High School Chemistry

Solutions and Solubility溶液与溶解度

Why does salt dissolve in water but not in oil? Why does adding antifreeze lower the freezing point of a car's coolant? Why do electrolyte drinks conduct electricity? Every answer comes back to solutions — homogeneous mixtures where a solute dissolves in a solvent. This guide covers the full Grade 11 solutions unit: terminology and the dissolving process, like-dissolves-like polarity reasoning, molarity and concentration calculations, the dilution equation $c_1V_1 = c_2V_2$, solubility curves and saturation, electrolytes and dissociation, and a first look at colligative properties. Worked examples and KaTeX formulas throughout.为什么食盐溶于水却不溶于油？为什么加入防冻液会降低汽车冷却液的冰点？为什么电解质饮料能导电？答案都回归到溶液（solution，溶液）——溶质（solute，溶质）溶解于溶剂（solvent，溶剂）所形成的均匀混合物。本指南覆盖 11 年级溶液单元的完整内容：术语与溶解过程、相似相溶的极性推理、摩尔浓度（molarity，摩尔浓度）与浓度（concentration，浓度）计算、稀释（dilution，稀释）方程 $c_1V_1 = c_2V_2$、溶解度（solubility，溶解度）曲线与饱和（saturated，饱和）、电解质（electrolyte，电解质）与解离，以及依数性（colligative properties，依数性）的初步介绍。全程使用例题与公式。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Feeder: IB Chemistry HL Reactivity 2衔接：IB Chemistry HL 反应性 2
Read me first先读这里

How to use this guide如何使用本指南

Solutions and solubility sit in Grade 11 chemistry across all four curricula we map to. The four programs agree on a robust core: what solutions are, the dissolving process, concentration and molarity, dilution calculations, solubility and saturation, and electrolytes. Ontario SCH3U Strand E and Alberta Chemistry 20 Unit C are the most detailed, providing verbatim content codes; BC Chemistry 11 organises content around the Big Idea "Solubility within a solution is determined by the nature of the solute and the solvent"; NGSS HS-PS1-3 frames the dissolving process through the lens of inter-particle electrical forces. All four agree that quantitative molarity calculations ($c = n/V$) and the dilution equation ($c_1V_1 = c_2V_2$) are Grade 11 content. Colligative properties (section 7) are introduced here but assessed in depth only at the IB or AP level.溶液与溶解度在我们所对照的四套大纲中均属于 11 年级化学内容。四套课程在核心内容上一致：溶液的定义、溶解过程、浓度与摩尔浓度、稀释计算、溶解度与饱和，以及电解质。安大略 SCH3U E 单元和阿尔伯塔 Chemistry 20 C 单元最为详细，提供了逐字的内容代码；BC Chemistry 11 围绕大概念"溶液中的溶解度由溶质和溶剂的性质决定"组织内容；NGSS HS-PS1-3 从颗粒间电力的角度理解溶解过程。四套大纲均认定摩尔浓度定量计算（$c = n/V$）和稀释方程（$c_1V_1 = c_2V_2$）为 11 年级内容。依数性（第 7 节）在此作简介，仅在 IB 或 AP 层次深入评估。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学 §1–§2 (solutions terminology, dissolving process) anchor on HS-PS1-3 (force-strength explains bulk behaviour). Molarity and dilution (§3–§4) are foundational skills; NGSS folds them into stoichiometry (HS-PS1-7).§1–§2（溶液术语、溶解过程）对应 HS-PS1-3（力的强度解释宏观行为）。摩尔浓度与稀释（§3–§4）是基础技能；NGSS 将其纳入化学计量学（HS-PS1-7）。 §7 (colligative properties): above the NGSS assessed floor. The Assessment Boundary for HS-PS1-3 explicitly excludes "Raoult's law calculations of vapor pressure."§7（依数性）：高于 NGSS 评估下限。HS-PS1-3 的评估边界明确排除"蒸气压的拉乌尔定律计算"。 NGSS HS-PS1 (Chemistry) — HS-PS1-3 PE + Clarification + Assessment Boundary— HS-PS1-3 表现期望 + 澄清说明 + 评估边界
🇨🇦 ON SCH3U (Grade 11)安大略 SCH3U（11 年级） §1–§6 in full. SCH3U Strand E (E2.1–E2.6, E3.1–E3.4) makes all seven core topics assessed content. Key quantitative items: molarity calculation (E2.2), dilution (E2.3), solubility tables and precipitates (E3.4).§1–§6 完整学习。SCH3U E 单元（E2.1–E2.6、E3.1–E3.4）将全部七个核心主题列为评估内容。关键定量项目：摩尔浓度计算（E2.2）、稀释（E2.3）、溶解度表与沉淀（E3.4）。 §7 (colligative properties): enters at SCH4U / IB depth. Mention briefly for completeness but not assessed in SCH3U.§7（依数性）：在 SCH4U / IB 深度引入。在 SCH3U 中仅简要提及，不作为评估内容。 Ontario SCH3U/4U Chemistry — SCH3U Strand E Overall Expectations E1–E3; Specific Expectations E2.1–E2.6, E3.1–E3.4— SCH3U E 单元总体期望 E1–E3；具体期望 E2.1–E2.6、E3.1–E3.4
🇨🇦 BC Chemistry 11BC Chemistry 11 §1–§6 in full. BC Chemistry 11 Big Idea "Solubility within a solution is determined by the nature of the solute and the solvent" directly maps all six sections as core. Molarity and dilution are elaborated as "stoichiometric calculations in aqueous solutions: molarity, dilution effect, concentration of ions in solution when two solutions are mixed."§1–§6 完整学习。BC Chemistry 11 大概念"溶液中的溶解度由溶质和溶剂的性质决定"将全部六节直接列为核心内容。摩尔浓度与稀释细化为"水溶液中的化学计量计算：摩尔浓度、稀释效应、两种溶液混合时溶液中离子浓度"。 §7 (colligative properties): not a named BC Chemistry 11 Content bullet; enters in Chemistry 12 context.§7（依数性）：不是 BC Chemistry 11 的命名内容；在 Chemistry 12 背景下引入。 BC Chemistry 11/12 — Chemistry 11 Big Idea "Solubility…"; Content "solubility of molecular and ionic compounds"; Elaboration "solubility: dissociation of ions, dissociation equation"; "stoichiometric calculations in aqueous solutions"— Chemistry 11 大概念"溶解度…"；内容"分子和离子化合物的溶解度"；细化"溶解度：离子解离、解离方程"；"水溶液中的化学计量计算"
🇨🇦 AB Chemistry 20阿尔伯塔 Chemistry 20 §1–§6 in full. Chemistry 20 Unit C GO1 covers all core solutions outcomes: dissolving as endo/exothermic, electrolytes vs nonelectrolytes, molarity, dilution, solubility, and saturation as equilibrium. Alberta explicitly names "electrolyte/nonelectrolyte" as a Key Concept.§1–§6 完整学习。Chemistry 20 C 单元 GO1 覆盖所有核心溶液结果：溶解的吸热/放热性、电解质与非电解质、摩尔浓度、稀释、溶解度，以及将饱和视为平衡。阿尔伯塔明确将"电解质/非电解质"列为关键概念。 §7 (colligative properties): not a Chemistry 20 knowledge outcome; enters in IB/AP-level work.§7（依数性）：不是 Chemistry 20 的知识结果；在 IB/AP 层次引入。 Alberta Chemistry 20/30 — Chemistry 20 Unit C GO1, Key Concepts, Knowledge outcome text— Chemistry 20 C 单元 GO1，关键概念，知识结果文本

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise five things: molarity $c = n/V$; dilution $c_1V_1 = c_2V_2$; like dissolves like (polar dissolves polar); saturated = dissolved rate equals crystallisation rate; strong electrolytes fully dissociate, weak electrolytes partially. Read every cram-cheat box first.背熟五件事：摩尔浓度 $c = n/V$；稀释 $c_1V_1 = c_2V_2$；相似相溶（极性溶极性）；饱和 = 溶解速率等于结晶速率；强电解质完全解离，弱电解质部分解离。先读每个速记框。

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If you are going for the top mark如果你目标顶分

Be precise about the thermodynamics of dissolving (enthalpy of solution, entropy drive); know why pressure affects gas solubility but not solid solubility; write dissociation equations for strong and weak electrolytes; explain each colligative property using the particle-count argument; and connect solubility equilibrium to the concept of $K_{sp}$ you will meet in IB/AP. ON SCH3U E2.5 adds net ionic equations for precipitation.精确掌握溶解热力学（溶解焓、熵驱动）；了解为何压力影响气体溶解度但不影响固体溶解度；为强弱电解质写出解离方程；用颗粒数量论证解释每种依数性；并将溶解度平衡与你将在 IB/AP 中遇到的 $K_{sp}$ 概念相联系。ON SCH3U E2.5 还增加了沉淀反应的净离子方程。

Section 1 · Solutions, solutes, solvents & terminology第 1 节 · 溶液、溶质、溶剂与术语

Solutions, Solutes, and Solvents溶液、溶质与溶剂

Three words, one definition — master these and everything else follows.三个词，一个定义 — 掌握这些，其余内容自然跟上。
  • Solution溶液 (solution) — a homogeneous (uniform throughout) mixture of two or more substances. Every sample you take from any part of the solution has the same composition. Examples: seawater, vinegar, antifreeze, a copper sulfate solution in a lab beaker.— 两种或多种物质形成的均匀（各处组成相同）混合物。从溶液任何部位取出的每个样品组成相同。示例：海水、醋、防冻液、实验室烧杯中的硫酸铜溶液。
  • Solute溶质 (solute) — the substance that is dissolved. There can be more than one solute. In salt water, NaCl is the solute. Usually the minor component, but not always.— 被溶解的物质。溶质可以不止一种。在盐水中，NaCl 是溶质。通常是次要组分，但并非总是如此。
  • Solvent溶剂 (solvent) — the substance that does the dissolving. In an aqueous solution, water is always the solvent. Water is the most common solvent in biology and chemistry; its polarity and hydrogen-bonding capacity make it exceptionally good at dissolving ionic and polar substances. 🇨🇦 SCH3U E3.1 / AB Chem 20 C-GO1— 起溶解作用的物质。在水溶液中，水始终是溶剂。水是生物学和化学中最常见的溶剂；其极性和氢键能力使其在溶解离子性和极性物质方面表现出色。🇨🇦 SCH3U E3.1 / AB Chem 20 C-GO1

Key vocabulary you must know:必须掌握的关键词汇：

  • Aqueous solution水溶液 — solvent is water; denoted (aq) in equations.— 溶剂为水；方程式中用 (aq) 标注。
  • Dilute solution稀溶液 — relatively small amount of solute per unit volume.— 每单位体积中溶质相对较少。
  • Concentrated solution浓溶液 — relatively large amount of solute per unit volume.— 每单位体积中溶质相对较多。
  • Miscible / Immiscible互溶 / 不互溶 — two liquids that mix in all proportions (miscible, e.g. ethanol + water) vs. liquids that do not mix (immiscible, e.g. oil + water). 🇨🇦 AB Chem 20 C-GO1— 以任意比例混合的两种液体（互溶，如乙醇 + 水）与不混合的液体（不互溶，如油 + 水）。🇨🇦 AB Chem 20 C-GO1
Worked Example 1 · Identifying components of a solution例题 1 · 识别溶液的组成

A student dissolves $5.85\ \mathrm{g}$ of table salt (NaCl) in $250\ \mathrm{mL}$ of water to make a saline solution for a lab. Identify (a) the solute, (b) the solvent, (c) whether the solution is aqueous, and (d) classify the solution as dilute or concentrated relative to seawater ($\approx 3.5\%$ NaCl by mass).一位同学将 $5.85\ \mathrm{g}$ 食盐（NaCl）溶于 $250\ \mathrm{mL}$ 水中，为实验配制生理盐水。请确认 (a) 溶质，(b) 溶剂，(c) 该溶液是否为水溶液，(d) 与海水（质量分数约 $3.5\%$ NaCl）相比，该溶液是稀溶液还是浓溶液。

