CH12 Unit 4: Solubility — 溶解度 完整解答¶
状态: ✅ 已完成
创建日期: 2026-02-18
最后更新: 2026-02-18
4.1 Solubility — 溶解度¶
Q1¶
EN (Original): What is the difference between a homogeneous mixture and a heterogeneous mixture?
中文翻译： 均匀混合物和非均匀混合物有什么区别？
Answer 答案:
EN: A homogeneous mixture has a uniform composition throughout — you cannot visually distinguish the different components. It appears as a single phase (e.g., salt dissolved in water). A heterogeneous mixture has a non-uniform composition — you can see distinct regions or phases (e.g., sand in water, oil and vinegar).
中文： 均匀混合物（homogeneous mixture）的组成在整体上是一致的——无法用肉眼区分不同成分，呈单一相态（如食盐溶于水）。非均匀混合物（heterogeneous mixture）的组成不均匀——可以看到明显的区域或相态（如沙子在水中、油和醋的混合物）。
Q2¶
EN (Original): What do you think will happen if more solute is added to a saturated solution in flask?
中文翻译： 如果向烧瓶中的饱和溶液再加入更多溶质，会发生什么？
Answer 答案:
EN: The additional solute will not dissolve. It will remain as undissolved solid at the bottom of the flask. A saturated solution already contains the maximum amount of dissolved solute at that temperature, so no more can dissolve.
中文： 多余的溶质不会溶解，将以固体形式沉在烧瓶底部。饱和溶液在该温度下已经溶解了最大量的溶质，无法再溶解更多。
Q3¶
EN (Original): What is the concentration we use to determine if a substance is soluble or insoluble in a solvent?
中文翻译： 我们用什么浓度标准来判断物质在溶剂中是可溶还是不可溶的？
Answer 答案:
EN: The threshold is 0.1 M. If a substance dissolves to give a concentration ≥ 0.1 M, it is considered soluble. If the concentration is < 0.1 M, it is considered insoluble (or "slightly soluble").
中文： 标准为 0.1 M。如果物质溶解后浓度 ≥ 0.1 M，则视为可溶；如果浓度 < 0.1 M，则视为不可溶（或"微溶"）。
Q4¶
EN (Original): What is the equation for solubility (s) in relation to Ksp for a 1:1:1 reaction? Rearrange this to determine the equation for Ksp in terms of solubility (s).
中文翻译： 对于 1:1:1 反应，溶解度 (s) 与 Ksp 的关系方程是什么？将其变形为用 s 表示 Ksp 的方程。
Answer 答案:
EN: For a 1:1:1 dissociation, e.g., AB(s) ⇌ A⁺(aq) + B⁻(aq):
$$K_{sp} = [\text{A}^+][\text{B}^-] = s \times s = s^2$$
$$\therefore\ s = \sqrt{K_{sp}} \qquad \text{and} \qquad K_{sp} = s^2$$
中文： 对于 1:1:1 解离，如 AB(s) ⇌ A⁺(aq) + B⁻(aq)：Ksp = s²，所以 s = √Ksp。
Q5¶
EN (Original): What is the equation for solubility (s) in relation to Ksp for a 1:2:1 or 1:1:2 reaction? Rearrange this to determine the equation for Ksp in terms of solubility (s).
中文翻译： 对于 1:2:1 或 1:1:2 反应，溶解度 (s) 与 Ksp 的关系方程是什么？
Answer 答案:
EN: For a 1:1:2 dissociation, e.g., AB₂(s) ⇌ A²⁺(aq) + 2B⁻(aq):
$$K_{sp} = [\text{A}^{2+}][\text{B}^-]^2 = (s)(2s)^2 = 4s^3$$
$$\therefore\ s = \sqrt[3]{\frac{K_{sp}}{4}} \qquad \text{and} \qquad K_{sp} = 4s^3$$
中文： 对于 1:1:2 解离：Ksp = s × (2s)² = 4s³，所以 s = ∛(Ksp/4)。
Q6¶
EN (Original): What is the equation for solubility (s) in relation to Ksp for a 1:1:3 or 1:3:1 reaction? Rearrange this to determine the equation for Ksp in terms of solubility (s).
中文翻译： 对于 1:1:3 或 1:3:1 反应，溶解度 (s) 与 Ksp 的关系方程是什么？
Answer 答案:
EN: For a 1:1:3 dissociation, e.g., AB₃(s) ⇌ A³⁺(aq) + 3B⁻(aq):
$$K_{sp} = [\text{A}^{3+}][\text{B}^-]^3 = (s)(3s)^3 = 27s^4$$
$$\therefore\ s = \sqrt[4]{\frac{K_{sp}}{27}} \qquad \text{and} \qquad K_{sp} = 27s^4$$
中文： 对于 1:1:3 解离：Ksp = s × (3s)³ = 27s⁴，所以 s = ⁴√(Ksp/27)。
Q7¶
EN (Original): Look up the Ksp of strontium sulphate in your data booklet. Write the dissociation equation for this compound, determine what ratio it is and then calculate its solubility.
