📚 AP Chemistry: Acid-Base Calculations | AP化学：酸碱计算专题解析
Acid-base chemistry lies at the heart of the AP Chemistry curriculum, combining equilibrium concepts, stoichiometry, and a deep understanding of molecular behavior in aqueous solutions. Mastering pH calculations, buffer systems, and titration curves is essential for success on the exam. This revision guide unpacks every major calculation type you will encounter, from strong acids to polyprotic systems, with clear step-by-step explanations and paired Chinese translations to reinforce key ideas.
酸碱化学是AP化学课程的核心内容，融合了平衡概念、化学计量学以及分子在水溶液中行为的深刻理解。掌握pH计算、缓冲系统和滴定曲线对于考试成功至关重要。本复习指南解析你将遇到的所有主要计算类型，从强酸到多元酸体系，并提供清晰的逐步解释和配对的中文翻译，以巩固关键概念。
1. The pH Scale and Water Autoionization | pH标度与水的自电离
The pH of a solution is defined as pH = −log[H₃O⁺], where [H₃O⁺] is the hydronium ion concentration in mol·L⁻¹. In pure water at 25 °C, a tiny fraction of molecules undergoes autoionization: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). The equilibrium constant for this process is the ion product of water, K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴.
溶液的pH定义为pH = −log[H₃O⁺]，其中[H₃O⁺]是水合氢离子浓度，单位为mol·L⁻¹。在25℃的纯水中，极少部分分子发生自电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。该过程的平衡常数是水的离子积，K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴。
Because the product of [H⁺] and [OH⁻] is constant, we can always find one if we know the other. The pOH scale is related by pOH = −log[OH⁻] and pH + pOH = 14.00 at 25 °C. Note that K_w increases with temperature, so at higher temperatures the neutral pH is below 7.
由于[H⁺]和[OH⁻]的乘积为常数，已知其一时必能求出另一者。pOH标度由pOH = −log[OH⁻]定义，且在25℃时满足pH + pOH = 14.00。注意K_w随温度升高而增大，因此在较高温度下，中性pH值低于7。
2. Strong Acids and Bases: Complete Dissociation | 强酸与强碱：完全解离
Strong acids such as HCl, HNO₃, and H₂SO₄ (first proton) ionize completely in water. For a monoprotic strong acid, [H₃O⁺] equals the initial acid concentration. Thus, the pH of 0.10 M HCl is simply −log(0.10) = 1.00.
强酸如HCl、HNO₃和H₂SO₄（第一级质子）在水中完全电离。对于一元强酸，[H₃O⁺]等于酸的初始浓度。因此，0.10 M HCl的pH值直接为−log(0.10) = 1.00。
Similarly, strong bases like NaOH and KOH dissociate fully to give OH⁻. For a 0.050 M NaOH solution, [OH⁻] = 0.050 M, pOH = −log(0.050) ≈ 1.30, and pH = 14.00 − 1.30 = 12.70. Always check the stoichiometry: Ba(OH)₂ provides two OH⁻ per formula unit, so [OH⁻] is twice the base concentration.
类似地，强碱如NaOH和KOH完全解离产生OH⁻。对于0.050 M NaOH溶液，[OH⁻] = 0.050 M，pOH = −log(0.050) ≈ 1.30，pH = 14.00 − 1.30 = 12.70。务必检查化学计量数：Ba(OH)₂每化学式提供两个OH⁻，因此[OH⁻]是碱浓度的两倍。
3. Weak Acids and Acid Dissociation Constant (Kₐ) | 弱酸与酸解离常数 (Kₐ)
A weak acid only partially ionizes in water. For a generic weak acid HA, the equilibrium is HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq). The acid dissociation constant is Kₐ = [H₃O⁺][A⁻]/[HA], where water is omitted from the expression because it is the solvent.
弱酸在水中仅部分电离。对于通式弱酸HA，平衡为HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)。酸解离常数表示为Kₐ = [H₃O⁺][A⁻]/[HA]，其中水因作为溶剂而从表达式中省略。
The smaller the Kₐ, the weaker the acid. For example, acetic acid (CH₃COOH) has Kₐ = 1.8 × 10⁻⁵, while hydrofluoric acid (HF) has Kₐ = 6.8 × 10⁻⁴. To compare acid strengths quickly, use pKₐ = −log Kₐ; the smaller the pKₐ, the stronger the acid.
Kₐ越小，酸越弱。例如，醋酸（CH₃COOH）的Kₐ = 1.8 × 10⁻⁵，而氢氟酸（HF）的Kₐ = 6.8 × 10⁻⁴。为快速比较酸强度，可使用pKₐ = −log Kₐ；pKₐ越小，酸越强。
4. Weak Bases and Base Dissociation Constant (Kb) | 弱碱与碱解离常数 (Kb)
Weak bases such as ammonia (NH₃) react with water to produce hydroxide ions: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). The base dissociation constant is Kb = [NH₄⁺][OH⁻]/[NH₃]. A larger Kb indicates a stronger weak base.
