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Solutions详解

Chemical Equilibrium · Solutions化学平衡 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 25 marksAP 选择题 + 安/卑省考短答 · 共 25 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 🇨🇦 ON安 AP-style MCQAP 风格选择题 §1 Dynamic equilibrium动态平衡 · HS-PS1-6 [3 marks][3 分]

For $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$ in a sealed flask, which statement correctly describes dynamic equilibrium?密封烧瓶中 $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$，哪项正确描述了动态平衡？

Answer:答案：  (C)  Forward and reverse rates are equal; concentrations remain constant.正向与逆向反应速率相等；浓度保持不变。

Identify the defining feature of dynamic equilibrium识别动态平衡的本质特征 M1·A1·A1

At dynamic equilibrium the forward reaction ($\text{N}_2\text{O}_4 \to 2\,\text{NO}_2$) and reverse reaction ($2\,\text{NO}_2 \to \text{N}_2\text{O}_4$) both continue to occur at the molecular level. Because the two rates are equal, the macroscopic concentrations of all species stay constant. This is option (C).动态平衡时，正向反应（$\text{N}_2\text{O}_4 \to 2\,\text{NO}_2$）和逆向反应（$2\,\text{NO}_2 \to \text{N}_2\text{O}_4$）在分子水平上仍在持续进行。由于两者速率相等，所有组分的宏观浓度保持不变。这对应选项 (C)。
Why the distractors fail.干扰项分析。
(A) The reaction has NOT stopped; dynamic equilibrium is an ongoing molecular process.反应并未停止；动态平衡是持续进行的分子过程。
(B) Equal concentrations of all species is not required; the concentrations are simply constant (whatever values the system settled at).并不要求各组分浓度相等，只要求浓度恒定（即体系稳定后各自的数值）。
(D) Both forward and reverse reactions continue; only the forward reaction is not the correct description.正向与逆向反应都在进行，仅有正向反应的说法不正确。
Dynamic equilibrium: constant concentrations do NOT mean the reaction stopped.动态平衡：浓度不变并不意味着反应停止。 The word "dynamic" is the key: it emphasises that both the forward and reverse reactions are still occurring simultaneously. The constant-concentration observation at the macroscopic level is a consequence of the two microscopic rates being equal, not of the reactions ceasing. On every exam, the most common distractor is "the reaction has stopped" (option A). A secondary trap is to confuse "equal concentrations" (the amounts of each species happen to be the same) with "constant concentrations" (the amounts do not change over time, but can be any value). The correct definition: rate$_{\text{fwd}}$ = rate$_{\text{rev}}$ leads to $d[\text{X}]/dt = 0$ for all species X."动态"二字是关键：它强调正向和逆向反应仍在同时发生。宏观层面的浓度不变是两个微观速率相等的结果，而非反应停止的结果。每次考试最常见的干扰项都是"反应已停止"（选项 A）。次级陷阱是将"浓度相等"（各组分的量恰好相同）与"浓度不变"（各量随时间不变，但可取任意值）混淆。正确定义：$\text{速率}_{\text{正}} = \text{速率}_{\text{逆}}$，从而使所有组分 X 满足 $d[\text{X}]/dt = 0$。
Q2EASY易 🇨🇦 ON安 🇨🇦 BC卑 AP-style MCQAP 风格选择题 §2 $K_{eq}$ expression$K_{eq}$ 表达式 · SCH4U E3 [4 marks][4 分]

Which expression correctly gives $K_{eq}$ for $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$?下列哪个表达式正确给出了 $\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$ 的 $K_{eq}$？

Answer:答案：  (A)  $K_{eq} = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

Apply the equilibrium constant expression rule套用平衡常数表达式规则 M1·A1·A1·A1

For a general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium expression is:对于一般反应 $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$，平衡表达式为： $$ K_{eq} = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}. $$ Here products = $\text{NH}_3$ (coefficient 2); reactants = $\text{N}_2$ (coefficient 1) and $\text{H}_2$ (coefficient 3). Substituting:此处产物为 $\text{NH}_3$（系数 2）；反应物为 $\text{N}_2$（系数 1）和 $\text{H}_2$（系数 3）。代入： $$ K_{eq} = \frac{[\text{NH}_3]^2}{[\text{N}_2]^1[\text{H}_2]^3}. $$ This matches option (A).这与选项 (A) 一致。
Why the distractors fail.干扰项分析。
(B) Reactants over products (inverted fraction). This gives $1/K_{eq}$, the expression for the reverse reaction.分子分母颠倒（反应物在上）。这给出的是逆反应的 $1/K_{eq}$。
(C) Missing the stoichiometric exponents entirely; $[\text{NH}_3]^1$ and $[\text{H}_2]^1$ are wrong.完全忽略了化学计量数作为指数；$[\text{NH}_3]^1$ 和 $[\text{H}_2]^1$ 均错误。
(D) Uses stoichiometric coefficients as multipliers rather than exponents, and adds reactant concentrations in the denominator. Both are common errors.将化学计量数作为乘数而非指数，且将反应物浓度相加。两者都是常见错误。
$K_{eq}$: products over reactants, stoichiometric coefficients become exponents.$K_{eq}$：产物除以反应物，化学计量数作为指数。 The two rules are non-negotiable: (1) products always go in the numerator; (2) each concentration is raised to the power of its stoichiometric coefficient. Pure solids and pure liquids are omitted (their activity = 1). For this Haber-process reaction, the large denominator $[\text{H}_2]^3$ means $K_{eq}$ is very sensitive to the H$_2$ concentration, which is why industrial plants maintain a large excess of H$_2$. Reversing the equation gives $K_{eq}' = 1/K_{eq}$; multiplying the equation by a factor $n$ gives $K_{eq}' = K_{eq}^n$. These scaling rules are standard exam questions.两条规则不容违反：(1) 产物始终在分子上；(2) 每种浓度的指数等于其化学计量数。纯固体和纯液体省略（活度 = 1）。对于此哈伯法反应，分母中的 $[\text{H}_2]^3$ 意味着 $K_{eq}$ 对 H$_2$ 浓度极为敏感，这也是工业装置维持大量过量 H$_2$ 的原因。将方程式颠倒得 $K_{eq}' = 1/K_{eq}$；将方程式各系数乘以因子 $n$ 得 $K_{eq}' = K_{eq}^n$。这些缩放规则是标准考题。
Q3MEDIUM中 🇨🇦 BC卑 🇨🇦 AB阿 BC Provincial-style卑诗省考风格 §3 Reaction quotient $Q$反应商 $Q$ · Chem 12 Honors荣誉级 [6 marks][6 分]

$\text{A(g)} \rightleftharpoons 2\,\text{B(g)}$, $K_{eq} = 4.00$. Initial: $[\text{A}] = 0.500\ \text{mol/L}$, $[\text{B}] = 0.600\ \text{mol/L}$. (a) Calculate $Q$. (b) Predict direction of shift. (c) State condition for $Q = K_{eq}$.$\text{A(g)} \rightleftharpoons 2\,\text{B(g)}$，$K_{eq} = 4.00$。初始：$[\text{A}] = 0.500\ \text{mol/L}$，$[\text{B}] = 0.600\ \text{mol/L}$。(a) 计算 $Q$。(b) 预测移动方向。(c) 说明 $Q = K_{eq}$ 的条件。

