IB Chemistry: pH Calculation Key Points | IB 化学：pH计算 考点精讲

📚 IB Chemistry: pH Calculation Key Points | IB 化学：pH计算 考点精讲

pH calculations form a core component of the IB Chemistry curriculum, appearing in topics on acids and bases, equilibrium, and analytical techniques. Mastering these calculations requires a solid understanding of the definitions of pH and pOH, strong and weak acids/bases, Ka and Kb, the ionic product of water Kw, buffer systems, and titration curves. This guide breaks down each key area with bilingual explanations and worked examples to help you ace your exams.

pH 计算是 IB 化学课程的核心组成部分，出现在酸碱、平衡和分析技术等主题中。要掌握这些计算，需要扎实理解 pH 和 pOH 的定义、强弱酸/碱、Ka 和 Kb、水的离子积 Kw、缓冲体系以及滴定曲线。本指南通过双语讲解和例题分析，分解各个关键领域，助你轻松应对考试。

1. Definition of pH and pOH | pH与pOH的定义

[OH⁻]. At 25°C, the ionic product of water Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴, so pH + pOH = 14.

溶液的 pH 定义为氢离子浓度的负对数（以 10 为底）：pH = −log[H⁺]。类似地，pOH = −log[OH⁻]。在 25°C 时，水的离子积 Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴，因此 pH + pOH = 14。

These logarithmic relationships mean that a change of one pH unit corresponds to a tenfold change in [H⁺]. For instance, a solution with pH 3 has 10 times the [H⁺] of a solution with pH 4.

这些对数关系意味着 pH 值每变化 1 个单位，[H⁺] 就变化十倍。例如，pH=3 的溶液的 [H⁺] 是 pH=4 溶液的 10 倍。

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2. Strong Acids and Bases | 强酸与强碱

centration (multiplied by the number of dissociable protons if diprotic). For a monoprotic strong acid of concentration c, pH = −log c.

强酸（例如 HCl、HNO₃ 和 H₂SO₄ 的第一步电离）在水溶液中完全解离。因此，[H⁺] 等于初始酸浓度（如果是二元酸则乘以可电离质子数）。对于浓度为 c 的一元强酸，pH = −log c。

from pH = 14 − pOH at 25°C. For Ba(OH)₂, remember each formula unit produces two OH⁻ ions.

强碱（如 NaOH 和 KOH）完全解离产生 OH⁻ 离子。[OH⁻] 等于碱浓度，pOH = −log[OH⁻]；在 25°C 时，pH = 14 − pOH。对于 Ba(OH)₂，注意每个单元产生两个 OH⁻ 离子。

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3. Weak Acids and the Acid Dissociation Constant, Kₐ | 弱酸与酸解离常数 Kₐ

Weak acids partially dissociate: HA ⇌ h initial concentration c, assuming the amount dissociated x is small (c/Kₐ > 100), [H⁺] ≈ √Kₐ·c.

弱酸部分解离：HA ⇌ H⁺ + A⁻。平衡常数 Kₐ = [H⁺][A⁻]/[HA]。对于初始浓度为 c 的弱酸，假设解离度 x 很小（c/Kₐ > 100），则 [H⁺] ≈ √Kₐ·c。

Always check the approximation: if [H⁺]/c < 5% se the quadratic equation: Kₐ = x²/(c − x), and solve for x = [H⁺]. The pH is then −log x.

务必检查近似条件：若 [H⁺]/c < 5%，近似有效。否则需使用二次方程：Kₐ = x²/(c − x)，求解 x = [H⁺]，然后 pH = −log x。

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4. Weak Bases and the Base Dissociation Constant, K_b | 弱碱与碱解离常数 K_b

Weak bases like NH₃ react with water: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. The base dissociation constant K_b = [NH₄⁺][OH⁻]/[NH₃]. Analogously, for a weak base of concentration c, [OH⁻] ≈ √K_b·c.

弱碱如 NH₃ 与水反应：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。碱解离常数 K_b = [NH₄⁺][OH⁻]/[NH₃]。类似地，对于浓度为 c 的弱碱，[OH⁻] ≈ √K_b·c。

Therefore, pOH = −log[OH⁻] and then pH = 14 − pOH. The same approximation criteria apply.

