High School Chemistry — Acids, Bases and pH — Solutions

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Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Acid/base definitions酸碱定义 · HS-PS1-6 [3 marks][3 分]

In $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+$, which species is the Bronsted-Lowry acid in the forward direction?在 $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+$ 中，正向反应的布朗斯特-劳里酸是哪种物质？

Answer:答案： (B) $\text{CH}_3\text{COOH}$

(a) Identify the proton donor in the forward reaction确定正向反应中的质子供体 M1·A1·A1

A Bronsted-Lowry acid is a proton ($\text{H}^+$) donor. In the forward reaction, $\text{CH}_3\text{COOH}$ transfers a proton to $\text{H}_2\text{O}$, producing $\text{H}_3\text{O}^+$ and the conjugate base $\text{CH}_3\text{COO}^-$. Therefore $\text{CH}_3\text{COOH}$ is the Bronsted-Lowry acid and $\text{H}_2\text{O}$ is the Bronsted-Lowry base.布朗斯特-劳里酸是质子（$\text{H}^+$）供体。正向反应中，$\text{CH}_3\text{COOH}$ 将质子转移给 $\text{H}_2\text{O}$，生成 $\text{H}_3\text{O}^+$ 和共轭碱 $\text{CH}_3\text{COO}^-$。因此 $\text{CH}_3\text{COOH}$ 是布朗斯特-劳里酸，$\text{H}_2\text{O}$ 是布朗斯特-劳里碱。

Why the distractors fail.干扰项分析。

(A) $\text{H}_2\text{O}$: water acts as the Bronsted-Lowry base here (proton acceptor), not the acid.水在此处是布朗斯特-劳里碱（质子受体），而非酸。

(C) $\text{CH}_3\text{COO}^-$: this is the conjugate base formed after $\text{CH}_3\text{COOH}$ donates its proton; it acts as a base in the reverse reaction.这是 $\text{CH}_3\text{COOH}$ 给出质子后形成的共轭碱，在逆反应中充当碱。

(D) $\text{H}_3\text{O}^+$: this is the conjugate acid formed when $\text{H}_2\text{O}$ accepts the proton; it is a product, not a reactant acid.这是 $\text{H}_2\text{O}$ 接受质子后形成的共轭酸，是产物而非反应物中的酸。

Bronsted-Lowry vs Arrhenius: broader definitions.布朗斯特-劳里与阿伦尼乌斯：更广泛的定义。 The Arrhenius definition limits acids to $\text{H}^+$ producers in water. The Bronsted-Lowry definition generalises this to any proton-transfer reaction: the acid donates $\text{H}^+$, the base accepts it. In every such reaction there is one conjugate acid-base pair on each side. Here the two pairs are $\text{CH}_3\text{COOH}/\text{CH}_3\text{COO}^-$ (acid/conjugate base) and $\text{H}_3\text{O}^+/\text{H}_2\text{O}$ (conjugate acid/base). On AP and provincial exams you will be asked to label all four species; practice doing so automatically.阿伦尼乌斯定义将酸限制为在水中产生 $\text{H}^+$ 的物质。布朗斯特-劳里定义将其推广至任何质子转移反应：酸给出 $\text{H}^+$，碱接受 $\text{H}^+$。每对反应物和产物各构成一个共轭酸碱对。此处两对为 $\text{CH}_3\text{COOH}/\text{CH}_3\text{COO}^-$（酸/共轭碱）和 $\text{H}_3\text{O}^+/\text{H}_2\text{O}$（共轭酸/碱）。AP 及省考常要求标注全部四种物质，请养成习惯。

Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §3 pH scalepH 标度 · HS-PS1-6 [3 marks][3 分]

A solution has $[\text{H}^+] = 0.010\ \text{mol/L}$ at 25 °C. What is the pH?某溶液在 25 °C 时 $[\text{H}^+] = 0.010\ \text{mol/L}$。pH 是多少？