中文翻译： 在数据手册中查找硫酸锶的 Ksp。写出解离方程式，确定比例类型，然后计算其溶解度。
Answer 答案:
EN:
$$\text{SrSO}_4(s) \rightleftharpoons \text{Sr}^{2+}(aq) + \text{SO}_4^{2-}(aq)$$
Ratio: 1:1:1 → Ksp = s²
From data booklet: Ksp = 3.4 × 10⁻⁷
$$s = \sqrt{K_{sp}} = \sqrt{3.4 \times 10^{-7}} = \boxed{5.8 \times 10^{-4}\ \text{M}}$$
中文： SrSO₄ 为 1:1:1 类型。Ksp = 3.4 × 10⁻⁷，溶解度 s = √(3.4 × 10⁻⁷) = 5.8 × 10⁻⁴ M。
✅ Answer Key 答案: 5.8×10⁻⁴ M — 一致 ✓
Q8¶
EN (Original): The Ksp of Sc(OH)₃ is 8.0×10⁻³¹. Write the dissociation equation for this compound, determine what ratio it is and then calculate its solubility.
中文翻译： Sc(OH)₃ 的 Ksp 为 8.0×10⁻³¹。写出解离方程式，确定比例类型，计算溶解度。
Answer 答案:
EN:
$$\text{Sc(OH)}_3(s) \rightleftharpoons \text{Sc}^{3+}(aq) + 3\text{OH}^-(aq)$$
Ratio: 1:1:3 → Ksp = 27s⁴
$$s^4 = \frac{8.0 \times 10^{-31}}{27} = 2.96 \times 10^{-32}$$
$$s = (2.96 \times 10^{-32})^{0.25} = \boxed{1.3 \times 10^{-8}\ \text{M}}$$
中文： Sc(OH)₃ 为 1:1:3 类型。Ksp = 27s⁴，s⁴ = 2.96 × 10⁻³²，s = 1.3 × 10⁻⁸ M。
✅ Answer Key 答案: 1.3×10⁻⁸ M — 一致 ✓
Q9¶
EN (Original): The solubility of NiS(s) is 5.5 × 10⁻¹⁰ M at 25°C. Write the dissociation equation and calculate Ksp.
中文翻译： NiS(s) 在 25°C 的溶解度为 5.5 × 10⁻¹⁰ M。写出解离方程式并计算 Ksp。
Answer 答案:
EN:
$$\text{NiS}(s) \rightleftharpoons \text{Ni}^{2+}(aq) + \text{S}^{2-}(aq)$$
Ratio: 1:1:1 → Ksp = s²
$$K_{sp} = (5.5 \times 10^{-10})^2 = \boxed{3.0 \times 10^{-19}}$$
中文： NiS 为 1:1:1 类型。Ksp = s² = (5.5 × 10⁻¹⁰)² = 3.0 × 10⁻¹⁹。
✅ Answer Key 答案: 3.0×10⁻¹⁹ — 一致 ✓
Q10¶
EN (Original): The solubility of copper (II) hydroxide is 1.8 × 10⁻⁷ M at 25°C. Write the dissociation equation and calculate Ksp.
中文翻译： 氢氧化铜 (II) 在 25°C 的溶解度为 1.8 × 10⁻⁷ M。写出解离方程式并计算 Ksp。
Answer 答案:
EN:
$$\text{Cu(OH)}_2(s) \rightleftharpoons \text{Cu}^{2+}(aq) + 2\text{OH}^-(aq)$$
Ratio: 1:1:2 → Ksp = 4s³
$$K_{sp} = 4(1.8 \times 10^{-7})^3 = 4(5.832 \times 10^{-21}) = \boxed{2.3 \times 10^{-20}}$$
中文： Cu(OH)₂ 为 1:1:2 类型。Ksp = 4s³ = 4 × (1.8 × 10⁻⁷)³ = 2.3 × 10⁻²⁰。
✅ Answer Key 答案: 2.3×10⁻²⁰ — 一致 ✓
Q11¶
EN (Original): A 250. L sample of saturated silver bromide is evaporated to dryness. What is the mass of the solid residue?
