📚 AQA 9620 CH04 Specimen Paper 2016 Question 2 Walkthrough | 2016年国际 A Level 化学样卷第2题精讲
The AQA 9620 CH04 International A Level Chemistry specimen paper is a key resource for understanding the depth of A2 chemistry assessment. Question 2 typically focuses on acid-base equilibria, including weak acid pH, Ka, pKa, buffer solutions, and titration curves. This walkthrough breaks down the exact skills, formulas, unit conventions, and mark-scheme habits you need in order to answer Question 2 confidently.
AQA 9620 CH04 国际 A Level 化学样卷是理解 A2 化学考查深度的重要资源。第2题通常聚焦酸碱平衡，包括弱酸 pH、Ka、pKa、缓冲溶液和滴定曲线。本篇精讲拆解回答第2题所需的全部核心技能、公式、单位规范与评分习惯，帮助你从容应对并在这一题上拿满分。
1. What Question 2 Assesses | 第2题考查内容
In the 2016 specimen paper, Question 2 is built around a weak acid such as ethanoic acid or a similar organic acid. You must move fluently between Ka, pKa, [H⁺], pH, and the concentrations of the conjugate base and the undissociated acid.
在 2016 年样卷中，第2题围绕弱酸（如乙酸或类似有机酸）展开。你必须在 Ka、pKa、[H⁺]、pH，以及共轭碱浓度和未解离酸浓度之间灵活转换。
Examiners often reward clear working, correct units, and sensible significant figures. A final pH should usually be quoted to 2 decimal places, and Ka values to 2 or 3 significant figures.
考官通常奖励清晰的演算步骤、正确的单位和合理的有效数字。最终 pH 通常保留两位小数，Ka 值保留两位或三位有效数字。
The question is rarely a single calculation. It is normally a chain of linked parts, so an error in part (a) can be carried forward into part (c). That is why showing your working matters so much.
这道题很少只考一个计算。它通常是一串相互关联的小问，所以第 (a) 问的错误可能被带入第 (c) 问。这正是写出完整步骤如此重要的原因。
- Weak acid and weak base pH calculations — 弱酸与弱碱的 pH 计算
- Ka, pKa and Kw conversions — Ka、pKa 与 Kw 的换算
- Buffer solutions, including buffer preparation and pH after adding acid or base — 缓冲溶液，包括缓冲溶液的配制及加酸加碱后的 pH
- Titration curves and the half-equivalence point — 滴定曲线与半当量点
- Units, significant figures and explanation marks — 单位、有效数字与解释性得分点
2. Extraction of Data | 数据提取
Start by listing every value given: the acid concentration, the value of Ka or pKa, and any added salt or base concentration. Convert all volumes to dm³ and all concentrations to mol dm⁻³ before substituting into any equation.
首先列出所有给定数值：酸浓度、Ka 或 pKa 的值，以及任何加入的盐或碱的浓度。在代入任何方程前，先将所有体积换算成 dm³，所有浓度换算成 mol dm⁻³。
If the question gives pKa, convert it immediately. If it gives Ka, you may need pKa instead. Keeping both numbers written down at the top of your answer saves you from re-deriving them later.
如果题目给出 pKa，立刻进行转换。如果给出 Ka，你可能需要的是 pKa。把两个数值都写在答题区上方，可以避免后面重复推导。
pKa = −log₁₀(Ka) and Ka = 10⁻ᵖᴷᵃ
lost. Remember that 1 dm³ = 1000 cm³, so 25.0 cm³ = 0.0250 dm³ and 12.5 cm³ = 0.01250 dm³.
体积换算是丢分最集中的地方。记住 1 dm³ = 1000 cm³，因此 25.0 cm³ = 0.0250 dm³，12.5 cm³ = 0.01250 dm³。
n = c × V (V in dm³) and 1 dm³ = 1000 cm³
Also note whether the acid is monoprotic or diprotic. A diprotic acid such as sulfuric acid can donate two protons, so its stoichiometry in a titration differs from that of ethanoic acid.
还要注意该酸是一元酸还是二元酸。像硫酸这样的二元酸可以给出两个质子，所以它在滴定中的化学计量关系与乙酸不同。
3. Building the Governing Equations | 建立核心方程
Every calculation in Question 2 rests on two equilibrium expressions: the acid dissociation constant Ka and the ionic product of water Kw.
第2题中的每一个计算都建立在两个平衡表达式之上：酸解离常数 Ka 和水的离子积 Kw。
Ka = [H⁺][A⁻] ÷ [HA]
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K
art from the tiny contribution from water itself. Combining this with the Ka expression gives the familiar square-root relationship.
