📚 pH Calculations | pH计算
Understanding pH calculations is fundamental to mastering acid–base chemistry at A-Level. From strong acids and bases to buffers and salt hydrolysis, this guide covers the essential principles, equations, and problem-solving strategies required for Cambridge A-Level Chemistry.
理解pH计算是掌握A-Level酸与碱化学的基础。从强酸强碱到缓冲溶液和盐的水解，本指南涵盖了剑桥A-Level化学所需的基本原理、方程式和解题策略。
1. The Definition of pH and pOH | pH与pOH的定义
xpressed in mol dm⁻³. The concept extends to pOH, where pOH = −log₁₀[OH⁻]. Together, they obey the relation pH + pOH = 14 (at 25 °C).
pH定义为氢离子浓度的负对数（以10为底）：pH = −log₁₀[H⁺]。在25 °C的水溶液中，[H⁺]通常以mol dm⁻³表示。这一概念扩展至pOH，pOH = −log₁₀[OH⁻]。它们共同满足关系式pH + pOH = 14（25 °C下）。
⁻³ has a pH of 3.0. Each unit change in pH represents a tenfold change in [H⁺], which means pH is a logarithmic scale that compresses a huge range of concentrations.
当[H⁺] = 1.0 × 10⁻³ mol dm⁻³时，pH为3.0。pH每变化1个单位，[H⁺]就变化10倍，这意味着pH是一个对数标度，浓缩了极宽范围的浓度。
2. The Ionic Product of Water, Kw | 水的离子积Kw
Water undergoes self-ionisation: H₂O ⇌ 1.0 × 10⁻¹⁴ mol² dm⁻⁶, so in pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³ and pH = 7.00.
水发生自偶电离：H₂O ⇌ H⁺ + OH⁻。这一可逆反应的平衡常数为Kw = [H⁺][OH⁻]。在298 K时，Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶，因此在纯水中[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，pH = 7.00。
10⁻¹² mol² dm⁻⁶, making neutral pH 6.00. This must be considered when measuring pH at non‑standard temperatures.
Kw随温度变化；温度升高Kw增大。在100 °C时，Kw ≈ 1 × 10⁻¹² mol² dm⁻⁶，此时中性pH为6.00。在非标准温度下测量pH时必须考虑这一点。
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (at 25 °C)
3. pH of Strong Acids | 强酸的pH
tration of the acid. Therefore, the calculation is direct: pH = −log₁₀(acid concentration).
强酸如HCl、H₂SO₄和HNO₃在水溶液中完全电离。对于一元强酸，[H⁺]等于酸的初始浓度。因此计算直接：pH = −log₁₀(酸的浓度)。
g₁₀(0.050) = 1.30. When the concentration is very low (e.g. ≤ 1 × 10⁻⁶ mol dm⁻³), the self‑ionisation of water contributes to [H⁺] and must be included.
示例：0.050 mol dm⁻³ HCl给出[H⁺] = 0.050 mol dm⁻³，因此pH = −log₁₀(0.050) = 1.30。当浓度极低时（如≤ 1 × 10⁻⁶ mol dm⁻³），水的自偶电离对[H⁺]的贡献不可忽略，必须考虑。
4. pH of Strong Bases | 强碱的pH
concentration, then determine pOH = −log₁₀[OH⁻], and finally use pH = 14 − pOH (at 25 °C).
强碱如NaOH和KOH也完全电离，释放OH⁻离子。为求pH，先由碱浓度计算[OH⁻]，再求pOH = −log₁₀[OH⁻]，最后利用pH = 14 − pOH（25 °C）。
r Group 2 metal hydroxides like Ba(OH)₂, remember that each mole of the base releases two moles of OH⁻.
示例：0.020 mol dm⁻³ NaOH给出[OH⁻] = 0.020 mol dm⁻³，pOH = −log₁₀(0.020) = 1.70，故pH = 14 − 1.70 = 12.30。对于Ba(OH)₂等第2族金属氢氧化物，注意每摩尔碱释放两摩尔OH⁻。
5. Weak Acids and Ka | 弱酸与Ka
A weak acid, HA, only partially dissociates in water: HA ⇌ s the extent of ionisation: Ka = [H⁺][A⁻] / [HA]. The smaller the Ka, the weaker the acid.
弱酸HA在水中仅部分电离：HA ⇌ H⁺ + A⁻。酸解离常数Ka定量描述了电离程度：Ka = [H⁺][A⁻] / [HA]。Ka越小，酸越弱。
s c and x is the amount dissociated, then at equilibrium [H⁺] = [A⁻] = x and [HA] = c − x.