Answer:答案：  (a) $Q = 0.720$  ·  (b) shift right (toward B)右移（向 B 方向）  ·  (c) system is already at equilibrium体系已处于平衡状态

(a) Calculate $Q$计算 $Q$ M1·A1

The reaction quotient $Q$ uses the same expression form as $K_{eq}$ but with non-equilibrium concentrations:反应商 $Q$ 与 $K_{eq}$ 的表达式形式相同，但代入非平衡浓度： $$ Q = \frac{[\text{B}]^2}{[\text{A}]} = \frac{(0.600)^2}{0.500} = \frac{0.360}{0.500} = 0.720. $$

(b) Compare $Q$ and $K_{eq}$; predict direction比较 $Q$ 与 $K_{eq}$；预测方向 M1·A1·A1

$Q = 0.720 < K_{eq} = 4.00$. The numerator (products) is too small relative to $K_{eq}$, so the reaction must proceed in the forward direction (to the right), producing more B, until $Q$ rises to equal $K_{eq}$.$Q = 0.720 < K_{eq} = 4.00$。分子（产物）相对 $K_{eq}$ 偏小，故反应须向正向（向右）进行，产生更多 B，直至 $Q$ 升高等于 $K_{eq}$。

(c) When does $Q = K_{eq}$?何时 $Q = K_{eq}$？ A1

$Q = K_{eq}$ when the system has reached chemical equilibrium. At that point, the concentrations of all species have their equilibrium values and no net change occurs.当体系达到化学平衡时，$Q = K_{eq}$。此时所有组分的浓度均为其平衡值，不再发生净变化。
$Q$ vs. $K_{eq}$: the three-case rule.$Q$ 与 $K_{eq}$ 的三种情形规律。 This comparison is among the most frequently tested equilibrium skills. $Q < K_{eq}$: the reaction shifts forward (right) to make more products. $Q > K_{eq}$: the reaction shifts in reverse (left) to make more reactants. $Q = K_{eq}$: equilibrium; no net shift. A useful memory device: $Q$ needs to "grow" toward $K_{eq}$ when $Q < K_{eq}$ (more products needed, reaction goes right), and "shrink" when $Q > K_{eq}$ (fewer products needed, reaction goes left). The calculation in (a) requires writing the $Q$ expression before substituting. Always write the expression first on any exam to secure the method mark.这一比较是最常考的平衡技能之一。$Q < K_{eq}$：反应向正向（右）移动，产生更多产物。$Q > K_{eq}$：反应向逆向（左）移动，产生更多反应物。$Q = K_{eq}$：平衡，无净移动。一个有用的记忆方法：当 $Q < K_{eq}$ 时，$Q$ 需要"增大"向 $K_{eq}$ 靠拢（需要更多产物，反应向右）；当 $Q > K_{eq}$ 时，$Q$ 需要"减小"（需要更少产物，反应向左）。(a) 的计算要求先写出 $Q$ 的表达式再代入数值。在任何考试中，务必先写表达式以确保获得方法分。
Q4MEDIUM中 🇺🇸 US美 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Le Chatelier's principle勒沙特列原理 · SCH4U E3 [6 marks][6 分]

$2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$, $\Delta H = -198\ \text{kJ/mol}$. Predict shift and effect on $[\text{SO}_3]$ for (a) adding SO$_2$, (b) decreasing volume, (c) increasing temperature.$2\,\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\,\text{SO}_3\text{(g)}$，$\Delta H = -198\ \text{kJ/mol}$。对(a)加入 SO$_2$，(b)减小体积，(c)升温，各预测平衡移动方向及对 $[\text{SO}_3]$ 的影响。

Answer:答案：  (a) right; $[\text{SO}_3]$ increases右移；$[\text{SO}_3]$ 增大  ·  (b) right; $[\text{SO}_3]$ increases右移；$[\text{SO}_3]$ 增大  ·  (c) left; $[\text{SO}_3]$ decreases左移；$[\text{SO}_3]$ 减小

(a) Adding $\text{SO}_2$ at constant volume and temperature恒容恒温下加入 $\text{SO}_2$ M1·A1

Adding a reactant increases its concentration, making $Q < K_{eq}$. Le Chatelier's principle predicts the system responds by shifting right (forward) to consume the added SO$_2$ and restore equilibrium. The net effect is an increase in $[\text{SO}_3]$.加入反应物使其浓度增大，导致 $Q < K_{eq}$。勒沙特列原理预测体系通过右移（正向）来消耗加入的 SO$_2$ 以恢复平衡。净效果是 $[\text{SO}_3]$ 增大。

(b) Decreasing volume at constant temperature (pressure increase)恒温下减小体积（增大压强） M1·A1

Count moles of gas: left side = 2 + 1 = 3 mol gas; right side = 2 mol gas. Decreasing volume increases the total pressure. Le Chatelier's principle predicts a shift toward the side with fewer moles of gas, which is the product side (right). The equilibrium shifts right, so $[\text{SO}_3]$ increases.计算气体物质的量：左侧 = 2 + 1 = 3 mol 气体；右侧 = 2 mol 气体。减小体积使总压升高。勒沙特列原理预测平衡向气体物质的量较少的一侧（右侧）移动。平衡右移，$[\text{SO}_3]$ 增大。

(c) Increasing temperature (exothermic reaction)升温（放热反应） M1·A1

The forward reaction is exothermic ($\Delta H = -198\ \text{kJ/mol}$), so heat is a product. Treating heat as a product, adding more heat (by raising temperature) shifts the equilibrium left (toward reactants) to absorb the excess heat. The net effect is a decrease in $[\text{SO}_3]$.正向反应为放热反应（$\Delta H = -198\ \text{kJ/mol}$），热量是产物之一。将热量视为产物，升温（增加热量输入）使平衡向左移动（向反应物方向）以吸收多余热量。净效果是 $[\text{SO}_3]$ 减小。
Le Chatelier summary: the system always counteracts the applied stress.勒沙特列原理总结：体系总是对抗所施加的扰动。 Three types of stress appear on virtually every equilibrium exam: (1) concentration change, (2) pressure/volume change, and (3) temperature change. For concentration: adding a substance shifts away from it; removing a substance shifts toward it. For pressure (by volume change): shift toward the side with fewer moles of gas; if moles are equal on both sides, no shift. For temperature: exothermic forward reaction means higher temperature shifts left; endothermic forward reaction means higher temperature shifts right. Note that a catalyst does NOT shift equilibrium (it speeds up both directions equally). Also note that adding an inert (non-reactive) gas at constant volume does NOT shift equilibrium because the partial pressures of the reacting species do not change.几乎每道平衡考题都会涉及三类扰动：(1) 浓度变化，(2) 压强/体积变化，(3) 温度变化。对于浓度：加入某物质使平衡离开该侧；移除某物质使平衡向该侧移动。对于压强（通过改变体积）：向气体物质的量较少的一侧移动；若两侧气体物质的量相等，则不移动。对于温度：正向放热反应升温后向左移动；正向吸热反应升温后向右移动。注意：催化剂不会使平衡移动（它对正逆两个方向的加速程度相同）。恒容条件下加入惰性气体也不会使平衡移动，因为各反应组分的分压不变。
Q5MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2 $K_{eq}$ calculation$K_{eq}$ 计算 · Chem 30 Unit D Honors荣誉级 [6 marks][6 分]