(a) Solute: NaCl(a) 溶质：NaCl — the substance that was dissolved into the water.— 溶入水中的物质。

(b) Solvent: water (H₂O)(b) 溶剂：水（H₂O） — the dissolving medium, present in much larger amount ($250\ \mathrm{g}$).— 溶解介质，用量远多于溶质（$250\ \mathrm{g}$）。

(c) Aqueous.(c) 水溶液。 Water is the solvent, so this is an aqueous NaCl solution, written $\mathrm{NaCl(aq)}$ in equations.水为溶剂，故这是水溶液，方程式中写作 $\mathrm{NaCl(aq)}$。

(d) Dilute vs seawater.(d) 与海水比较。 Mass of solute / total mass $\approx 5.85/(5.85+250) \approx 2.3\%$. Seawater is $\approx 3.5\%$, so this solution is more dilute than seawater.溶质质量/总质量 $\approx 5.85/(5.85+250) \approx 2.3\%$。海水约为 $3.5\%$，故该溶液比海水更稀。

In a solution of copper sulfate ($\mathrm{CuSO_4}$) dissolved in water, what is the solvent?在硫酸铜（$\mathrm{CuSO_4}$）溶于水的溶液中，溶剂是什么？
§1 · Q1
Copper sulfate ($\mathrm{CuSO_4}$)硫酸铜（$\mathrm{CuSO_4}$）
Water ($\mathrm{H_2O}$)水（$\mathrm{H_2O}$）
Copper ions ($\mathrm{Cu^{2+}}$)铜离子（$\mathrm{Cu^{2+}}$）
The solution itself溶液本身
The solvent is the dissolving medium — water. $\mathrm{CuSO_4}$ is the solute (the substance dissolved). Since water is the solvent, this is an aqueous solution.溶剂是溶解介质——水。$\mathrm{CuSO_4}$ 是溶质（被溶解的物质）。由于水是溶剂，这是一种水溶液。
The solute is what gets dissolved ($\mathrm{CuSO_4}$); the solvent does the dissolving (water). The solution is the result — the homogeneous mixture of both.溶质是被溶解的物质（$\mathrm{CuSO_4}$）；溶剂是起溶解作用的物质（水）。溶液是结果——两者形成的均匀混合物。
Which of the following is the best definition of a solution?以下哪项是溶液的最佳定义？
§1 · Q2
A homogeneous mixture of two or more substances两种或多种物质形成的均匀混合物
Any liquid that contains dissolved solids任何含有溶解固体的液体
A pure substance with a fixed composition具有固定组成的纯物质
A heterogeneous mixture that can be filtered可以过滤的非均匀混合物
A solution is a homogeneous (uniform) mixture. It can exist as a solid, liquid, or gas — not just a liquid with dissolved solids. Air is a gaseous solution; alloys (e.g. brass) are solid solutions.溶液是均匀的混合物。它可以是固态、液态或气态——不仅仅是含有溶解固体的液体。空气是气态溶液；合金（如黄铜）是固态溶液。
Solutions can be solid, liquid, or gas. They are mixtures (not pure substances), and they are homogeneous — not heterogeneous, and not filterable to separate components.溶液可以是固态、液态或气态。它们是混合物（不是纯物质），且是均匀的——不是非均匀的，也不能通过过滤来分离组分。
Going deeper — why water is such an extraordinary solvent深入 — 为什么水是如此非凡的溶剂

Water's ability to dissolve so many substances comes from two structural features. First, the $\mathrm{H_2O}$ molecule is bent (bond angle $\approx 104.5°$) and the oxygen end carries a partial negative charge ($\delta^-$) while the hydrogen ends carry partial positive charges ($\delta^+$) — water is a polar molecule. Second, the $\delta^+$ hydrogens of one water molecule are attracted to the $\delta^-$ oxygens of neighbouring molecules, forming hydrogen bonds (approximately $20\ \mathrm{kJ/mol}$ each). Together, these features let water interact strongly with both ions (via ion-dipole forces) and polar molecules (via dipole-dipole and hydrogen bonding). SCH3U E3.1 specifically names "polarity" and "hydrogen bonding" as the properties explaining water's solvent behaviour. NGSS HS-PS1-3 frames this as "infer the strength of electrical forces between particles."水溶解如此多物质的能力来自两个结构特征。第一，$\mathrm{H_2O}$ 分子是弯曲的（键角 $\approx 104.5°$），氧端带部分负电荷（$\delta^-$），而氢端带部分正电荷（$\delta^+$）——水是极性分子。第二，一个水分子的 $\delta^+$ 氢被相邻分子的 $\delta^-$ 氧所吸引，形成氢键（每个约 $20\ \mathrm{kJ/mol}$）。这两个特征合在一起，使水既能与离子强烈相互作用（通过离子-偶极力），也能与极性分子相互作用（通过偶极-偶极力和氢键）。SCH3U E3.1 明确将"极性"和"氢键"列为解释水的溶剂行为的性质。NGSS HS-PS1-3 将其框架为"推断颗粒间电力的强度"。

Section 2 · The dissolving process & "like dissolves like"第 2 节 · 溶解过程与相似相溶

The Dissolving Process and "Like Dissolves Like"溶解过程与相似相溶

Dissolving is a competition: solute-solute attractions vs solvent-solute attractions.溶解是一场竞争：溶质-溶质引力 vs 溶剂-溶质引力。
  • Dissolution of an ionic compound (e.g. NaCl in water)离子化合物的溶解（如 NaCl 溶于水）
    1. Water molecules, attracted by their dipoles, surround ions at the crystal surface.水分子因偶极子的吸引而包围晶体表面的离子。
    2. Ion-dipole forces overcome the lattice energy, pulling ions free.离子-偶极力克服晶格能，将离子拉出。
    3. Ions are surrounded by a hydration shell of water molecules and disperse through the solution. 🇨🇦 SCH3U E3.2 / AB Chem 20 C-GO1离子被水分子形成的水化壳包围，并分散到溶液中。🇨🇦 SCH3U E3.2 / AB Chem 20 C-GO1
  • Enthalpy of dissolving溶解焓 = energy to break solute-solute bonds − energy released forming solute-solvent bonds. Net result can be endothermic (e.g. NH₄NO₃ dissolving — cold packs) or exothermic (e.g. NaOH dissolving — gets hot). 🇨🇦 AB Chem 20 C-GO1= 断裂溶质-溶质键的能量 − 形成溶质-溶剂键释放的能量。净结果可以是吸热的（如 NH₄NO₃ 溶解——冰袋）或放热的（如 NaOH 溶解——发热）。🇨🇦 AB Chem 20 C-GO1

"Like dissolves like" rule:相似相溶规则：

  • Polar solvents (e.g. water) dissolve polar solutes (e.g. glucose, ethanol) and ionic compounds (e.g. NaCl, KNO₃).极性溶剂（如水）溶解极性溶质（如葡萄糖、乙醇）和离子化合物（如 NaCl、KNO₃）。
  • Nonpolar solvents (e.g. hexane, oils) dissolve nonpolar solutes (e.g. grease, fats, iodine). Water and oil do not mix because there is no strong attraction between polar water molecules and nonpolar oil molecules. 🇺🇸 NGSS HS-PS1-3 / 🇨🇦 BC Chem 11非极性溶剂（如己烷、油）溶解非极性溶质（如油脂、脂肪、碘）。水与油不混溶，因为极性水分子与非极性油分子之间没有强烈吸引力。🇺🇸 NGSS HS-PS1-3 / 🇨🇦 BC Chem 11
Worked Example 2 · Predicting solubility using "like dissolves like"例题 2 · 用相似相溶预测溶解性

Predict whether each substance dissolves readily in water (polar) or in hexane (nonpolar): (a) $\mathrm{NaCl}$, (b) $\mathrm{I_2}$, (c) $\mathrm{CH_3OH}$ (methanol), (d) motor oil (long nonpolar hydrocarbon chains). Give a brief reason for each.预测以下各物质是否易溶于水（极性）或己烷（非极性）：(a) $\mathrm{NaCl}$，(b) $\mathrm{I_2}$，(c) $\mathrm{CH_3OH}$（甲醇），(d) 机油（长链非极性烃）。每种给出简要理由。

(a) $\mathrm{NaCl}$ → Water.(a) $\mathrm{NaCl}$ → 水。 Ionic compound; strong ion-dipole forces with polar water overcome the lattice energy.离子化合物；与极性水形成强离子-偶极力，克服晶格能。

(b) $\mathrm{I_2}$ → Hexane.(b) $\mathrm{I_2}$ → 己烷。 Nonpolar molecule; London dispersion forces with nonpolar hexane are sufficient. $\mathrm{I_2}$ is nearly insoluble in water (only $\approx 0.3\ \mathrm{g/L}$).非极性分子；与非极性己烷形成的伦敦色散力已足够。$\mathrm{I_2}$ 几乎不溶于水（仅约 $0.3\ \mathrm{g/L}$）。

(c) $\mathrm{CH_3OH}$ → Water (miscible in all proportions).(c) $\mathrm{CH_3OH}$ → 水（以任意比例互溶）。 Polar; $\mathrm{O-H}$ group allows hydrogen bonding with water molecules.极性；$\mathrm{O-H}$ 基团可与水分子形成氢键。

(d) Motor oil → Hexane.(d) 机油 → 己烷。 Long nonpolar chains; only London forces between chains and nonpolar solvent. Immiscible with water — this is why oil and water form separate layers.长链非极性结构；链与非极性溶剂之间仅有伦敦力。与水不互溶——这就是油水分层的原因。

A student wants to remove a grease stain from fabric. Based on "like dissolves like," which solvent is most effective?一位同学想去除织物上的油脂污渍。根据相似相溶原则，哪种溶剂最有效？
§2 · Q1
Water水
Salt water盐水
A nonpolar solvent (e.g. dry-cleaning fluid)非极性溶剂（如干洗液）
Vinegar (dilute acetic acid)醋（稀乙酸）
Grease is a nonpolar substance. "Like dissolves like" predicts that a nonpolar solvent will dissolve grease effectively. Dry-cleaning solvents (e.g. perchloroethylene) are nonpolar for exactly this reason. Water, salt water, and vinegar are all polar or ionic — they do not dissolve nonpolar grease well.油脂是非极性物质。相似相溶预测非极性溶剂能有效溶解油脂。干洗溶剂（如四氯乙烯）正是因此而为非极性。水、盐水和醋都是极性或离子性的——它们不能很好地溶解非极性油脂。
Grease is nonpolar. Water, salt water, and vinegar are polar — they cannot break the nonpolar grease-fabric interactions. A nonpolar solvent interacts with grease via London dispersion forces and dissolves it.油脂是非极性的。水、盐水和醋是极性的——它们无法破坏非极性油脂与织物之间的相互作用。非极性溶剂通过伦敦色散力与油脂相互作用并将其溶解。
When NaCl dissolves in water, what is the immediate role of the water molecules around a sodium ion?当 NaCl 溶于水时，围绕钠离子的水分子的直接作用是什么？
§2 · Q2
They react chemically with Na⁺ to form a new compound它们与 Na⁺ 发生化学反应形成新化合物
They push Na⁺ away from Cl⁻ using kinetic energy它们用动能将 Na⁺ 推离 Cl⁻
They form covalent bonds with Na⁺它们与 Na⁺ 形成共价键
They surround Na⁺ with their negative (oxygen) ends, stabilising it in solution via ion-dipole attraction它们用负极端（氧端）包围 Na⁺，通过离子-偶极引力将其稳定在溶液中
Na⁺ is a cation (positive). The oxygen ends ($\delta^-$) of water molecules are attracted to it, forming an ion-dipole interaction called a hydration shell. This stabilises the ion in solution and provides the energy that compensates for breaking the crystal lattice.Na⁺ 是阳离子（正离子）。水分子的氧端（$\delta^-$）被其吸引，形成称为水化壳的离子-偶极相互作用。这使离子在溶液中稳定，并提供补偿晶格键断裂所需的能量。
Dissolving is a physical process (no new bonds formed, no chemical reaction). Water molecules use electrostatic attraction (ion-dipole forces) to stabilise the ions — not kinetic pushing or covalent bonding.溶解是物理过程（不形成新键，不发生化学反应）。水分子通过静电引力（离子-偶极力）稳定离子——而非动能推动或共价键合。
Section 3 · Concentration & molarity第 3 节 · 浓度与摩尔浓度

