A-Level化学 酸碱平衡 缓冲溶液 pH计算

A-Level化学 酸碱平衡 缓冲溶液 pH计算

1. Bronsted-Lowry Acid-Base Theory 布朗斯特酸碱理论

In the Bronsted-Lowry framework, an acid is a proton (H+) donor and a base is a proton acceptor. When an acid donates a proton, it becomes its conjugate base; when a base accepts a proton, it becomes its conjugate acid. This conjugate pair relationship is fundamental to understanding all acid-base equilibria.

在布朗斯特-劳里理论中，酸是质子(H+)供体，碱是质子受体。酸失去一个质子后变成其共轭碱；碱得到一个质子后变成其共轭酸。这种共轭对关系是理解所有酸碱平衡的基础。

For example, when HCl dissolves in water: HCl + H2O → H3O+ + Cl-. Here HCl is the acid (proton donor), H2O acts as the base (proton acceptor), H3O+ is the conjugate acid of water, and Cl- is the conjugate base of HCl. Water is amphoteric (can act as either acid or base), which is why it serves as the universal solvent for acid-base chemistry.

例如，HCl溶于水：HCl + H2O → H3O+ + Cl-。HCl是酸（质子供体），H2O作为碱（质子受体），H3O+是水的共轭酸，Cl-是HCl的共轭碱。水是两性的（既可作酸也可作碱），这是它成为酸碱化学通用溶剂的原因。

2. The Ionic Product of Water and pH Scale 水的离子积和pH标度

Pure water undergoes autoprotolysis: 2H2O ⇌ H3O+ + OH-. The equilibrium constant for this process is Kw = [H+][OH-] = 1.0 × 10^-14 mol² dm^-6 at 298K. In pure water, [H+] = [OH-] = 1.0 × 10^-7 mol dm^-3, giving pH = 7 at 25°C. Kw is temperature-dependent and increases with temperature because autoprotolysis is endothermic.

纯水发生自质子解离：2H2O ⇌ H3O+ + OH-。该过程的平衡常数 Kw = [H+][OH-] = 1.0 × 10^-14 mol² dm^-6（298K）。纯水中[H+] = [OH-] = 1.0 × 10^-7 mol dm^-3，因此25°C时pH = 7。Kw与温度有关，随温度升高而增大，因为自解离是吸热过程。

The pH scale is defined as pH = -log₁₀[H+]. Similarly, pOH = -log₁₀[OH-], and pKw = pH + pOH →14 at 298K. A solution with pH < 7 is acidic, pH = 7 is neutral, and pH > 7 is alkaline. The logarithmic nature means each unit change in pH represents a tenfold change in [H+].

pH的定义是pH = -log₁₀[H+]。类似地，pOH = -log₁₀[OH-]，且pKw = pH + pOH →14（298K）。pH < 7为酸性溶液，pH = 7为中性，pH > 7为碱性。对数性质意味着pH每变化1个单位，[H+]变化10倍。

3. Strong and Weak Acids and Bases 强酸强碱与弱酸弱碱

Strong acids (HCl, H₂SO₄, HNO₃) and strong bases (NaOH, KOH) dissociate completely in aqueous solution. For a monoprotic strong acid of concentration c, [H+] = c, so pH = -log₁₀(c). Weak acids (CH₃COOH, HF) and weak bases (NH₃, CH₃NH₂) only partially dissociate, establishing an equilibrium between the undissociated form and the ions.

强酸（HCl、H₂SO₄、HNO₃）和强碱（NaOH、KOH）在水溶液中完全解离。对于浓度为c的一元强酸，[H+] = c，因此pH = -log₁₀(c)。弱酸（CH₃COOH、HF）和弱碱（NH₃、CH₃NH₂）只部分解离，在未解离形式和离子之间建立平衡。

For a weak acid HA: HA ⇌ H+ + A-, the acid dissociation constant Ka = [H+][A-]/[HA]. A larger Ka indicates a stronger acid (more dissociation). The pKa = -log₁₀(Ka) provides a convenient scale: lower pKa means stronger acid. For a weak base B: B + H₂O ⇌ BH+ + OH-, the base dissociation constant Kb = [BH+][OH-]/[B].