对于水中纯弱酸 HA，除水本身贡献的极小量外，电荷守恒要求 [H⁺] = [A⁻]。把它与 Ka 表达式结合，就得到熟悉的平方根关系式。
[H⁺] = √Ka × [HA]
The square-root form is only valid when the degree of dissociation is small, usually taken as less than about 5 %. If the acid is relatively strong or very dilute, you must solve the full quadratic instead.
平方根形式只在解离度很小时成立，通常取小于约 5 %。如果酸相对较强或溶液非常稀，就必须解完整的二次方程。
4. Weak Acid pH by the Square-Root Method | 用平方根法求弱酸 pH
l dm⁻³. Set out the calculation in three clean lines so the examiner can follow your reasoning.
假设题目给出 0.100 mol dm⁻³ 的乙酸，Ka = 1.74 × 10⁻⁵ mol dm⁻³。用三行清晰的计算呈现，让考官能跟上你的推理。
Line one: state the assumption that [HA] at equilibrium is approximately equal to the initial concentration, 0.100 mol dm⁻³. Line two: substitute into the square-root expression. Line three: convert [H⁺] into pH.
第一行：说明假设——平衡时 [HA] 近似等于初始浓度 0.100 mol dm⁻³。第二行：代入平方根表达式。第三行：把 [H⁺] 换算为 pH。
[H⁺] = √1.74 × 10⁻⁵ × 0.100 = √1.74 × 10⁻⁶ = 1.32 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.32 × 10⁻³) = 2.88
Notice that the units work out correctly: Ka has units of mol dm⁻³, multiplying by another mol dm⁻³ gives mol² dm⁻⁶, and the square root returns mol dm⁻³. A quick unit check catches most algebra slips.
注意单位是正确的：Ka 的单位是 mol dm⁻³，再乘以 mol dm⁻³ 得到 mol² dm⁻⁶，开平方后回到 mol dm⁻³。快速检查单位可以抓出大多数代数错误。
5. Checking the Approximation | 检验近似是否成立
After obtaining [H⁺], always test the assumption you made. Divide [H⁺] by the initial acid concentration and express the result as a percentage.
得到 [H⁺] 之后，一定要检验你做过的假设。用 [H⁺] 除以酸的初始浓度，并把结果表示为百分比。
percentage dissociation = ([H⁺] ÷ [HA]initial) × 100 %
In the ethanoic acid example this gives (1.32 × 10⁻³ ÷ 0.100) × 100 % = 1.32 %, which is well below 5 %, so the approximation is justified and the answer stands.
在乙酸的例子中，这给出 (1.32 × 10⁻³ ÷ 0.100) × 100 % = 1.32 %，远低于 5 %，所以近似成立，答案有效。
If the percentage exceeds about 5 %, the equilibrium concentration of HA is noticeably smaller than the initial value. You must then let the extent of dissociation be x and solve the resulting quadratic equation.
如果百分比超过约 5 %，平衡时 HA 的浓度就明显小于初始值。此时你必须设解离量为 x，并解由此得到的二次方程。
Ka = x² ÷ (c − x) → x² + Ka·x − Ka·c = 0
Very dilute weak acids also break the approximation for a different reason: the autoionisation of water becomes a significant source of H⁺, and the simple treatment underestimates the true [H⁺].
极稀的弱酸也会因另一个原因破坏该近似：水的自偶电离成为 H⁺ 的重要来源，简单处理会低估真实的 [H⁺]。
6. Buffer Solutions and the Henderson-Hasselbalch Equation | 缓冲溶液与亨德森-哈塞尔巴赫方程
A buffer contains a weak acid and its conjugate base in comparable amounts. Rearranging the Ka expression and taking logarithms gives the working equation for buffer pH.
缓冲溶液含有浓度相当的弱酸及其共轭碱。对 Ka 表达式变形并取对数，就得到计算缓冲溶液 pH 的实用方程。
pH = pKa + log₁₀([A⁻] ÷ [HA])
ol dm⁻³ in both ethanoic acid and sodium ethanoate, the ratio is 1, the logarithm is 0, and the pH is exactly 4.76.
对乙酸来说，pKa = −log₁₀(1.74 × 10⁻⁵) = 4.76。如果溶液中的乙酸和乙酸钠浓度都是 0.100 mol dm⁻³，比值为 1，对数为 0，pH 恰好为 4.76。
A buffer works best when the pH required is within about one unit of the pKa of the acid used. Outside that window the buffer capacity falls sharply and the pH changes markedly on adding acid or base.
当所需 pH 落在所用酸 pKa 的约一个单位范围内时，缓冲效果最佳。超出这个区间，缓冲容量会急剧下降，加入酸或碱后 pH 变化明显。
Buffer capacity also increases with total concentration. A 1.0 mol dm⁻³ buffer resists pH change far better than a 0.010 mol dm⁻³ buffer with the same ratio.
