📚 IB CIE Chemistry: pH Calculations Key Points | IB CIE化学：pH计算考点精讲
Mastering pH calculations is a fundamental skill for IB and CIE Chemistry students, forming the backbone of acid-base equilibria, titration analysis, and buffer design. This guide distills the essential concepts, formulas, and exam shortcuts you need to tackle any pH problem with confidence. From strong acid fundamentals to the nuances of Henderson-Hasselbalch and salt hydrolysis, every key point is explained with clarity.
掌握pH计算是IB与CIE化学学生的基本功，它构成了酸碱平衡、滴定分析及缓冲体系设计的基础。本指南提炼了必备概念、公式与应试技巧，助你自信应对各类pH题型。从强酸基础到亨德森-哈塞尔巴尔赫方程的微妙之处，再到盐类水解，每个考点都清晰讲解。
1. Definition of pH and the Ionic Product of Water, Kw | pH定义与水的离子积Kw
neutral pH of 7. The ionic product of water, Kw = [H⁺][OH⁻], equals 1.0 × 10⁻¹⁴ at this temperature and increases with rising temperature because the auto-ionisation of water is endothermic.
pH标度是对氢离子浓度的对数度量：pH = -log₁₀[H⁺]。在298 K的纯水中，[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³，中性pH为7。水的离子积Kw = [H⁺][OH⁻]在此温度下为1.0 × 10⁻¹⁴，并随温度升高而增大，因为水的自解离是吸热过程。
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K
pKw = -log₁₀Kw = 14 at 298 K
You should also be comfortable with the relationship pOH = -log₁₀[OH⁻] and pH + pOH = pKw.
在任何水溶液中，Kw始终成立，即若[H⁺]上升，[OH⁻]必定相应下降。也应熟练掌握pOH = -log₁₀[OH⁻]以及pH + pOH = pKw 的关系。
2. pH of Strong Acids | 强酸的pH计算
tion, provided the acid is not too dilute (typically [acid] > 10⁻⁶ mol dm⁻³, otherwise the auto-ionisation of water contributes significantly). Thus, pH = -log₁₀[acid].
强酸如HCl、HNO₃ 和 H₂SO₄（一级电离）在水中完全解离。对于一元强酸，只要浓度不是极稀（通常[acid] > 10⁻⁶ mol dm⁻³，否则水的自解离贡献不可忽略），[H⁺]就等于酸的浓度。因此，pH = -log₁₀[acid]。
g acids such as H₂SO₄, the first proton is fully dissociated, and the second proton (HSO₄⁻ ⇌ alculations, you may be instructed to treat only the first dissociation as strong, or use the given Ka₂ to work out total [H⁺].
对于二元强酸如H₂SO₄，第一个质子完全解离，而第二个质子（HSO₄⁻ ⇌ H⁺ + SO₄²⁻）部分解离，Ka₂ 约为 1.0 × 10⁻² mol dm⁻³。考试计算中，可能要求只把第一步解离视为强酸，或使用给出的Ka₂求得总[H⁺]。
- Always write the dissociation equation and assume complete ionisation for strong acids.
- 永远写出解离方程式，并假设强酸完全电离。
- Check whether water's contribution to [H⁺] matters when the acid concentration is below 10⁻⁶ mol dm⁻³.
- 当酸浓度低于10⁻⁶ mol dm⁻³时，检查水的[H⁺]贡献是否重要。
3. pH of Strong Bases | 强碱的pH计算
ion. Then use Kw to find [H⁺] = Kw / [OH⁻] and compute pH = -log₁₀[H⁺], or simply find pOH first and subtract from 14 at 298 K.
强碱如NaOH、KOH和Ba(OH)₂完全解离释放OH⁻离子。对于一元强碱，[OH⁻]等于碱的浓度。然后利用Kw求得[H⁺] = Kw / [OH⁻]并计算pH = -log₁₀[H⁺]，或先求pOH，在298 K时用14减去pOH。
ry dilute, the auto-ionisation of water must be considered, leading to solving a quadratic equation involving Kw.
