📚 Solutions and Concentration | 溶液与浓度
In A-Level Chemistry, solutions are central to quantitative analysis, titrations, and equilibrium studies. A clear understanding of concentration units, conversions, and dilution is essential for success in paper 1, paper 2, and practical assessments.
在 A-Level 化学中，溶液是定量分析、滴定和平衡研究的核心。清楚理解浓度单位、换算和稀释对于在 Paper 1、Paper 2 以及实验考核中取得成功至关重要。
1. The Nature of Solutions | 溶液的基本性质
A solution is a homogeneous mixture formed when a solute dissolves in a solvent. The solvent is usually the component present in the largest quantity, and in aqueous solutions water is the solvent.
溶液是溶质溶解在溶剂中形成的均相混合物。溶剂通常是含量最多的组分，在水溶液中水是溶剂。
The solute can be a solid, liquid, or gas. For example, in brine, sodium chloride is the solute and water is the solvent; in carbonated drinks, carbon dioxide gas is the solute.
溶质可以是固体、液体或气体。例如，在盐水中氯化钠是溶质、水是溶剂；在碳酸饮料中，二氧化碳气体是溶质。
A saturated solution is one in which no more solute can dissolve at a given temperature. Undissolved solute remains in equilibrium with dissolved solute, so the rate of dissolving equals the rate of crystallisation.
饱和溶液是指在某一温度下不能再溶解更多溶质的溶液。未溶解的溶质与已溶解的溶质保持平衡，因此溶解速率等于结晶速率。
2. Concentration Units | 浓度单位
Concentration is the amount of solute dissolved in a unit volume of solution. In A-Level Chemistry, the two most important measures are mass concentration and molar concentration.
浓度是单位体积溶液中所含溶质的量。在 A-Level 化学中，最重要的两种浓度表示是质量浓度和摩尔浓度。
Mass concentration is expressed in grams per cubic decimetre (g dm⁻³). It is calculated by dividing the mass of solute in grams by the volume of solution in dm³.
质量浓度以克每立方分米 (g dm⁻³) 表示，由溶质质量（g）除以溶液体积（dm³）得到。
mass concentration = mass of solute (g) ÷ volume of solution (dm³)
Molar concentration, often called molarity, is expressed in moles per cubic decimetre (mol dm⁻³). It is calculated by dividing the amount of solute in moles by the volume of solution in dm³.
摩尔浓度，通常称为物质的量浓度，以摩尔每立方分米 (mol dm⁻³) 表示，由溶质的物质的量（mol）除以溶液体积（dm³）得到。
c = n ÷ V
where n is amount of solute in mol, V is volume of solution in dm³, and c is molar concentration in mol dm⁻³.
其中 n 为溶质的物质的量（mol），V 为溶液体积（dm³），c 为摩尔浓度（mol dm⁻³）。
dm³ = 1000 cm³ = 1 L. This conversion is needed in almost every concentration calculation.
请始终记住：1 dm³ = 1000 cm³ = 1 L。几乎每个浓度计算都需要用到这一换算。
3. Converting Mass Concentration to Molar Concentration | 质量浓度与摩尔浓度的换算
ass concentration by the molar mass M of the solute. This follows directly from n = m ÷ M.
将质量浓度转换为摩尔浓度时，用质量浓度除以溶质的摩尔质量 M。这直接来自 n = m ÷ M。
c (mol dm⁻³) = mass concentration (g dm⁻³) ÷ M (g mol⁻¹)
mass of NaCl is 58.5 g mol⁻¹, so the molar concentration is 11.7 ÷ 58.5 = 0.200 mol dm⁻³.
例如，某溶液含有 11.7 g dm⁻³ 的 NaCl。NaCl 的摩尔质量为 58.5 g mol⁻¹，因此摩尔浓度为 11.7 ÷ 58.5 = 0.200 mol dm⁻³。
concentration = c × M. This is useful when preparing a solution from a required molarity.
将摩尔浓度转换为质量浓度时，乘以摩尔质量：质量浓度 = c × M。这在根据所需摩尔浓度配制溶液时非常有用。
If the solute is hydrated, such as CuSO₄·5H₂O, you must include the water molecules in the molar mass calculation.
如果溶质是水合物，如 CuSO₄·5H₂O，计算摩尔质量时必须包含结晶水分子。
4. Moles in Solution and Volume Conversions | 溶液中物质的量与体积换算
ged to n = c × V. This is used to find the amount of solute in a given volume of solution.
关键公式 c = n ÷ V 可变形为 n = c × V，用于计算一定体积溶液中的溶质物质的量。
n = c × V
efore substituting. If the volume is given in cm³, divide by 1000: V(dm³) = V(cm³) ÷ 1000.