中文翻译： 将 250. L 饱和溴化银溶液蒸干，固体残留物的质量是多少？
Answer 答案:
EN:
AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq), Ksp = 5.4 × 10⁻¹³ (1:1:1)
$$s = \sqrt{5.4 \times 10^{-13}} = 7.35 \times 10^{-7}\ \text{M}$$
$$\text{Moles AgBr} = 7.35 \times 10^{-7} \times 250. = 1.84 \times 10^{-4}\ \text{mol}$$
$$M_{\text{AgBr}} = 107.87 + 79.90 = 187.77\ \text{g/mol}$$
$$\text{Mass} = 1.84 \times 10^{-4} \times 187.77 = \boxed{0.035\ \text{g}}$$
中文： s = √(5.4 × 10⁻¹³) = 7.35 × 10⁻⁷ M。在 250 L 中：mol = 1.84 × 10⁻⁴。质量 = 1.84 × 10⁻⁴ × 187.77 = 0.035 g。
✅ Answer Key 答案: 0.035 g AgBr — 一致 ✓
Q12¶
EN (Original): A 250 L sample of saturated copper(II) iodate is evaporated to dryness. What is the mass of the solid residue?
中文翻译： 将 250 L 饱和碘酸铜 (II) 溶液蒸干，固体残留物的质量是多少？
Answer 答案:
EN:
Cu(IO₃)₂(s) ⇌ Cu²⁺(aq) + 2IO₃⁻(aq), Ksp = 6.9 × 10⁻⁸ (1:1:2)
$$s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{6.9 \times 10^{-8}}{4}} = \sqrt[3]{1.725 \times 10^{-8}} = 2.59 \times 10^{-3}\ \text{M}$$
$$\text{Moles} = 2.59 \times 10^{-3} \times 250 = 0.647\ \text{mol}$$
$$M_{\text{Cu(IO₃)₂}} = 63.55 + 2(126.90 + 48.00) = 63.55 + 349.80 = 413.35\ \text{g/mol}$$
$$\text{Mass} = 0.647 \times 413.35 = \boxed{270\ \text{g}}$$
中文： s = ∛(6.9 × 10⁻⁸ / 4) = 2.59 × 10⁻³ M。在 250 L 中：mol = 0.647。质量 = 0.647 × 413.35 ≈ 270 g。
✅ Answer Key 答案: 270 g — 一致 ✓
Q13¶
EN (Original): A solution which contains only one of the following anions: I⁻, SO₄²⁻, or OH⁻ is tested with various reagents and the following results are obtained:
| Reagent | Result |
|---|---|
| 0.2 M AgNO₃ | Precipitate |
| 0.2 M Fe(NO₃)₂ | Precipitate |
| 0.2 M Sr(NO₃)₂ | No Precipitate |
a) Which anion does the solution contain? b) Explain why each of the other two ions is not the ion in the solution.
中文翻译： 一种仅含有 I⁻、SO₄²⁻ 或 OH⁻ 其中之一的溶液，用各种试剂测试后得到上述结果。a) 溶液中含有哪种阴离子？b) 解释为什么另外两种离子不是溶液中的离子。
Answer 答案:
a) OH⁻
EN — Reasoning:
| Anion | With Ag⁺ | With Fe²⁺ | With Sr²⁺ | Match? |
|---|---|---|---|---|
| I⁻ | AgI — ppt ✓ | FeI₂ — soluble ✗ | SrI₂ — soluble ✓ | ✗ |
| SO₄²⁻ | Ag₂SO₄ — low solubility, ppt ✓ | FeSO₄ — soluble ✗ | SrSO₄ — ppt ✗ | ✗ |
| OH⁻ | AgOH — ppt ✓ | Fe(OH)₂ — ppt ✓ | Sr(OH)₂ — soluble ✓ | ✓ |
b)
- I⁻ is eliminated: FeI₂ is soluble (not in Ksp table), so I⁻ would NOT precipitate with Fe²⁺. But the test shows a precipitate with Fe(NO₃)₂.
- SO₄²⁻ is eliminated: SrSO₄ is insoluble (Ksp = 3.4 × 10⁻⁷, in Ksp table), so SO₄²⁻ WOULD precipitate with Sr²⁺. But the test shows no precipitate with Sr(NO₃)₂. Also, FeSO₄ is soluble, contradicting the precipitate with Fe²⁺.
中文：
a) OH⁻
b) I⁻ 被排除： FeI₂ 是可溶的（不在 Ksp 表中），所以 I⁻ 不会与 Fe²⁺ 沉淀，但测试显示有沉淀。SO₄²⁻ 被排除： SrSO₄ 是不可溶的（Ksp = 3.4 × 10⁻⁷），所以 SO₄²⁻ 会与 Sr²⁺ 沉淀，但测试显示无沉淀。而且 FeSO₄ 可溶，与 Fe²⁺ 产生沉淀的结果矛盾。
✅ Answer Key 答案: a) OH⁻ — 一致 ✓
Q14¶
EN (Original): A solution which contains only one of the following cations: Ag⁺, Pb²⁺, or Cu⁺ is tested:
| Reagent | Result |
|---|---|
| 0.2 M NaCl | Precipitate |
| 0.2 M Na₂SO₄ | No Precipitate |
| 0.2 M NaNO₃ | No Precipitate |
a) Which cation does the solution contain? b) Explain why each of the other two ions is not the ion in the solution.
