Solutions and Concentration | 溶液与浓度

📚 Solutions and Concentration | 溶液与浓度

A solution is a homogeneous mixture composed of a solute dissolved in a solvent. In A-level chemistry, aqueous solutions are by far the most common, where water acts as the solvent. Uniform distribution of particles ensures that any portion of the solution has identical composition and properties. Mastering the concept of concentration – how much solute is present per unit volume or mass – is essential for quantitative analysis, reaction kinetics, and equilibrium studies.

溶液是由溶质溶解在溶剂中形成的均匀混合物。在A-level化学中，水溶液最为常见，其中水作为溶剂。溶质粒子均匀分布，确保溶液中任意部分的组成和性质相同。掌握浓度的概念——单位体积或单位质量中溶质的含量——对于定量分析、反应动力学和平衡研究至关重要。

1. Introduction to Solutions | 溶液简介

A solution consists of a solute and a solvent. The solute is the substance being dissolved, typically present in a smaller amount, while the solvent is the dissolving medium. Solutions can exist in gas, liquid, or solid phases, but liquid solutions dominate laboratory work. The attraction between solute and solvent particles (solvation) drives the dissolving process. For aqueous solutions, water molecules surround ions or polar molecules, stabilising them in solution.

溶液由溶质和溶剂组成。溶质是被溶解的物质，通常量较少；溶剂是溶解介质。溶液可以以气态、液态或固态存在，但实验室中以液态溶液为主。溶质与溶剂粒子之间的吸引力（溶剂化）推动溶解过程。对于水溶液，水分子包围离子或极性分子，使它们在溶液中稳定存在。

Concentration is a measure of ‘how much solute’ is contained in a given quantity of solvent or solution. It can be expressed in many ways, each suitable for specific contexts – from molarity in titration to percentage purity in pharmaceuticals. Understanding these different expressions and being able to convert between them is a fundamental skill in physical chemistry.

浓度是衡量“有多少溶质”存在于一定量溶剂或溶液中的量度。它可以用多种方式表示，每种方式适用于特定情境——从滴定中的摩尔浓度到制药中的百分比纯度。理解这些不同的表达方式并能在它们之间进行转换，是物理化学中的一项基本技能。

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2. Expressing Concentration: Molarity | 浓度表示：摩尔浓度

Molarity, symbol c, is the number of moles of solute dissolved per cubic decimetre (dm³) of solution. The unit is mol dm⁻³, often shortened to M. It is the most widely used concentration term in volumetric analysis because it directly relates to the mole concept and the balanced equation.

摩尔浓度，符号c，是每立方分米（dm³）溶液中所含溶质的物质的量。单位是mol dm⁻³，常缩写为M。这是容量分析中使用最广泛的浓度术语，因为它直接与物质的量概念和配平的化学方程式相关。

c = n / V

where n is the amount of solute in moles and V is the volume of solution in dm³. For example, if 0.50 mol of NaCl is dissolved to give 2.00 dm³ of solution, the molarity is c = 0.50 mol / 2.00 dm³ = 0.25 mol dm⁻³. Remember to convert cm³ to dm³ by dividing by 1000 before substituting into the formula.

其中n是溶质的物质的量（mol），V是溶液的体积（dm³）。例如，如果0.50 mol的NaCl溶解后得到2.00 dm³的溶液，则摩尔浓度为c = 0.50 mol / 2.00 dm³ = 0.25 mol dm⁻³。代入公式前，请记得将cm³转换为dm³（除以1000）。

Molarity depends on temperature because the volume of a liquid changes with temperature. For precise work, solutions are often prepared at a controlled temperature, normally 20 °C. In calculations involving reacting quantities, we combine c = n / V with the mole ratio from the chemical equation to determine unknown concentrations or volumes.