缓冲容量还随总浓度增大而提高。总浓度 1.0 mol dm⁻³、比例相同的缓冲液，其抗 pH 变化能力远强于 0.010 mol dm⁻³ 的缓冲液。
7. Buffer Preparation and Ratio Reasoning | 缓冲溶液的配制与比例推理
Some parts of Question 2 ask you to work backwards: given a target pH, find the required ratio of conjugate base to acid. Rearranging the buffer equation gives the ratio directly.
第2题有时会要求你反向推算：给定目标 pH，求共轭碱与酸所需的比例。将缓冲方程变形即可直接得到该比例。
[A⁻] ÷ [HA] = 10⁽ᵖᴴ ⁻ ᵖᴷᵃ⁾
0⁰·³⁰ = 2.0. You therefore need twice as many moles of ethanoate ions as of ethanoic acid.
以目标 pH 5.06、乙酸 pKa = 4.76 为例。指数为 0.30，所以比值为 10⁰·³⁰ = 2.0。因此你需要乙酸根离子的物质的量是乙酸的两倍。
Buffers are often made in the laboratory by partially neutralising a weak acid with a strong base. If half of the acid is neutralised, [A⁻] equals [HA] and the resulting pH is the pKa.
实验室常用强碱部分中和弱酸来配制缓冲液。如果中和掉一半的酸，则 [A⁻] 等于 [HA]，所得 pH 就是 pKa。
Work in moles rather than concentrations whenever you mix two solutions. Because both species end up in the same total volume, the volume cancels out of the ratio, which simplifies the arithmetic considerably.
当两种溶液混合时，用物质的量而不是浓度来思考。由于两种物种最终处于同一总体积中，体积在比例中会约掉，这能大大简化计算。
10⁻³, moles of OH⁻ = 1.00 × 10⁻³, so [A⁻] = 1.00 × 10⁻³ and remaining [HA] = 1.50 × 10⁻³.
一个实例：25.0 cm³ 的 0.100 mol dm⁻³ 乙酸加 10.0 cm³ 的 0.100 mol dm⁻³ NaOH。酸的物质的量 = 2.50 × 10⁻³，OH⁻ 的物质的量 = 1.00 × 10⁻³，因此 [A⁻] = 1.00 × 10⁻³，剩余 [HA] = 1.50 × 10⁻³。
pH = 4.76 + log₁₀(1.00 ÷ 1.50) = 4.76 − 0.18 = 4.58
8. The Half-Equivalence Point and the Titration Curve | 半当量点与滴定曲线
When a weak acid is titrated with a strong base, the pH rises slowly at first, then sharply through the equivalence point. The curve has three regions the examiner expects you to comment on.
用强碱滴定弱酸时，pH 先缓慢上升，然后在当量点附近急剧变化。这条曲线有三个区域是考官希望你加以描述的。
Before the equivalence point the mixture is a buffer: unreacted HA plus the A⁻ formed. The pH is calculated with the Henderson-Hasselbalch equation, not with the square-root method.
当量点之前，混合物是一种缓冲体系：未反应的 HA 加上生成的 A⁻。此时 pH 用亨德森-哈塞尔巴赫方程计算，而不是平方根法。
ogarithmic term becomes zero. This is the neatest way to read a pKa off a titration curve.
在半当量点，恰好一半的酸转化为共轭碱，因此 [HA] = [A⁻]，对数项为零。这是从滴定曲线读取 pKa 最简洁的方法。
at half-equivalence: pH = pKa, since log₁₀(1) = 0
At the equivalence point all the weak acid has been converted into its conjugate base, which hydrolyses to give an alkaline solution. The pH at equivalence is therefore above 7, not exactly 7.
在当量点，全部弱酸都转化为其共轭碱，后者水解使溶液呈碱性。因此当量点的 pH 大于 7，而不是恰好等于 7。
[OH⁻] = √Kw × [A⁻] ÷ Ka then pH = 14.00 − pOH
0.100 mol dm⁻³ ethanoate at 298 K: Kb = 1.00 × 10⁻¹⁴ ÷ 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰, [OH⁻] = √5.75 × 10⁻¹⁰ × 0.100 = 7.58 × 10⁻⁶ mol dm⁻³, pOH = 5.12 and pH = 8.88.
对 298 K 下 0.100 mol dm⁻³ 的乙酸根：Kb = 1.00 × 10⁻¹⁴ ÷ 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰，[OH⁻] = √5.75 × 10⁻¹⁰ × 0.100 = 7.58 × 10⁻⁶ mol dm⁻³，pOH = 5.12，pH = 8.88。
9. Reading Key Points of the Curve | 滴定曲线关键点的计算路径
A useful habit is to label each point on the curve with the species that dominates there, then pick the equation that matches. The table below summarises the standard route for a weak acid titrated with a strong base.
一个有用的习惯是标出曲线上每个点的主要物种，再选择与之匹配
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