在弱酸溶液中，平衡浓度通过解离的化学计量关系相联系。若初始浓度为c，电离量为x，则平衡时[H⁺] = [A⁻] = x，[HA] = c − x。
Ka = [H⁺][A⁻] / [HA]
6. Calculating pH of Weak Acids: Approximations | 弱酸pH计算：近似处理
For many weak acids, the degree of dissociation is very small, so x ≪ pproximation [HA] ≈ c. Substituting into the Ka expression gives Ka ≈ x² / c, hence [H⁺] ≈ √Ka × c. This is valid when c / Ka > 1000 or the dissociation is less than 5%.
对于许多弱酸，电离度非常小，因此x ≪ c，可使用近似[HA] ≈ c。代入Ka表达式得Ka ≈ x² / c，从而[H⁺] ≈ √Ka × c。当c / Ka > 1000或电离度小于5%时，该近似成立。
ample: Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵ mol dm⁻³). [H⁺] = √1.8 × 10⁻⁵ × 0.100 = −log₁₀(1.34 × 10⁻³) ≈ 2.87. Always check that x / c < 0.05 to verify the approximation.
示例：计算0.100 mol dm⁻³乙酸（Ka = 1.8 × 10⁻⁵ mol dm⁻³）的pH。[H⁺] = √1.8 × 10⁻⁵ × 0.100 = 1.34 × 10⁻³ mol dm⁻³，pH = −log₁₀(1.34 × 10⁻³) ≈ 2.87。务必验证x / c < 0.05以确认近似合理。
7. pKa and Acid Strength | pKa与酸强度
mpare acid strengths. A smaller pKa indicates a stronger acid. For example, pKa of ethanoic acid is 4.76, while chloroethanoic acid has a pKa around 2.86 due to the electron‑withdrawing chlorine atom.
正如pH是一个方便的对数标度，pKa = −log₁₀(Ka)被用来比较酸强度。pKa越小，酸性越强。例如，乙酸的pKa为4.76，而氯乙酸由于吸电子的氯原子，其pKa约为2.86。
The relationship pH = pKa + log₁₀([A⁻]/[HA]) is known as the Henderson– hows that when [A⁻] = [HA], pH = pKa. This is particularly useful for buffer calculations.
关系式pH = pKa + log₁₀([A⁻]/[HA])被称为Henderson–Hasselbalch方程。它表明当[A⁻] = [HA]时，pH = pKa。这对缓冲溶液的计算特别有用。
8. Weak Bases and Kb | 弱碱与Kb
Weak bases such as ammonia, NH₃, react partially with water: NH₃ + H₂O ⇌ [NH₄⁺][OH⁻] / [NH₃]. The same approximation techniques used for weak acids apply: [OH⁻] ≈ √Kb × c.
弱碱如氨NH₃部分与水反应：NH₃ + H₂O ⇌ NH₄⁺ + OH⁻。碱解离常数Kb定义为Kb = [NH₄⁺][OH⁻] / [NH₃]。用于弱酸的相同近似方法在此适用：[OH⁻] ≈ √Kb × c。
on with Kb = 1.8 × 10⁻⁵ mol dm⁻³, [OH⁻] ≈ 1.34 × 10⁻³ mol dm⁻³, pOH = 2.87, so pH = 11.13.
要求pH，先由[OH⁻]计算pOH，再用pH = 14 − pOH。对于0.10 mol dm⁻³ NH₃溶液，Kb = 1.8 × 10⁻⁵ mol dm⁻³，[OH⁻] ≈ 1.34 × 10⁻³ mol dm⁻³，pOH = 2.87，故pH = 11.13。
9. Buffer Solutions | 缓冲溶液
and its conjugate acid). The pH of an acidic buffer can be calculated using the Henderson–Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]).
缓冲溶液在加入少量酸或碱时能够抵抗pH变化。它们由弱酸及其共轭碱（或弱碱及其共轭酸）组成。酸性缓冲液的pH可用Henderson–Hasselbalch方程计算：pH = pKa + log₁₀([A⁻]/[HA])。
s chemists to design buffers with a specific pH by choosing an acid with a pKa close to the desired value.
当弱酸与其盐的浓度相等时，对数项为零，pH = pKa。这允许化学家通过选择pKa接近目标值的酸来设计特定pH的缓冲液。
dding a small amount of strong acid converts some A⁻ to HA, but the ratio [A⁻]/[HA] changes only slightly, so the pH remains nearly constant.