摩尔浓度依赖于温度，因为液体的体积随温度变化。为了精确工作，溶液通常在控温（通常是20 °C）下配制。在涉及反应量的计算中，我们结合c = n / V和化学方程式中的摩尔比来确定未知浓度或体积。

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3. Mass Concentration | 质量浓度

Mass concentration, often given the symbol ρ or simply ‘concentration in g dm⁻³’, is the mass of solute dissolved per unit volume of solution. The unit is g dm⁻³. It is particularly useful when the molar mass of the solute is unknown or when preparing solutions from hydrated salts.

质量浓度，常用符号ρ或直接称为“以g dm⁻³表示的浓度”，是指单位体积溶液中溶解的溶质质量。单位是g dm⁻³。当溶质的摩尔质量未知，或使用水合盐配制溶液时，质量浓度特别有用。

ρ = m / V

where m is the mass of solute in grams and V is the volume of solution in dm³. For instance, dissolving 10.0 g of NaOH in water to make 500 cm³ (0.500 dm³) of solution yields a mass concentration of 10.0 g / 0.500 dm³ = 20.0 g dm⁻³.

其中m是溶质的质量（g），V是溶液的体积（dm³）。例如，将10.0 g NaOH溶于水，配制成500 cm³ (0.500 dm³)的溶液，质量浓度为10.0 g / 0.500 dm³ = 20.0 g dm⁻³。

The relationship between molarity and mass concentration is given by c = ρ / M, where M is the molar mass of the solute. This equation allows easy conversion: a 20.0 g dm⁻³ NaOH solution (M = 40.0 g mol⁻¹) has a molarity of 20.0 / 40.0 = 0.500 mol dm⁻³. Such interconversions are essential when results of an experiment are reported in different concentration units.

摩尔浓度与质量浓度之间的关系由c = ρ / M给出，其中M是溶质的摩尔质量。这个公式可以轻松转换：20.0 g dm⁻³的NaOH溶液（M = 40.0 g mol⁻¹）的摩尔浓度为20.0 / 40.0 = 0.500 mol dm⁻³。当实验结果以不同浓度单位报告时，这种互相转换是必不可少的。

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4. Parts Per Million (ppm) | 百万分率

For very dilute solutions, using molarity or mass concentration becomes impractical. Parts per million (ppm) is a convenient way to express very low concentrations. It represents the mass of solute per million mass units of solution. In aqueous systems where the density of the solution is approximately 1.0 g cm⁻³, 1 ppm is equivalent to 1 mg of solute per dm³ of solution (1 mg dm⁻³). This approximation holds well for dilute aqueous solutions.

对于非常稀的溶液，使用摩尔浓度或质量浓度变得不切实际。百万分率 (ppm) 是表示极低浓度的一种便捷方式。它代表每百万质量单位溶液中所含溶质的质量单位。在水溶液中，当溶液密度约为1.0 g cm⁻³时，1 ppm相当于每dm³溶液中含有1 mg溶质 (1 mg dm⁻³)。这个近似值对于稀水溶液来说很适用。

ppm = (mass of solute / mass of solution) × 10⁶

For example, a sample of drinking water containing 0.0025 g of lead in 1000 g of water has a lead concentration of (0.0025 g / 1000 g) × 10⁶ = 2.5 ppm. In air or other fluids, ppm by volume is also used. Laboratory specifications for impurity levels are often given in ppm, and analytical techniques such as atomic absorption spectroscopy are calibrated using ppm standards.

例如，一份饮用水样品在1000 g水中含有0.0025 g铅，则铅浓度为(0.0025 g / 1000 g) × 10⁶ = 2.5 ppm。在空气或其他流体中，也使用体积ppm表示。实验室对杂质水平的规格通常以ppm给出，像原子吸收光谱等分析技术要用ppm标准来校准。

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5. Preparing Standard Solutions | 配制标准溶液

A standard solution is one of accurately known concentration. It is prepared using a primary standard – a substance of high purity, known stoichiometry and stability in air. The preparation involves a series of critical steps that must be performed with precision to minimise errors.