请记住，Ba(OH)₂每摩尔提供两摩尔OH⁻，故[OH⁻] = 2 × [Ba(OH)₂]。当碱极度稀释时，必须考虑水的自解离，这需要求解涉及Kw的二次方程。
pH = 14 - pOH = 14 + log₁₀[OH⁻] (at 298 K)
4. Weak Acids and Ka | 弱酸与酸离解常数Ka
Weak acids partially dissociate in equilibrium: HA ⇌ olution, assume that the amount dissociated, x, is very small compared to the initial concentration, c, so that [HA] ≈ c. Then [H⁺] = √Ka × c and pH = -log₁₀[H⁺].
弱酸在水中部分解离，存在平衡：HA ⇌ H⁺ + A⁻。酸离解常数Ka定义为Ka = [H⁺][A⁻] / [HA]。Ka越小，酸越弱。计算弱酸溶液的pH时，假设解离量x远小于初始浓度c，因此[HA] ≈ c。于是[H⁺] = √Ka × c，进而pH = -log₁₀[H⁺]。
and CIE exams often provide Ka values and expect you to use the simplified expression where appropriate.
当c / Ka > 1000 时，近似[HA] ≈ c 成立；否则需解二次方程Ka = x² / (c - x)。IB与CIE考试常给出Ka值，并期望在适当情况下使用简化式。
- Write the ICE table (Initial, Change, Equilibrium) to set up the Ka expression.
- 绘制ICE表格（初始、变化、平衡），建立Ka表达式。
- Always check the approximation before using the simplified formula; if invalid, solve the full equation.
- 使用简化公式前务必验证近似条件；若不满足，求解完整方程。
5. Weak Bases and Kb | 弱碱与碱离解常数Kb
Weak bases such as NH₃ and amines accept protons: B + H₂O ⇌ ⁺ + OH⁻. The base dissociation constant is Kb = [BH⁺][OH⁻] / [B]. For a weak base, [OH⁻] = √Kb × c and pH can be calculated via pOH and Kw, or directly by using Ka of the conjugate acid if pKa is known.
弱碱如NH₃和胺类接受质子：B + H₂O ⇌ BH⁺ + OH⁻。碱离解常数Kb = [BH⁺][OH⁻] / [B]。对于弱碱，[OH⁻] = √Kb × c，然后通过pOH和Kw计算pH，或者若已知共轭酸的pKa，也可直接利用pKa换算。
find the pH of a weak base from a given pKa value. Convert: pKa + pKb = pKw (14 at 298 K).
共轭酸碱对的Ka与Kb关系：Ka × Kb = Kw。当需要从已知pKa值求弱碱的pH时，此关系尤为有用。换算公式：pKa + pKb = pKw（298 K时为14）。
pKa + pKb = 14
6. pH of Buffer Solutions | 缓冲溶液的pH计算
d its conjugate acid). The pH of an acidic buffer can be calculated from the equilibrium expression: Ka = [H⁺][A⁻] / [HA], which rearranges to [H⁺] = Ka × [HA] / [A⁻]. Taking negative logarithms gives the Henderson-Hasselbalch equation.
缓冲溶液能在加入少量酸或碱时抵制pH变化。它由弱酸及其共轭碱（或弱碱及其共轭酸）组成。酸性缓冲液的pH可从平衡表达式推出：Ka = [H⁺][A⁻] / [HA]，重排得[H⁺] = Ka × [HA] / [A⁻]。取负对数即得亨德森-哈塞尔巴尔赫方程。
Key points: the buffer' er does not change its pH significantly because the ratio [HA]/[A⁻] remains constant. Always use equilibrium concentrations; in buffer calculations, the initial amounts can often be used because the dissociation is suppressed by the common ion effect.
要点：当酸与其共轭碱浓度相等时（pH = pKa），缓冲能力最强。稀释缓冲液时，因[HA]/[A⁻]比值不变，pH基本不变。必须使用平衡浓度计算；但在缓冲计算中，通常可用初始物质的量，因为同离子效应抑制了解离。
7. Henderson-Hasselbalch Equation | 亨德森-哈塞尔巴尔赫方程
The Henderson-Hasselbalch equation provides a direct route to buffer pH:
pH = pKa + log₁₀([A⁻] / [HA])
This equation is valid when the concentrations of acid and conjugate base are sufficiently high, and the assumptions of the buffer model hold. In IB and CIE exams, you'll often be asked to calculate the ratio of salt to acid needed to achieve a target pH, or to determine pH after adding small amounts of strong acid or base to a buffer.