中文翻译： 一种仅含有 Ag⁺、Pb²⁺ 或 Cu⁺ 其中之一的溶液，测试结果如上。a) 含有哪种阳离子？b) 解释排除其他两种的原因。
Answer 答案:
a) Cu⁺
EN — Reasoning:
| Cation | With Cl⁻ | With SO₄²⁻ | With NO₃⁻ | Match? |
|---|---|---|---|---|
| Ag⁺ | AgCl — low solubility, ppt ✓ | Ag₂SO₄ — low solubility, ppt ✗ | AgNO₃ — soluble ✓ | ✗ |
| Pb²⁺ | PbCl₂ — low solubility, ppt ✓ | PbSO₄ — low solubility, ppt ✗ | Pb(NO₃)₂ — soluble ✓ | ✗ |
| Cu⁺ | CuCl — low solubility, ppt ✓ | Cu₂SO₄ — soluble ✓ | CuNO₃ — soluble ✓ | ✓ |
b)
- Ag⁺ is eliminated: According to the solubility chart, Ag₂SO₄ has low solubility, so Ag⁺ would precipitate with Na₂SO₄. But the test shows no precipitate.
- Pb²⁺ is eliminated: PbSO₄ has low solubility, so Pb²⁺ would precipitate with Na₂SO₄. But the test shows no precipitate.
中文：
a) Cu⁺
b) Ag⁺ 被排除： 根据公式表第 4 页的溶解度表，Ag₂SO₄ 属于低溶解度，因此 Ag⁺ 会与 Na₂SO₄ 产生沉淀，但题目显示无沉淀。Pb²⁺ 被排除： PbSO₄ 也是低溶解度，Pb²⁺ 会与 Na₂SO₄ 产生沉淀，但题目显示无沉淀。
ℹ️ 答案页未提供此题答案。
Q15¶
EN (Original): If 120. mL of 0.00125 M AgNO₃ and 70.0 mL of 0.00250 M K₂CO₃ are combined, will a precipitate form?
中文翻译： 若将 120. mL 的 0.00125 M AgNO₃ 与 70.0 mL 的 0.00250 M K₂CO₃ 混合，是否会形成沉淀？
Answer 答案:
EN:
Possible precipitate: Ag₂CO₃
$$\text{Ag}2\text{CO}_3(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CO}_3^{2-}(aq) \qquad K$$} = 8.5 \times 10^{-12
Step 1 — Find concentrations after mixing:
Total volume = 120. + 70.0 = 190. mL
$$[\text{Ag}^+] = \frac{0.00125 \times 120.}{190.} = 7.89 \times 10^{-4}\ \text{M}$$
$$[\text{CO}_3^{2-}] = \frac{0.00250 \times 70.0}{190.} = 9.21 \times 10^{-4}\ \text{M}$$
Step 2 — Calculate Trial Ksp:
$$\text{Trial}\ K_{sp} = [\text{Ag}^+]^2[\text{CO}_3^{2-}] = (7.89 \times 10^{-4})^2(9.21 \times 10^{-4})$$
$$= (6.23 \times 10^{-7})(9.21 \times 10^{-4}) = \boxed{5.73 \times 10^{-10}}$$
Step 3 — Compare:
Trial Ksp (5.73 × 10⁻¹⁰) > Ksp (8.5 × 10⁻¹²) → YES, a precipitate WILL form.
中文： 可能沉淀为 Ag₂CO₃（Ksp = 8.5 × 10⁻¹²）。混合后 [Ag⁺] = 7.89 × 10⁻⁴ M，[CO₃²⁻] = 9.21 × 10⁻⁴ M。Trial Ksp = 5.73 × 10⁻¹⁰ > Ksp → 会形成沉淀。
✅ Answer Key 答案: Trial Ksp = 5.73×10⁻¹⁰ — 一致 ✓
Q16¶
EN (Original): Given a 0.00500 M solution of KOH, what concentration of Mg(NO₃)₂ will be needed to just start precipitation of Mg(OH)₂?
中文翻译： 在 0.00500 M KOH 溶液中，需要多少浓度的 Mg(NO₃)₂ 才能刚好开始沉淀 Mg(OH)₂？
Answer 答案:
EN:
$$\text{Mg(OH)}2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \qquad K$$} = 5.6 \times 10^{-12
KOH is a strong base: [OH⁻] = 0.00500 M
At the point of precipitation (Trial Ksp = Ksp):
$$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$$
$$5.6 \times 10^{-12} = [\text{Mg}^{2+}]\,(0.00500)^2$$
$$[\text{Mg}^{2+}] = \frac{5.6 \times 10^{-12}}{2.5 \times 10^{-5}} = \boxed{2.24 \times 10^{-7}\ \text{M}}$$
Since each Mg(NO₃)₂ gives one Mg²⁺: [Mg(NO₃)₂] = 2.24 × 10⁻⁷ M
中文： Ksp = [Mg²⁺][OH⁻]²。代入 [OH⁻] = 0.00500 M：[Mg²⁺] = 5.6 × 10⁻¹² / (0.00500)² = 2.24 × 10⁻⁷ M。
✅ Answer Key 答案: 2.24×10⁻⁷ M — 一致 ✓
Q17¶
EN (Original): A dissociation equation may be exothermic or endothermic. This determines how a temperature change will affect the solubility of a chemical. Fill in the table.