Concentration and Molarity浓度与摩尔浓度

Molarity $c$ = moles of solute per litre of solution.摩尔浓度 $c$ = 每升溶液中溶质的摩尔数。 $$ c = \frac{n}{V} $$
  • $c$ = molar concentration / molarity $[\mathrm{mol\,L^{-1}}]$ or $[\mathrm{M}]$$c$ = 摩尔浓度（摩尔/升） $[\mathrm{mol\,L^{-1}}]$ 或 $[\mathrm{M}]$
  • $n$ = amount of solute in moles $[\mathrm{mol}]$$n$ = 溶质的摩尔数 $[\mathrm{mol}]$
  • $V$ = volume of solution in litres $[\mathrm{L}]$$V$ = 溶液的体积（升） $[\mathrm{L}]$

Common concentration units (ON SCH3U E2.2 / AB Chem 20 C-GO1):常见浓度单位（ON SCH3U E2.2 / AB Chem 20 C-GO1）：

  • Molarity (mol/L or M) — most common in stoichiometry calculations.摩尔浓度（mol/L 或 M）——化学计量计算中最常用。
  • % by mass — mass of solute / total mass of solution × 100%. Used on labels (e.g. 3% H₂O₂).质量分数——溶质质量/溶液总质量 × 100%。用于标签（如 3% H₂O₂）。
  • ppm / ppb — parts per million / billion. Used for trace concentrations in water-quality analysis.ppm / ppb——百万分比 / 十亿分比。用于水质分析中的痕量浓度。
Converting between units: use $n = m/M_r$ (where $M_r$ is molar mass) to get moles from grams, then $c = n/V$. 🇺🇸 NGSS HS-PS1-3 / 🇨🇦 SCH3U E2.2 / BC Chem 11 / AB Chem 20 C-GO1单位换算：用 $n = m/M_r$（$M_r$ 为摩尔质量）将克转换为摩尔，再用 $c = n/V$。🇺🇸 NGSS HS-PS1-3 / 🇨🇦 SCH3U E2.2 / BC Chem 11 / AB Chem 20 C-GO1
Worked Example 3 · Calculating molarity例题 3 · 计算摩尔浓度

A student dissolves $14.6\ \mathrm{g}$ of NaCl ($M_r = 58.44\ \mathrm{g/mol}$) in enough water to make $500.0\ \mathrm{mL}$ of solution. Calculate the molar concentration of the solution.一位同学将 $14.6\ \mathrm{g}$ 的 NaCl（$M_r = 58.44\ \mathrm{g/mol}$）溶于足量水中，配制 $500.0\ \mathrm{mL}$ 溶液。计算该溶液的摩尔浓度。

Step 1 — Convert mass to moles.第一步 — 将质量转换为摩尔数。

$$ n = \frac{m}{M_r} = \frac{14.6\ \mathrm{g}}{58.44\ \mathrm{g/mol}} = 0.2498\ \mathrm{mol}. $$

Step 2 — Convert volume to litres.第二步 — 将体积转换为升。

$$ V = 500.0\ \mathrm{mL} = 0.5000\ \mathrm{L}. $$

Step 3 — Apply $c = n/V$.第三步 — 应用 $c = n/V$。

$$ c = \frac{0.2498\ \mathrm{mol}}{0.5000\ \mathrm{L}} = 0.500\ \mathrm{mol/L}\ (0.500\ \mathrm{M}). $$

The NaCl solution has a molar concentration of $0.500\ \mathrm{M}$. This is called "0.500 molar NaCl" and written $[\mathrm{NaCl}] = 0.500\ \mathrm{mol/L}$. ✓该 NaCl 溶液的摩尔浓度为 $0.500\ \mathrm{M}$。称为"0.500 摩尔 NaCl 溶液"，写作 $[\mathrm{NaCl}] = 0.500\ \mathrm{mol/L}$。✓

How many moles of $\mathrm{HCl}$ are present in $250\ \mathrm{mL}$ of a $2.0\ \mathrm{mol/L}$ $\mathrm{HCl}$ solution?$250\ \mathrm{mL}$ 的 $2.0\ \mathrm{mol/L}$ $\mathrm{HCl}$ 溶液中含有多少摩尔 $\mathrm{HCl}$？
§3 · Q1
$2.0\ \mathrm{mol}$$2.0\ \mathrm{mol}$
$0.125\ \mathrm{mol}$$0.125\ \mathrm{mol}$
$0.50\ \mathrm{mol}$$0.50\ \mathrm{mol}$
$8.0\ \mathrm{mol}$$8.0\ \mathrm{mol}$
Rearrange $c = n/V$ to $n = cV$. Convert volume: $250\ \mathrm{mL} = 0.250\ \mathrm{L}$. Then $n = 2.0\ \mathrm{mol/L} \times 0.250\ \mathrm{L} = 0.50\ \mathrm{mol}$.将 $c = n/V$ 变形为 $n = cV$。换算体积：$250\ \mathrm{mL} = 0.250\ \mathrm{L}$。则 $n = 2.0\ \mathrm{mol/L} \times 0.250\ \mathrm{L} = 0.50\ \mathrm{mol}$。
Use $n = cV$. Always convert $\mathrm{mL}$ to $\mathrm{L}$ first: $250\ \mathrm{mL} = 0.250\ \mathrm{L}$. Then $n = 2.0 \times 0.250 = 0.50\ \mathrm{mol}$.使用 $n = cV$。始终先将 $\mathrm{mL}$ 换算为 $\mathrm{L}$：$250\ \mathrm{mL} = 0.250\ \mathrm{L}$。则 $n = 2.0 \times 0.250 = 0.50\ \mathrm{mol}$。
What volume (in mL) of a $0.40\ \mathrm{mol/L}$ $\mathrm{NaOH}$ solution contains $0.0200\ \mathrm{mol}$ of $\mathrm{NaOH}$?$0.40\ \mathrm{mol/L}$ 的 $\mathrm{NaOH}$ 溶液中，哪个体积（mL）含有 $0.0200\ \mathrm{mol}$ 的 $\mathrm{NaOH}$？
§3 · Q2
$8.0\ \mathrm{mL}$$8.0\ \mathrm{mL}$
$50\ \mathrm{mL}$$50\ \mathrm{mL}$
$0.050\ \mathrm{mL}$$0.050\ \mathrm{mL}$
$200\ \mathrm{mL}$$200\ \mathrm{mL}$
$V = n/c = 0.0200\ \mathrm{mol} / 0.40\ \mathrm{mol/L} = 0.050\ \mathrm{L} = 50\ \mathrm{mL}$. Always convert litres back to millilitres if the question asks for mL.$V = n/c = 0.0200\ \mathrm{mol} / 0.40\ \mathrm{mol/L} = 0.050\ \mathrm{L} = 50\ \mathrm{mL}$。如果题目要求 mL，始终将升换算回毫升。
Rearrange $c = n/V$ to $V = n/c$: $V = 0.0200/0.40 = 0.050\ \mathrm{L} = 50\ \mathrm{mL}$.将 $c = n/V$ 变形为 $V = n/c$：$V = 0.0200/0.40 = 0.050\ \mathrm{L} = 50\ \mathrm{mL}$。
Section 4 · Dilution第 4 节 · 稀释

Dilution稀释

Dilution: add solvent, moles of solute stay the same.稀释：加入溶剂，溶质的摩尔数不变。

When you dilute a solution, you increase the volume of solvent while keeping the number of moles of solute constant. Because $n = cV$ and $n$ is constant:稀释溶液时，增加溶剂体积，而溶质的摩尔数保持不变。由于 $n = cV$ 且 $n$ 为常数：

$$ c_1 V_1 = c_2 V_2 $$
  • $c_1$, $V_1$ = concentration and volume before dilution (the stock solution)$c_1$、$V_1$ = 稀释前的浓度和体积（储备液）
  • $c_2$, $V_2$ = concentration and volume after dilution (the diluted solution)$c_2$、$V_2$ = 稀释后的浓度和体积（稀溶液）
  • Units of $c$ must match ($\mathrm{mol/L}$ or $\mathrm{M}$); units of $V$ must match ($\mathrm{mL}$ or $\mathrm{L}$, as long as both sides are the same).$c$ 的单位必须一致（$\mathrm{mol/L}$ 或 $\mathrm{M}$）；$V$ 的单位必须一致（$\mathrm{mL}$ 或 $\mathrm{L}$，只要两边相同即可）。

Laboratory procedure (SCH3U E2.3 / AB Chem 20 C-GO1): measure the required volume $V_1$ of stock solution, transfer it to a volumetric flask, and add solvent up to the $V_2$ mark. Never add the concentrated acid or base to a small amount of water first — always add concentrated solute to the larger volume of solvent.实验室操作（SCH3U E2.3 / AB Chem 20 C-GO1）：量取所需体积 $V_1$ 的储备液，转移至容量瓶，加溶剂至 $V_2$ 刻度线。切勿先将浓酸或浓碱加入少量水中——始终将浓溶质加入较大体积的溶剂中。

Worked Example 4 · Preparing a dilute solution例题 4 · 配制稀溶液

A lab has a stock solution of $\mathrm{HCl}$ at $12.0\ \mathrm{mol/L}$. What volume of this stock solution is needed to prepare $500\ \mathrm{mL}$ of $0.600\ \mathrm{mol/L}$ $\mathrm{HCl}$?实验室有 $12.0\ \mathrm{mol/L}$ 的 $\mathrm{HCl}$ 储备液。需要取多少体积的该储备液，才能配制 $500\ \mathrm{mL}$ 的 $0.600\ \mathrm{mol/L}$ $\mathrm{HCl}$？

Identify variables.确定变量。

$$ c_1 = 12.0\ \mathrm{mol/L}, \quad V_1 = ?, \quad c_2 = 0.600\ \mathrm{mol/L}, \quad V_2 = 500\ \mathrm{mL}. $$

Apply $c_1V_1 = c_2V_2$ and solve for $V_1$.应用 $c_1V_1 = c_2V_2$ 并解出 $V_1$。

$$ V_1 = \frac{c_2 V_2}{c_1} = \frac{0.600\ \mathrm{mol/L} \times 500\ \mathrm{mL}}{12.0\ \mathrm{mol/L}} = 25.0\ \mathrm{mL}. $$

Interpret.解读。 Measure $25.0\ \mathrm{mL}$ of the $12.0\ \mathrm{mol/L}$ stock, transfer to a $500\ \mathrm{mL}$ volumetric flask, and dilute to the mark with distilled water. The resulting solution is $0.600\ \mathrm{mol/L}$ HCl. Note the units of $V$ cancelled on both sides (both in mL). ✓量取 $25.0\ \mathrm{mL}$ 的 $12.0\ \mathrm{mol/L}$ 储备液，转移至 $500\ \mathrm{mL}$ 容量瓶，用蒸馏水稀释至刻度线。所得溶液为 $0.600\ \mathrm{mol/L}$ HCl。注意两边的 $V$ 单位抵消（均为 mL）。✓