亨德森-哈塞尔巴尔赫方程为缓冲液pH提供了直接计算途径。当酸与共轭碱的浓度足够高，且缓冲模型成立时，该方程准确。在IB与CIE考试中，常要求计算达到目标pH所需的盐与酸的比例，或计算向缓冲液中添加少量强酸、强碱后的pH。
Remember that adding a strong acid to a buffer converts some A⁻ to HA, while adding a strong base converts some HA to A⁻. Use stoichiometry first, then apply the Henderson-Hasselbalch equation with the new amounts.
记住，向缓冲液中加强酸会将部分A⁻转化为HA，加强碱则把部分HA转化为A⁻。先进行化学计量计算，再用新物质的量代入亨德森-哈塞尔巴尔赫方程。
8. Acid-Base Titrations and pH Curves | 酸碱滴定与pH曲线
Titration pH curves plot pH against volume of titrant added. Key features include the initial pH, the buffering region, the equivalence point, and the final pH. For a strong acid–strong base titration, the equivalence point is at pH 7 (298 K). For a weak acid–strong base titration, the equivalence point occurs at a pH > 7 because the conjugate base hydrolyses water to produce OH⁻. The opposite is true for a strong acid–weak base titration.
滴定pH曲线以pH对加入的滴定剂体积作图。关键特征包括初始pH、缓冲区域、等当点和最终pH。强酸-强碱滴定的等当点pH=7（298 K）。弱酸-强碱滴定的等当点pH>7，因为共轭碱水解生成OH⁻。强酸-弱碱滴定则相反。
At the half-equivalence point of a weak acid– so pH = pKa. This is a crucial concept for determining pKa experimentally from a pH curve.
在弱酸-强碱滴定的半等当点处，[HA] = [A⁻]，故pH = pKa。该概念是通过pH曲线实验测定pKa的关键。
- Sketching and interpreting titration curves is a common exam task; practise labeling significant points.
- 绘制并解读滴定曲线是常见考题；练习标记各重要点位。
- Use the correct species to calculate pH at the equivalence point: hydrolysis of the salt formed.
- 计算等当点pH时，需使用生成的盐的水解平衡。
9. Indicators and pH | 指示剂与pH
equal to pKᵢₙ ± 1. Choose an indicator whose pH range lies entirely within the steep portion of the titration curve; for a strong acid–g base titration, both methyl orange and phenolphthalein are suitable, but for a weak acid–strong base titration, phenolphthalein (range 8.3–10.0) is ideal, while for a strong acid–weak base, methyl orange (range 3.1–4.4) is appropriate.
酸碱指示剂是自身为弱酸或弱碱、共轭形态颜色不同的物质。颜色变化发生在约pKᵢₙ ± 1的pH区间内。应选择变色范围完全落在滴定曲线陡峭部分的指示剂；强酸-强碱滴定可选用甲基橙或酚酞，弱酸-强碱滴定宜用酚酞（范围8.3–10.0），强酸-弱碱滴定则适用甲基橙（范围3.1–4.4）。
ch the two coloured forms are present in equal concentrations. Understanding the indicator's equilibrium: HIn ⇌ H⁺ + In⁻, and applying pKᵢₙ = pH, is often tested.
指示剂的pKᵢₙ 是两种颜色形态浓度相等时的pH值。常考查指示剂平衡 HIn ⇌ H⁺ + In⁻ 及 pKᵢₙ = pH 的理解。
10. Dilution and pH Changes | 稀释与pH变化
Diluting a strong acid decreases [H⁺] proportionally, and pH rises by 1 unit for every 10-fold dilution (provided concentrations stay above ~10⁻⁶ mol dm⁻³). For a weak acid, dilution shifts the equilibrium to increase the degree of dissociation, so the pH change is less dramatic. For a buffer, dilution essentially leaves pH unchanged due to the constant ratio [HA]/[A⁻]. In extreme dilution, the pH of any solution approaches 7 because water's auto-ionisation dominates.
