PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 24 marksAP 风格选择题 + 安/卑省考短答 · 共 24 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter and show enough working in the margin. For short-answer items, state units where applicable. Use $K_w = 1.0 \times 10^{-14}$ at 25 °C and $\text{pH} + \text{pOH} = 14.00$ throughout. Calculator permitted on Q3-Q5; Q1-Q2 no calculator.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写出足够的步骤。短答题需注明单位。全卷取 $K_w = 1.0 \times 10^{-14}$（25 °C），$\text{pH} + \text{pOH} = 14.00$。Q3-Q5 可用计算器；Q1-Q2 不可用。
In the reaction $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+$, which species is the Bronsted-Lowry acid in the forward direction?在反应 $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+$ 中，正向反应的布朗斯特-劳里酸是哪种物质？
A solution has $[\text{H}^+] = 0.010\ \text{mol/L}$ at 25 °C. What is the pH of this solution?某溶液在 25 °C 时 $[\text{H}^+] = 0.010\ \text{mol/L}$。该溶液的 pH 是多少？
Two solutions have the same molar concentration: $0.10\ \text{mol/L}\ \text{HCl}$ and $0.10\ \text{mol/L}\ \text{CH}_3\text{COOH}$ (acetic acid, $K_a = 1.8 \times 10^{-5}$).两种溶液的摩尔浓度相同：$0.10\ \text{mol/L}\ \text{HCl}$ 和 $0.10\ \text{mol/L}\ \text{CH}_3\text{COOH}$（乙酸，$K_a = 1.8 \times 10^{-5}$）。
(a)State the approximate $[\text{H}^+]$ in each solution and explain which has the lower pH.分别写出两种溶液中 $[\text{H}^+]$ 的近似值，并解释哪种溶液的 pH 更低。[2]
(b)Explain why HCl is classified as a strong acid while $\text{CH}_3\text{COOH}$ is classified as a weak acid.解释为什么 HCl 被归类为强酸，而 $\text{CH}_3\text{COOH}$ 被归类为弱酸。[2]
A student titrates a $25.00\ \text{mL}$ sample of hydrochloric acid with $0.150\ \text{mol/L}$ NaOH. The endpoint is reached after adding $31.50\ \text{mL}$ of NaOH.一名学生用 $0.150\ \text{mol/L}$ NaOH 滴定 $25.00\ \text{mL}$ 的盐酸样品。加入 $31.50\ \text{mL}$ NaOH 后到达终点。
(a)Calculate the moles of NaOH used.计算所用 NaOH 的物质的量。[2]
(b)Calculate the concentration of HCl in the original sample.计算原样品中 HCl 的浓度。[2]
(c)State the pH at the equivalence point and identify the appropriate indicator class (colour change at pH near 7).写出终点处的 pH，并指明应选用的指示剂类别（在 pH 接近 7 时变色）。[2]
(d)Identify one source of error in this titration and state how it would affect the calculated HCl concentration.指出该滴定中一个误差来源，并说明其如何影响计算所得的 HCl 浓度。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 33 marksAP 衔接简答题 + 荣誉级 · 共 33 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of your reasoning. Define all symbols and state units in every final answer. Use $K_w = 1.0 \times 10^{-14}$ at 25 °C. For logarithm calculations, retain at least 2 decimal places. Calculator permitted throughout Part II.每一步推理都要写出。定义所有符号并在最终答案中写出单位。取 $K_w = 1.0 \times 10^{-14}$（25 °C）。对数计算保留至少 2 位小数。第二部分全程可用计算器。
Answer each part independently. All solutions are at 25 °C.各小问独立作答，均在 25 °C 下。
(a)A solution has $[\text{OH}^-] = 5.0 \times 10^{-3}\ \text{mol/L}$. Find the pOH, the pH, and classify the solution as acidic, basic, or neutral.某溶液 $[\text{OH}^-] = 5.0 \times 10^{-3}\ \text{mol/L}$。求 pOH、pH，并判断溶液的酸碱性。[3]