标准溶液是指浓度准确已知的溶液。它使用基准物质配制——这种物质纯度高、化学计量明确、在空气中稳定。配制过程涉及一系列关键步骤，必须精确操作以最大程度减少误差。

The typical procedure is: (1) Calculate the mass of solute required to achieve the desired concentration and volume. (2) Weigh the solute accurately on a balance, using a weighing boat and recording the mass to the required number of significant figures. (3) Transfer the solid into a clean beaker, dissolve it in a small volume of deionised water, and stir. (4) Carefully pour the solution into a volumetric flask of the chosen volume via a funnel. Rinse the beaker and funnel several times with deionised water, transferring all washings to the flask. (5) Fill the flask with deionised water until the bottom of the meniscus aligns with the calibration mark on the neck. Stopper the flask and invert it several times to ensure homogeneity.

典型步骤是：(1) 计算达到所需浓度和体积所需的溶质质量。(2) 使用天平准确称量溶质，用称量舟并记录质量至所需的有效数字位数。(3) 将固体转移至干净的烧杯中，用少量去离子水溶解并搅拌。(4) 通过漏斗小心地将溶液倒入选定体积的容量瓶中。用去离子水冲洗烧杯和漏斗数次，将所有洗涤液转移至容量瓶中。(5) 继续加入去离子水，直到弯月面底部与容量瓶颈部的刻度线对齐。塞紧瓶塞，倒转摇晃数次以确保均匀。

Common errors include loss of solute during transfer, insufficient rinsing, parallax error when reading the meniscus, and failing to mix thoroughly. Always use the solvent at room temperature and re-invert the flask after temperature equilibration.

常见误差包括转移过程中溶质损失、冲洗不充分、读取弯月面时的视差错误以及混合不充分。务必在室温下使用溶剂，并在温度平衡后再次倒转容量瓶。

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6. Dilution of Solutions | 溶液的稀释

Dilution is the process of reducing the concentration of a solution by adding more solvent. The amount of solute remains constant, so we can write: c₁V₁ = c₂V₂, where c₁ and V₁ are the initial concentration and volume, and c₂ and V₂ are the final concentration and volume. This equation works as long as the same volume unit is used for both V₁ and V₂ and the concentration units match.

稀释是通过添加更多溶剂来降低溶液浓度的过程。溶质的物质的量保持不变，因此可以写成：c₁V₁ = c₂V₂，其中c₁和V₁是初始浓度和体积，c₂和V₂是最终浓度和体积。只要V₁和V₂使用相同的体积单位，并且浓度单位一致，这个公式就成立。

For instance, to prepare 250 cm³ of 0.100 mol dm⁻³ HCl from a stock solution of 2.00 mol dm⁻³, we calculate the required volume of stock: V₁ = (c₂V₂)/c₁ = (0.100 mol dm⁻³ × 0.250 dm³) / 2.00 mol dm⁻³ = 0.0125 dm³ = 12.5 cm³. A pipette is used to measure 12.5 cm³ of the concentrated acid, which is then transferred to a 250 cm³ volumetric flask and made up to the mark with deionised water.

例如，要从2.00 mol dm⁻³的浓盐酸储备液配制250 cm³、0.100 mol dm⁻³的HCl溶液，计算所需储备液体积：V₁ = (c₂V₂)/c₁ = (0.100 mol dm⁻³ × 0.250 dm³) / 2.00 mol dm⁻³ = 0.0125 dm³ = 12.5 cm³。使用移液管量取12.5 cm³的浓酸，然后转移至250 cm³容量瓶中，用去离子水定容至刻度。

When diluting concentrated acids, always add acid to water, not the reverse, to dissipate heat safely. The dilution equation can also be expressed in terms of mass concentration: ρ₁V₁ = ρ₂V₂, provided the solute is the same.

稀释浓酸时，永远是将酸加入水中，而不是反过来，以便安全散热。稀释公式也可以用质量浓度表示：ρ₁V₁ = ρ₂V₂，前提是溶质相同。

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7. Volumetric Analysis and Titration Calculations | 容量分析与滴定计算

Titration is an analytical technique in which a solution of known concentration (the titrant) is reacted with a known volume of a solution of unknown concentration (the analyte) until the reaction reaches its endpoint. Acid–base titrations are the classic example. The mole ratio from the balanced equation is central to finding the unknown concentration.