$\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$ in $2.00\ \text{L}$. At equilibrium: $0.240\ \text{mol H}_2$, $0.240\ \text{mol I}_2$, $1.92\ \text{mol HI}$. (a) Equilibrium concentrations. (b) Calculate $K_{eq}$. (c) Interpret magnitude.$2.00\ \text{L}$ 容器中 $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\,\text{HI(g)}$。平衡时：$0.240\ \text{mol H}_2$，$0.240\ \text{mol I}_2$，$1.92\ \text{mol HI}$。(a) 平衡浓度。(b) 计算 $K_{eq}$。(c) 解读大小。

Answer:答案：  (a) $[\text{H}_2]=[\text{I}_2]=0.120\ \text{mol/L}$; $[\text{HI}]=0.960\ \text{mol/L}$  ·  (b) $K_{eq} = 64.0$  ·  (c) strongly favours products强烈偏向产物

(a) Calculate equilibrium concentrations计算平衡浓度 M1·A1

$$ [\text{H}_2] \;=\; \frac{0.240\ \text{mol}}{2.00\ \text{L}} \;=\; 0.120\ \text{mol/L}; \quad [\text{I}_2] \;=\; 0.120\ \text{mol/L}; \quad [\text{HI}] \;=\; \frac{1.92\ \text{mol}}{2.00\ \text{L}} \;=\; 0.960\ \text{mol/L.} $$

(b) Write the $K_{eq}$ expression and substitute写出 $K_{eq}$ 表达式并代入 M1·A1

$$ K_{eq} \;=\; \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \;=\; \frac{(0.960)^2}{(0.120)(0.120)} \;=\; \frac{0.9216}{0.0144} \;=\; 64.0. $$

(c) Interpret $K_{eq} = 64.0$解读 $K_{eq} = 64.0$ A1·A1

$K_{eq} = 64.0 \gg 1$ means the equilibrium strongly favours the products (HI). At equilibrium, the concentration of HI is far greater than those of H$_2$ and I$_2$. The reaction proceeds nearly to completion but not entirely, because $K_{eq}$ is large but finite.$K_{eq} = 64.0 \gg 1$ 意味着平衡强烈偏向产物（HI）。平衡时，HI 的浓度远大于 H$_2$ 和 I$_2$ 的浓度。反应接近完全进行，但非完全，因为 $K_{eq}$ 虽大但有限。
Interpreting $K_{eq}$: magnitude tells you which side is favoured at equilibrium.解读 $K_{eq}$：大小告诉你平衡偏向哪一侧。 The scale is roughly: $K_{eq} \gg 1$ (e.g., $> 10^3$) means essentially complete reaction; $K_{eq} \approx 1$ means significant concentrations of both reactants and products; $K_{eq} \ll 1$ (e.g., $< 10^{-3}$) means the reaction barely proceeds. $K_{eq} = 64.0$ sits in the "products strongly favoured" range but is not so large that the reverse reaction is negligible. Note: $K_{eq}$ is a dimensionless number (concentrations are divided by the standard state $1\ \text{mol/L}$), though in introductory courses the units are often omitted by convention. Also: $K_{eq}$ depends only on temperature, not on the initial amounts or the volume of the container.粗略的判断尺度：$K_{eq} \gg 1$（例如 $> 10^3$）意味着反应几乎完全进行；$K_{eq} \approx 1$ 意味着反应物和产物均有可观浓度；$K_{eq} \ll 1$（例如 $< 10^{-3}$）意味着反应几乎不进行。$K_{eq} = 64.0$ 处于"产物强烈被偏向"的范围，但不大到逆反应可以忽略的程度。注意：$K_{eq}$ 是无量纲数（浓度除以标准态 $1\ \text{mol/L}$），但在入门课程中惯例上省略单位。另外：$K_{eq}$ 仅取决于温度，与初始量或容器体积无关。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §3 + §4 $Q$ and Le Chatelier$Q$ 与勒沙特列 · HS-PS1-6 Honors荣誉级 [8 marks][8 分]

$\text{CO(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$, $K_{eq} = 3.92$ at $800\ \text{K}$. Initial: $[\text{CO}] = [\text{H}_2] = 0.200\ \text{mol/L}$, $[\text{CH}_4] = [\text{H}_2\text{O}] = 0.100\ \text{mol/L}$. (a) $K_{eq}$ expression. (b) Calculate $Q$. (c) Direction of shift. (d) Effect of halving the volume.$\text{CO(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$，$800\ \text{K}$ 时 $K_{eq} = 3.92$。初始：$[\text{CO}] = [\text{H}_2] = 0.200\ \text{mol/L}$，$[\text{CH}_4] = [\text{H}_2\text{O}] = 0.100\ \text{mol/L}$。(a) $K_{eq}$ 表达式。(b) 计算 $Q$。(c) 移动方向。(d) 体积减半的影响。

Answer:答案：  (a) $K_{eq}=\dfrac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$  ·  (b) $Q = 6.25$  ·  (c) shift left左移  ·  (d) shift right (toward fewer gas moles)右移（向气体物质的量较少一侧）

(a) $K_{eq}$ expression$K_{eq}$ 表达式 A1

$$ K_{eq} \;=\; \frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}. $$

(b) Calculate $Q$计算 $Q$ M1·A1·A1

$$ Q \;=\; \frac{(0.100)(0.100)}{(0.200)(0.200)^3} \;=\; \frac{0.0100}{(0.200)(0.00800)} \;=\; \frac{0.0100}{0.00160} \;=\; 6.25. $$

(c) Predict direction by comparing $Q$ and $K_{eq}$通过比较 $Q$ 与 $K_{eq}$ 预测方向 M1·A1

$Q = 6.25 > K_{eq} = 3.92$. The reaction quotient exceeds the equilibrium constant, meaning there are too many products relative to equilibrium. The reaction must shift left (reverse) to consume products and form more reactants until $Q$ decreases to $3.92$.$Q = 6.25 > K_{eq} = 3.92$。反应商超过平衡常数，说明相对于平衡态产物过多。反应须左移（逆向）消耗产物并形成更多反应物，直到 $Q$ 降至 $3.92$。

