📚 Acids, Alkalis and pH: Calculation Questions | 酸、碱和pH：计算题型
Understanding acids, alkalis and pH is fundamental to A-level Chemistry, and mastering the associated calculations is essential for exam success. This revision guide focuses exclusively on the calculation skills required for strong and weak acids, strong bases, neutralisation, dilution, and buffer solutions, providing clear methods, worked examples, and tips to avoid common errors. Whether you are calculating pH from [H⁺], using the ionic product of water Kw, or handling more advanced Ka expressions, this article equips you with the tools to tackle numerical problems confidently.
理解酸、碱和pH是A-level化学的基础，而掌握相关的计算对于考试成功至关重要。本复习指南专注于强酸、弱酸、强碱、中和、稀释和缓冲溶液所需的计算技能，提供清晰的方法、解题示例以及避免常见错误的技巧。无论是由[H⁺]计算pH、运用水的离子积Kw，还是处理更复杂的Ka表达式，这篇文章都将为你提供信心满满地攻克计算题的工具。
1. Understanding the pH Scale and [H⁺] | 理解pH标度和[H⁺]
The pH of a solution is defined as the negative logarithm to base 10 of the hydrogen ion concentration: pH = –log₁₀[H⁺]. Conversely, if you know the pH, you can find [H⁺] using [H⁺] = 10⁻ᵖᴴ. These equations work for any aqueous solution at around 25°C. Remember that [H⁺] must be in mol dm⁻³, and a difference of 1 pH unit represents a ten-fold change in [H⁺].
溶液的pH定义为氢离子浓度的以10为底的负对数：pH = –log₁₀[H⁺]。反之，如果已知pH，可通过[H⁺] = 10⁻ᵖᴴ求得[H⁺]。这些方程适用于25°C左右的任何水溶液。务必记住[H⁺]的单位是mol dm⁻³，且pH值相差1代表[H⁺]改变10倍。
For example, if [H⁺] = 0.001 mol dm⁻³, then pH = – If pH = 5.30, [H⁺] = 10⁻⁵·³⁰ = 5.01 × 10⁻⁶ mol dm⁻³. Always check your calculator: use the 'log' button for pH and '10ˣ' or 'shift log' for [H⁺].
例如，若[H⁺] = 0.001 mol dm⁻³，则pH = –log(1 × 10⁻³) = 3。若pH = 5.30，[H⁺] = 10⁻⁵·³⁰ = 5.01 × 10⁻⁶ mol dm⁻³。务必检查计算器：使用“log”键计算pH，使用“10ˣ”或“shift log”键计算[H⁺]。
2. Calculating pH of Strong Acids | 强酸的pH计算
= 2 × [H₂SO₄], provided the second ionisation is also complete at the given concentration.
强酸（如HCl、HNO₃和H₂SO₄）在水中完全电离。对于一元强酸，[H⁺]就等于酸的浓度。因此，0.10 mol dm⁻³ HCl的[H⁺] = 0.10 mol dm⁻³，pH = 1.00。对于二元酸H₂SO₄，每摩尔放出两摩尔H⁺，因此若浓度较高时二级电离也完全，则[H⁺] = 2 × [H₂SO₄]。
H of 0.050 mol dm⁻³ HNO₃. Since HNO₃ is monoprotic and strong, [H⁺] = 0.050 mol dm⁻³. pH = –log(0.050) = 1.30. Always express pH to 2 decimal places unless otherwise specified.
解题示例：计算0.050 mol dm⁻³ HNO₃的pH。因HNO₃为一元强酸，[H⁺] = 0.050 mol dm⁻³。pH = –log(0.050) = 1.30。除非另有要求，pH通常保留两位小数。
3. The Ionic Product of Water, Kw | 水的离子积Kw
Water undergoes slight self-ionisation: 2H₂O ⇌ H₃O⁺ + OH⁻, often simplified as H₂O ⇌ ⁻⁶. This constant is crucial for linking [H⁺] and [OH⁻] in aqueous solutions, allowing us to calculate pH of strong bases and mixtures.
水会发生微弱自电离：2H₂O ⇌ H₃O⁺ + OH⁻，常简写为H₂O ⇌ H⁺ + OH⁻。水的离子积Kw = [H⁺][OH⁻]，在25°C时，Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶。这个常数对于关联水溶液中的[H⁺]和[OH⁻]至关重要，使我们能够计算强碱和混合物的pH。
you know [OH⁻], you can find [H⁺] by [H⁺] = Kw/[OH⁻], then calculate pH. Similarly, pOH = – + pOH = 14 at 25°C (since pKw = 14). This is often the quickest route for basic solutions.