对于弱酸HA：HA ⇌ H+ + A-，酸解离常数Ka = [H+][A-]/[HA]。Ka越大表示酸越强（解离程度越大）。pKa = -log₁₀(Ka)提供了方便的标度：pKa越低，酸越强。对于弱碱B：B + H₂O ⇌ BH+ + OH-，碱解离常数Kb = [BH+][OH-]/[B]。

4. Ka, pKa, Kb, and pKb Relationships 酸碱解离常数关系

For a conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10^-14 at 298K. Taking negative logarithms: pKa + pKb = pKw = 14 at 298K. This relationship is crucial: if you know the Ka of a weak acid, you can immediately calculate the Kb of its conjugate base, and vice versa. For example, ethanoic acid has pKa = 4.76, so its conjugate base (ethanoate ion) has pKb = 14 – 4.76 = 9.24.

对于共轭酸碱对，Ka × Kb = Kw = 1.0 × 10^-14（298K）。取负对数：pKa + pKb = pKw = 14（298K）。这个关系至关重要：如果知道弱酸的Ka，可以立即计算其共轭碱的Kb，反之亦然。例如，乙酸的pKa = 4.76，其共轭碱（乙酸根离子）的pKb = 14 – 4.76 = 9.24。

5. pH Calculations for Weak Acids and Bases 弱酸弱碱的pH计算

For a weak acid HA with concentration c and dissociation constant Ka: [H+] = √(Ka × c), provided c >> Ka and the approximation that [HA]eq ≈ c is valid (less than 5% dissociation). The pH is then pH = -log₁₀(√(Ka × c)) = ½(pKa – log₁₀c). Always check your assumption: if [H+]/c > 0.05, the approximation fails and the quadratic equation must be solved.

对于浓度为c、解离常数为Ka的弱酸HA：[H+] = √(Ka × c)，前提是c >> Ka且[HA]eq ≈ c的近似成立（解离度小于5%）。此时pH = -log₁₀(√(Ka × c)) = ½(pKa – log₁₀c)。务必检验假设：如果[H+]/c > 0.05，近似不成立，必须求解二次方程。

Worked example: Calculate the pH of 0.10 mol dm^-3 CH₃COOH (Ka = 1.74 × 10^-5). [H+] = √(1.74 × 10^-5 × 0.10) = √(1.74 × 10^-6) = 1.32 × 10^-3 mol dm^-3. Check: 1.32 × 10^-3 / 0.10 = 1.32% < 5%, so the approximation is valid. pH = -log₁₀(1.32 × 10^-3) = 2.88.

计算示例：计算0.10 mol dm^-3 CH₃COOH (Ka = 1.74 × 10^-5) 的pH。[H+] = √(1.74 × 10^-5 × 0.10) = √(1.74 × 10^-6) = 1.32 × 10^-3 mol dm^-3。检验：1.32 × 10^-3 / 0.10 = 1.32% < 5%，近似成立。pH = -log₁₀(1.32 × 10^-3) = 2.88。

6. Buffer Solutions 缓冲溶液

A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (acidic buffer), or a weak base and its conjugate acid (basic buffer). Common examples include ethanoic acid/sodium ethanoate and ammonia/ammonium chloride. Buffers are essential in biological systems: blood pH is maintained at 7.35-7.45 by the carbonic acid/hydrogencarbonate buffer system.

缓冲溶液能在加入少量酸或碱时抵抗pH变化。它由弱酸及其共轭碱（酸性缓冲液）或弱碱及其共轭酸（碱性缓冲液）组成。常见例子包括乙酸/乙酸钠和氨/氯化铵。缓冲液在生物系统中至关重要：血液pH通过碳酸/碳酸氢盐缓冲系统维持在7.35-7.45。

The Henderson-Hasselbalch equation relates the pH of a buffer to the pKa of the weak acid and the ratio of conjugate base to acid concentrations: pH = pKa + log₁₀([A-]/[HA]). This is the most important equation in buffer chemistry. When [A-] = [HA], pH = pKa, and the buffer has its maximum buffering capacity. The effective buffering range is approximately pKa ± 1.