(d) Effect of halving the volume at constant temperature恒温下体积减半的影响 M1·A1

Counting moles of gas: left side has $1\ (\text{CO}) + 3\ (\text{H}_2) = 4\ \text{mol gas}$; right side has $1\ (\text{CH}_4) + 1\ (\text{H}_2\text{O}) = 2\ \text{mol gas}$. Halving the volume doubles the total pressure. Le Chatelier's principle predicts the system shifts toward the side with fewer moles of gas, which is the right (product) side. The equilibrium shifts right.计算气体物质的量：左侧 $1\ (\text{CO}) + 3\ (\text{H}_2) = 4\ \text{mol}$ 气体；右侧 $1\ (\text{CH}_4) + 1\ (\text{H}_2\text{O}) = 2\ \text{mol}$ 气体。体积减半使总压加倍。勒沙特列原理预测体系向气体物质的量较少一侧（右侧，即产物侧）移动。平衡右移。
The $[\text{H}_2]^3$ term dominates $Q$: small changes in $[\text{H}_2]$ have a large effect.$[\text{H}_2]^3$ 项主导 $Q$：$[\text{H}_2]$ 的小幅变化对 $Q$ 影响很大。 Notice that the denominator contains $[\text{H}_2]^3$. This means the reaction quotient is extremely sensitive to changes in the hydrogen concentration. Doubling $[\text{H}_2]$ would decrease $Q$ by a factor of $2^3 = 8$. This is why industrial methane-reforming plants carefully control H$_2$ partial pressure. For part (d), always count the moles of gas on each side of the balanced equation before applying the pressure-volume stress: it is the difference in moles of gas (not total moles) that determines the direction of the pressure shift. If both sides had equal moles of gas, a pressure change at constant temperature would cause no net shift (though concentrations would change).注意分母含有 $[\text{H}_2]^3$，这意味着反应商对氢气浓度的变化极为敏感。$[\text{H}_2]$ 加倍将使 $Q$ 缩小 $2^3 = 8$ 倍。这就是工业甲烷重整装置需要精确控制 H$_2$ 分压的原因。对于 (d)，在施加压强-体积扰动之前，务必先统计配平方程式两侧的气体物质的量：决定压强移动方向的是两侧气体物质的量之差（而非总物质的量）。若两侧气体物质的量相等，恒温下的压强变化不会引起净移动（尽管浓度会改变）。
Q7MEDIUM中 🇨🇦 ON安 🇨🇦 BC卑 ON Provincial-style安大略省考风格 §5 Solubility equilibria $K_{sp}$溶解平衡 $K_{sp}$ · SCH4U E3 Honors荣誉级 [7 marks][7 分]

$\text{BaSO}_4\text{(s)} \rightleftharpoons \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$, $K_{sp} = 1.1 \times 10^{-10}$. (a) $K_{sp}$ expression. (b) Molar solubility in pure water. (c) Molar solubility in $0.100\ \text{mol/L}\ \text{Na}_2\text{SO}_4$.$\text{BaSO}_4\text{(s)} \rightleftharpoons \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$，$K_{sp} = 1.1 \times 10^{-10}$。(a) $K_{sp}$ 表达式。(b) 纯水中摩尔溶解度。(c) 在 $0.100\ \text{mol/L}\ \text{Na}_2\text{SO}_4$ 中的摩尔溶解度。

Answer:答案：  (a) $K_{sp}=[\text{Ba}^{2+}][\text{SO}_4^{2-}]$  ·  (b) $s = 1.05 \times 10^{-5}\ \text{mol/L}$  ·  (c) less than (b); common-ion effect suppresses solubility小于 (b)；同离子效应抑制溶解度

(a) $K_{sp}$ expression$K_{sp}$ 表达式 A1

Pure solid BaSO$_4$ is omitted (activity = 1):纯固体 BaSO$_4$ 省略（活度 = 1）： $$ K_{sp} \;=\; [\text{Ba}^{2+}][\text{SO}_4^{2-}]. $$

(b) Molar solubility in pure water纯水中的摩尔溶解度 M1·A1·A1

Let $s$ = mol/L of BaSO$_4$ that dissolves. Each formula unit produces one Ba$^{2+}$ and one SO$_4^{2-}$, so $[\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = s$.设 $s$ = BaSO$_4$ 溶解量（mol/L）。每个化学式单元产生一个 Ba$^{2+}$ 和一个 SO$_4^{2-}$，故 $[\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = s$。 $$ s^2 \;=\; K_{sp} \;=\; 1.1 \times 10^{-10} \;\Longrightarrow\; s \;=\; \sqrt{1.1 \times 10^{-10}} \;=\; 1.05 \times 10^{-5}\ \text{mol/L.} $$

(c) Molar solubility in $0.100\ \text{mol/L}\ \text{Na}_2\text{SO}_4$在 $0.100\ \text{mol/L}\ \text{Na}_2\text{SO}_4$ 中的摩尔溶解度 M1·A1·A1

Na$_2$SO$_4$ dissolves completely, providing $[\text{SO}_4^{2-}]_{\text{initial}} = 0.100\ \text{mol/L}$. This is a common ion (SO$_4^{2-}$). By Le Chatelier's principle, the existing SO$_4^{2-}$ shifts the dissolving equilibrium to the left (toward undissolved BaSO$_4$), suppressing solubility. Therefore the molar solubility of BaSO$_4$ in this solution is less than in pure water ($1.05 \times 10^{-5}\ \text{mol/L}$). (Quantitatively: with $[\text{SO}_4^{2-}] \approx 0.100$, $s' = K_{sp}/0.100 = 1.1 \times 10^{-9}\ \text{mol/L}$, much smaller.)Na$_2$SO$_4$ 完全溶解，提供 $[\text{SO}_4^{2-}]_{\text{初}} = 0.100\ \text{mol/L}$。这是一个同离子（SO$_4^{2-}$）。根据勒沙特列原理，已有的 SO$_4^{2-}$ 使溶解平衡向左移动（趋向未溶解的 BaSO$_4$），抑制了溶解度。因此 BaSO$_4$ 在该溶液中的摩尔溶解度小于在纯水中的值（$1.05 \times 10^{-5}\ \text{mol/L}$）。（定量地：$[\text{SO}_4^{2-}] \approx 0.100$ 时，$s' = K_{sp}/0.100 = 1.1 \times 10^{-9}\ \text{mol/L}$，小得多。）
The common-ion effect: any shared ion suppresses solubility via Le Chatelier.同离子效应：任何共同离子都通过勒沙特列原理抑制溶解度。 When a sparingly soluble salt is dissolved in a solution already containing one of its ions, the solubility is dramatically reduced. This is the common-ion effect and it is a direct application of Le Chatelier's principle: the excess common ion shifts the dissolution equilibrium toward the undissolved solid. The quantitative drop is large: here from $1.05 \times 10^{-5}$ to $1.1 \times 10^{-9}\ \text{mol/L}$, a factor of roughly 10,000. This effect is used in analytical chemistry to drive precipitate formation to near-completion, and in water treatment to remove metal ions. The approximation $[\text{SO}_4^{2-}] \approx 0.100\ \text{mol/L}$ holds because $s' = 1.1 \times 10^{-9} \ll 0.100$.当微溶盐溶于已含有其某种离子的溶液时，溶解度大幅降低。这就是同离子效应，是勒沙特列原理的直接应用：过量的同离子使溶解平衡向未溶解固体方向移动。定量降幅很大：此处从 $1.05 \times 10^{-5}$ 降至 $1.1 \times 10^{-9}\ \text{mol/L}$，约下降了 10000 倍。该效应在分析化学中用于使沉淀几乎完全析出，在水处理中用于去除金属离子。因为 $s' = 1.1 \times 10^{-9} \ll 0.100$，近似 $[\text{SO}_4^{2-}] \approx 0.100\ \text{mol/L}$ 成立。
Q8HARD难Honors荣誉级 🇨🇦 ON安 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6 ICE tableICE 表格 · SCH4U E3 [8 marks][8 分]