滴定是一种分析技术，用已知浓度的溶液（滴定剂）与已知体积的未知浓度溶液（分析物）反应，直到反应达到终点。酸碱滴定是最经典的例子。配平方程式中的摩尔比是求得未知浓度的关键。

In a typical acid–base titration, the average titre V_titre (in dm³) of standard solution of concentration c_standard is recorded. Using n = cV, the amount of standard used is calculated. From the stoichiometric ratio, the amount of analyte is determined, and then its concentration is found by dividing by the pipetted volume V_analyte.

在典型的酸碱滴定中，记录标准溶液的平均滴定体积V_titre (dm³) 及其浓度c_standard。利用n = cV，计算所用的标准的物质的量。根据化学计量比，确定分析物的物质的量，然后除以移取的体积V_analyte得到其浓度。

Example: 25.0 cm³ of NaOH solution is titrated with 0.100 mol dm⁻³ HCl. The average titre is 22.40 cm³. Reaction: NaOH + HCl → NaCl + H₂O (1:1 ratio). n_HCl = 0.100 mol dm⁻³ × 0.02240 dm³ = 2.24 × 10⁻³ mol. Since the mole ratio is 1:1, n_NaOH = 2.24 × 10⁻³ mol. c_NaOH = n_NaOH / V_NaOH = 2.24 × 10⁻³ mol / 0.0250 dm³ = 0.0896 mol dm⁻³. Concordant titres within 0.10 cm³ should be obtained for reliable results.

例如：用0.100 mol dm⁻³ HCl滴定25.0 cm³ NaOH溶液，平均滴定体积为22.40 cm³。反应：NaOH + HCl → NaCl + H₂O（1:1摩尔比）。n_HCl = 0.100 mol dm⁻³ × 0.02240 dm³ = 2.24 × 10⁻³ mol。由于摩尔比为1:1，n_NaOH = 2.24 × 10⁻³ mol。c_NaOH = n_NaOH / V_NaOH = 2.24 × 10⁻³ mol / 0.0250 dm³ = 0.0896 mol dm⁻³。为得到可靠结果，应获得相差在0.10 cm³以内的吻合滴定体积。

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8. Back Titration and Indirect Analysis | 返滴定与间接分析

When direct titration is not feasible – for example, if the analyte is insoluble, volatile, or reacts slowly – a back titration is employed. In this method, an excess measured amount of a standard reagent is added to the sample. After reaction, the remaining unreacted reagent is titrated with a second standard solution. The amount that reacted with the sample is found by subtraction.

当直接滴定不可行时——例如分析物不溶、易挥发或反应缓慢——就采用返滴定。在这种方法中，向样品中加入过量且计量准确的标准试剂。反应完成后，未反应的试剂用第二种标准溶液滴定。通过减法求得与样品反应的量。

A classic example is the determination of calcium carbonate in limestone. A known mass of limestone is reacted with excess standard HCl. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. The unreacted HCl is then titrated with standard NaOH. The difference between the initial moles of HCl and the moles titrated gives the moles of HCl that reacted with CaCO₃, and hence the mass or purity of CaCO₃ can be calculated.

一个经典例子是测定石灰石中碳酸钙的含量。称取已知质量的石灰石，与过量标准HCl反应。CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O。未反应的HCl再用标准NaOH滴定。初始HCl的物质的量与滴定的物质的量之差，即为与CaCO₃反应的HCl的物质的量，进而可计算出CaCO₃的质量或纯度。

Back titrations are also useful for determining the active ingredient in indigestion tablets or the ammonia content in ammonium salts. The key is to ensure that the first reagent is genuinely in excess and that the two titrations are carried out under the same conditions.