弱碱如氨（NH₃）与水反应产生氢氧根离子：NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)。碱解离常数Kb = [NH₄⁺][OH⁻]/[NH₃]。Kb越大，表示弱碱越强。
Common weak bases include amines, carbonates, and the conjugate bases of weak acids. For instance, the acetate ion (CH₃COO⁻) is a weak base with Kb = K_w/Kₐ of its conjugate acid. Understanding Kb allows you to calculate the pH of a solution of a weak base using an ICE table, just as for weak acids.
常见的弱碱包括胺类、碳酸盐以及弱酸的共轭碱。例如，醋酸根离子（CH₃COO⁻）是一种弱碱，其Kb = K_w / Kₐ(共轭酸）。理解Kb后，你可以像处理弱酸一样，使用ICE表计算弱碱溶液的pH。
5. pH Calculations for Weak Acids: ICE Tables | 弱酸pH计算：ICE表
To find the pH of a weak acid solution, set up an ICE (Initial, Change, Equilibrium) table. Suppose we have 0.10 M CH₃COOH with Kₐ = 1.8 × 10⁻⁵. The initial concentrations are [CH₃COOH] = 0.10 M, [H₃O⁺] ≈ 0, [CH₃COO⁻] = 0. Let x be the amount that dissociates. At equilibrium, [H₃O⁺] = [CH₃COO⁻] = x, and [CH₃COOH] = 0.10 − x.
为求弱酸溶液的pH，需建立ICE（初始、变化、平衡）表。假设有0.10 M CH₃COOH，Kₐ = 1.8 × 10⁻⁵。初始浓度为[CH₃COOH] = 0.10 M，[H₃O⁺] ≈ 0，[CH₃COO⁻] = 0。设解离的量为x。平衡时，[H₃O⁺] = [CH₃COO⁻] = x，[CH₃COOH] = 0.10 − x。
Substituting into the Kₐ expression gives Kₐ = x²/(0.10 − x) = 1.8 × 10⁻⁵. Since Kₐ is very small, we can approximate 0.10 − x ≈ 0.10. Then x² = (1.8 × 10⁻⁵)(0.10) → x = √(1.8 × 10⁻⁶) = 1.3 × 10⁻³ M. Checking the 5% rule: (1.3×10⁻³ / 0.10)×100% = 1.3% < 5%, so the approximation is valid. Hence pH = −log(1.3×10⁻³) ≈ 2.89.
代入Kₐ表达式得Kₐ = x²/(0.10 − x) = 1.8 × 10⁻⁵。由于Kₐ很小，可近似0.10 − x ≈ 0.10。则x² = (1.8 × 10⁻⁵)(0.10) → x = √(1.8 × 10⁻⁶) = 1.3 × 10⁻³ M。检查5%规则：(1.3×10⁻³ / 0.10)×100% = 1.3% < 5%，近似成立。因此pH = −log(1.3×10⁻³) ≈ 2.89。
If the approximation fails (x is more than 5% of the initial concentration), you must solve the quadratic equation: x²/(C₀ − x) = Kₐ. The AP exam may require you to set up the quadratic and simplify using the approximation, but always check validity.
如果近似失败（x超过初始浓度的5%），则必须求解二次方程：x²/(C₀ − x) = Kₐ。AP考试可能要求你建立二次方程并通过近似简化，但务必检查其有效性。
6. pH Calculations for Weak Bases | 弱碱pH计算
For a weak base like 0.20 M NH₃ (Kb = 1.8 × 10⁻⁵), we again use an ICE table. The reaction is NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Let x = [OH⁻] at equilibrium. Then Kb = x²/(0.20 − x) ≈ x²/0.20, giving x = √(1.8×10⁻⁵ × 0.20) = 1.9 × 10⁻³ M. Check approximation: 1.9×10⁻³/0.20 = 0.95% < 5%, valid. Then pOH = −log(1.9×10⁻³) ≈ 2.72, and pH = 14.00 − 2.72 = 11.28.
对于弱碱如0.20 M NH₃（Kb = 1.8 × 10⁻⁵），我们同样使用ICE表。反应为NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。设x = [OH⁻]平衡浓度。则Kb = x²/(0.20 − x) ≈ x²/0.20，得出x = √(1.8×10⁻⁵ × 0.20) = 1.9 × 10⁻³ M。检查近似：1.9×10⁻³/0.20 = 0.95% < 5%，有效。然后pOH = −log(1.9×10⁻³) ≈ 2.72，pH = 14.00 − 2.72 = 11.28。
Always remember to obtain pH from pOH at the end. The same 5% rule and quadratic approach apply to weak base calculations. For the conjugate base of a weak acid, first determine Kb from Kₐ, then proceed with the ICE table.