$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$, $K_{eq} = 0.360$. $1.00\ \text{L}$ flask, initially $1.00\ \text{mol}$ pure $\text{N}_2\text{O}_4$. (a) ICE table. (b) Solve quadratic for $x$. (c) State equilibrium concentrations and verify.$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}$，$K_{eq} = 0.360$。$1.00\ \text{L}$ 烧瓶，初始 $1.00\ \text{mol}$ 纯 $\text{N}_2\text{O}_4$。(a) ICE 表格。(b) 求解 $x$ 的一元二次方程。(c) 写出平衡浓度并验证。

Answer:答案：  (b) $x \approx 0.258\ \text{mol/L}$  ·  (c) $[\text{N}_2\text{O}_4] = 0.742\ \text{mol/L}$; $[\text{NO}_2] = 0.516\ \text{mol/L}$; verified $K_{eq} \approx 0.360$

(a) Complete ICE table完整 ICE 表格 M1·A1

Initial concentrations: $[\text{N}_2\text{O}_4]_0 = 1.00/1.00 = 1.00\ \text{mol/L}$; $[\text{NO}_2]_0 = 0$. Let $x$ = mol/L of N$_2$O$_4$ that decomposes.初始浓度：$[\text{N}_2\text{O}_4]_0 = 1.00/1.00 = 1.00\ \text{mol/L}$；$[\text{NO}_2]_0 = 0$。设 $x$ = 分解的 N$_2$O$_4$ 浓度（mol/L）。
N$_2$O$_4$NO$_2$
I1.000
C$-x$$+2x$
E$1.00-x$$2x$

(b) Write and solve the quadratic写出并求解一元二次方程 M1·A1·A1·A1

$$ K_{eq} \;=\; \frac{(2x)^2}{1.00-x} \;=\; \frac{4x^2}{1.00-x} \;=\; 0.360. $$ $$ 4x^2 \;=\; 0.360(1.00-x) \;=\; 0.360 - 0.360x. $$ $$ 4x^2 + 0.360x - 0.360 \;=\; 0. $$ Applying the quadratic formula with $a = 4$, $b = 0.360$, $c = -0.360$:用求根公式，$a = 4$，$b = 0.360$，$c = -0.360$： $$ x \;=\; \frac{-0.360 \pm \sqrt{(0.360)^2 - 4(4)(-0.360)}}{2(4)} \;=\; \frac{-0.360 \pm \sqrt{0.1296 + 5.760}}{8} \;=\; \frac{-0.360 \pm \sqrt{5.8896}}{8}. $$ $$ \sqrt{5.8896} \;\approx\; 2.4269. $$ $$ x \;=\; \frac{-0.360 + 2.4269}{8} \;=\; \frac{2.0669}{8} \;\approx\; 0.2584\ \text{mol/L.} $$ (The negative root $x = (-0.360 - 2.4269)/8 < 0$ is rejected as unphysical.)（负根 $x = (-0.360 - 2.4269)/8 < 0$ 不合物理意义，舍去。）

(c) Equilibrium concentrations and verification平衡浓度及验证 M1·A1

$$ [\text{N}_2\text{O}_4]_{eq} \;=\; 1.00 - 0.2584 \;=\; 0.742\ \text{mol/L}. $$ $$ [\text{NO}_2]_{eq} \;=\; 2(0.2584) \;=\; 0.517\ \text{mol/L.} $$ Verify:验证： $$ K_{eq} \;=\; \frac{(0.517)^2}{0.742} \;=\; \frac{0.2673}{0.742} \;\approx\; 0.360. \checkmark $$
ICE tables + quadratic formula: the standard method for any equilibrium calculation from scratch.ICE 表格 + 求根公式：从初始条件计算平衡的标准方法。 The ICE table is the scaffold: it systematically tracks the change in each species. The key sign-check: N$_2$O$_4$ decreases ($-x$) and NO$_2$ increases ($+2x$) in a 1:2 stoichiometric ratio, matching the balanced equation. Always reject the negative root for $x$ because a negative concentration change would mean producing reactants before any product exists, which violates the physical setup. After solving, verify by back-substitution: plugging the equilibrium values back into the $K_{eq}$ expression should return the given $K_{eq}$ to within rounding. A small discrepancy (here $0.360$ vs. $0.360$) is acceptable; a discrepancy of more than 1-2% indicates an algebra error.ICE 表格是支架：它系统地追踪每种组分的变化。关键符号检查：N$_2$O$_4$ 减少（$-x$），NO$_2$ 增加（$+2x$），符合 1:2 的化学计量比，与配平方程式一致。务必舍去 $x$ 的负根，因为负的浓度变化意味着在产物存在之前反应物反而增加，不符合物理情形。求解后用回代验证：将平衡值代回 $K_{eq}$ 表达式应还原给定的 $K_{eq}$（在取整范围内）。小的偏差（此处 $0.360$ vs. $0.360$）可接受；超过 1-2% 的偏差提示代数运算有误。
Q9HARD难Honors荣誉级 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §5 + §3 $K_{sp}$ and precipitation$K_{sp}$ 与沉淀判断 · Chem 30 Unit D [7 marks][7 分]

$\text{PbCl}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$, $K_{sp} = 1.7 \times 10^{-5}$. (a) $K_{sp}$ expression. (b) Molar solubility. (c) Mix equal volumes of $0.100\ \text{mol/L}\ \text{Pb(NO}_3)_2$ and $0.060\ \text{mol/L}\ \text{NaCl}$; does PbCl$_2$ precipitate?$\text{PbCl}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\,\text{Cl}^-\text{(aq)}$，$K_{sp} = 1.7 \times 10^{-5}$。(a) $K_{sp}$ 表达式。(b) 摩尔溶解度。(c) 等体积混合 $0.100\ \text{mol/L}\ \text{Pb(NO}_3)_2$ 与 $0.060\ \text{mol/L}\ \text{NaCl}$；PbCl$_2$ 是否沉淀？

Answer:答案：  (a) $K_{sp}=[\text{Pb}^{2+}][\text{Cl}^-]^2$  ·  (b) $s \approx 0.0162\ \text{mol/L}$  ·  (c) $Q_{sp} = 4.5 \times 10^{-5} > K_{sp}$; yes, PbCl$_2$ precipitates是，PbCl$_2$ 会沉淀

(a) $K_{sp}$ expression$K_{sp}$ 表达式 A1

$$ K_{sp} \;=\; [\text{Pb}^{2+}][\text{Cl}^-]^2. $$

(b) Molar solubility of $\text{PbCl}_2$ in pure waterPbCl$_2$ 在纯水中的摩尔溶解度 M1·A1·A1