中文翻译： 解离方程可以是放热或吸热的。温度变化如何影响溶解度？填写表格。
Answer 答案:
| Dissociation Type 解离类型 | Temperature Increases 温度升高 | Temperature Decreases 温度降低 |
|---|---|---|
| Exothermic 放热 | Decreased solubility 溶解度降低 | Increased solubility 溶解度升高 |
| Endothermic 吸热 | Increased solubility 溶解度升高 | Decreased solubility 溶解度降低 |
EN — Explanation: By Le Chatelier's Principle: adding heat shifts equilibrium away from the exothermic direction. For an exothermic dissolution (heat is a product), raising temperature shifts left → less dissolves. For an endothermic dissolution (heat is a reactant), raising temperature shifts right → more dissolves.
中文： 根据勒夏特列原理：加热使平衡向吸热方向移动。放热溶解中，升温使平衡左移→溶解度降低。吸热溶解中，升温使平衡右移→溶解度升高。
ℹ️ 答案页未提供此题答案。
Q18¶
EN (Original): For the equilibrium CuCl₂(s) ⇌ Cu²⁺(aq) + 2Cl⁻(aq), determine whether adding each compound will increase solubility, decrease solubility or have no effect.
中文翻译： 对于平衡 CuCl₂(s) ⇌ Cu²⁺(aq) + 2Cl⁻(aq)，判断加入以下化合物对溶解度的影响。
Answer 答案:
| Compound Added 加入的化合物 | Effect 影响 | Reason 原因 |
|---|---|---|
| NaCl | Decreased 降低 | Cl⁻ is a common ion → shifts equilibrium left 同离子效应 |
| Cu(NO₃)₂ | Decreased 降低 | Cu²⁺ is a common ion → shifts equilibrium left 同离子效应 |
| CuCl₂ | No effect 无影响 | Adding more solid doesn't change Ksp (solid concentration is constant) 固体浓度恒定 |
| Pb(NO₃)₂ | Increased 升高 | Pb²⁺ reacts with Cl⁻ → PbCl₂ ppt (Ksp = 1.2×10⁻⁵) removes Cl⁻ → shifts right |
中文说明： - NaCl 和 Cu(NO₃)₂ 提供同离子（Cl⁻ 和 Cu²⁺），使平衡左移，溶解度降低。 - CuCl₂ 为固体，加入更多固体不改变 Ksp 表达式中的浓度。 - Pb(NO₃)₂ 提供的 Pb²⁺ 与 Cl⁻ 反应生成 PbCl₂ 沉淀，移除 Cl⁻，使平衡右移，溶解度升高。
ℹ️ 答案页未提供此题答案。
4.2 Saturated Solutions — 饱和溶液¶
Q19¶
EN (Original): Explain why a saturated solution with NO solid present is NOT at equilibrium.
中文翻译： 解释为什么没有固体存在的饱和溶液不处于平衡状态。
Answer 答案:
EN: Equilibrium is a dynamic process requiring both the forward reaction (dissolution) and the reverse reaction (crystallization/precipitation) to occur simultaneously at equal rates. Without solid present:
- There is no undissolved solid to continue dissolving (forward reaction stops)
- There is no solid surface for dissolved ions to crystallize onto (reverse reaction stops)
- Since neither process is occurring, there is no dynamic equilibrium
A saturated solution with solid present IS at equilibrium because ions are continuously dissolving and re-crystallizing at the same rate.
中文： 平衡是一个动态过程，要求正向反应（溶解）和逆向反应（结晶/沉淀）同时以相等速率进行。没有固体时：正向反应无法进行（没有固体可溶解），逆向反应也无法进行（没有固体表面供离子结晶）。两个过程都停止，不存在动态平衡。有固体存在的饱和溶液才是真正的平衡。
ℹ️ 答案页未提供此题答案。
Q20¶
EN (Original): Which of the following saturated solutions will have the lowest [Ba²⁺]? Explain. Barium carbonate Barium chromate Barium sulphate
中文翻译： 以下哪种饱和溶液的 [Ba²⁺] 最低？硫酸钡、碳酸钡、铬酸钡。请解释。
Answer 答案:
EN: All three are 1:1:1 dissociations, so [Ba²⁺] = s = √Ksp.
| Compound | Ksp | s = √Ksp |
|---|---|---|
| BaCO₃ | 2.6 × 10⁻⁹ | 5.1 × 10⁻⁵ M |
| BaCrO₄ | 1.2 × 10⁻¹⁰ | 1.1 × 10⁻⁵ M |
| BaSO₄ | 1.1 × 10⁻¹⁰ | 1.05 × 10⁻⁵ M |
Barium sulphate has the lowest Ksp and therefore the lowest [Ba²⁺].