利用Kw，若已知[OH⁻]，可由[H⁺] = Kw/[OH⁻]求出[H⁺]，再计算pH。类似地，pOH = –log[OH⁻]，且在25°C时pH + pOH = 14（因为pKw = 14）。这通常是处理碱性溶液最快的方法。
4. Calculating pH of Strong Bases | 强碱的pH计算
example, 0.20 mol dm⁻³ NaOH gives [OH⁻] = 0.20 mol dm⁻³. To find pH, first calculate pOH = –log(0.20) = 0.70, then pH = 14 – Alternatively, use [H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴ / 0.20 = 5.0 × 10⁻¹⁴ mol dm⁻³, then pH = –log(5.0 × 10⁻¹⁴) = 13.30.
像NaOH和KOH这样的强碱在水中完全电离。对于第一主族金属氢氧化物，[OH⁻]就等于碱的浓度。例如，0.20 mol dm⁻³ NaOH的[OH⁻] = 0.20 mol dm⁻³。求pH时，先计算pOH = –log(0.20) = 0.70，然后pH = 14 – 0.70 = 13.30。也可使用[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴ / 0.20 = 5.0 × 10⁻¹⁴ mol dm⁻³，再算pH = –log(5.0 × 10⁻¹⁴) = 13.30。
rite the dissociation equation first to ensure you have the correct stoichiometric factor.
对于第二主族氢氧化物如Ba(OH)₂，每个单元提供两个OH⁻离子。因此，0.10 mol dm⁻³ Ba(OH)₂的[OH⁻] = 0.20 mol dm⁻³。务必先写出电离方程式，确保使用正确的化学计量系数。
5. Dilution and pH Changes | 稀释与pH变化
tremely dilute (below 10⁻⁶ mol dm⁻³), the contribution of H⁺ from water’s autoionisation becomes significant, and simple dilution formulas no longer apply.
用水稀释酸会降低[H⁺]并使pH升高。对强酸而言，稀释10倍pH增加1个单位。例如，将0.1 mol dm⁻³ HCl (pH=1) 稀释10倍至0.01 mol dm⁻³，pH变为2。然而，当极度稀释时（低于10⁻⁶ mol dm⁻³），来自水自电离的H⁺贡献变得不可忽略，简单的稀释公式不再适用。
fter dilution. A 100-fold dilution of 0.05 mol dm⁻³ NaOH gives 0.0005 mol dm⁻³ NaOH, pOH = –log(0.0005) = 3.30, pH = 10.70.
类似地，稀释强碱会降低[OH⁻]并使pH下降。借助Kw计算稀释后的新[OH⁻]或[H⁺]。将0.05 mol dm⁻³ NaOH稀释100倍，得到0.0005 mol dm⁻³ NaOH，pOH = –log(0.0005) = 3.30，pH = 10.70。
6. Mixing Acids and Bases: Neutralisation Calculations | 酸碱混合：中和计算
OH⁻ in the total volume. If neither is in excess, the solution is neutral (pH=7 at 25°C).
当酸与碱混合时，对于一元酸和一元碱，它们以1:1的摩尔比反应：H⁺ + OH⁻ → H₂O。关键是判断哪种反应物过量，并计算在总体积中过量的H⁺或OH⁻浓度。若均无过量，溶液为中性（25°C时pH=7）。
Example: 0 mol. Total volume = 45.0 cm³ = 0.0450 dm³. [H⁺] = 0.00050/0.0450 = 0.0111 mol dm⁻³. pH = –log(0.0111) ≈ 1.95.
示例：将25.0 cm³ 0.10 mol dm⁻³ HCl加到20.0 cm³ 0.10 mol dm⁻³ NaOH中。H⁺的物质的量 = (25.0/1000)×0.10 = 0.00250 mol。OH⁻的物质的量 = (20.0/1000)×0.10 = 0.00200 mol。H⁺过量0.00050 mol。总体积 = 45.0 cm³ = 0.0450 dm³。[H⁺] = 0.00050/0.0450 = 0.0111 mol dm⁻³。pH = –log(0.0111) ≈ 1.95。
7. pH of Weak Acids: Introducing Ka | 弱酸的pH：引入Ka
Weak acids such as CH₃COOH do not fully dissociate. Their strength is described by the acid dissociation constant Ka. For a general weak acid HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA] ⁻] and that the amount dissociated is negligible compared to initial concentration, giving [H⁺] = √Ka × C, where C is the initial concentration of the acid.