Let $s$ = mol/L dissolved. Each formula unit gives one Pb$^{2+}$ and two Cl$^-$, so $[\text{Pb}^{2+}] = s$ and $[\text{Cl}^-] = 2s$.设 $s$ = 溶解量（mol/L）。每个化学式单元产生一个 Pb$^{2+}$ 和两个 Cl$^-$，故 $[\text{Pb}^{2+}] = s$，$[\text{Cl}^-] = 2s$。 $$ K_{sp} \;=\; (s)(2s)^2 \;=\; 4s^3 \;=\; 1.7 \times 10^{-5}. $$ $$ s^3 \;=\; \frac{1.7 \times 10^{-5}}{4} \;=\; 4.25 \times 10^{-6}. $$ $$ s \;=\; \sqrt[3]{4.25 \times 10^{-6}} \;\approx\; 0.01624 \;\approx\; 0.0162\ \text{mol/L.} $$

(c) Mixing equal volumes: calculate $Q_{sp}$ and compare to $K_{sp}$等体积混合：计算 $Q_{sp}$ 并与 $K_{sp}$ 比较 M1·A1·A1

When equal volumes are mixed, each concentration is halved:等体积混合时，每种浓度减半： $$ [\text{Pb}^{2+}]_{\text{mix}} \;=\; \frac{0.100}{2} \;=\; 0.0500\ \text{mol/L}; \qquad [\text{Cl}^-]_{\text{mix}} \;=\; \frac{0.060}{2} \;=\; 0.030\ \text{mol/L.} $$ $$ Q_{sp} \;=\; [\text{Pb}^{2+}][\text{Cl}^-]^2 \;=\; (0.0500)(0.030)^2 \;=\; (0.0500)(9.0 \times 10^{-4}) \;=\; 4.5 \times 10^{-5}. $$ Since $Q_{sp} = 4.5 \times 10^{-5} > K_{sp} = 1.7 \times 10^{-5}$, the ion product exceeds the solubility limit. PbCl$_2$ precipitates from the mixed solution.因为 $Q_{sp} = 4.5 \times 10^{-5} > K_{sp} = 1.7 \times 10^{-5}$，离子积超过溶解度上限。PbCl$_2$ 从混合溶液中沉淀析出。
Precipitation rule: if $Q_{sp} > K_{sp}$, the solution is supersaturated and precipitation occurs.沉淀规则：若 $Q_{sp} > K_{sp}$，溶液过饱和，沉淀析出。 The ion product $Q_{sp}$ plays the same role as the reaction quotient $Q$: compare it to $K_{sp}$ to determine the direction. Three outcomes: $Q_{sp} < K_{sp}$: unsaturated, more solid can dissolve; $Q_{sp} = K_{sp}$: exactly saturated, equilibrium; $Q_{sp} > K_{sp}$: supersaturated, precipitation occurs until $Q_{sp}$ drops to $K_{sp}$. The factor-of-2 dilution when mixing equal volumes is the most frequently forgotten step on diploma exams. The Cl$^-$ exponent of 2 in $K_{sp}$ is critical: even a small increase in $[\text{Cl}^-]$ has a large effect because it appears squared. This is why sea water (high Cl$^-$) has very low dissolved lead concentrations.离子积 $Q_{sp}$ 与反应商 $Q$ 的作用相同：与 $K_{sp}$ 比较来判断方向。三种结果：$Q_{sp} < K_{sp}$：不饱和，可以溶解更多固体；$Q_{sp} = K_{sp}$：恰好饱和，处于平衡；$Q_{sp} > K_{sp}$：过饱和，沉淀析出直至 $Q_{sp}$ 降至 $K_{sp}$。等体积混合时浓度减半是毕业考上最常被遗忘的步骤。$K_{sp}$ 中 Cl$^-$ 的指数为 2 非常关键：即使 $[\text{Cl}^-]$ 小幅增大，效果也很大，因为它以平方出现。这就是为什么海水（高 Cl$^-$）中溶解的铅浓度极低。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解Haber process + ICE calculations · 26 marks哈伯法 + ICE 定量计算 · 共 26 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇺🇸 US美 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §7 Haber process哈伯法 · Chem 30 Unit D [8 marks][8 分]

$\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$, $\Delta H = -92\ \text{kJ/mol}$. Industrial plants: $400-500\ ^\circ\text{C}$, $150-300\ \text{atm}$, Fe catalyst. (a) Effect of increasing pressure on NH$_3$ yield. (b) Why high temperature despite being exothermic. (c) Why catalyst does not change $K_{eq}$.$\text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)}$，$\Delta H = -92\ \text{kJ/mol}$。工业条件：$400-500\ ^\circ\text{C}$，$150-300\ \text{atm}$，铁催化剂。(a) 增压对 NH$_3$ 产率的影响。(b) 放热反应为何在高温下进行。(c) 催化剂为何不改变 $K_{eq}$。

Answer:答案：  (a) yield increases (shift right)产率增大（右移）  ·  (b) kinetics: too slow at room temperature; high $T$ is a rate-yield compromise动力学：室温下反应太慢；高温是速率与产率的折中  ·  (c) catalyst lowers activation energy equally for both directions; thermodynamics unchanged催化剂对正逆两个方向同等降低活化能；热力学不变

(a) Effect of increasing pressure增大压强的影响 M1·A1·A1

Count moles of gas: left side = $1\ (\text{N}_2) + 3\ (\text{H}_2) = 4\ \text{mol gas}$; right side = $2\ (\text{NH}_3) = 2\ \text{mol gas}$. Increasing pressure shifts the equilibrium toward the side with fewer moles of gas, which is the right (product) side. Therefore, increasing pressure increases the yield of NH$_3$. This is why industrial Haber plants operate at 150-300 atm.计算气体物质的量：左侧 = $1\ (\text{N}_2) + 3\ (\text{H}_2) = 4\ \text{mol}$ 气体；右侧 = $2\ (\text{NH}_3) = 2\ \text{mol}$ 气体。增大压强使平衡向气体物质的量较少一侧（右侧，即产物侧）移动。因此，增大压强可提高 NH$_3$ 的产率。这就是工业哈伯装置在 150-300 atm 下运行的原因。

(b) Thermodynamics vs. kinetics: the compromise temperature热力学与动力学的矛盾：折中温度 M1·A1·A1

Thermodynamic argument: the forward reaction is exothermic, so lower temperature favours the products (Le Chatelier shifts right). At room temperature, $K_{eq}$ is very large and the equilibrium strongly favours NH$_3$. Kinetic argument: at low temperatures, the activation energy barrier is rarely overcome, and the reaction rate is extremely slow. In practice, it would take years to reach equilibrium at room temperature even with a catalyst. The industrial compromise of $400-500\ ^\circ\text{C}$ gives a rate fast enough to be commercially viable (equilibrium reached in minutes to hours) while still producing an acceptable NH$_3$ yield (typically 15-25%). Higher temperature gives faster rate but lower yield; lower temperature gives higher yield but impractically slow rate.热力学角度：正向反应为放热反应，较低温度有利于产物（勒沙特列向右移动）。在室温下，$K_{eq}$ 非常大，平衡强烈偏向 NH$_3$。动力学角度：温度低时，活化能壁垒极少被克服，反应速率极慢。实际上，即使有催化剂，室温下达到平衡可能需要数年。工业折中条件 $400-500\ ^\circ\text{C}$ 提供了足够快的速率（几分钟到几小时内达到平衡），同时仍能产生可接受的 NH$_3$ 产率（通常为 15-25%）。温度越高速率越快但产率越低；温度越低产率越高但速率低得无法实用。