中文： 三者均为 1:1:1 解离，[Ba²⁺] = √Ksp。BaSO₄ 的 Ksp 最小（1.1 × 10⁻¹⁰），因此 [Ba²⁺] 最低。
ℹ️ 答案页未提供此题答案。
Q21¶
EN (Original): When do you stop adding solution from your burette when doing a titration?
中文翻译： 做滴定时，什么时候停止从滴定管中加入溶液？
Answer 答案:
EN: You stop adding solution when you observe the first permanent color change of the indicator that does not disappear upon swirling. This is called the endpoint of the titration.
中文： 当观察到指示剂出现第一次持久的颜色变化（摇晃后颜色不消失）时停止加入溶液。这称为滴定的终点（endpoint）。
ℹ️ 答案页未提供此题答案。
4.3 Ions — 离子¶
Q22¶
EN (Original): When you are separating ions using filtration, what is the name of the solution that passes through the filter?
中文翻译： 用过滤法分离离子时，通过滤纸的溶液叫什么？
Answer 答案:
EN: The solution that passes through the filter is called the filtrate (滤液).
中文： 通过滤纸的溶液称为滤液（filtrate）。
ℹ️ 答案页未提供此题答案。
Q23¶
EN (Original): For each pair of ions, circle an ion which could be added to allow separation.
中文翻译： 对于每对离子，选择一种可以实现分离的离子。
Answer 答案:
Pair 1: Cu⁺ and Na⁺ → Cl⁻
| Added Ion | With Cu⁺ | With Na⁺ | Separates? |
|---|---|---|---|
| Cl⁻ | CuCl — insoluble (ppt) | NaCl — soluble | Yes ✓ |
| SO₄²⁻ | Cu₂SO₄ — soluble | Na₂SO₄ — soluble | No |
| NO₃⁻ | CuNO₃ — soluble | NaNO₃ — soluble | No |
Pair 2: Sr²⁺ and Pb²⁺ → SO₄²⁻
| Added Ion | With Sr²⁺ | With Pb²⁺ | Separates? |
|---|---|---|---|
| PO₄³⁻ | Sr₃(PO₄)₂ — insoluble | Pb₃(PO₄)₂ — insoluble | No (both ppt) |
| SO₄²⁻ | SrSO₄ (Ksp = 3.4×10⁻⁷) | PbSO₄ (Ksp = 1.8×10⁻⁸) | Yes (selective) |
| OH⁻ | Sr(OH)₂ — soluble | Pb(OH)₂ — insoluble | Also works |
With careful addition of SO₄²⁻, PbSO₄ (lower Ksp) precipitates first while SrSO₄ (higher Ksp) remains dissolved → selective precipitation.
Pair 3: S²⁻ and OH⁻ → Mg²⁺
| Added Ion | With S²⁻ | With OH⁻ | Separates? |
|---|---|---|---|
| Sr²⁺ | SrS — soluble | Sr(OH)₂ — soluble | No |
| Mg²⁺ | MgS — soluble | Mg(OH)₂ (Ksp = 5.6×10⁻¹²) — ppt | Yes ✓ |
| Ag⁺ | Ag₂S — insoluble | AgOH — unstable | Also works |
中文： 1. Cu⁺ 和 Na⁺ → 加 Cl⁻：CuCl 不可溶（沉淀），NaCl 可溶（留在溶液中）。 2. Sr²⁺ 和 Pb²⁺ → 加 SO₄²⁻：PbSO₄（Ksp 更小）先沉淀，SrSO₄ 留在溶液中（选择性沉淀）。 3. S²⁻ 和 OH⁻ → 加 Mg²⁺：Mg(OH)₂ 沉淀（在 Ksp 表中），MgS 可溶。
ℹ️ 答案页未提供此题答案。
Q24¶
EN (Original): Write an equilibrium reaction for each precipitate that forms in question 23.
中文翻译： 为 Q23 中形成的每种沉淀写出平衡反应。
Answer 答案:
Pair 1:
$$\text{CuCl}(s) \rightleftharpoons \text{Cu}^+(aq) + \text{Cl}^-(aq)$$
Pair 2:
$$\text{PbSO}4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \qquad K$$} = 1.8 \times 10^{-8
Pair 3:
$$\text{Mg(OH)}2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \qquad K$$} = 5.6 \times 10^{-12
ℹ️ 答案页未提供此题答案。
Q25¶
EN (Original): What concentration of silver bromate can dissociate in 1.6 M sodium bromate?