于是，pOH = −log[OH⁻]，然后 pH = 14 − pOH。相同的近似条件适用。

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5. The Autoionization of Water and K_w | 水的自解离与 K_w

Water undergoes autoionization: 2H₂O ⇌ H₃O⁺ + OH⁻. The equilibrium constant K_w ] = 1.0 × 10⁻¹⁴ at 25°C. In pure water, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol·dm⁻³, giving pH = 7.

水存在自解离：2H₂O ⇌ H₃O⁺ + OH⁻。在 25°C 时，平衡常数 K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴。纯水中 [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol·dm⁻³，pH=7。

K_w is temperature-dependent; it increases with temperature. In IB questions, unless otherwise stated, assume 25°C and K_w = 10⁻¹⁴.

K_w 随温度变化；温度升高而增大。IB 考题中，除非另有说明，默认 25°C 且 K_w = 10⁻¹⁴。

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6. pH of Salt Solutions | 盐溶液的 pH

. A salt from a strong acid and weak base (e.g., NH₄Cl) produces an acidic solution: NH₄⁺ hydrolyzes to give H⁺. A salt from a weak acid and strong base (e.g., CH₃COONa) gives a basic solution: CH₃COO⁻ hydrolyzes to produce OH⁻.

盐类因水解可产生酸性或碱性溶液。强酸强碱盐（如 NaCl）溶液呈中性（pH=7）。强酸弱碱盐（如 NH₄Cl）呈酸性：NH₄⁺ 水解产生 H⁺。弱酸强碱盐（如 CH₃COONa）呈碱性：CH₃COO⁻ 水解产生 OH⁻。

For a salt of a weak acid and weak base, the pH depends on the relative strengths: if Kₐ (of the conjugate acid) > K_b (of the conjugate base), the solution is acidic, and vice versa.

对于弱酸弱碱盐，pH 值取决于相对强度：若共轭酸的 Kₐ > 共轭碱的 K_b，溶液呈酸性，反之呈碱性。

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7. Buffer Solutions and the Henderson–Hasselbalch Equation | 缓冲溶液与 Henderson–Hasselbalch 方程

base (or a weak base and its conjugate acid). The pH of a buffer is given by the Henderson–Hasselbalch equation: pH = pKₐ + log([A⁻]/[HA]), where pKₐ = −log Kₐ.

缓冲溶液能抵抗少量酸碱加入引起的 pH 变化。它由弱酸及其共轭碱（或弱碱及其共轭酸）组成。缓冲液的 pH 值由 Henderson–Hasselbalch 方程给出：pH = pKₐ + log([A⁻]/[HA])，其中 pKₐ = −log Kₐ。

Buffer capacity is highest when the ratio is close to 1 and when concentrations are high.

当 [A⁻] = [HA] 时，pH = pKₐ。当比率接近 1 且浓度较高时，缓冲能力最强。

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8. pH Curves and Titration Calculations | pH 曲线与滴定计算

In an acid–base titration, the pH changes gradually, then rapidly near the equivalence point. For a strong acid–strong base titration, the equivalence point is at pH 7. For a weak acid–strong base titration, the equivalence point is >7; for a strong acid–weak base, it is <7.

在酸碱滴定中，pH 先缓慢变化，在等当点附近急剧变化。强酸-强碱滴定的等当点 pH=7。弱酸-强碱滴定等当点 >7；强酸-弱碱滴定等当点 <7。

ful: at this point, [HA] = [A⁻], so pH = pKₐ. The buffer region extends around this point.

半等当点十分有用：此时 [HA] = [A⁻]，因此 pH = pKₐ。缓冲区域围绕该点展开。

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9. Common Pitfalls in pH Calculations | pH 计算中的常见误区

Students often forget to check the approximation for weak acids/bases, leading to inaccurate pH values, especially at low concentrations. Always verify the 5% rule.

学生常忘记检查弱酸/弱碱的近似条件，导致 pH 值不准确，尤其在浓度很低时。务必验证 5% 规则。

Another mistake is using moles instead of concentrations in the Henderson–Hasselbalch equation. Since the volume is the same for both species, you can use mole ratios, but the equation strictly requires concentrations. Be careful with dilutions.

另一个错误是在 Henderson–Hasselbalch 方程中使用物质的量而非浓度。由于两者体积相同，可以用摩尔比

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