(c) Why a catalyst does not change $K_{eq}$催化剂为何不改变 $K_{eq}$ M1·A1

A catalyst provides an alternative reaction pathway with a lower activation energy. Critically, it lowers the activation energy for both the forward and the reverse reactions by the same amount. The thermodynamic quantities ($\Delta G$, $\Delta H$, $\Delta S$) and the equilibrium position are unchanged. Since $K_{eq}$ is determined by thermodynamics (the relative stability of reactants and products at a given temperature), not by kinetics, the catalyst has no effect on $K_{eq}$ or on the equilibrium concentrations. It only determines how quickly equilibrium is reached.催化剂提供了一条活化能较低的替代反应路径。关键在于，它对正向和逆向反应的活化能降低量相同。热力学量（$\Delta G$、$\Delta H$、$\Delta S$）和平衡位置不变。由于 $K_{eq}$ 由热力学（给定温度下反应物和产物的相对稳定性）决定，而非由动力学决定，催化剂对 $K_{eq}$ 或平衡浓度没有影响。它只决定达到平衡的快慢。
The Haber process is the classic example of a thermodynamics-kinetics trade-off in industrial chemistry.哈伯法是工业化学中热力学与动力学权衡的经典案例。 Three levers are pulled simultaneously: (1) very high pressure (150-300 atm) to exploit the mole-imbalance and maximise yield; (2) moderate temperature (400-500 C) as a compromise between thermodynamic yield (favours low T) and reaction rate (favours high T); (3) Fe catalyst to achieve an acceptable rate at the chosen temperature. The catalyst is recycled because it is not consumed. This process produces about 150 million tonnes of ammonia per year for agricultural fertilisers, feeding roughly half the world's population. The Haber process consumes about 1-2% of global energy output. Understanding the interplay of $K_{eq}$, rate, and industrial constraints is central to AP and IB exam questions on industrial equilibrium.同时调节三个杠杆：(1) 极高压强（150-300 atm）以利用物质的量不平衡并最大化产率；(2) 适中温度（400-500 C）作为热力学产率（偏好低温）与反应速率（偏好高温）的折中；(3) 铁催化剂以在选定温度下实现可接受的反应速率。催化剂可循环使用，因为它不被消耗。该工艺每年生产约 1.5 亿吨氨用于农业化肥，养活了全球约一半的人口。哈伯法消耗全球约 1-2% 的能源。理解 $K_{eq}$、速率和工业约束条件之间的相互作用是 AP 和 IB 工业平衡考题的核心。
Q11MEDIUM中 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §6 ICE table (weak acid)ICE 表格（弱酸） · SCH4U E3 Honors荣誉级 [9 marks][9 分]

$\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$, $K_a = 1.8 \times 10^{-5}$. $0.100\ \text{mol/L}$ solution. (a) ICE table. (b) Solve for $[\text{H}^+]$ using approximation. (c) Calculate pH. (d) Validate approximation.$\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$，$K_a = 1.8 \times 10^{-5}$。$0.100\ \text{mol/L}$ 溶液。(a) ICE 表格。(b) 用近似求 $[\text{H}^+]$。(c) 计算 pH。(d) 验证近似。

Answer:答案：  (b) $[\text{H}^+] = 1.34 \times 10^{-3}\ \text{mol/L}$  ·  (c) $\text{pH} \approx 2.87$  ·  (d) $1.34\%\,<\,5\%$; valid$1.34\%\,<\,5\%$；近似有效

(a) ICE table for acetic acid ionization乙酸电离的 ICE 表格 M1·A1

Let $x = [\text{H}^+]$ formed at equilibrium.设 $x = $ 平衡时生成的 $[\text{H}^+]$。
CH$_3$COOHH$^+$CH$_3$COO$^-$
I0.10000
C$-x$$+x$$+x$
E$0.100-x$$x$$x$

(b) Apply approximation and solve for $x$应用近似并求解 $x$ M1·A1·A1

$$ K_a \;=\; \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \;=\; \frac{x \cdot x}{0.100 - x}. $$ Approximation: since $K_a = 1.8 \times 10^{-5} \ll 0.100$, we assume $x \ll 0.100$, so $0.100 - x \approx 0.100$:近似：因为 $K_a = 1.8 \times 10^{-5} \ll 0.100$，假设 $x \ll 0.100$，故 $0.100 - x \approx 0.100$： $$ \frac{x^2}{0.100} \;\approx\; 1.8 \times 10^{-5} \;\Longrightarrow\; x^2 \;=\; 1.8 \times 10^{-6} \;\Longrightarrow\; x \;=\; \sqrt{1.8 \times 10^{-6}} \;=\; 1.342 \times 10^{-3}\ \text{mol/L.} $$

(c) Calculate pH计算 pH M1·A1

$$ \text{pH} \;=\; -\log[\text{H}^+] \;=\; -\log(1.342 \times 10^{-3}) \;=\; 3 - \log(1.342) \;=\; 3 - 0.1277 \;\approx\; 2.87. $$

(d) Validate the approximation验证近似 M1·A1

$$ \frac{x}{[\text{HA}]_0} \times 100\% \;=\; \frac{1.342 \times 10^{-3}}{0.100} \times 100\% \;=\; 1.34\%. $$ Since $1.34\% < 5\%$, the approximation is valid. The error introduced by assuming $0.100 - x \approx 0.100$ is less than 5%, which is acceptable for a two-significant-figure result.因为 $1.34\% < 5\%$，近似有效。假设 $0.100 - x \approx 0.100$ 引入的误差小于 5%，对于两位有效数字的结果可以接受。
Weak-acid ICE: the 5% approximation test is mandatory, not optional.弱酸 ICE：5% 近似检验是必要步骤，不可省略。 Every weak-acid pH calculation proceeds in the same four steps: (1) write the ICE table; (2) write the $K_a$ expression; (3) apply the approximation $[\text{HA}]_0 - x \approx [\text{HA}]_0$ to get $x = \sqrt{K_a \cdot C_0}$; (4) validate using the 5% rule. If the approximation fails (percent ionisation $> 5\%$), you must solve the full quadratic. For acetic acid at $0.100\ \text{mol/L}$, the 1.34% ionisation confirms the approximation is excellent. Also note: a weak acid at $\text{pH} = 2.87$ is much less acidic than a strong acid at the same concentration (pH $= 1.00$ for $0.100\ \text{mol/L}$ HCl). This pH difference ($1.87$ units, a factor of $74\times$ in $[\text{H}^+]$) is the quantitative definition of weak vs. strong acid at this concentration.每道弱酸 pH 计算都按相同的四步进行：(1) 写 ICE 表格；(2) 写 $K_a$ 表达式；(3) 用近似 $[\text{HA}]_0 - x \approx [\text{HA}]_0$ 得 $x = \sqrt{K_a \cdot C_0}$；(4) 用 5% 规则验证。若近似不成立（电离度 $> 5\%$），则必须解完整的一元二次方程。对于 $0.100\ \text{mol/L}$ 乙酸，$1.34\%$ 的电离度证明近似非常精确。另外注意：pH = 2.87 的弱酸远没有同浓度的强酸酸性强（$0.100\ \text{mol/L}$ HCl 的 pH = 1.00）。这一 pH 差值（$1.87$ 个单位，$[\text{H}^+]$ 相差 $74$ 倍）正是在该浓度下弱酸与强酸的定量区别。
Q12HARD难Honors荣誉级 🇨🇦 BC卑 🇨🇦 AB阿 BC Provincial-style卑诗省考风格 §5 $K_{sp}$ + common-ion effect$K_{sp}$ + 同离子效应 · Chem 12 [9 marks][9 分]