A $2.00\ \mathrm{mol/L}$ solution of $\mathrm{NaCl}$ is diluted from $100\ \mathrm{mL}$ to $400\ \mathrm{mL}$. What is the new concentration?将 $100\ \mathrm{mL}$ 的 $2.00\ \mathrm{mol/L}$ $\mathrm{NaCl}$ 溶液稀释至 $400\ \mathrm{mL}$。新浓度是多少？
§4 · Q1
$8.00\ \mathrm{mol/L}$$8.00\ \mathrm{mol/L}$
$2.00\ \mathrm{mol/L}$$2.00\ \mathrm{mol/L}$
$1.00\ \mathrm{mol/L}$$1.00\ \mathrm{mol/L}$
$0.500\ \mathrm{mol/L}$$0.500\ \mathrm{mol/L}$
$c_2 = c_1V_1/V_2 = (2.00)(100)/(400) = 0.500\ \mathrm{mol/L}$. The volume quadrupled, so the concentration is reduced to one quarter.$c_2 = c_1V_1/V_2 = (2.00)(100)/(400) = 0.500\ \mathrm{mol/L}$。体积扩大了四倍，所以浓度降至四分之一。
$c_1V_1 = c_2V_2 \Rightarrow c_2 = c_1V_1/V_2 = (2.00 \times 100)/400 = 0.500\ \mathrm{mol/L}$. Diluting quadruples the volume, so the concentration becomes $1/4$ of the original.$c_1V_1 = c_2V_2 \Rightarrow c_2 = c_1V_1/V_2 = (2.00 \times 100)/400 = 0.500\ \mathrm{mol/L}$。稀释使体积变为四倍，浓度变为原来的 $1/4$。
When you dilute a solution, which of the following stays constant?稀释溶液时，以下哪项保持不变？
§4 · Q2
The concentration of the solute溶质的浓度
The number of moles of solute溶质的摩尔数
The volume of the solution溶液的体积
The mass of the solvent溶剂的质量
Dilution means adding more solvent. No solute is added or removed, so the moles of solute ($n = cV$) remain constant. The concentration decreases because the same $n$ is now spread over a larger volume.稀释意味着加入更多溶剂。不添加也不去除溶质，所以溶质的摩尔数（$n = cV$）保持不变。浓度降低，因为相同的 $n$ 现在分散在更大的体积中。
Adding solvent increases the volume and decreases the concentration. The solute mass (and moles) stays fixed. $c_1V_1 = n = c_2V_2$ — moles are conserved in dilution.加入溶剂增大体积并降低浓度。溶质的质量（和摩尔数）保持固定。$c_1V_1 = n = c_2V_2$——稀释中摩尔数守恒。
Section 5 · Solubility curves & saturation第 5 节 · 溶解度曲线与饱和

Solubility Curves and Saturation溶解度曲线与饱和

Solubility: maximum grams of solute that dissolve in 100 g of solvent at a given temperature.溶解度：在给定温度下，每 100 g 溶剂中能溶解的溶质最大克数。
  • Saturated solution饱和溶液 — contains the maximum amount of dissolved solute at a given temperature; undissolved solute is in equilibrium with dissolved solute (rate of dissolving = rate of crystallisation). 🇨🇦 AB Chem 20 C-GO1— 在给定温度下含有最大量溶解溶质；未溶解的溶质与溶解的溶质处于平衡（溶解速率 = 结晶速率）。🇨🇦 AB Chem 20 C-GO1
  • Unsaturated solution不饱和溶液 — can dissolve more solute; below the solubility curve.— 还能溶解更多溶质；在溶解度曲线以下。
  • Supersaturated solution过饱和溶液 — contains more dissolved solute than the equilibrium solubility at that temperature; unstable — a seed crystal or disturbance triggers rapid crystallisation.— 溶解的溶质超过该温度下的平衡溶解度；不稳定——晶种或扰动会触发快速结晶。

Effect of temperature and pressure on solubility (SCH3U E3.3 / AB Chem 20 C-GO1):温度和压力对溶解度的影响（SCH3U E3.3 / AB Chem 20 C-GO1）：

  • Solids in liquids: solubility generally increases with increasing temperature (e.g. KNO₃ in water). Most solubility curves slope upward with temperature.固体溶于液体：溶解度通常随温度升高而增大（如 KNO₃ 溶于水）。大多数溶解度曲线随温度升高而向上倾斜。
  • Gases in liquids: solubility decreases with increasing temperature (O₂, CO₂). This is why warm soda goes flat and why dissolved oxygen in lakes drops in summer (affects aquatic life).气体溶于液体：溶解度随温度升高而降低（O₂、CO₂）。这就是为什么温热的碳酸饮料会变平，以及为什么夏季湖中溶解氧会下降（影响水生生物）。
  • Pressure: affects only gas solubility. Henry's law: at constant temperature, the solubility of a gas is proportional to its partial pressure above the solution. Pressure does not significantly affect the solubility of solids or liquids.压力：仅影响气体溶解度。亨利定律：在恒温下，气体的溶解度与溶液上方的分压成正比。压力对固体或液体的溶解度没有显著影响。
Worked Example 5 · Reading a solubility curve例题 5 · 读取溶解度曲线

At $40\ ^\circ\mathrm{C}$, the solubility of $\mathrm{KNO_3}$ in water is $65\ \mathrm{g}$ per $100\ \mathrm{g}$ of water. A student dissolves $50\ \mathrm{g}$ of $\mathrm{KNO_3}$ in $100\ \mathrm{g}$ of water at $40\ ^\circ\mathrm{C}$. (a) Is the solution saturated, unsaturated, or supersaturated? (b) If the solution is cooled to $20\ ^\circ\mathrm{C}$ where the solubility is $32\ \mathrm{g}/100\ \mathrm{g}$, how much $\mathrm{KNO_3}$ crystallises out?在 $40\ ^\circ\mathrm{C}$ 时，$\mathrm{KNO_3}$ 在水中的溶解度为每 $100\ \mathrm{g}$ 水 $65\ \mathrm{g}$。一位同学在 $40\ ^\circ\mathrm{C}$ 下将 $50\ \mathrm{g}$ $\mathrm{KNO_3}$ 溶于 $100\ \mathrm{g}$ 水中。(a) 该溶液是饱和、不饱和还是过饱和溶液？(b) 若将溶液冷却至 $20\ ^\circ\mathrm{C}$（此时溶解度为 $32\ \mathrm{g}/100\ \mathrm{g}$），有多少 $\mathrm{KNO_3}$ 结晶析出？

(a)(a) Dissolved: $50\ \mathrm{g}$. Maximum at $40\ ^\circ\mathrm{C}$: $65\ \mathrm{g}$. Since $50 < 65$, the solution is unsaturated — more $\mathrm{KNO_3}$ could still dissolve. ✓溶解量：$50\ \mathrm{g}$。$40\ ^\circ\mathrm{C}$ 的最大溶解度：$65\ \mathrm{g}$。由于 $50 < 65$，溶液为不饱和——还能继续溶解更多 $\mathrm{KNO_3}$。✓

(b)(b) At $20\ ^\circ\mathrm{C}$, maximum solubility is $32\ \mathrm{g}$ per $100\ \mathrm{g}$ water. Amount that crystallises = dissolved amount $-$ new solubility limit $= 50 - 32 = 18\ \mathrm{g}$ of $\mathrm{KNO_3}$ crystallises out. ✓在 $20\ ^\circ\mathrm{C}$ 时，每 $100\ \mathrm{g}$ 水的最大溶解度为 $32\ \mathrm{g}$。结晶析出量 = 溶解量 $-$ 新溶解度上限 $= 50 - 32 = 18\ \mathrm{g}$ $\mathrm{KNO_3}$ 结晶析出。✓

A sample of water at $25\ ^\circ\mathrm{C}$ is gently heated. What happens to the concentration of dissolved oxygen ($\mathrm{O_2}$)?在 $25\ ^\circ\mathrm{C}$ 下轻轻加热一份水样。溶解氧（$\mathrm{O_2}$）的浓度会发生什么变化？
§5 · Q1
It increases, because higher temperature speeds up dissolving增加，因为较高温度加速溶解
It stays the same, because temperature does not affect gas solubility不变，因为温度不影响气体溶解度
It decreases, because gas solubility decreases with increasing temperature减少，因为气体溶解度随温度升高而降低
It increases, because the solution becomes more dilute增加，因为溶液变得更稀
Gas solubility in liquids decreases with increasing temperature. As water warms, dissolved $\mathrm{O_2}$ is released. This is why warm lakes have lower dissolved oxygen, threatening fish that need oxygen-rich water.气体在液体中的溶解度随温度升高而降低。当水变暖时，溶解的 $\mathrm{O_2}$ 会释放出来。这就是为什么温暖的湖泊溶解氧较低，威胁到需要富氧水的鱼类。
For gases dissolved in liquids, the trend is the opposite of solids: gas solubility decreases as temperature increases. Higher kinetic energy allows gas molecules to escape the solution more easily.对于溶解在液体中的气体，趋势与固体相反：气体溶解度随温度升高而降低。更高的动能使气体分子更容易逸出溶液。
A saturated solution is best defined as one that:饱和溶液最准确的定义是：
§5 · Q2
Contains the maximum amount of dissolved solute at a given temperature, with dissolving and crystallisation rates equal在给定温度下含有最大量溶解溶质，溶解速率与结晶速率相等
Contains so much solute that no more water can be added含有如此多的溶质以至于不能再加入更多水
Can dissolve more solute if it is stirred如果搅拌可以溶解更多溶质
Contains more dissolved solute than the equilibrium value allows含有的溶解溶质超过平衡值
A saturated solution is at equilibrium: the rate at which solute dissolves equals the rate at which it crystallises. This is the maximum dissolved amount at that temperature — adding more solute would leave it undissolved. Alberta explicitly phrases this as "equal rates of dissolving and crystallization."饱和溶液处于平衡状态：溶质溶解的速率等于其结晶的速率。这是该温度下最大溶解量——加入更多溶质会使其未溶解。阿尔伯塔明确将此表述为"溶解和结晶速率相等"。
Supersaturated (option D) exceeds the equilibrium value and is unstable. Unsaturated (option C) can dissolve more. A saturated solution is at the equilibrium maximum — dissolving rate = crystallisation rate.过饱和溶液（选项D）超过平衡值，不稳定。不饱和溶液（选项C）还能溶解更多。饱和溶液处于平衡最大值——溶解速率 = 结晶速率。
Section 6 · Electrolytes & dissociation第 6 节 · 电解质与解离

Electrolytes and Dissociation电解质与解离

Electrolytes: dissolved substances that produce ions and conduct electricity.电解质：溶解后产生离子并导电的物质。
  • Strong electrolyte强电解质 — dissociates (or ionises) completely in water. Produces a maximum number of ions; aqueous solution conducts electricity well. Examples: all soluble ionic salts (NaCl, KNO₃), strong acids (HCl, HNO₃, H₂SO₄), strong bases (NaOH, KOH). 🇨🇦 AB Chem 20 C-GO1— 在水中完全解离（或电离）。产生最大数量的离子；水溶液导电性好。示例：所有可溶性离子盐（NaCl、KNO₃）、强酸（HCl、HNO₃、H₂SO₄）、强碱（NaOH、KOH）。🇨🇦 AB Chem 20 C-GO1
  • Weak electrolyte弱电解质 — ionises partially; most dissolved particles remain as molecules. Conducts weakly. Examples: weak acids (CH₃COOH — acetic acid), weak bases (NH₃).— 部分电离；大多数溶解的颗粒仍以分子形式存在。导电性弱。示例：弱酸（CH₃COOH——乙酸）、弱碱（NH₃）。
  • Nonelectrolyte非电解质 — does not produce ions in solution; does not conduct electricity. Examples: glucose (C₆H₁₂O₆), sucrose, ethanol (C₂H₅OH). 🇨🇦 AB Chem 20 C-GO1 / SCH3U E3.2— 溶于水后不产生离子；不导电。示例：葡萄糖（C₆H₁₂O₆）、蔗糖、乙醇（C₂H₅OH）。🇨🇦 AB Chem 20 C-GO1 / SCH3U E3.2