Answer:答案： (A) $2$

(a) Apply the pH definition运用 pH 定义 M1·A1·A1

$$ \text{pH} = -\log[\text{H}^+] = -\log(0.010) = -\log(10^{-2}) = 2. $$ The solution is acidic (pH $< 7$). Option (A).该溶液为酸性（pH $< 7$）。选 (A)。

Why the distractors fail.干扰项分析。

(B) $7$: pH 7 is neutral ($[\text{H}^+] = 10^{-7}$); $0.010\ \text{mol/L}$ is far more acidic.pH 7 是中性（$[\text{H}^+] = 10^{-7}$）；$0.010\ \text{mol/L}$ 远比这更酸。

(D) $-2$: arises from forgetting the negative sign in the definition: $\log(0.010) = -2$, so $\text{pH} = -(-2) = +2$.忘记定义中的负号，直接取 $\log(0.010) = -2$，实际 pH $= -(-2) = +2$。

The pH scale is logarithmic: each unit represents a 10-fold change in $[\text{H}^+]$.pH 标度是对数标度：每差 1 个单位，$[\text{H}^+]$ 相差 10 倍。 $\text{pH} = -\log[\text{H}^+]$ means that a solution with pH 2 has ten times more $[\text{H}^+]$ than one with pH 3. For concentrations that are exact powers of 10, the calculation is mental arithmetic: $[\text{H}^+] = 10^{-n} \Rightarrow \text{pH} = n$. Memorise that pH $< 7$ is acidic, pH $= 7$ is neutral, and pH $> 7$ is basic at 25 °C. Below pH 0 is possible in concentrated strong-acid solutions.$\text{pH} = -\log[\text{H}^+]$ 意味着 pH 2 的溶液比 pH 3 的溶液 $[\text{H}^+]$ 高 10 倍。当浓度恰好为 10 的幂次时，可心算：$[\text{H}^+] = 10^{-n} \Rightarrow \text{pH} = n$。记住：25 °C 下 pH $< 7$ 为酸性，pH $= 7$ 为中性，pH $> 7$ 为碱性。在浓强酸溶液中 pH 可低于 0。

Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Strong vs weak强弱酸碱 · SCH3U E3.5 [4 marks][4 分]

$0.10\ \text{mol/L}\ \text{HCl}$ vs $0.10\ \text{mol/L}\ \text{CH}_3\text{COOH}$ ($K_a = 1.8 \times 10^{-5}$). (a) $[\text{H}^+]$ in each; which has lower pH? (b) Why is HCl strong and $\text{CH}_3\text{COOH}$ weak?$0.10\ \text{mol/L}\ \text{HCl}$ 与 $0.10\ \text{mol/L}\ \text{CH}_3\text{COOH}$（$K_a = 1.8 \times 10^{-5}$）。(a) 各自的 $[\text{H}^+]$ 及哪个 pH 更低？(b) 为何 HCl 是强酸而 $\text{CH}_3\text{COOH}$ 是弱酸？

Answer:答案： (a) HCl: $[\text{H}^+] = 0.10\ \text{mol/L}$; acetic acid: $[\text{H}^+] \approx 1.3 \times 10^{-3}\ \text{mol/L}$; HCl has the lower pH.HCl：$[\text{H}^+] = 0.10\ \text{mol/L}$；乙酸：$[\text{H}^+] \approx 1.3 \times 10^{-3}\ \text{mol/L}$；HCl 的 pH 更低。 (b) HCl fully dissociates; $\text{CH}_3\text{COOH}$ partially ionizes.HCl 完全解离；$\text{CH}_3\text{COOH}$ 部分电离。