$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$, $K_{sp} = 1.8 \times 10^{-10}$. (a) Molar solubility in pure water. (b) Molar solubility in $0.0200\ \text{mol/L}\ \text{AgNO}_3$. (c) Compare and name the phenomenon.$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$，$K_{sp} = 1.8 \times 10^{-10}$。(a) 纯水中摩尔溶解度。(b) 在 $0.0200\ \text{mol/L}\ \text{AgNO}_3$ 中的摩尔溶解度。(c) 比较并命名该现象。

Answer:答案：  (a) $s = 1.34 \times 10^{-5}\ \text{mol/L}$  ·  (b) $s' = 9.0 \times 10^{-9}\ \text{mol/L}$  ·  (c) common-ion effect; Le Chatelier's principle同离子效应；勒沙特列原理

(a) Molar solubility of AgCl in pure waterAgCl 在纯水中的摩尔溶解度 M1·A1

Let $s$ = mol/L of AgCl that dissolves. $[\text{Ag}^+] = [\text{Cl}^-] = s$.设 $s$ = AgCl 溶解量（mol/L）。$[\text{Ag}^+] = [\text{Cl}^-] = s$。 $$ K_{sp} \;=\; s^2 \;=\; 1.8 \times 10^{-10} \;\Longrightarrow\; s \;=\; \sqrt{1.8 \times 10^{-10}} \;=\; 1.34 \times 10^{-5}\ \text{mol/L.} $$

(b) Molar solubility of AgCl in $0.0200\ \text{mol/L}\ \text{AgNO}_3$AgCl 在 $0.0200\ \text{mol/L}\ \text{AgNO}_3$ 中的摩尔溶解度 M1·A1·A1·A1

AgNO$_3$ dissociates completely: $[\text{Ag}^+]_{\text{initial}} = 0.0200\ \text{mol/L}$. Let $s'$ = mol/L of AgCl that dissolves. Then $[\text{Ag}^+] = 0.0200 + s'$ and $[\text{Cl}^-] = s'$.AgNO$_3$ 完全电离：$[\text{Ag}^+]_{\text{初}} = 0.0200\ \text{mol/L}$。设 $s'$ = AgCl 溶解量（mol/L）。则 $[\text{Ag}^+] = 0.0200 + s'$，$[\text{Cl}^-] = s'$。 $$ K_{sp} \;=\; (0.0200 + s') \cdot s' \;=\; 1.8 \times 10^{-10}. $$ Approximation: $s' \ll 0.0200$ (justified below), so $0.0200 + s' \approx 0.0200$:近似：$s' \ll 0.0200$（见下方验证），故 $0.0200 + s' \approx 0.0200$： $$ 0.0200 \cdot s' \;\approx\; 1.8 \times 10^{-10} \;\Longrightarrow\; s' \;=\; \frac{1.8 \times 10^{-10}}{0.0200} \;=\; 9.0 \times 10^{-9}\ \text{mol/L.} $$ Justification: $s' = 9.0 \times 10^{-9} \ll 0.0200$, so the approximation is valid.验证：$s' = 9.0 \times 10^{-9} \ll 0.0200$，近似有效。

(c) Compare solubilities; name and explain the phenomenon比较溶解度；命名并解释该现象 M1·A1·A1

Pure water: $s = 1.34 \times 10^{-5}\ \text{mol/L}$. AgNO$_3$ solution: $s' = 9.0 \times 10^{-9}\ \text{mol/L}$. The solubility decreased by a factor of about 1500. This illustrates the common-ion effect. The general principle it follows is Le Chatelier's principle: the added Ag$^+$ (a common ion) shifts the dissolving equilibrium $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ to the left, suppressing dissolution and drastically reducing solubility.纯水中：$s = 1.34 \times 10^{-5}\ \text{mol/L}$。AgNO$_3$ 溶液中：$s' = 9.0 \times 10^{-9}\ \text{mol/L}$。溶解度降低了约 1500 倍。这说明了同离子效应。它遵循的一般性原理是勒沙特列原理：加入的 Ag$^+$（同离子）使溶解平衡 $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ 向左移动，抑制溶解，使溶解度大幅降低。
Common-ion effect: the quantitative face of Le Chatelier applied to solubility.同离子效应：勒沙特列原理在溶解度上的定量体现。 The result here is striking: a factor of about 1500 reduction in solubility from adding only 0.0200 mol/L of a common ion. This is possible because $K_{sp}$ is a product of ion concentrations, so the pre-existing $[\text{Ag}^+]$ of 0.0200 mol/L forces $[\text{Cl}^-]$ to be tiny to keep the product at $1.8 \times 10^{-10}$. The common-ion effect has major practical applications: (1) in analytical chemistry, adding excess precipitating agent drives precipitation to near-completion for gravimetric analysis; (2) in pharmacology, the solubility of ionic drugs changes in biological fluids containing the same ions; (3) in water treatment, controlling ion concentrations prevents unwanted precipitation or dissolves existing scale. The key formula to remember: when the common ion has initial concentration $c_0 \gg s'$, then $s' \approx K_{sp}/c_0$.此处结果令人印象深刻：仅加入 $0.0200\ \text{mol/L}$ 的同离子就使溶解度降低了约 1500 倍。这是可能的，因为 $K_{sp}$ 是离子浓度的乘积：已有的 $[\text{Ag}^+] = 0.0200\ \text{mol/L}$ 迫使 $[\text{Cl}^-]$ 极小以维持乘积等于 $1.8 \times 10^{-10}$。同离子效应有重要的实际应用：(1) 在分析化学中，加入过量沉淀剂使沉淀接近完全，用于重量分析；(2) 在药学中，离子型药物在含有相同离子的生物流体中溶解度会改变；(3) 在水处理中，控制离子浓度可防止不需要的沉淀或溶解已有的水垢。需记住的关键公式：当同离子的初始浓度 $c_0 \gg s'$ 时，$s' \approx K_{sp}/c_0$。