Writing dissociation equations (BC Chem 11 elaboration "dissociation of ions, dissociation equation"):书写解离方程（BC Chem 11 细化"离子解离、解离方程"）：

$$ \mathrm{NaCl(s)} \xrightarrow{\mathrm{H_2O}} \mathrm{Na^+(aq)} + \mathrm{Cl^-(aq)} $$ $$ \mathrm{MgCl_2(s)} \xrightarrow{\mathrm{H_2O}} \mathrm{Mg^{2+}(aq)} + 2\,\mathrm{Cl^-(aq)} $$ Note: the coefficient in front of each ion tells you the concentration of that ion relative to the original compound. $1.0\ \mathrm{mol/L}$ $\mathrm{MgCl_2}$ produces $1.0\ \mathrm{mol/L}$ $\mathrm{Mg^{2+}}$ AND $2.0\ \mathrm{mol/L}$ $\mathrm{Cl^-}$.注意：每种离子前的系数告诉你该离子相对于原化合物的浓度。$1.0\ \mathrm{mol/L}$ $\mathrm{MgCl_2}$ 产生 $1.0\ \mathrm{mol/L}$ $\mathrm{Mg^{2+}}$ 和 $2.0\ \mathrm{mol/L}$ $\mathrm{Cl^-}$。
Worked Example 6 · Ion concentrations from dissociation例题 6 · 由解离计算离子浓度

A solution of calcium chloride $\mathrm{CaCl_2}$ has a concentration of $0.150\ \mathrm{mol/L}$. Assuming complete dissociation, calculate (a) the concentration of $\mathrm{Ca^{2+}}$ ions and (b) the concentration of $\mathrm{Cl^-}$ ions in solution.氯化钙 $\mathrm{CaCl_2}$ 溶液的浓度为 $0.150\ \mathrm{mol/L}$。假设完全解离，计算 (a) 溶液中 $\mathrm{Ca^{2+}}$ 离子的浓度和 (b) $\mathrm{Cl^-}$ 离子的浓度。

Write the dissociation equation.写出解离方程。

$$ \mathrm{CaCl_2(s)} \xrightarrow{\mathrm{H_2O}} \mathrm{Ca^{2+}(aq)} + 2\,\mathrm{Cl^-(aq)} $$

(a) $[\mathrm{Ca^{2+}}]$.(a) $[\mathrm{Ca^{2+}}]$。 One formula unit of $\mathrm{CaCl_2}$ gives one $\mathrm{Ca^{2+}}$: $[\mathrm{Ca^{2+}}] = 0.150\ \mathrm{mol/L}$. ✓一个 $\mathrm{CaCl_2}$ 式量单元给出一个 $\mathrm{Ca^{2+}}$：$[\mathrm{Ca^{2+}}] = 0.150\ \mathrm{mol/L}$。✓

(b) $[\mathrm{Cl^-}]$.(b) $[\mathrm{Cl^-}]$。 One formula unit gives two $\mathrm{Cl^-}$: $[\mathrm{Cl^-}] = 2 \times 0.150 = 0.300\ \mathrm{mol/L}$. ✓一个式量单元给出两个 $\mathrm{Cl^-}$：$[\mathrm{Cl^-}] = 2 \times 0.150 = 0.300\ \mathrm{mol/L}$。✓

Total ion concentration $= 0.150 + 0.300 = 0.450\ \mathrm{mol/L}$ — this matters for colligative properties (§7).总离子浓度 $= 0.150 + 0.300 = 0.450\ \mathrm{mol/L}$——这对依数性（§7）很重要。

Which of the following is classified as a strong electrolyte?以下哪种物质被归类为强电解质？
§6 · Q1
Glucose ($\mathrm{C_6H_{12}O_6}$)葡萄糖（$\mathrm{C_6H_{12}O_6}$）
Acetic acid ($\mathrm{CH_3COOH}$)乙酸（$\mathrm{CH_3COOH}$）
Potassium nitrate ($\mathrm{KNO_3}$)硝酸钾（$\mathrm{KNO_3}$）
Ethanol ($\mathrm{C_2H_5OH}$)乙醇（$\mathrm{C_2H_5OH}$）
$\mathrm{KNO_3}$ is a soluble ionic salt — it dissociates completely in water into $\mathrm{K^+(aq)}$ and $\mathrm{NO_3^-(aq)}$. Glucose and ethanol are nonelectrolytes; acetic acid is a weak electrolyte (partially ionises).$\mathrm{KNO_3}$ 是可溶性离子盐——它在水中完全解离为 $\mathrm{K^+(aq)}$ 和 $\mathrm{NO_3^-(aq)}$。葡萄糖和乙醇是非电解质；乙酸是弱电解质（部分电离）。
Ionic compounds (like $\mathrm{KNO_3}$) are strong electrolytes — they fully dissociate. Molecular compounds like glucose and ethanol do not produce ions (nonelectrolytes). Acetic acid only partially ionises (weak electrolyte).离子化合物（如 $\mathrm{KNO_3}$）是强电解质——它们完全解离。葡萄糖和乙醇等分子化合物不产生离子（非电解质）。乙酸只部分电离（弱电解质）。
What is the $[\mathrm{Cl^-}]$ in a $0.200\ \mathrm{mol/L}$ solution of $\mathrm{AlCl_3}$ (assuming complete dissociation)?在 $0.200\ \mathrm{mol/L}$ 的 $\mathrm{AlCl_3}$ 溶液中（假设完全解离），$[\mathrm{Cl^-}]$ 是多少？
§6 · Q2
$0.600\ \mathrm{mol/L}$$0.600\ \mathrm{mol/L}$
$0.200\ \mathrm{mol/L}$$0.200\ \mathrm{mol/L}$
$0.400\ \mathrm{mol/L}$$0.400\ \mathrm{mol/L}$
$0.067\ \mathrm{mol/L}$$0.067\ \mathrm{mol/L}$
$\mathrm{AlCl_3 \to Al^{3+} + 3\,Cl^-}$. Each mole of $\mathrm{AlCl_3}$ produces 3 moles of $\mathrm{Cl^-}$. So $[\mathrm{Cl^-}] = 3 \times 0.200 = 0.600\ \mathrm{mol/L}$.$\mathrm{AlCl_3 \to Al^{3+} + 3\,Cl^-}$。每摩尔 $\mathrm{AlCl_3}$ 产生 3 摩尔 $\mathrm{Cl^-}$。所以 $[\mathrm{Cl^-}] = 3 \times 0.200 = 0.600\ \mathrm{mol/L}$。
Write the dissociation equation: $\mathrm{AlCl_3 \to Al^{3+} + 3\,Cl^-}$. The coefficient 3 means each formula unit releases 3 chloride ions. $[\mathrm{Cl^-}] = 3 \times [\mathrm{AlCl_3}] = 3 \times 0.200 = 0.600\ \mathrm{mol/L}$.写出解离方程：$\mathrm{AlCl_3 \to Al^{3+} + 3\,Cl^-}$。系数 3 表示每个式量单元释放 3 个氯离子。$[\mathrm{Cl^-}] = 3 \times [\mathrm{AlCl_3}] = 3 \times 0.200 = 0.600\ \mathrm{mol/L}$。
Section 7 · Colligative properties (intro)第 7 节 · 依数性（简介）

Colligative Properties: Introduction依数性：简介

Curriculum note.课纲提示。 Colligative properties are not a named content outcome in NGSS HS-PS1-3, Ontario SCH3U Strand E, BC Chemistry 11, or Alberta Chemistry 20 Unit C. This section introduces the concept as a logical extension of the electrolyte/dissociation work in §6, and because NGSS HS-PS1-3 explicitly names "vapor pressure" and "boiling point" as bulk properties to link to inter-particle force strength. The full quantitative treatment (Raoult's law, molality-based $\Delta T_f$ / $\Delta T_b$, van't Hoff factor) is IB Chemistry HL and AP Chemistry depth — it is introduced here without derivation so students heading to those programs are not surprised.依数性不是 NGSS HS-PS1-3、安大略 SCH3U E 单元、BC Chemistry 11 或阿尔伯塔 Chemistry 20 C 单元中的命名内容结果。本节将该概念作为 §6 电解质/解离内容的逻辑延伸引入，并且是因为 NGSS HS-PS1-3 明确将"蒸气压"和"沸点"列为需联系颗粒间力强度的宏观性质。完整的定量处理（拉乌尔定律、基于重量摩尔浓度的 $\Delta T_f$ / $\Delta T_b$、范托夫因子）是 IB Chemistry HL 和 AP Chemistry 的深度内容——这里仅作不含推导的介绍，以免面向这些课程的学生感到意外。
Colligative properties depend only on the number of dissolved particles, not on their chemical identity.依数性仅取决于溶解颗粒的数量，而与其化学性质无关。
  • Boiling-point elevation沸点升高 — dissolved particles interfere with solvent molecules escaping to the gas phase, so the solution boils at a higher temperature than the pure solvent. Example: salted pasta water boils above $100\ ^\circ\mathrm{C}$ (though the effect is tiny at kitchen concentrations).— 溶解颗粒干扰溶剂分子逸出到气相，因此溶液的沸点高于纯溶剂。示例：加盐的煮面水沸点高于 $100\ ^\circ\mathrm{C}$（虽然在厨房浓度下效果很小）。
  • Freezing-point depression凝固点降低 — dissolved particles disrupt the formation of the ordered solid lattice, so the solution freezes at a lower temperature than the pure solvent. Application: antifreeze (ethylene glycol or salt) in cold climates; de-icing roads in winter; why seawater does not freeze at $0\ ^\circ\mathrm{C}$.— 溶解颗粒破坏有序固态晶格的形成，因此溶液的凝固点低于纯溶剂。应用：寒冷气候中的防冻液（乙二醇或盐）；冬季道路除冰；为什么海水不在 $0\ ^\circ\mathrm{C}$ 结冰。
  • Osmotic pressure渗透压 — the pressure required to stop solvent molecules from moving through a semipermeable membrane from a dilute solution to a concentrated one. Important in biology (cell turgor, kidney function, IV drips).— 阻止溶剂分子通过半透膜从稀溶液向浓溶液移动所需的压力。在生物学中很重要（细胞膨压、肾功能、静脉注射）。
  • Particle count rule颗粒数量规则 — electrolytes dissociate into multiple ions, so they produce a greater colligative effect per mole of formula unit than nonelectrolytes. $1\ \mathrm{mol}$ NaCl $\to$ $2\ \mathrm{mol}$ particles; $1\ \mathrm{mol}$ MgCl₂ $\to$ $3\ \mathrm{mol}$ particles; $1\ \mathrm{mol}$ glucose $\to$ $1\ \mathrm{mol}$ particles.— 电解质解离成多种离子，因此每摩尔式量单元产生比非电解质更大的依数性效应。$1\ \mathrm{mol}$ NaCl $\to$ $2\ \mathrm{mol}$ 颗粒；$1\ \mathrm{mol}$ MgCl₂ $\to$ $3\ \mathrm{mol}$ 颗粒；$1\ \mathrm{mol}$ 葡萄糖 $\to$ $1\ \mathrm{mol}$ 颗粒。
Worked Example 7 · Comparing colligative effects of electrolytes and nonelectrolytes例题 7 · 比较电解质与非电解质的依数性效应

Three solutions are each prepared at a concentration of $0.10\ \mathrm{mol/L}$: (A) glucose ($\mathrm{C_6H_{12}O_6}$), (B) NaCl, (C) $\mathrm{MgCl_2}$. Rank them in order of increasing boiling-point elevation (smallest effect first). Explain using particle counts.以 $0.10\ \mathrm{mol/L}$ 浓度分别配制三种溶液：(A) 葡萄糖（$\mathrm{C_6H_{12}O_6}$），(B) NaCl，(C) $\mathrm{MgCl_2}$。按沸点升高从小到大排列（效果最小的在前）。用颗粒数量解释。