示例：含有0.20 mol dm⁻³乙酸和0.20 mol dm⁻³乙酸钠的缓冲液pH = pKa = 4.76。加入少量强酸会将部分A⁻转化为HA，但[A⁻]/[HA]的比值变化很小，因此pH几乎保持恒定。
10. pH of Salt Solutions (Hydrolysis) | 盐溶液的pH（水解）
from a weak acid and a strong base (e.g. CH₃COONa) give basic solutions because the anion hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
强酸与强碱形成的盐（如NaCl）产生中性溶液（pH = 7）。然而，弱酸与强碱形成的盐（如CH₃COONa）由于阴离子水解呈碱性：CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻。
estimated using the hydrolysis constant Kh = Kw / Ka. For 0.10 mol dm⁻³ CH₃COONa, [OH⁻] ≈ √Kh × c and then pH = 14 − pOH. This calculation often yields a mildly alkaline pH around 8–9.
此类盐溶液的pH可用水解常数Kh = Kw / Ka估算。对于0.10 mol dm⁻³ CH₃COONa，[OH⁻] ≈ √Kh × c，然后求得pH = 14 − pOH。计算结果通常是pH在8–9之间的弱碱性。
a weak base (e.g. NH₄Cl) give acidic solutions due to hydrolysis of the cation: NH₄⁺ + H₂O ⇌ + H₃O⁺. The acid dissociation constant for the conjugate acid is Ka(NH₄⁺) = Kw / Kb(NH₃).
相反，强酸与弱碱形成的盐（如NH₄Cl）因阳离子水解而呈酸性：NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺。其共轭酸的酸解离常数为Ka(NH₄⁺) = Kw / Kb(NH₃)。
11. pH Curves and Indicator Selection | pH曲线与指示剂选择
During an acid–base titration, the pH changes in a characteristic way, producing a curve. For a strong acid–strong base titration, the equivalence point is at pH 7, and the vertical region is steep. For a weak acid–strong base titration, the equivalence point is above pH 7.
在酸碱滴定过程中，pH以特征方式变化，形成曲线。强酸-强碱滴定的等当点在pH 7，且垂直区域陡峭。弱酸-强碱滴定的等当点高于pH 7。
An appropriate indicator must change colour within the vertical section of the pH curve. Phenolphthalein (range 8.2–10.0) is suitable for strong base titrations, while methyl orange (3.1–4.4) fits strong acid titrations. Choosing the wrong indicator leads to a gradual, unclear endpoint.
合适的指示剂必须在pH曲线垂直部分内变色。酚酞（范围8.2–10.0）适用于强碱滴定，而甲基橙（3.1–4.4）适用于强酸滴定。选择错误的指示剂会导致终点渐变、不明确。
Half‑equivalence points are used to determine pKa: at the half‑ int of a weak acid, pH = pKa. This is an important practical application of buffer theory.
半等当点可用于测定pKa：在弱酸的半中和点，pH = pKa。这是缓冲理论的重要实际应用。
12. Exam Tips and Common Pitfalls | 考试技巧与常见陷阱
Always state your assumptions clearly – at the autoprotolysis of water is negligible. Check the validity of these assumptions at the end of the calculation. Marks are often awarded for showing the verification step.
始终清楚地陈述你的假设——例如近似[HA] ≈ c以及水的自偶电离可忽略。在计算结束时检验这些假设的合理性。展示验证步骤通常有分。
Common mistakes include forgetting to square root when solving for [H⁺] in weak acids, using the wrong unit for volume, and confusing Ka with Kb. When mixing solutions, account for the dilution effect on all species before substituting into equilibrium expressions.
常见错误包括求解弱酸[H⁺]时忘记开平方根、使用错误的体积单位、混淆Ka和Kb。在混合溶液时，代入平衡表达式前要考虑所有物种的稀释效应。
For buffer pH changes, always calculate the new moles of acid and conjugate base after the addition of strong acid or base, then apply Henderson–Hasselbalch. Do not simply add the volumes without accounting for the reaction stoichiometry.
对于缓冲液pH变化，始终先计算加入强酸或强碱后酸和共轭碱的新物质的量，再应用Henderson–Hasselbalch方程。不能简单地加合体积而不考虑反应的化学计量关系。
Finally, remember that Kw and therefore neutral pH change with temperature. In exam questions, if the temperature is not 25 °C, be prepared to use the given Kw value to determine [H⁺] and pH.
最后，记住Kw及其中性pH随温度变化。在考试题中，如果温度不是25 °C，要准备使用给定的Kw值来确定[H⁺]和pH。
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