中文翻译： 在 1.6 M 溴酸钠溶液中，溴酸银的溶解度是多少？
Answer 答案:
EN:
$$\text{AgBrO}3(s) \rightleftharpoons \text{Ag}^+(aq) + \text{BrO}_3^-(aq) \qquad K$$} = 5.3 \times 10^{-5
NaBrO₃ provides 1.6 M BrO₃⁻ (common ion). Let s = solubility of AgBrO₃.
[Ag⁺] = s, [BrO₃⁻] = 1.6 + s ≈ 1.6 M (since s ≪ 1.6)
$$K_{sp} = s \times 1.6$$
$$s = \frac{5.3 \times 10^{-5}}{1.6} = \boxed{3.3 \times 10^{-5}\ \text{M}}$$
中文： NaBrO₃ 提供 1.6 M BrO₃⁻（同离子效应）。Ksp = s × 1.6，s = 5.3 × 10⁻⁵ / 1.6 = 3.3 × 10⁻⁵ M。
✅ Answer Key 答案: 3.3×10⁻⁵ M — 一致 ✓
Q26¶
EN (Original): What concentration of lead (II) iodide can dissociate in 3.00 M potassium iodide?
中文翻译： 在 3.00 M 碘化钾溶液中，碘化铅 (II) 的溶解度是多少？
Answer 答案:
EN:
$$\text{PbI}2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \qquad K$$} = 8.5 \times 10^{-9
KI provides 3.00 M I⁻. Let s = solubility.
[Pb²⁺] = s, [I⁻] = 3.00 + 2s ≈ 3.00 M
$$K_{sp} = s(3.00)^2 = 9.00s$$
$$s = \frac{8.5 \times 10^{-9}}{9.00} = \boxed{9.4 \times 10^{-10}\ \text{M}}$$
中文： KI 提供 3.00 M I⁻。Ksp = s × (3.00)² = 9.00s，s = 8.5 × 10⁻⁹ / 9.00 = 9.4 × 10⁻¹⁰ M。
✅ Answer Key 答案: 9.4×10⁻¹⁰ M — 一致 ✓
Q27¶
EN (Original): What concentration of iron (III) hydroxide can dissociate in 4.0 × 10⁻⁵ M potassium hydroxide?
中文翻译： 在 4.0 × 10⁻⁵ M 氢氧化钾溶液中，氢氧化铁 (III) 的溶解度是多少？
Answer 答案:
EN:
$$\text{Fe(OH)}3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \qquad K$$} = 2.6 \times 10^{-39
KOH provides 4.0 × 10⁻⁵ M OH⁻. Let s = solubility.
[Fe³⁺] = s, [OH⁻] = 4.0 × 10⁻⁵ + 3s ≈ 4.0 × 10⁻⁵ M
$$K_{sp} = s(4.0 \times 10^{-5})^3 = s(6.4 \times 10^{-14})$$
$$s = \frac{2.6 \times 10^{-39}}{6.4 \times 10^{-14}} = \boxed{4.1 \times 10^{-26}\ \text{M}}$$
中文： KOH 提供 4.0 × 10⁻⁵ M OH⁻。Ksp = s × (4.0 × 10⁻⁵)³ = 6.4 × 10⁻¹⁴ s，s = 4.1 × 10⁻²⁶ M。
✅ Answer Key 答案: 4.1×10⁻²⁶ — 一致 ✓
Q28¶
EN (Original): A large vessel contains 250. L of 0.00030 M barium sulfide (BaS). What mass of solid sodium carbonate (Na₂CO₃) would be needed to just start precipitation of barium carbonate (BaCO₃)?
中文翻译： 一个大容器中有 250. L 的 0.00030 M 硫化钡 (BaS) 溶液。需要多少质量的固体碳酸钠 (Na₂CO₃) 才能刚好开始沉淀碳酸钡 (BaCO₃)？
Answer 答案:
EN:
$$\text{BaCO}3(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{CO}_3^{2-}(aq) \qquad K$$} = 2.6 \times 10^{-9
BaS → Ba²⁺ + S²⁻, so [Ba²⁺] = 0.00030 M
At the point of precipitation:
$$K_{sp} = [\text{Ba}^{2+}][\text{CO}_3^{2-}]$$
$$[\text{CO}_3^{2-}] = \frac{2.6 \times 10^{-9}}{0.00030} = 8.67 \times 10^{-6}\ \text{M}$$
$$\text{Moles CO}_3^{2-} = 8.67 \times 10^{-6} \times 250. = 2.17 \times 10^{-3}\ \text{mol}$$
$$M_{\text{Na₂CO₃}} = 2(22.99) + 12.01 + 3(16.00) = 106.0\ \text{g/mol}$$
$$\text{Mass} = 2.17 \times 10^{-3} \times 106.0 = \boxed{0.23\ \text{g}}$$
中文： [Ba²⁺] = 0.00030 M。需要 [CO₃²⁻] = Ksp / [Ba²⁺] = 8.67 × 10⁻⁶ M。在 250 L 中需要 2.17 × 10⁻³ mol Na₂CO₃，质量 = 0.23 g。
✅ Answer Key 答案: 0.23 g — 一致 ✓
Q29¶
EN (Original): A large vessel contains 500. L of 0.00050 M silver (I) nitrate (AgNO₃). What mass of solid ammonium chromate ((NH₄)₂CrO₄) would be needed to just start precipitation of silver (I) chromate (Ag₂CrO₄)?