务必记住最后从pOH求出pH。相同的5%规则和二次方程方法也适用于弱碱计算。对于弱酸的共轭碱，先由Kₐ求出Kb，然后使用ICE表计算。
7. The Relationship between Kₐ and Kb: Kₐ × Kb = K_w | Kₐ 与 Kb 的关系
For any conjugate acid-base pair, the product of the acid dissociation constant of the acid and the base dissociation constant of its conjugate base equals the ion product of water: Kₐ × Kb = K_w = 1.0 × 10⁻¹⁴ at 25 °C. This relationship arises from the simultaneous equilibria: HA ⇌ H⁺ + A⁻ and A⁻ + H₂O ⇌ HA + OH⁻; adding these gives 2H₂O ⇌ H₃O⁺ + OH⁻, equilibrium constant K_w.
对于任何共轭酸碱对，酸的酸解离常数与其共轭碱的碱解离常数的乘积等于水的离子积：Kₐ × Kb = K_w = 1.0 × 10⁻¹⁴（25℃）。此关系源自同时存在的平衡：HA ⇌ H⁺ + A⁻ 与 A⁻ + H₂O ⇌ HA + OH⁻；两者相加得 2H₂O ⇌ H₃O⁺ + OH⁻，平衡常数为K_w。
This allows you to find the Kb of a conjugate base if you know the Kₐ of the parent weak acid. For example, the Kb of the acetate ion is K_w/Kₐ(CH₃COOH) = 1.0×10⁻¹⁴ / 1.8×10⁻⁵ = 5.6×10⁻¹⁰. The weaker the acid, the stronger its conjugate base, and vice versa.
这使得你能在知道母体弱酸Kₐ的情况下求出其共轭碱的Kb。例如，醋酸根离子的Kb = K_w / Kₐ(CH₃COOH) = 1.0×10⁻¹⁴ / 1.8×10⁻⁵ = 5.6×10⁻¹⁰。酸越弱，其共轭碱越强，反之亦然。
8. Salt Hydrolysis and pH of Salt Solutions | 盐的水解与盐溶液pH
When a salt dissolves, its ions may react with water (hydrolysis), affecting pH. The salt of a strong acid and strong base (e.g., NaCl) produces a neutral solution (pH = 7). Salts from a strong base and weak acid (e.g., CH₃COONa) yield basic solutions because the anion is the conjugate base of a weak acid and accepts a proton from water.
盐溶解时，其离子可能与水反应（水解），影响pH。强酸强碱盐（如NaCl）产生中性溶液（pH = 7）。强碱弱酸盐（如CH₃COONa）产生碱性溶液，因为阴离子是弱酸的共轭碱，能从水中夺取质子。
For CH₃COONa, the acetate ion hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. The pH is calculated using the Kb of acetate. A salt from a weak base and strong acid (e.g., NH₄Cl) is acidic because NH₄⁺ donates a proton: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. Use the Kₐ of NH₄⁺ (Kₐ = K_w/Kb(NH₃) = 5.6×10⁻¹⁰) in an ICE table.
对于CH₃COONa，醋酸根离子水解：CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻。使用醋酸根的Kb计算pH。弱碱强酸盐（如NH₄Cl）呈酸性，因为NH₄⁺提供质子：NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺。使用NH₄⁺的Kₐ（Kₐ = K_w/Kb(NH₃) = 5.6×10⁻¹⁰）建立ICE表计算。
Salts from weak acids and weak bases require comparing Kₐ and Kb of the ions. If Kₐ(ion) > Kb(ion), the solution is acidic; if Kb > Kₐ, basic; if Kₐ ≈ Kb, near neutral. The pH is calculated from the relevant equilibrium.
弱酸弱碱盐需要比较离子的Kₐ和Kb。若Kₐ(离子) > Kb(离子)，溶液呈酸性；若Kb > Kₐ，呈碱性；若Kₐ ≈ Kb，接近中性。通过相关平衡计算pH。
9. Buffer Solutions and the Henderson-Hasselbalch Equation | 缓冲溶液与Henderson-Hasselbalch方程
A buffer is a solution that resists pH change when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in appreciable concentrations. The pH of a buffer is given by the Henderson-Hasselbalch equation:
pH = pKₐ + log ([A⁻]/[HA])
缓冲溶液是一种在加入少量酸或碱时能抵抗pH变化的溶液。它由浓度可观的弱酸及其共轭碱（或弱碱及其共轭酸）组成。缓冲溶液的pH由Henderson-Hasselbalch方程给出：
pH = pKₐ + log ([A⁻]/[HA])
Here [A⁻] is the concentration of the conjugate base and [HA] the concentration of the weak acid. This equation assumes that the dissociation of the acid is negligible and that the buffer components do not react with water significantly. The buffer is most effective when the ratio [A⁻]/[HA] is between 0.1 and 10, i.e., pH = pKₐ ± 1.