亨德森-哈塞尔巴尔赫方程将缓冲液的pH与弱酸的pKa及共轭碱与酸的浓度比联系起来：pH = pKa + log₁₀([A-]/[HA])。这是缓冲化学中最重要的方程。当[A-] = [HA]时，pH = pKa，此时缓冲容量最大。有效缓冲范围约为pKa ± 1。

Worked example: A buffer contains 0.20 mol CH₃COOH and 0.15 mol CH₃COONa in 1.0 dm³. Calculate its pH (pKa of CH₃COOH = 4.76). pH = 4.76 + log₁₀(0.15/0.20) = 4.76 + log₁₀(0.75) = 4.76 + (-0.125) = 4.64. If 0.010 mol of HCl is added to 1.0 dm³ of this buffer, the added H+ reacts with CH₃COO- to form more CH₃COOH. New [CH₃COO-] = 0.15 – 0.010 = 0.14 mol dm^-3; new [CH₃COOH] = 0.20 + 0.010 = 0.21 mol dm^-3. New pH = 4.76 + log₁₀(0.14/0.21) = 4.76 + log₁₀(0.667) = 4.76 – 0.176 = 4.58. The pH changes by only 0.06 units, demonstrating the buffering action.

计算示例：缓冲液含0.20 mol CH₃COOH和0.15 mol CH₃COONa溶于1.0 dm³。计算其pH（CH₃COOH的pKa = 4.76）。pH = 4.76 + log₁₀(0.15/0.20) = 4.76 + log₁₀(0.75) = 4.76 + (-0.125) = 4.64。如果向1.0 dm³缓冲液中加入0.010 mol HCl，加入的H+与CH₃COO-反应生成更多CH₃COOH。新的[CH₃COO-] = 0.15 – 0.010 = 0.14 mol dm^-3；新的[CH₃COOH] = 0.20 + 0.010 = 0.21 mol dm^-3。新pH = 4.76 + log₁₀(0.14/0.21) = 4.76 + log₁₀(0.667) = 4.76 – 0.176 = 4.58。pH仅变化0.06个单位，展示了缓冲作用。

7. Titration Curves 滴定曲线

A titration curve plots pH against the volume of titrant added. The shape of the curve reveals key information about the acid and base involved. Four main types exist: strong acid-strong base (sharp endpoint at pH 7), strong acid-weak base (endpoint below pH 7), weak acid-strong base (endpoint above pH 7), and weak acid-weak base (no sharp endpoint). The equivalence point is where the amount of acid equals the amount of base (in moles), while the endpoint is where the indicator changes colour.

滴定曲线绘制pH随滴定剂加入体积的变化。曲线形状揭示了酸和碱的关键信息。存在四种主要类型：强酸-强碱（pH 7处有尖锐终点）、强酸-弱碱（终点pH < 7）、弱酸-强碱（终点pH > 7）和弱酸-弱碱（无明显终点）。当量点是酸的物质的量等于碱的物质的量的点，而终点是指示剂变色的点。

For a weak acid-strong base titration, the curve has several important features. The initial pH is higher than for a strong acid of the same concentration. Before the equivalence point, the mixture is a buffer so pH rises slowly. At half-neutralisation (halfway to the equivalence point), [HA] = [A-] so pH = pKa. At the equivalence point, pH > 7 because the conjugate base hydrolyses water: A- + H₂O ⇌ HA + OH-. After the equivalence point, pH is determined by excess strong base.

对于弱酸-强碱滴定，曲线有几个重要特征。初始pH高于同浓度强酸。在当量点之前，混合物是缓冲液，pH缓慢上升。在半中和点（当量点的一半路程），[HA] = [A-]，因此pH = pKa。在当量点，pH > 7，因为共轭碱水解水：A- + H₂O ⇌ HA + OH-。当量点之后，pH由过量强碱决定。

8. Acid-Base Indicators 酸碱指示剂

Indicators are weak acids or bases whose conjugate forms have different colours. For a weak acid indicator HIn: HIn ⇌ H+ + In-, where HIn and In- have different colours. The colour change occurs over a range of approximately pKIn ± 1. An indicator is suitable for a titration if its pKIn falls within the steep vertical region of the titration curve. Methyl orange (pKIn ≈ 3.7) is used for strong acid-strong base and strong acid-weak base titrations; phenolphthalein (pKIn ≈ 9.3) is used for strong acid-strong base and weak acid-strong base titrations.