返滴定也用于测定胃药片中的有效成分或铵盐中的氨含量。关键是确保第一种试剂确实过量，并且两次滴定在相同条件下进行。

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9. Percentage Concentration Expressions | 百分比浓度表示

Percentage concentrations are common in commercial and pharmaceutical contexts. Three main types are used: % w/w (mass/mass), % w/v (mass/volume), and % v/v (volume/volume). Each represents the number of grams or cm³ of solute per 100 g or 100 cm³ of solution. For instance, a 5% w/v NaCl solution contains 5 g of NaCl per 100 cm³ of solution.

百分比浓度在商业和制药领域很常见。主要使用三种类型：% w/w (质量/质量)、% w/v (质量/体积) 和 % v/v (体积/体积)。每种表示每100 g或100 cm³溶液中含有的溶质克数或cm³数。例如，5% w/v NaCl溶液表示每100 cm³溶液中含有5 g NaCl。

Conversion between % w/v and molarity requires the solute’s molar mass. For a 5% w/v NaCl, mass concentration is 5 g per 100 cm³ = 50 g dm⁻³. With M(NaCl)=58.5 g mol⁻¹, c = 50 / 58.5 ≈ 0.855 mol dm⁻³. Similarly, % v/v is used for liquid solutes; a 70% v/v ethanol solution contains 70 cm³ of pure ethanol per 100 cm³ of solution. These expressions are not temperature-dependent in the same way molarity is, but care must be taken to specify the basis (w/w, w/v, v/v).

将% w/v转换为摩尔浓度需要溶质的摩尔质量。对于5% w/v NaCl，质量浓度为5 g/100 cm³ = 50 g dm⁻³。M(NaCl)=58.5 g mol⁻¹，c = 50 / 58.5 ≈ 0.855 mol dm⁻³。类似地，% v/v用于液体溶质；70% v/v乙醇溶液表示每100 cm³溶液中含有70 cm³纯乙醇。这些表示法不像摩尔浓度那样依赖于温度，但必须明确其基准 (w/w, w/v, v/v)。

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10. Common Errors and Accuracy in Solution Preparation | 配制溶液的常见误差与准确度

Achieving accurate concentration requires recognising both systematic and random errors. Systematic errors include using an uncalibrated balance, moisture-sensitive primary standards absorbing water, or failing to rinse glassware. Random errors arise from estimating the meniscus, temperature fluctuations, or incomplete mixing.

要使浓度准确，需要识别系统误差和随机误差。系统误差包括使用未校准的天平、对水分敏感的基准物质吸水、或未冲洗玻璃器皿。随机误差源于估读弯月面、温度波动或混合不充分。

Error source	Effect on concentration	Prevention
Weighing by difference not done properly	Mass recorded higher than true, concentration too high	Tare balance, handle with care
Transfer losses: solid left in beaker	Actual solute mass lower, concentration too low	Rinse beaker and funnel thoroughly
Overfilling volumetric flask above mark	Volume too large, concentration too low	Use dropper for final addition, read meniscus at eye level
Air bubbles in pipette or burette	Delivered volume inaccurate	Inspect and remove bubbles before use

Calibration of volumetric glassware (Class A or B) plays a significant role. Class A apparatus is manufactured to tighter tolerances and is preferred for analytical work. Temperature also affects the volume of both the solution and the glassware; working at the calibration temperature (usually 20 °C) minimises this effect. Repeating the preparation and calculating the mean concentration with standard deviation gives a measure of precision.

容量玻璃器皿的校准（A级或B级）起着重要作用。A级器皿的公差更严格，是分析工作的首选。温度也会影响溶液和玻璃器皿的体积；在校准温度（通常20 °C）下操作可最大程度减少这种影响。重复配制并计算平均浓度和标准差，可评估精密度。

In summary, mastering solutions and concentration is not just about plugging numbers into formulas; it involves a deep understanding of the underlying principles, careful experimental technique, and the ability to critically evaluate the reliability of the resulting data.

总之，掌握溶液与浓度不仅仅是把数字代入公式；它涉及对基本原理的深刻理解、严谨的实验技术，以及批判性地评估所得数据可靠性的能力。

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