式中[A⁻]为共轭碱浓度，[HA]为弱酸浓度。此方程假设酸的电离可忽略，且缓冲组分不与水发生显著反应。当比例[A⁻]/[HA]在0.1到10之间，即pH = pKₐ ± 1 时，缓冲效率最高。
Example: A buffer made by mixing 0.50 M CH₃COOH and 0.30 M CH₃COONa has pH = −log(1.8×10⁻⁵) + log(0.30/0.50) = 4.74 + (−0.22) = 4.52. Changing the ratio changes the pH predictably. The AP exam often asks for the pH of a buffer after adding a small amount of strong acid or base; use stoichiometry to adjust the concentrations of HA and A⁻, then apply the equation.
示例：由0.50 M CH₃COOH和0.30 M CH₃COONa混合制成的缓冲液，其pH = −log(1.8×10⁻⁵) + log(0.30/0.50) = 4.74 + (−0.22) = 4.52。改变比例可预测地改变pH。AP考试常要求计算加入少量强酸或强碱后缓冲液的pH；先用化学计量法调整HA和A⁻的浓度，再应用方程。
10. Buffer Capacity and Preparing Buffers | 缓冲容量与缓冲液的配制
Buffer capacity refers to the amount of acid or base a buffer can neutralize before the pH changes significantly. It depends on the absolute concentrations of the buffer components; higher concentrations yield higher capacity. A buffer is exhausted when one component is completely consumed.
缓冲容量是指缓冲液在pH发生显著变化前能中和的酸或碱的量。它取决于缓冲组分的绝对浓度；浓度越高，容量越大。当某一组分被完全消耗时，缓冲作用失效。
To prepare a buffer at a desired pH, choose a weak acid whose pKₐ is close to the target pH, then adjust the ratio of conjugate base to acid. For example, to make a buffer at pH 5.0, acetic acid (pKₐ 4.74) is a good choice. Using the Henderson-Hasselbalch equation: 5.0 = 4.74 + log([A⁻]/[HA]) → [A⁻]/[HA] = 10^(0.26) ≈ 1.8. Mixing 0.18 M sodium acetate with 0.10 M acetic acid would achieve this.
要配制特定pH的缓冲液，选择pKₐ接近目标pH的弱酸，然后调整共轭碱与酸的比例。例如，配制pH 5.0的缓冲液，醋酸（pKₐ 4.74）是不错的选择。使用Henderson-Hasselbalch方程：5.0 = 4.74 + log([A⁻]/[HA]) → [A⁻]/[HA] = 10^(0.26) ≈ 1.8。将0.18 M醋酸钠与0.10 M醋酸混合即可实现。
Always check that the chosen concentrations are practical and that the buffer capacity is sufficient for the intended use. In the lab, buffers are often made by partially neutralizing a weak acid with a strong base, directly yielding the required ratio.
务必检查所选浓度是否可行，以及缓冲容量是否满足使用需求。在实验室中，常通过用强碱部分中和弱酸来直接制得所需比例的缓冲液。
11. Acid-Base Titrations: Strong Acid – Strong Base | 酸碱滴定：强酸-强碱
In a strong acid–strong base titration, the pH starts low, rises slowly at first, then sharply near the equivalence point, and finally levels off at high pH. The equivalence point occurs when stoichiometrically equal amounts of acid and base have reacted, and the pH is exactly 7.00 at 25 °C because the product is a neutral salt.
在强酸-强碱滴定中，pH起始较低，起初缓慢上升，随后在等当点附近急剧变化，最后在高pH处趋于平缓。等当点出现在酸和碱按化学计量完全反应之时，由于产物为中性盐，25℃时pH恰为7.00。
Before the equivalence point, the pH is determined by the excess strong acid or base. For example, if titrating 25.0 mL of 0.10 M HCl with 0.10 M NaOH, after adding 10.0 mL of NaOH, unreacted HCl remains. The total volume is 35.0 mL, and moles of excess H⁺ = (0.10 M × 25.0 mL) − (0.10 M × 10.0 mL) = 1.5 mmol. [H⁺] = 1.5 mmol / 35.0 mL = 0.043 M, pH ≈ 1.37. After the
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