指示剂是弱酸或弱碱，其共轭形式具有不同颜色。对于弱酸指示剂HIn：HIn ⇌ H+ + In-，其中HIn和In-颜色不同。颜色变化范围约为pKIn ± 1。如果指示剂的pKIn落在滴定曲线陡峭上升区间内，则该指示剂适合该滴定。甲基橙（pKIn ≈ 3.7）用于强酸-强碱和强酸-弱碱滴定；酚酞（pKIn ≈ 9.3）用于强酸-强碱和弱酸-强碱滴定。

9. Common Ion Effect 同离子效应

The common ion effect describes the suppression of the dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion. For example, adding sodium ethanoate (CH₃COONa) to a solution of ethanoic acid (CH₃COOH) introduces extra CH₃COO- ions. By Le Chatelier’s principle, the equilibrium CH₃COOH ⇌ H+ + CH₃COO- shifts left, reducing [H+] and raising the pH. This is the principle behind preparing buffer solutions.

同离子效应描述了通过加入含有共同离子的强电解质来抑制弱电解质解离的现象。例如，向乙酸(CH₃COOH)溶液中加入乙酸钠(CH₃COONa)会引入额外的CH₃COO-离子。根据勒夏特列原理，平衡CH₃COOH ⇌ H+ + CH₃COO-向左移动，降低[H+]并升高pH。这正是制备缓冲溶液背后的原理。

10. Salts and Salt Hydrolysis 盐类和水解

Salts formed from strong acid + strong base (e.g., NaCl, KNO₃) produce neutral solutions because neither ion hydrolyses. Salts from strong acid + weak base (e.g., NH₄Cl) are acidic because the cation hydrolyses: NH₄+ + H₂O ⇌ NH₃ + H₃O+. Salts from weak acid + strong base (e.g., CH₃COONa) are alkaline because the anion hydrolyses: CH₃COO- + H₂O ⇌ CH₃COOH + OH-. For salts from weak acid + weak base, both ions hydrolyse and the pH depends on the relative strengths of Ka and Kb.

由强酸+强碱形成的盐（如NaCl、KNO₃）产生中性溶液，因为两种离子都不水解。强酸+弱碱的盐（如NH₄Cl）呈酸性，因为阳离子水解：NH₄+ + H₂O ⇌ NH₃ + H₃O+。弱酸+强碱的盐（如CH₃COONa）呈碱性，因为阴离子水解：CH₃COO- + H₂O ⇌ CH₃COOH + OH-。对于弱酸+弱碱的盐，两种离子都水解，pH取决于Ka和Kb的相对强度。

11. Exam Tips 考试技巧

Always write the balanced equation for the acid-base equilibrium before attempting any calculation. This helps you identify which species are present and ensures you apply the correct Ka or Kb expression. When solving weak acid pH problems, always check your approximation: calculate [H+]/c × 100% and verify it is below 5%. If not, you must solve the quadratic equation. For buffer calculations, the Henderson-Hasselbalch equation is your primary tool ： memorise it and practice applying it in both directions.

在进行任何计算之前，务必先写出酸碱平衡的配平方程式。这有助于识别存在的物种，确保使用正确的Ka或Kb表达式。解答弱酸pH问题时，务必检验近似条件：计算[H+]/c × 100%，验证其低于5%。如果不满足，必须求解二次方程。对于缓冲液计算，亨德森-哈塞尔巴尔赫方程是主要工具：熟记并在两个方向上练习应用。

On titration curve questions, be able to identify the type of titration from the curve shape, state which indicator is suitable and why, and explain the pH at the equivalence point in terms of salt hydrolysis. For buffer questions, expect to calculate the new pH after adding a small amount of strong acid or base ： always determine which species is consumed and which is produced, then recalculate the [A-]/[HA] ratio before applying the Henderson-Hasselbalch equation.

对于滴定曲线题目，要能根据曲线形状判断滴定类型，说明哪种指示剂合适及原因，并从盐水解角度解释当量点的pH。对于缓冲液题目，准备计算加入少量强酸或强碱后的新pH：始终确定哪种物质被消耗、哪种物质产生，然后重新计算[A-]/[HA]比值，再应用亨德森-哈塞尔巴尔赫方程。