稀释强酸会使[H⁺]成比例下降，每稀释10倍pH升高约1个单位（前提是浓度保持在约10⁻⁶ mol dm⁻³以上）。对弱酸，稀释促使平衡向增大解离度方向移动，因此pH变化不那么显著。对缓冲液，因[HA]/[A⁻]比值不变，稀释基本不改变pH。在极度稀释时，任何溶液的pH都趋近7，因为水的自解离占主导。
Examiners may ask you to predict the effect of dilution on pH and justify your answer using Le Chatelier's principle or equilibrium arguments. Always compare the initial and final molar concentrations carefully.
考官可能会要求预测稀释对pH的影响，并用勒夏特列原理或平衡移动加以解释。务必仔细比较初始与最终的摩尔浓度。
11. pH of Salts (Hydrolysis) | 盐类水解的pH
ng base and a weak acid (e.g., CH₃COONa) hydrolyse to give basic solutions because the anion acts as a conjugate base: A⁻ + H₂O ⇌ parent weak acid. The [OH⁻] and then pH are found with the usual weak base approximation.
强酸强碱生成的盐（如NaCl）溶液呈中性（pH = 7）。强碱弱酸盐（如CH₃COONa）水解呈碱性，因为阴离子作为共轭碱：A⁻ + H₂O ⇌ HA + OH⁻。可利用阴离子的Kb进行计算，其中Kb = Kw / Ka(母体弱酸)。然后用弱碱的常规近似求得[OH⁻]和pH。
Salts of a weak base and a strong acid (e.g., NH₄Cl) give acidic solutions: BH⁺ + H₂O ⇌ [H⁺] using the Ka expression and the salt concentration. Salts of a weak acid and a weak base require comparing Ka and Kb; the pH may be acidic, basic or neutral.
弱碱强酸盐（如NH₄Cl）溶液呈酸性：BH⁺ + H₂O ⇌ B + H₃O⁺。共轭酸的Ka = Kw / Kb(弱碱)。利用Ka表达式及盐浓度求出[H⁺]。弱酸弱碱盐则需要比较Ka与Kb，溶液可呈酸性、碱性或中性。
12. Common Exam Mistakes & Tips | 常见考试错误与技巧
- Kw. Always check the given temperature; if not stated, assume 298 K and Kw = 1.0 × 10⁻¹⁴.
- 忘记换算与温度相关的Kw。务必确认给定温度；若未说明，假设298 K，Kw = 1.0 × 10⁻¹⁴。
- Applying the weak acid approximation without verifying c / Ka > 1000. This leads to significant pH errors.
- 未验证c / Ka > 1000就使用弱酸近似，导致pH显著偏差。
- Mixing up concentration and amount of substance in Henderson-Hasselbalch; since the volume cancels, you may use moles directly, but be consistent.
- 在亨德森-哈塞尔巴尔赫方程中混淆浓度与物质的量；因体积可约去，可直接使用摩尔数，但需保持单位一致。
- Ignoring the contribution of the auto-ionisation of water in very dilute solutions.
- 在极稀溶液中忽视水自解离的贡献。
- Using pKa instead of Ka when setting up the equilibrium expression, or vice versa.
- 建立平衡表达式时混淆使用pKa与Ka。
- Forgetting that diprotic acids have two dissociation steps and exam questions often ask for total [H⁺] considering both.
- 忘记二元酸有二级解离，考题常要求计算包含两步在内的总[H⁺]。
- Memorise: pH = pKa at half-equivalence; buffer pH ≈ pKa when [acid] = [conjugate base].
- 记忆：半等当点时pH = pKa；当[酸]=[共轭碱]时，缓冲液pH ≈ pKa。
- Practise drawing titration curves with correct axes labels and clearly showing the buffer region and steep rise.
- 练习绘制滴定曲线，正确标注坐标轴，清晰显示缓冲区域和陡升段。
Published by TutorHao | IB CIE Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图