Count particles per 0.10 mol/L formula unit.计算每 0.10 mol/L 式量单元的颗粒数。

  • (A) Glucose: nonelectrolyte, no dissociation. Particles $= 0.10\ \mathrm{mol/L}$.(A) 葡萄糖：非电解质，不解离。颗粒数 $= 0.10\ \mathrm{mol/L}$。
  • (B) NaCl $\to$ Na⁺ + Cl⁻: 2 particles per formula unit. Particles $= 0.20\ \mathrm{mol/L}$.(B) NaCl $\to$ Na⁺ + Cl⁻：每个式量单元 2 个颗粒。颗粒数 $= 0.20\ \mathrm{mol/L}$。
  • (C) $\mathrm{MgCl_2}$ $\to$ Mg²⁺ + 2Cl⁻: 3 particles per formula unit. Particles $= 0.30\ \mathrm{mol/L}$.(C) $\mathrm{MgCl_2}$ $\to$ Mg²⁺ + 2Cl⁻：每个式量单元 3 个颗粒。颗粒数 $= 0.30\ \mathrm{mol/L}$。

Ranking (smallest to largest boiling-point elevation):排名（沸点升高从小到大）：

A (glucose) < B (NaCl) < C ($\mathrm{MgCl_2}$). More particles $=$ greater boiling-point elevation (and greater freezing-point depression). ✓A（葡萄糖）< B（NaCl）< C（$\mathrm{MgCl_2}$）。颗粒越多 = 沸点升高越大（凝固点降低越大）。✓

Why does adding salt to water lower its freezing point?为什么向水中加盐会降低其冰点？
§7 · Q1
Salt reacts chemically with ice to release heat盐与冰发生化学反应释放热量
Salt increases the density of water, making it harder to freeze盐增加了水的密度，使其更难结冰
Salt absorbs heat from the surroundings as it dissolves盐在溶解时从周围吸收热量
Dissolved ions disrupt the formation of the ordered ice lattice, requiring a lower temperature to freeze溶解的离子破坏有序冰晶格的形成，需要更低的温度才能结冰
Freezing-point depression is a colligative property. The dissolved Na⁺ and Cl⁻ ions interfere with water molecules aligning into the ordered crystal structure of ice. More energy (lower temperature) is needed to overcome this interference. This is why salt is spread on icy roads and added to antifreeze.凝固点降低是一种依数性。溶解的 Na⁺ 和 Cl⁻ 离子干扰水分子排列成冰的有序晶体结构。需要更多能量（更低温度）来克服这种干扰。这就是为什么在结冰的道路上撒盐以及在防冻液中加盐的原因。
Freezing-point depression is physical, not chemical. Salt doesn't react with ice or absorb heat to cause the effect. The reason is structural: dissolved particles disrupt ice-crystal formation, lowering the temperature at which the solution solidifies.凝固点降低是物理现象，不是化学反应。盐不与冰反应，也不通过吸热来产生效果。原因是结构性的：溶解的颗粒破坏冰晶的形成，降低溶液凝固的温度。
A $0.10\ \mathrm{mol/L}$ solution of $\mathrm{CaCl_2}$ will have a boiling-point elevation approximately how many times larger than a $0.10\ \mathrm{mol/L}$ glucose solution?$0.10\ \mathrm{mol/L}$ 的 $\mathrm{CaCl_2}$ 溶液的沸点升高大约是 $0.10\ \mathrm{mol/L}$ 葡萄糖溶液的几倍？
§7 · Q2
1 times (the same)1 倍（相同）
2 times2 倍
3 times3 倍
4 times4 倍
$\mathrm{CaCl_2 \to Ca^{2+} + 2\,Cl^-}$: 3 particles per formula unit. Glucose: 1 particle per formula unit. At the same formula-unit concentration, $\mathrm{CaCl_2}$ produces 3 times as many particles, so the boiling-point elevation is approximately 3 times larger.$\mathrm{CaCl_2 \to Ca^{2+} + 2\,Cl^-}$：每个式量单元 3 个颗粒。葡萄糖：每个式量单元 1 个颗粒。在相同的式量浓度下，$\mathrm{CaCl_2}$ 产生的颗粒数是葡萄糖的 3 倍，因此沸点升高大约是 3 倍。
Colligative effects depend on particle count. $\mathrm{CaCl_2}$ gives 3 ions per formula unit; glucose gives 1 particle. At equal formula-unit concentrations, the ratio of particles is 3:1, so the colligative effect is $\approx 3\times$ larger for $\mathrm{CaCl_2}$.依数性取决于颗粒数量。$\mathrm{CaCl_2}$ 每个式量单元给出 3 个离子；葡萄糖给出 1 个颗粒。在相等的式量浓度下，颗粒数之比为 3:1，所以 $\mathrm{CaCl_2}$ 的依数性效应约大 3 倍。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline for molarity and dilution questions (§3–§4)摩尔浓度和稀释题的解题纪律（§3–§4）
  • Convert mL to L before substituting.代入前将 mL 换算为 L。 The formula $c = n/V$ requires $V$ in litres. $1\ \mathrm{mL} = 0.001\ \mathrm{L}$. The most common error is using millilitres without converting — you will be off by a factor of 1000.公式 $c = n/V$ 要求 $V$ 以升为单位。$1\ \mathrm{mL} = 0.001\ \mathrm{L}$。最常见的错误是不换算就使用毫升——结果会差 1000 倍。
  • For dilution: identify $c_1$, $V_1$, $c_2$, $V_2$ before you compute.稀释题：计算前确定 $c_1$、$V_1$、$c_2$、$V_2$。 Write out $c_1V_1 = c_2V_2$ with the known values substituted. Solve for the unknown. Always check: is the final solution more dilute or more concentrated than the original? The answer should make sense.将已知值代入 $c_1V_1 = c_2V_2$ 并写出。解出未知量。始终检查：最终溶液比原溶液更稀还是更浓？答案应该合理。
  • Significant figures.有效数字。 Match the fewest significant figures in the given data. Molarity answers typically need 3 sig figs; dilution results often need 3. Never give more figures than justified.与给定数据中最少的有效数字位数一致。摩尔浓度答案通常需要 3 位有效数字；稀释结果通常也需要 3 位。不要给出超过合理位数的数字。
Like dissolves like and electrolytes (§2, §6)相似相溶与电解质（§2、§6）
  • State polarity before predicting solubility.预测溶解性前先判断极性。 Is the solute ionic, polar covalent, or nonpolar? Is the solvent polar (water) or nonpolar (hexane)? Ionic and polar $\to$ dissolves in water. Nonpolar $\to$ dissolves in nonpolar solvent.溶质是离子型、极性共价型还是非极性型？溶剂是极性（水）还是非极性（己烷）？离子型和极性 $\to$ 溶于水。非极性 $\to$ 溶于非极性溶剂。
  • Ion concentration multiplies by the dissociation stoichiometry.离子浓度等于摩尔浓度乘以解离化学计量数。 $\mathrm{Al_2(SO_4)_3}$ dissociates into 2 Al³⁺ and 3 SO₄²⁻ — so a $0.10\ \mathrm{mol/L}$ solution gives $0.20\ \mathrm{mol/L}$ Al³⁺ and $0.30\ \mathrm{mol/L}$ SO₄²⁻. Write the dissociation equation first, then read off the coefficients.$\mathrm{Al_2(SO_4)_3}$ 解离成 2 个 Al³⁺ 和 3 个 SO₄²⁻——所以 $0.10\ \mathrm{mol/L}$ 的溶液产生 $0.20\ \mathrm{mol/L}$ Al³⁺ 和 $0.30\ \mathrm{mol/L}$ SO₄²⁻。先写出解离方程，再读取系数。
Solubility and temperature (§5)溶解度与温度（§5）
  • Solids vs gases — opposite trends.固体与气体——相反趋势。 Most solid solutes become more soluble at higher temperatures; gases become less soluble. Pressure only matters for gases (Henry's law), not for solids or liquids.大多数固体溶质在高温下溶解度更大；气体溶解度更小。压力仅对气体有影响（亨利定律），对固体或液体无显著影响。
  • Reading a solubility curve: three regions.读取溶解度曲线：三个区域。 Below the curve = unsaturated (can dissolve more). On the curve = saturated (at equilibrium). Above the curve = supersaturated (unstable, will crystallise).曲线以下 = 不饱和（还能溶解更多）。曲线上 = 饱和（处于平衡）。曲线以上 = 过饱和（不稳定，会结晶）。
  • Mass of crystals on cooling = original dissolved mass $-$ new solubility limit.冷却后结晶质量 = 原溶解质量 $-$ 新溶解度上限。 State the temperature, the solubility at that temperature, and the amount already dissolved. Then subtract.说明温度、该温度下的溶解度以及已溶解的量，然后相减。
Colligative properties (§7) and general answer hygiene依数性（§7）与通用作答规范
  • Always count particles, not moles of formula unit.始终计算颗粒数，而非式量摩尔数。 The colligative effect is proportional to total solute particles in solution. An electrolyte produces more particles per formula unit than a nonelectrolyte at the same concentration.依数性与溶液中溶质的总颗粒数成正比。在相同浓度下，电解质比非电解质每个式量单元产生更多颗粒。
  • Unit labels on every number.每个数字都要标注单位。 Write mol/L (or M) for concentrations, L for volumes, g/mol for molar mass, g for mass. Drop a unit and you will make a dimensional error.浓度写 mol/L（或 M），体积写 L，摩尔质量写 g/mol，质量写 g。漏掉单位会导致量纲错误。
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Solution?溶液 (solution)？
Homogeneous mixture of solute dissolved in solvent. Uniform composition throughout.溶质溶于溶剂形成的均匀混合物。各处组成相同。
Molarity formula?摩尔浓度公式 (molarity formula)？
$$c = \tfrac{n}{V}$$ $c$ in mol/L; $n$ in mol; $V$ in L.$c$ 单位 mol/L；$n$ 单位 mol；$V$ 单位 L。
Dilution equation?稀释方程 (dilution equation)？
$$c_1 V_1 = c_2 V_2$$ Moles of solute are conserved; only volume changes.溶质摩尔数守恒；仅体积改变。
Like dissolves like?相似相溶 (like dissolves like)？
Polar solvents dissolve polar/ionic solutes. Nonpolar solvents dissolve nonpolar solutes.极性溶剂溶解极性/离子型溶质。非极性溶剂溶解非极性溶质。
Saturated solution?饱和溶液 (saturated solution)？
Maximum dissolved solute at that temperature. Rate of dissolving = rate of crystallisation.该温度下溶解溶质的最大量。溶解速率 = 结晶速率。
Strong electrolyte?强电解质 (strong electrolyte)？
Dissociates completely in water. Soluble ionic salts, strong acids and bases. Conducts well.在水中完全解离。可溶性离子盐、强酸和强碱。导电性好。
Nonelectrolyte?非电解质 (nonelectrolyte)？
Dissolves but produces no ions. Does not conduct electricity. Examples: glucose, sucrose, ethanol.溶于水但不产生离子。不导电。示例：葡萄糖、蔗糖、乙醇。
Solubility of gas vs temperature?气体溶解度 vs 温度 (gas solubility vs temperature)？
Decreases as temperature increases. Opposite of most solids.随温度升高而降低。与大多数固体相反。
Colligative properties?依数性 (colligative properties)？
Depend on number of dissolved particles, not their identity. Include boiling-point elevation, freezing-point depression, osmotic pressure.取决于溶解颗粒的数量，而非其种类。包括沸点升高、凝固点降低、渗透压。
Why does NaCl lower freezing point more than glucose (at equal mol/L)?为何等摩尔浓度下 NaCl 比葡萄糖更能降低冰点？
NaCl $\to$ 2 particles (Na⁺ + Cl⁻); glucose $\to$ 1 particle. More particles = larger colligative effect.NaCl $\to$ 2 个颗粒（Na⁺ + Cl⁻）；葡萄糖 $\to$ 1 个颗粒。颗粒越多 = 依数性效应越大。
Ion-dipole force in dissolving?溶解中的离子-偶极力 (ion-dipole force)?
Attraction between an ion and the partial charge on a polar solvent molecule. Drives ionic dissolution in water. Forms hydration shell.离子与极性溶剂分子部分电荷之间的吸引力。驱动离子型物质在水中溶解。形成水化壳。
Supersaturated solution?过饱和溶液 (supersaturated solution)？
Contains more dissolved solute than the equilibrium limit. Unstable — a seed crystal triggers rapid crystallisation.含有超过平衡极限的溶解溶质。不稳定——晶种触发快速结晶。
Concentration of Cl⁻ in 0.5 mol/L CaCl₂?0.5 mol/L CaCl₂ 中 Cl⁻ 的浓度？
$\mathrm{CaCl_2 \to Ca^{2+} + 2\,Cl^-}$; $[\mathrm{Cl^-}] = 2 \times 0.5 = 1.0\ \mathrm{mol/L}$.$\mathrm{CaCl_2 \to Ca^{2+} + 2\,Cl^-}$；$[\mathrm{Cl^-}] = 2 \times 0.5 = 1.0\ \mathrm{mol/L}$。
Moles of solute in 300 mL of 2.0 mol/L NaOH?300 mL 的 2.0 mol/L NaOH 中溶质的摩尔数？
$n = cV = 2.0 \times 0.300 = 0.60\ \mathrm{mol}$.$n = cV = 2.0 \times 0.300 = 0.60\ \mathrm{mol}$。
Mixed Practice综合练习