(b)A solution has $\text{pH} = 4.50$. Find $[\text{H}^+]$ and $[\text{OH}^-]$.某溶液 $\text{pH} = 4.50$。求 $[\text{H}^+]$ 和 $[\text{OH}^-]$。[2]
(c)A solution has $[\text{H}^+] = 2.5 \times 10^{-11}\ \text{mol/L}$. Find the pH and pOH and state whether the solution is acidic, basic, or neutral.某溶液 $[\text{H}^+] = 2.5 \times 10^{-11}\ \text{mol/L}$。求 pH 和 pOH，并判断酸碱性。[3]
A student titrates $20.00\ \text{mL}$ of $0.250\ \text{mol/L}\ \text{NaOH}$ with $0.100\ \text{mol/L}\ \text{HCl}$.一名学生用 $0.100\ \text{mol/L}\ \text{HCl}$ 滴定 $20.00\ \text{mL}$ 的 $0.250\ \text{mol/L}\ \text{NaOH}$。
(a)Calculate the moles of NaOH in the flask.计算锥形瓶中 NaOH 的物质的量。[1]
(b)Calculate the volume of HCl solution required to reach the equivalence point.计算到达等当点所需 HCl 溶液的体积。[3]
(c)State the pH at the equivalence point and explain why.写出等当点的 pH 并解释原因。[2]
(d)If only $40.00\ \text{mL}$ of HCl is added (not the full equivalence volume), describe what is in solution and whether the pH is above, below, or equal to 7.若仅加入 $40.00\ \text{mL}$ HCl（未到等当点体积），描述溶液中存在什么物质，并说明 pH 高于、低于还是等于 7。[2]
A buffer solution contains $0.10\ \text{mol/L}$ acetic acid $(\text{CH}_3\text{COOH})$ and $0.15\ \text{mol/L}$ sodium acetate $(\text{CH}_3\text{COONa})$. The $K_a$ of acetic acid is $1.8 \times 10^{-5}$ ($\text{p}K_a = 4.74$).某缓冲溶液含 $0.10\ \text{mol/L}$ 乙酸 $(\text{CH}_3\text{COOH})$ 和 $0.15\ \text{mol/L}$ 乙酸钠 $(\text{CH}_3\text{COONa})$。乙酸的 $K_a = 1.8 \times 10^{-5}$（$\text{p}K_a = 4.74$）。
(a)Using the Henderson-Hasselbalch equation, calculate the pH of the buffer.用 Henderson-Hasselbalch 方程计算该缓冲溶液的 pH。[3]
(b)If $0.010\ \text{mol}$ of HCl is added to $1.00\ \text{L}$ of this buffer, calculate the new pH. (Assume the volume change is negligible.)若将 $0.010\ \text{mol}$ HCl 加入 $1.00\ \text{L}$ 该缓冲溶液中，计算新的 pH（假设体积变化可忽略）。[4]
(c)Explain in terms of Le Chatelier's principle why the buffer resists the pH change when HCl is added.用勒夏特列原理解释为什么加入 HCl 后缓冲溶液能抵抗 pH 变化。[2]
Q9HARD难Honors荣誉级🇺🇸 US美AP-feeder FRQAP 衔接简答题§7 Ka / Kb and conjugate pairsKa / Kb 与共轭酸碱对 · HS-PS1-6 (Honors)（荣誉）[8 marks][8 分]
A weak acid HA has $K_a = 2.0 \times 10^{-5}$ at 25 °C.弱酸 HA 在 25 °C 时 $K_a = 2.0 \times 10^{-5}$。
(a)Write the ionization equilibrium expression for HA in water and the expression for $K_a$.写出 HA 在水中的电离平衡方程式及 $K_a$ 的表达式。[2]
(b)Calculate $K_b$ for the conjugate base $\text{A}^-$ and find p$K_b$.计算共轭碱 $\text{A}^-$ 的 $K_b$，并求 p$K_b$。[3]
(c)Explain what it means for a weak acid to have a larger $K_a$ value, and state whether $\text{A}^-$ is a stronger or weaker base compared with a conjugate base of an acid with $K_a = 1.0 \times 10^{-3}$.解释弱酸 $K_a$ 较大意味着什么，并说明与 $K_a = 1.0 \times 10^{-3}$ 的酸对应的共轭碱相比，$\text{A}^-$ 是更强还是更弱的碱。[3]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all symbols with units at the start of each question. Show all equilibrium or stoichiometry steps clearly. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义所有符号（含单位）。清楚写出全部平衡或化学计量步骤。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
