic Buffers | 酸性缓冲液的亨德森-哈塞尔巴尔赫方程
The pH of an acidic buffer can be calculated using the Henderson-Hasselbalch equation:
pH = pKₐ + log₁₀([A⁻] / [HA])
tration of the conjugate base (salt) and [HA] is the concentration of the weak acid. pKₐ = –log₁₀(Kₐ).
酸性缓冲液的pH可以使用亨德森-哈塞尔巴尔赫方程计算：pH = pKₐ + log₁₀([A⁻] / [HA])，其中[A⁻]是共轭碱（盐）的浓度，[HA]是弱酸的浓度。pKₐ = –log₁₀(Kₐ)。
most effective buffer region. The equation helps in designing buffers with a specific pH.
当[A⁻] = [HA]时，pH = pKₐ。这是最有效的缓冲区域。该方程有助于设计具有特定pH的缓冲液。
ffer, a similar equation applies: pOH = pK_b + log₁₀([conjugate acid]/[base]), and pH = 14 – asic buffer is often written as pH = pKₐ(conjugate acid) + log₁₀([base]/[conjugate acid]).
对于碱性缓冲液，类似的方程适用：pOH = pK_b + log₁₀([共轭酸]/[碱])，且pH = 14 – pOH。然而，碱性缓冲液的亨德森-哈塞尔巴尔赫形式常写作pH = pKₐ(共轭酸) + log₁₀([碱]/[共轭酸])。
6. Preparing Buffer Solutions | 缓冲溶液的配制
Buffers can be prepared by mixing a weak acid with its salt, e.g., adding ethanoic acid to sodium ethanoate solution. Alternatively, a buffer can be made by partially neutralising a weak acid with a strong base, or a weak base with a strong acid. For example, adding sodium hydroxide solution to an excess of ethanoic acid produces the buffer in situ.
缓冲液可以通过将弱酸与其盐混合来制备，例如将乙酸加入乙酸钠溶液中。或者，可以通过用强碱部分中和弱酸，或用强酸部分中和弱碱来制备缓冲液。例如，将氢氧化钠溶液加入过量的乙酸中可原位生成缓冲液。
the Henderson-Hasselbalch equation. Often a target pH ±1 unit around the pKₐ is practical.
为了获得所需的pH，可以根据亨德森-哈塞尔巴尔赫方程调整盐与酸的比例。通常在pKₐ附近±1个pH单位是可行的。
7. Buffer Capacity | 缓冲容量
e ratio [A⁻]/[HA]. The greater the concentrations, the greater the buffer capacity. Maximum buffer capacity occurs when [A⁻] = [HA], i.e., when pH = pKₐ.
缓冲容量是衡量缓冲液抵抗pH变化能力的指标。它取决于缓冲组分的总浓度以及[A⁻]/[HA]之比。浓度越大，缓冲容量越大。当[A⁻] = [HA]即pH = pKₐ时，缓冲容量最大。
When too much acid or base is added, the buffer becomes exhausted, and the pH changes sharply. A buffer works effectively only when the added amount of H⁺ or OH⁻ is small compared to the amounts of buffer components present.
当加入过多的酸或碱时，缓冲液被耗尽，pH会发生急剧变化。仅当加入的H⁺或OH⁻的量与存在的缓冲组分量相比较小时，缓冲液才有效工作。
8. Calculating pH Changes in Buffers | 计算缓冲溶液中pH的变化
When a small amount of strong acid is added to an acidic buffer, the added H⁺ reacts with the conjugate base A⁻ to form HA. The moles of A⁻ decrease and moles of HA increase. The new pH can be calculated using the new ratio in the Henderson-Hasselbalch equation, assuming the volume change is negligible or accounted for.
当向酸性缓冲液中加入少量强酸时，加入的H⁺与共轭碱A⁻反应生成HA。A⁻的物质的量减少，HA的物质的量增加。新的pH可以使用亨德森-哈塞尔巴尔赫方程中的新比值计算，假设体积变化可忽略或已考虑。
of HCl is added to 1 dm³ of buffer, the added H⁺ reacts with CH₃COO⁻. New [CH₃COO⁻] = 0.50 – mol dm⁻³, [CH₃COOH] = 0.50 + 0.10 = 0.60 mol dm⁻³. New pH = 4.74 + log₁₀(0.40/0.60) = 4.74 – 0.18 = 4.56. The change is small compared to adding acid to water.
示例：缓冲液含0.50 mol dm⁻³ CH₃COOH和0.50 mol dm⁻³ CH₃COONa。Kₐ = 1.8 × 10⁻⁵ mol dm⁻³，pKₐ = 4.74。初始pH = 4.74。若向1 dm³缓冲液中加入0.10 mol HCl，加入的H⁺与CH₃COO⁻反应。新的[CH₃COO⁻] = 0.50 – 0.10 = 0.40 mol dm⁻³，[CH₃COOH] = 0.50 + 0.10 = 0.60 mol dm⁻³。新pH = 4.74 + log₁₀(0.40/0.60) = 4.74 – 0.18 = 4.56。与向水中加酸相比变化很小。
Similarly, adding a strong base consumes HA, increasing A⁻, and the pH rises slightly. These stepwise calculations are typical in A-Level examination questions.
类似地，加入强碱会消耗HA，增加A⁻，pH略微上升。这种逐步计算在A-Level考试题中很常见。
9. The Blood Buffer System – Carbonate-Bicarbonate | 血液缓冲系统 – 碳酸/碳酸氢盐
Human blood maintains a pH around 7.40. The main buffer is the carbonic acid – hydrogencarbonate system: H₂CO₃ ⇌ H⁺ + HCO₃⁻. Dissolved CO₂ forms carbonic acid: CO₂ + H₂O ⇌ H₂CO₃. The equilibrium couples with respiration. When blood pH falls (acidic), the equilibrium shifts left, and breathing rate increases to expel CO₂, lowering H₂CO₃ concentration. When blood pH
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