Practice Quiz综合测验

A student dissolves $10.6\ \mathrm{g}$ of $\mathrm{Na_2CO_3}$ ($M_r = 106\ \mathrm{g/mol}$) in water to make $200\ \mathrm{mL}$ of solution. What is the molar concentration? 🇨🇦 SCH3U E2.2 / AB Chem 20 C-GO1一位同学将 $10.6\ \mathrm{g}$ 的 $\mathrm{Na_2CO_3}$（$M_r = 106\ \mathrm{g/mol}$）溶于水中，配制 $200\ \mathrm{mL}$ 的溶液。摩尔浓度是多少？🇨🇦 SCH3U E2.2 / AB Chem 20 C-GO1
Q1
$1.06\ \mathrm{mol/L}$$1.06\ \mathrm{mol/L}$
$0.100\ \mathrm{mol/L}$$0.100\ \mathrm{mol/L}$
$0.500\ \mathrm{mol/L}$$0.500\ \mathrm{mol/L}$
$0.053\ \mathrm{mol/L}$$0.053\ \mathrm{mol/L}$
$n = m/M_r = 10.6/106 = 0.100\ \mathrm{mol}$. $V = 200\ \mathrm{mL} = 0.200\ \mathrm{L}$. $c = n/V = 0.100/0.200 = 0.500\ \mathrm{mol/L}$.$n = m/M_r = 10.6/106 = 0.100\ \mathrm{mol}$。$V = 200\ \mathrm{mL} = 0.200\ \mathrm{L}$。$c = n/V = 0.100/0.200 = 0.500\ \mathrm{mol/L}$。
Step 1: $n = m/M_r = 10.6/106 = 0.100\ \mathrm{mol}$. Step 2: $V = 0.200\ \mathrm{L}$. Step 3: $c = 0.100/0.200 = 0.500\ \mathrm{mol/L}$.第一步：$n = m/M_r = 10.6/106 = 0.100\ \mathrm{mol}$。第二步：$V = 0.200\ \mathrm{L}$。第三步：$c = 0.100/0.200 = 0.500\ \mathrm{mol/L}$。
What volume of $6.0\ \mathrm{mol/L}$ $\mathrm{H_2SO_4}$ is needed to prepare $1.50\ \mathrm{L}$ of $0.400\ \mathrm{mol/L}$ $\mathrm{H_2SO_4}$? 🇨🇦 SCH3U E2.3 / BC Chem 11需要多少体积的 $6.0\ \mathrm{mol/L}$ $\mathrm{H_2SO_4}$ 来配制 $1.50\ \mathrm{L}$ 的 $0.400\ \mathrm{mol/L}$ $\mathrm{H_2SO_4}$？🇨🇦 SCH3U E2.3 / BC Chem 11
Q2
$225\ \mathrm{mL}$$225\ \mathrm{mL}$
$100\ \mathrm{mL}$$100\ \mathrm{mL}$
$22.5\ \mathrm{mL}$$22.5\ \mathrm{mL}$
$400\ \mathrm{mL}$$400\ \mathrm{mL}$
$c_1V_1 = c_2V_2 \Rightarrow V_1 = c_2V_2/c_1 = (0.400 \times 1.50)/6.0 = 0.100\ \mathrm{L} = 100\ \mathrm{mL}$.$c_1V_1 = c_2V_2 \Rightarrow V_1 = c_2V_2/c_1 = (0.400 \times 1.50)/6.0 = 0.100\ \mathrm{L} = 100\ \mathrm{mL}$。
$V_1 = c_2V_2/c_1 = (0.400)(1.50)/6.0 = 0.60/6.0 = 0.100\ \mathrm{L} = 100\ \mathrm{mL}$.$V_1 = c_2V_2/c_1 = (0.400)(1.50)/6.0 = 0.60/6.0 = 0.100\ \mathrm{L} = 100\ \mathrm{mL}$。
Which pair of substances would be expected to be miscible (mix in all proportions)? 🇺🇸 NGSS HS-PS1-3 / 🇨🇦 SCH3U E3.2以下哪对物质预期可以互溶（以任意比例混合）？🇺🇸 NGSS HS-PS1-3 / 🇨🇦 SCH3U E3.2
Q3
Water and motor oil水和机油
Hexane and water己烷和水
Iodine and water碘和水
Ethanol and water乙醇和水
Ethanol ($\mathrm{C_2H_5OH}$) is polar and can hydrogen-bond with water — they are miscible in all proportions. Oil (nonpolar), hexane (nonpolar), and iodine (weakly polar) are all either immiscible with water or only sparingly soluble.乙醇（$\mathrm{C_2H_5OH}$）是极性的，可以与水形成氢键——它们以任意比例互溶。机油（非极性）、己烷（非极性）和碘（弱极性）都与水不互溶或只有微弱溶解性。
Miscibility requires similar polarity. Ethanol is polar with an O-H group — compatible with water. Motor oil and hexane are nonpolar; iodine is nearly nonpolar; none of these mix with polar water.互溶需要相似的极性。乙醇是带有 O-H 基团的极性物质——与水相容。机油和己烷是非极性的；碘几乎是非极性的；这些都不与极性水混合。
At $60\ ^\circ\mathrm{C}$ the solubility of $\mathrm{KNO_3}$ is $110\ \mathrm{g}/100\ \mathrm{g}$ water. A solution at this temperature contains $80\ \mathrm{g}$ of $\mathrm{KNO_3}$ per $100\ \mathrm{g}$ water. If the solution is cooled to $20\ ^\circ\mathrm{C}$ where the solubility is $32\ \mathrm{g}/100\ \mathrm{g}$, how much $\mathrm{KNO_3}$ crystallises? 🇨🇦 SCH3U E3.3 / AB Chem 20 C-GO1在 $60\ ^\circ\mathrm{C}$ 时，$\mathrm{KNO_3}$ 的溶解度为 $110\ \mathrm{g}/100\ \mathrm{g}$ 水。该温度下某溶液每 $100\ \mathrm{g}$ 水含有 $80\ \mathrm{g}$ $\mathrm{KNO_3}$。若溶液冷却至 $20\ ^\circ\mathrm{C}$（此时溶解度为 $32\ \mathrm{g}/100\ \mathrm{g}$），有多少 $\mathrm{KNO_3}$ 结晶析出？🇨🇦 SCH3U E3.3 / AB Chem 20 C-GO1
Q4
$48\ \mathrm{g}$$48\ \mathrm{g}$
$32\ \mathrm{g}$$32\ \mathrm{g}$
$80\ \mathrm{g}$$80\ \mathrm{g}$
$110\ \mathrm{g}$$110\ \mathrm{g}$
Amount crystallised = dissolved $-$ new solubility = $80 - 32 = 48\ \mathrm{g}$. Only $32\ \mathrm{g}$ can remain dissolved at $20\ ^\circ\mathrm{C}$; the excess $48\ \mathrm{g}$ comes out as crystals.结晶析出量 = 溶解量 $-$ 新溶解度 = $80 - 32 = 48\ \mathrm{g}$。在 $20\ ^\circ\mathrm{C}$ 下只有 $32\ \mathrm{g}$ 能保持溶解；多余的 $48\ \mathrm{g}$ 以晶体形式析出。
Crystals = (amount dissolved at high T) $-$ (solubility at low T) $= 80 - 32 = 48\ \mathrm{g}$. The solution was unsaturated at $60\ ^\circ\mathrm{C}$ (80 < 110), so no crystals formed yet. On cooling, the maximum drops to 32 g.结晶量 = （高温溶解量）$-$（低温溶解度）$= 80 - 32 = 48\ \mathrm{g}$。在 $60\ ^\circ\mathrm{C}$ 时溶液未饱和（80 < 110），所以尚未有晶体析出。冷却后最大值降至 32 g。
Which solution contains the highest total ion concentration? All are $0.10\ \mathrm{mol/L}$ formula unit. 🇨🇦 BC Chem 11 / AB Chem 20 C-GO1以下哪种溶液的总离子浓度最高？均为 $0.10\ \mathrm{mol/L}$ 式量浓度。🇨🇦 BC Chem 11 / AB Chem 20 C-GO1
Q5
$\mathrm{NaCl}$ (sodium chloride)$\mathrm{NaCl}$（氯化钠）
$\mathrm{Al_2(SO_4)_3}$ (aluminium sulfate)$\mathrm{Al_2(SO_4)_3}$（硫酸铝）
$\mathrm{MgCl_2}$ (magnesium chloride)$\mathrm{MgCl_2}$（氯化镁）
Glucose $\mathrm{C_6H_{12}O_6}$葡萄糖 $\mathrm{C_6H_{12}O_6}$
$\mathrm{Al_2(SO_4)_3 \to 2\,Al^{3+} + 3\,SO_4^{2-}}$: 5 ions per formula unit $\Rightarrow$ total ion concentration $= 5 \times 0.10 = 0.50\ \mathrm{mol/L}$. $\mathrm{NaCl}$: 2 ions ($0.20\ \mathrm{mol/L}$). $\mathrm{MgCl_2}$: 3 ions ($0.30\ \mathrm{mol/L}$). Glucose: 0 ions.$\mathrm{Al_2(SO_4)_3 \to 2\,Al^{3+} + 3\,SO_4^{2-}}$：每个式量单元 5 个离子 $\Rightarrow$ 总离子浓度 $= 5 \times 0.10 = 0.50\ \mathrm{mol/L}$。$\mathrm{NaCl}$：2 个离子（$0.20\ \mathrm{mol/L}$）。$\mathrm{MgCl_2}$：3 个离子（$0.30\ \mathrm{mol/L}$）。葡萄糖：0 个离子。
Write each dissociation: NaCl (2 ions), MgCl₂ (3 ions), Al₂(SO₄)₃ (5 ions), glucose (0 ions). Total particle concentration = coefficient × formula-unit concentration. Al₂(SO₄)₃ gives the most.写出每种解离方程：NaCl（2 个离子），MgCl₂（3 个离子），Al₂(SO₄)₃（5 个离子），葡萄糖（0 个离子）。总颗粒浓度 = 系数 × 式量浓度。Al₂(SO₄)₃ 产生最多。
An aqueous solution of $\mathrm{KOH}$ conducts electricity well. This is because $\mathrm{KOH}$ is: 🇨🇦 AB Chem 20 C-GO1 / SCH3U E3.2$\mathrm{KOH}$ 的水溶液导电性好。这是因为 $\mathrm{KOH}$ 是：🇨🇦 AB Chem 20 C-GO1 / SCH3U E3.2
Q6
A nonelectrolyte that does not produce ions不产生离子的非电解质
A weak electrolyte that partially ionises部分电离的弱电解质
A strong electrolyte that fully dissociates into $\mathrm{K^+}$ and $\mathrm{OH^-}$完全解离为 $\mathrm{K^+}$ 和 $\mathrm{OH^-}$ 的强电解质
A covalent compound that forms a network solid形成网状固体的共价化合物
$\mathrm{KOH}$ is a strong base and an ionic compound. It dissociates completely: $\mathrm{KOH(s) \to K^+(aq) + OH^-(aq)}$. The high ion concentration allows strong electrical conduction.$\mathrm{KOH}$ 是强碱和离子化合物。它完全解离：$\mathrm{KOH(s) \to K^+(aq) + OH^-(aq)}$。高离子浓度允许强导电性。
KOH is an ionic compound (strong base) — it fully dissociates in water. Nonelectrolytes produce no ions; weak electrolytes only partially ionise. KOH does neither — it is a strong electrolyte.KOH 是离子化合物（强碱）——在水中完全解离。非电解质不产生离子；弱电解质只部分电离。KOH 都不是——它是强电解质。
Why does a carbonated drink go flat more quickly when left open in a warm room than in a refrigerator? 🇺🇸 NGSS HS-PS1-3 / 🇨🇦 SCH3U E3.3为什么碳酸饮料在温暖的室温下开盖后比在冰箱中更快变平？🇺🇸 NGSS HS-PS1-3 / 🇨🇦 SCH3U E3.3
Q7
The CO₂ reacts chemically with the drink at higher temperaturesCO₂ 在较高温度下与饮料发生化学反应
Warm temperatures increase the pressure above the drink, forcing CO₂ out较高温度增加饮料上方的压力，将 CO₂ 逼出
Cold slows down the evaporation of water, keeping the CO₂ in低温减慢水的蒸发，将 CO₂ 保留在内
Gas solubility decreases with increasing temperature, so more CO₂ escapes from the warm solution气体溶解度随温度升高而降低，因此更多 CO₂ 从温热溶液中逸出
The solubility of a gas in a liquid decreases as temperature increases. At a higher temperature, $\mathrm{CO_2}$ has a lower equilibrium solubility in the drink, so more escapes as bubbles. In the refrigerator, the lower temperature keeps more $\mathrm{CO_2}$ dissolved.气体在液体中的溶解度随温度升高而降低。在较高温度下，$\mathrm{CO_2}$ 在饮料中的平衡溶解度更低，因此更多 $\mathrm{CO_2}$ 以气泡形式逸出。在冰箱中，较低的温度使更多 $\mathrm{CO_2}$ 保持溶解状态。
CO₂ fizzing is physical, not chemical. Temperature doesn't "create pressure" — it affects solubility. Gas solubility decreases with temperature (Henry's law direction), so warm drinks lose CO₂ faster.CO₂ 冒泡是物理现象，不是化学反应。温度不会"产生压力"——它影响溶解度。气体溶解度随温度降低（亨利定律方向），所以温热饮料失去 CO₂ 更快。
A $0.050\ \mathrm{mol/L}$ solution of $\mathrm{Na_3PO_4}$ (sodium phosphate, assume complete dissociation) has a total ion concentration of: 🇨🇦 BC Chem 11 elab "concentration of ions…" / AB Chem 20 C-GO1$0.050\ \mathrm{mol/L}$ 的 $\mathrm{Na_3PO_4}$（磷酸钠，假设完全解离）溶液的总离子浓度为：🇨🇦 BC Chem 11 elab / AB Chem 20 C-GO1
Q8
$0.200\ \mathrm{mol/L}$$0.200\ \mathrm{mol/L}$
$0.050\ \mathrm{mol/L}$$0.050\ \mathrm{mol/L}$
$0.150\ \mathrm{mol/L}$$0.150\ \mathrm{mol/L}$
$0.100\ \mathrm{mol/L}$$0.100\ \mathrm{mol/L}$
$\mathrm{Na_3PO_4 \to 3\,Na^+ + PO_4^{3-}}$: 4 ions per formula unit. Total $[\mathrm{ions}] = 4 \times 0.050 = 0.200\ \mathrm{mol/L}$. ($[\mathrm{Na^+}] = 0.150\ \mathrm{mol/L}$; $[\mathrm{PO_4^{3-}}] = 0.050\ \mathrm{mol/L}$.)$\mathrm{Na_3PO_4 \to 3\,Na^+ + PO_4^{3-}}$：每个式量单元 4 个离子。总 $[\mathrm{离子}] = 4 \times 0.050 = 0.200\ \mathrm{mol/L}$。（$[\mathrm{Na^+}] = 0.150\ \mathrm{mol/L}$；$[\mathrm{PO_4^{3-}}] = 0.050\ \mathrm{mol/L}$。）
Dissociation: $\mathrm{Na_3PO_4 \to 3\,Na^+ + PO_4^{3-}}$ gives 4 total ions. $[\mathrm{total\ ions}] = 4 \times 0.050 = 0.200\ \mathrm{mol/L}$.解离方程：$\mathrm{Na_3PO_4 \to 3\,Na^+ + PO_4^{3-}}$ 共给出 4 个离子。$[\mathrm{总离子}] = 4 \times 0.050 = 0.200\ \mathrm{mol/L}$。
Antifreeze contains ethylene glycol dissolved in water. This lowers the freezing point of the coolant. Which best explains this effect?防冻液含有溶于水中的乙二醇。这降低了冷却液的凝固点。哪项最能解释这种效应？
Q9
Ethylene glycol has a lower freezing point than water乙二醇的凝固点低于水
Ethylene glycol reacts chemically with water molecules at low temperature乙二醇在低温下与水分子发生化学反应
Dissolved ethylene glycol molecules disrupt the formation of ice crystals, requiring a lower temperature to freeze溶解的乙二醇分子破坏冰晶的形成，需要更低的温度才能结冰
Ethylene glycol increases the density of water, lowering the freezing point乙二醇增加了水的密度，降低冰点
Freezing-point depression is a colligative property. The dissolved molecules disrupt the formation of the ice crystal lattice. More energy must be removed (lower temperature) to force the solution to freeze. This is why the molar quantity of solute particles matters, not their identity.凝固点降低是依数性。溶解的分子破坏冰晶格的形成。必须移除更多能量（更低温度）才能使溶液结冰。这就是为什么溶质颗粒的摩尔量比其种类更重要。
Freezing-point depression is a colligative property arising from dissolved particles disrupting crystallisation. It's not a reaction, not a density effect, and not simply about the solute's own freezing point.凝固点降低是由溶解颗粒破坏结晶引起的依数性。这不是反应，不是密度效应，也不仅仅与溶质自身的凝固点有关。
How many moles of solute are in $750\ \mathrm{mL}$ of $1.20\ \mathrm{mol/L}$ $\mathrm{HNO_3}$? 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U E2.2$750\ \mathrm{mL}$ 的 $1.20\ \mathrm{mol/L}$ $\mathrm{HNO_3}$ 中含有多少摩尔溶质？🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U E2.2
Q10
$1.20\ \mathrm{mol}$$1.20\ \mathrm{mol}$
$0.900\ \mathrm{mol}$$0.900\ \mathrm{mol}$
$0.750\ \mathrm{mol}$$0.750\ \mathrm{mol}$
$1.60\ \mathrm{mol}$$1.60\ \mathrm{mol}$
$n = cV = 1.20\ \mathrm{mol/L} \times 0.750\ \mathrm{L} = 0.900\ \mathrm{mol}$.$n = cV = 1.20\ \mathrm{mol/L} \times 0.750\ \mathrm{L} = 0.900\ \mathrm{mol}$。
$n = cV$. Convert $750\ \mathrm{mL}$ to $0.750\ \mathrm{L}$, then $n = 1.20 \times 0.750 = 0.900\ \mathrm{mol}$.$n = cV$。将 $750\ \mathrm{mL}$ 换算为 $0.750\ \mathrm{L}$，然后 $n = 1.20 \times 0.750 = 0.900\ \mathrm{mol}$。
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Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Solutions and solubility is one of the most directly applicable units in high school chemistry — its concepts appear in every subsequent quantitative topic. The molarity and dilution skills from this unit underpin acid-base titrations (Unit 9), the stoichiometry of reactions in solution (Unit 5/6 revisited), and equilibrium calculations (Unit 12). The cross-references below point at the college-credit feeder and the adjacent High School Chemistry units.溶液与溶解度是高中化学中应用最直接的单元之一——其概念出现在每个后续定量主题中。本单元的摩尔浓度和稀释技能支撑着酸碱滴定（第 9 单元）、溶液中的反应化学计量（重温第 5/6 单元）和平衡计算（第 12 单元）。以下链接指向大学学分衔接课程和相邻的高中化学单元。