A chemist mixes $50.0\ \text{mL}$ of $0.200\ \text{mol/L}\ \text{H}_2\text{SO}_4$ (a strong diprotic acid) with $50.0\ \text{mL}$ of $0.150\ \text{mol/L}\ \text{NaOH}$.一名化学师将 $50.0\ \text{mL}$ 的 $0.200\ \text{mol/L}\ \text{H}_2\text{SO}_4$（强二元酸）与 $50.0\ \text{mL}$ 的 $0.150\ \text{mol/L}\ \text{NaOH}$ 混合。
(a)Calculate the total moles of $\text{H}^+$ from $\text{H}_2\text{SO}_4$ and the moles of $\text{OH}^-$ from NaOH.计算 $\text{H}_2\text{SO}_4$ 提供的 $\text{H}^+$ 总物质的量和 NaOH 提供的 $\text{OH}^-$ 物质的量。[2]
(b)Determine the excess reagent and the moles in excess after the neutralization.确定过量试剂及中和后的过量物质的量。[2]
(c)Calculate $[\text{H}^+]$ in the final mixture and hence calculate the pH.计算最终混合溶液中的 $[\text{H}^+]$，进而计算 pH。[3]
(d)State one-sentence conclusion about whether the mixture is acidic, basic, or neutral.用一句话说明混合溶液是酸性、碱性还是中性。[1]
Acetic acid $(\text{CH}_3\text{COOH},\ K_a = 1.8 \times 10^{-5})$ is dissolved in water to give a $0.100\ \text{mol/L}$ solution at 25 °C.乙酸 $(\text{CH}_3\text{COOH},\ K_a = 1.8 \times 10^{-5})$ 溶于水，在 25 °C 时配制成 $0.100\ \text{mol/L}$ 溶液。
(a)Write the ICE table for the ionization equilibrium of acetic acid in water.写出乙酸在水中电离平衡的 ICE 表格。[2]
(b)Using the approximation $0.100 - x \approx 0.100$, calculate $[\text{H}^+]$ and the pH of the solution.用近似 $0.100 - x \approx 0.100$，计算 $[\text{H}^+]$ 和溶液的 pH。[3]
(c)Verify the approximation is valid by checking that the percent ionization is less than 5%.通过验证电离百分数小于 5% 来确认近似的有效性。[2]
(d)Compare the pH of this acetic acid solution to that of a $0.100\ \text{mol/L}$ HCl solution and explain the difference in one sentence.将该乙酸溶液的 pH 与 $0.100\ \text{mol/L}$ HCl 溶液的 pH 进行比较，并用一句话解释差异。[2]
Lactic acid $(\text{CH}_3\text{CH(OH)COOH})$ has $K_a = 1.4 \times 10^{-4}$. A buffer is prepared with $0.200\ \text{mol/L}$ lactic acid and $0.200\ \text{mol/L}$ sodium lactate.乳酸 $(\text{CH}_3\text{CH(OH)COOH})$ 的 $K_a = 1.4 \times 10^{-4}$。用 $0.200\ \text{mol/L}$ 乳酸和 $0.200\ \text{mol/L}$ 乳酸钠配制缓冲溶液。
(a)Calculate $\text{p}K_a$ for lactic acid and hence find the pH of the buffer.计算乳酸的 $\text{p}K_a$，并求该缓冲溶液的 pH。[3]
(b)Calculate $K_b$ for the lactate ion $(\text{CH}_3\text{CH(OH)COO}^-)$.计算乳酸根离子 $(\text{CH}_3\text{CH(OH)COO}^-)$ 的 $K_b$。[2]
(c)Explain in one or two sentences why this buffer is effective near its calculated pH and what range of pH values it would buffer effectively.用一至两句话解释为什么该缓冲溶液在其计算 pH 附近有效，以及它能有效缓冲的 pH 范围。[3]
🇨🇦 Alberta阿尔伯塔Chem 20 Unit C(acids/bases, neutralization); （酸碱、中和）；Chem 30 Unit D(pH, Ka/Kb, buffers, titration)（pH、Ka/Kb、缓冲液、滴定）
Full Syllabus Map in the companion Study Guide. Note: NGSS assesses Arrhenius definitions, strong vs weak, and the pH scale; titration, buffers, and Ka/Kb equilibria are Honors level for US and ON SCH3U students but core for BC Chemistry 12 and AB Chem 30.完整大纲对照表见配套学习指南。注：NGSS 仅考查阿伦尼乌斯定义、强弱酸碱及 pH 标度；滴定、缓冲液及 Ka/Kb 平衡为荣誉级（适合美国及安大略 SCH3U 学生），但为卑诗化学 12 和阿省化学 30 的核心内容。