弱酸（如CH₃COOH）不会完全电离。它们的强度由酸解离常数Ka描述。对于一般弱酸HA ⇌ H⁺ + A⁻，Ka = [H⁺][A⁻]/[HA]（单位mol dm⁻³）。求pH时，假设平衡时[H⁺] ≈ [A⁻]，且解离部分与初始浓度相比可忽略，从而得到[H⁺] = √Ka × C，C为酸的初始浓度。
For instance, calculate the pH of 0.20 mol dm⁻³ CH₃COOH (Ka = 1.8 × 10⁻⁵). [H⁺] = √1.8 × 10⁻⁵ × 0.20 = √3.6 × 10⁻⁶ = 1.90 × 10⁻³ mol dm⁻³. pH = –log(1.90 × 10⁻³) = 2.72. Always check the simplification: approximate if [H⁺] < 5% of initial C; here 1.9×10⁻³ / 0.20 = 0.95% — valid.
例如，计算0.20 mol dm⁻³ CH₃COOH（Ka = 1.8 × 10⁻⁵）的pH。[H⁺] = √1.8 × 10⁻⁵ × 0.20 = √3.6 × 10⁻⁶ = 1.90 × 10⁻³ mol dm⁻³。pH = –log(1.90 × 10⁻³) = 2.72。务必检查简化条件：若[H⁺]小于初始浓度C的5%则成立；此处1.9×10⁻³ / 0.20 = 0.95% — 有效。
8. Calculating pH of Weak Bases | 弱碱的pH计算
t protons. Their strength is expressed by the base dissociation constant Kb. For NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, Kb = [NH₄⁺][OH⁻]/[NH₃]. Using the same approximation, [OH⁻] = √Kb × C, then pOH = –log[OH⁻] and pH = 14 – pOH. Alternatively, you can find [H⁺] via Kw.
弱碱如NH₃部分接受质子。它们的强度用碱解离常数Kb表示。对于NH₃ + H₂O ⇌ NH₄⁺ + OH⁻，Kb = [NH₄⁺][OH⁻]/[NH₃]。采用相同近似，[OH⁻] = √Kb × C，然后pOH = –log[OH⁻]，pH = 14 – pOH。也可通过Kw求出[H⁺]。
Example: 0.10 mol dm⁻³ NH₃, Kb = 1.8 × 10⁻⁵. [OH⁻] = √1.8 × 10⁻⁵ × 0.10 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³. pOH = –log(1.34 × 10⁻³) = 2.87. pH = 14 – 2.87 = 11.13.
示例：0.10 mol dm⁻³ NH₃，Kb = 1.8 × 10⁻⁵。[OH⁻] = √1.8 × 10⁻⁵ × 0.10 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³。pOH = –log(1.34 × 10⁻³) = 2.87。pH = 14 – 2.87 = 11.13。
9. Buffer Solutions and pH Calculation | 缓冲溶液与pH计算
A buffer resists changes in pH and is typically a mixture of a weak acid and its conjugate base. The pH of an acidic buffer can be calculated using the Henderson–Hasselbalch equation: pH = pKa + log₁₀([salt]/[acid]) or pH = pKa + log₁₀([A⁻]/[HA]). pKa is –log₁₀(Ka). This equation assumes equal volumes so the ratio of concentrations equals the ratio of moles.
缓冲溶液能抵抗pH变化，通常由弱酸及其共轭碱组成。酸性缓冲的pH可用Henderson–Hasselbalch方程计算：pH = pKa + log₁₀([盐]/[酸]) 或 pH = pKa + log₁₀([A⁻]/[HA])。pKa = –log₁₀(Ka)。该方程假设等体积，因此浓度比等于物质的量比。
m³ of 0.10 mol dm⁻³ CH₃COOH (Ka = 1.8 × 10⁻⁵) with 25 cm³ of 0.10 mol dm⁻³ CH₃COONa. pKa = – = 0.0050, moles of salt = 0.0025. Ratio = 0.0025/0.0050 = 0.5. pH = 4.74 + log(0.5) = 4.74 – 0.30 = 4.44.