Within High School Chemistry.在 HS Chemistry 内部。

Unit 9 (Acids, Bases and pH) builds directly on the solution skills here: molarity ($c = n/V$) is the tool for calculating acid and base concentrations; dilution ($c_1V_1 = c_2V_2$) is used in preparing buffer solutions and standard solutions for titrations; and the ion-dissociation framework from §6 is the foundation for strong/weak acid and base behaviour. Unit 5 (The Mole and Stoichiometry) uses the same concentration formulas to solve solution-stoichiometry problems involving limiting reagents. Unit 12 (Chemical Equilibrium) extends the saturated-solution idea (§5) into the formal solubility product constant $K_{sp}$.第 9 单元（酸、碱与 pH）直接建立在本单元的溶液技能上：摩尔浓度（$c = n/V$）是计算酸碱浓度的工具；稀释（$c_1V_1 = c_2V_2$）用于配制缓冲液和滴定标准溶液；§6 的离子解离框架是强弱酸碱行为的基础。第 5 单元（摩尔与化学计量）使用相同的浓度公式来解决涉及限量试剂的溶液化学计量问题。第 12 单元（化学平衡）将 §5 的饱和溶液概念延伸到正式的溶度积常数 $K_{sp}$。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Reactivity 2: How Much, How Fast, How Far (the college-credit feeder — extends solution stoichiometry into $K_{sp}$, titration curves, ICE tables, and quantitative equilibrium at IB depth)IB Chemistry HL · 反应性 2：多少、多快、多远（大学学分衔接——将溶液化学计量延伸到 $K_{sp}$、滴定曲线、ICE 表格和 IB 深度下的定量平衡）

If you are aiming for IB Chemistry HL or AP Chemistry, the concentration, dilution, and electrolyte material here is assumed from the first week of the college-credit course. IB Chemistry HL Reactivity 2 extends the ideas of this unit into quantitative equilibrium: the saturated-solution concept becomes the solubility product $K_{sp}$; the ion-dissociation equations become ICE table inputs for pH and buffer calculations; and the colligative-properties idea (§7) is elaborated with Raoult's law and the van't Hoff factor. AP Chemistry Unit 4 (Chemical Reactions) and Unit 7 (Equilibrium) assume fluent molarity and dissociation skills from this unit.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的浓度、稀释和电解质材料从大学学分课程的第一周就被默认掌握。IB Chemistry HL 反应性 2 将本单元的概念延伸到定量平衡：饱和溶液概念变为溶度积 $K_{sp}$；离子解离方程成为 pH 和缓冲液计算的 ICE 表格输入；依数性概念（§7）通过拉乌尔定律和范托夫因子进一步阐述。AP Chemistry 第 4 单元（化学反应）和第 7 单元（平衡）都假定你熟练掌握本单元的摩尔浓度和解离技能。

DINGRUI SCHOLARS

High School Chemistry · Solutions and Solubility · v1高中化学 · 溶液与溶解度 · v1