代入前务必先将体积换算为 dm³。若体积以 cm³ 给出，则除以 1000：V(dm³) = V(cm³) ÷ 1000。
For example, how many moles of HCl are present in 25.0 cm³ of 0.200 mol dm⁻³ HCl? First convert 25.0 cm³ to 0.0250 dm³, then n = 0.200 × 0.0250 = 0.00500 mol.
例如，25.0 cm³ 0.200 mol dm⁻³ 的 HCl 中含有多少摩尔 HCl？先将 25.0 cm³ 换算为 0.0250 dm³，然后 n = 0.200 × 0.0250 = 0.00500 mol。
. This is useful when you know the moles of solute needed and the concentration available.
也可以变形求体积：V = n ÷ c。当已知所需溶质的物质的量和可用浓度时，这一公式非常有用。
5. Preparing a Standard Solution | 配制标准溶液
A standard solution is one whose concentration is known accurately. It is usually prepared in a volumetric flask with a precisely known volume, such as 100 cm³ or 250 cm³.
标准溶液是浓度精确已知的溶液，通常在体积精确已知的容量瓶中配制，如 100 cm³ 或 250 cm³。
0.250 = 0.0250 mol. M(Na₂CO₃) = 106.0 g mol⁻¹, so mass = n × M = 0.0250 × 106.0 = 2.65 g.
配制 250 cm³ 0.100 mol dm⁻³ Na₂CO₃ 时，先计算所需质量：n = c × V = 0.100 × 0.250 = 0.0250 mol。M(Na₂CO₃) = 106.0 g mol⁻¹，因此质量 = n × M = 0.0250 × 106.0 = 2.65 g。
The solid is weighed accurately, dissolved in a beaker with distilled water, and then transferred to the volumetric flask. The beaker and funnel are rinsed with distilled water, and the washings are added to the flask.
准确称取固体，在烧杯中用蒸馏水溶解，然后转移至容量瓶中。用蒸馏水冲洗烧杯和漏斗，并将洗涤液加入容量瓶。
Distilled water is added until the meniscus reaches the graduation mark. The flask is stoppered and inverted several times to ensure a uniform concentration.
继续加入蒸馏水直至弯月面到达刻度线。塞紧瓶塞并倒置数次，以确保浓度均匀。
6. Dilution Calculations | 稀释计算
When a solution is diluted, the number of moles of solute stays the same. Only the volume and concentration change, so we can use the dilution equation:
溶液稀释时，溶质的物质的量保持不变，只有体积和浓度发生变化，因此可以使用稀释公式：
c₁V₁ = c₂V₂
Here c₁ and V₁ are the initial concentration and volume, and c₂ and V₂ are the final concentration and volume. Both volumes must be in the same units.
其中 c₁ 和 V₁ 为初始浓度和体积，c₂ 和 V₂ 为最终浓度和体积。两个体积必须使用相同单位。
example, what volume of 2.0 mol dm⁻³ HCl is needed to prepare 500 cm³ of 0.10 mol dm⁻³ HCl? V₁ = c₂V₂ ÷ c₁ = 0.10 × 0.500 ÷ 2.0 = 0.025 dm³ = 25 cm³.
例如，配制 500 cm³ 0.10 mol dm⁻³ HCl 需要多少体积的 2.0 mol dm⁻³ HCl？V₁ = c₂V₂ ÷ c₁ = 0.10 × 0.500 ÷ 2.0 = 0.025 dm³ = 25 cm³。
In the laboratory, you would measure 25 cm³ of the concentrated solution with a pipette and add distilled water until the total volume reaches 500 cm³.
在实验室中，你会用移液管量取 25 cm³ 浓溶液，然后加蒸馏水直至总体积达到 500 cm³。
7. Ionic Concentrations in Solution | 溶液中离子浓度
Ionic compounds dissociate in water. The concentration of each ion depends on the formula of the compound and the number of each ion released per formula unit.
离子化合物在水中发生解离，每种离子的浓度取决于化合物的化学式以及每个化学式单元释放的离子数目。
For a 0.2 mol dm⁻³ solution of CaCl₂, the calcium ion concentration is 0.2 mol dm⁻³, while the chloride ion concentration is 0.4 mol dm⁻³ because each formula unit produces two Cl⁻ ions.
对于 0.2 mol dm⁻³ 的 CaCl₂ 溶液，钙离子浓度为 0.2 mol dm⁻³，而氯离子浓度为 0.4 mol dm⁻³，因为每个化学式单元产生两个 Cl⁻ 离子。
is is often written using square brackets: [Ca²⁺] = 0.2 mol dm⁻³ and [Cl⁻] = 0.4 mol dm⁻³.
这通常用方括号
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