中文翻译： 一个大容器中有 500. L 的 0.00050 M 硝酸银 (AgNO₃) 溶液。需要多少质量的固体铬酸铵才能刚好开始沉淀铬酸银 (Ag₂CrO₄)？
Answer 答案:
EN:
$$\text{Ag}2\text{CrO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq) \qquad K$$} = 1.1 \times 10^{-12
AgNO₃ → Ag⁺ + NO₃⁻, so [Ag⁺] = 0.00050 M
$$K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$$
$$[\text{CrO}_4^{2-}] = \frac{1.1 \times 10^{-12}}{(0.00050)^2} = \frac{1.1 \times 10^{-12}}{2.5 \times 10^{-7}} = 4.4 \times 10^{-6}\ \text{M}$$
$$\text{Moles CrO}_4^{2-} = 4.4 \times 10^{-6} \times 500. = 2.2 \times 10^{-3}\ \text{mol}$$
$$M_{(\text{NH}_4)_2\text{CrO}_4} = 2(18.04) + 52.00 + 4(16.00) = 152.1\ \text{g/mol}$$
$$\text{Mass} = 2.2 \times 10^{-3} \times 152.1 = \boxed{0.33\ \text{g}}$$
中文： [Ag⁺] = 0.00050 M。需要 [CrO₄²⁻] = 4.4 × 10⁻⁶ M。在 500 L 中需要 2.2 × 10⁻³ mol (NH₄)₂CrO₄，质量 = 0.33 g。
✅ Answer Key 答案: 0.33 g — 一致 ✓
术语表 | Terminology¶
| English 英文 | Chinese 中文 | Definition 定义 |
|---|---|---|
| Solubility Product Constant (Ksp) | 溶度积常数 | Equilibrium constant for a solid dissolving into ions |
| Saturated Solution | 饱和溶液 | Solution containing the maximum amount of dissolved solute |
| Common Ion Effect | 同离子效应 | Decrease in solubility when a common ion is already present |
| Trial Ksp | 试算 Ksp | Ion product calculated to predict if precipitation will occur |
| Precipitate | 沉淀 | Insoluble solid formed when Trial Ksp > Ksp |
| Filtrate | 滤液 | Solution that passes through a filter |
| Selective Precipitation | 选择性沉淀 | Using Ksp differences to precipitate one ion while another stays dissolved |
| Dissociation | 解离 | Solid breaking apart into individual ions in solution |
| Endpoint | 终点 | Point at which indicator changes color during titration |
答案核对总表 | Answer Key Verification Summary¶
| Question | Our Answer | Answer Key | Status |
|---|---|---|---|
| Q7 | 5.8 × 10⁻⁴ M | 5.8 × 10⁻⁴ M | ✅ |
| Q8 | 1.3 × 10⁻⁸ M | 1.3 × 10⁻⁸ M | ✅ |
| Q9 | 3.0 × 10⁻¹⁹ | 3.0 × 10⁻¹⁹ | ✅ |
| Q10 | 2.3 × 10⁻²⁰ | 2.3 × 10⁻²⁰ | ✅ |
| Q11 | 0.035 g | 0.035 g | ✅ |
| Q12 | 270 g | 270 g | ✅ |
| Q13a | OH⁻ | OH⁻ | ✅ |
| Q15 | Trial Ksp = 5.73 × 10⁻¹⁰ | 5.73 × 10⁻¹⁰ | ✅ |
| Q16 | 2.24 × 10⁻⁷ M | 2.24 × 10⁻⁷ M | ✅ |
| Q25 | 3.3 × 10⁻⁵ M | 3.3 × 10⁻⁵ M | ✅ |
| Q26 | 9.4 × 10⁻¹⁰ M | 9.4 × 10⁻¹⁰ M | ✅ |
| Q27 | 4.1 × 10⁻²⁶ | 4.1 × 10⁻²⁶ | ✅ |
| Q28 | 0.23 g | 0.23 g | ✅ |
| Q29 | 0.33 g | 0.33 g | ✅ |
Result: 14/14 answers verified — 100% match ✅