示例：将50 cm³ 0.10 mol dm⁻³ CH₃COOH（Ka = 1.8 × 10⁻⁵）与25 cm³ 0.10 mol dm⁻³ CH₃COONa混合制备的缓冲液。pKa = –log(1.8×10⁻⁵) ≈ 4.74。酸的物质的量 = 0.0050，盐的物质的量 = 0.0025。比值 = 0.0025/0.0050 = 0.5。pH = 4.74 + log(0.5) = 4.74 – 0.30 = 4.44。
10. Titration Curves and Key pH Points | 滴定曲线与关键pH点
Titration curves show pH versus volume of titrant added. Understanding the initial pH, equivalence point, and half-equivalence point is vital for calculations. For a strong acid–strong base titration, the equivalence point is pH 7. For a weak acid–strong base, the pH at equivalence is >7 due to the formation of the conjugate base which hydrolyses.
滴定曲线显示了pH随滴定剂加入体积的变化。理解初始pH、等当点和半等当点对计算至关重要。强酸-强碱滴定中，等当点为pH 7。而弱酸-强碱滴定时，由于生成共轭碱发生水解，等当点的pH大于7。
pKa. This is extremely useful for determining Ka from experimental data. Simply read the pH at the volume half that of the equivalence point; that pH equals the pKa of the weak acid.
在弱酸滴定的半等当点处，[HA] = [A⁻]，因此pH = pKa。这对于通过实验数据确定Ka极为有用。只需读出等当点一半体积时的pH值，该pH值就等于弱酸的pKa。
11. Exam-Style Calculation Questions | 考试风格计算题
Question 1: A student dissolves 1.26 g of HNO₃ (molar mass 63.0 g mol⁻¹) in water to make 250 cm³ of solution. Calculate the pH.
解答1： mol. Concentration = 0.0200/0.250 = 0.0800 mol dm⁻³. As a strong acid, [H⁺] = 0.0800, pH = –log(0.0800) = 1.10.
Question 2: 10.0 cm³ of 0.50 mol dm⁻³ NaOH is added to 40.0 cm³ of water. Find the pH of the resulting solution.
解答2： 0/1000)×0.50 = 0.00500 mol. New [OH⁻] = 0.00500/0.0500 = 0.100 mol dm⁻³. pOH = 1, pH = 13.
Question 3: mol dm⁻³ CH₃COOH with 200 cm³ of 0.10 mol dm⁻³ CH₃COONa. Calculate the pH. (Ka = 1.8×10⁻⁵)
解答3：Moles acid = 0.020, moles salt = 0.020. Ratio = 1; pH = pKa = 4.74.
These exam-style questions reinforce the step-by-step approach: convert to moles, find excess or ratio, calculate concentrations, and apply the appropriate formula.
这些考试风格的题目强化了分步解题的方法：转换为物质的量，判断过量或比值，计算浓度，并应用合适的公式。
12. Common Mistakes and Tips | 常见错误与技巧
- Forgetting to convert cm³ to dm³. Always divide by 1000. | 忘记将cm³转换为dm³。务必除以1000。
- Ignoring stoichiometry of diprotic acids or Group 2 hydroxides. Write the dissociation equation. | 忽略二元酸或第二主族氢氧化物的化学计量。要写电离方程式。
- Using Kw outside 25°C without the given value. Kw varies with temperature. | 在非25°C时使用默认Kw值。Kw随温度变化。
- Applying the weak acid approximation [H⁺] = √Ka C without checking the assumption. Ensure percent dissociation < 5%. | 未检查假设就使用弱酸近似式[H⁺] = √Ka C。确保解离度小于5%。
- Confusing pKa with Ka: pKa = –log Ka, and a smaller pKa indicates a stronger acid. | 混淆pKa与Ka：pKa = –log Ka，且pKa越小酸性越强。
- For buffers, using concentrations rather than moles in the log term is fine if volumes cancel; otherwise, use moles ratio directly to save time. | 对于缓冲溶液，若体积相同可在log项中使用浓度比；若不同，直接用物质的量比可节省时间。
Always write down what you know and what you need to find. Organise your working clearly: equation → moles → excess → new concentration → pH. With practice, these calculations become routine.
始终写下已知量和待求量。清晰地组织解题过程：方程式 → 物质的量 → 过量 → 新浓度 → pH。通过练习，这些计算将变得常规化。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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