(a) $[\text{H}^+]$ in each solution各溶液中的 $[\text{H}^+]$ M1·A1

HCl (strong acid): dissociates 100%, so $[\text{H}^+] = 0.10\ \text{mol/L}$, pH $= -\log(0.10) = 1.00$.HCl（强酸）：100% 解离，故 $[\text{H}^+] = 0.10\ \text{mol/L}$，pH $= -\log(0.10) = 1.00$。
$\text{CH}_3\text{COOH}$ (weak acid): using the approximation $[\text{H}^+] \approx \sqrt{K_a \times C}$:$\text{CH}_3\text{COOH}$（弱酸）：使用近似 $[\text{H}^+] \approx \sqrt{K_a \times C}$： $$ [\text{H}^+] \approx \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} \approx 1.3 \times 10^{-3}\ \text{mol/L.} $$ pH $= -\log(1.3 \times 10^{-3}) \approx 2.87$. HCl has the lower pH (1.00 vs 2.87).pH $= -\log(1.3 \times 10^{-3}) \approx 2.87$。HCl 的 pH 更低（1.00 vs 2.87）。

(b) Strong vs weak acid classification强酸与弱酸的分类 M1·A1

HCl is a strong acid because it dissociates completely in water ($K_a \gg 1$): every HCl molecule produces one $\text{H}^+$ and one $\text{Cl}^-$. Acetic acid is a weak acid because it only partially ionizes ($K_a = 1.8 \times 10^{-5} \ll 1$): at equilibrium, most $\text{CH}_3\text{COOH}$ molecules remain intact.HCl 是强酸，因为它在水中完全解离（$K_a \gg 1$）：每个 HCl 分子均产生一个 $\text{H}^+$ 和一个 $\text{Cl}^-$。乙酸是弱酸，因为它只部分电离（$K_a = 1.8 \times 10^{-5} \ll 1$）：平衡时大多数 $\text{CH}_3\text{COOH}$ 分子保持完整。

Same concentration does not mean same $[\text{H}^+]$: only strong acids give $[\text{H}^+] = C$.浓度相同不等于 $[\text{H}^+]$ 相同：只有强酸才有 $[\text{H}^+] = C$。 A classic exam trap is to assume pH depends only on concentration. It depends on both concentration and degree of ionization. The approximate formula $[\text{H}^+] \approx \sqrt{K_a C}$ applies when percent ionization is below 5% (check: $1.3 \times 10^{-3} / 0.10 \times 100\% = 1.3\%$, valid). The percent ionization of a weak acid increases as the solution is diluted, a counterintuitive but testable result.经典考题陷阱是认为 pH 只取决于浓度，实际上 pH 同时取决于浓度和电离程度。近似式 $[\text{H}^+] \approx \sqrt{K_a C}$ 在电离百分数低于 5% 时有效（验证：$1.3 \times 10^{-3} / 0.10 \times 100\% = 1.3\%$，有效）。弱酸的电离百分数随稀释而增大，这是一个反直觉但常考的结论。

Q4MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Neutralization中和反应 · Chemistry 11 [6 marks][6 分]

$40.0\ \text{mL}$ of $0.300\ \text{mol/L}\ \text{HCl}$ mixed with $0.200\ \text{mol/L}\ \text{NaOH}$. (a) Moles of HCl. (b) Volume of NaOH for complete neutralization. (c) Complete ionic equation.$40.0\ \text{mL}$ 的 $0.300\ \text{mol/L}\ \text{HCl}$ 与 $0.200\ \text{mol/L}\ \text{NaOH}$ 混合。(a) HCl 的物质的量。(b) 完全中和所需 NaOH 的体积。(c) 完整离子方程式。

Answer:答案： (a) $n(\text{HCl}) = 0.0120\ \text{mol}$ · (b) $V(\text{NaOH}) = 60.0\ \text{mL}$ · (c) $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\ell)$

(a) Moles of HClHCl 的物质的量 A1

$$ n(\text{HCl}) = C \times V = 0.300\ \text{mol/L} \times 0.0400\ \text{L} = 0.0120\ \text{mol.} $$

(b) Volume of NaOH required所需 NaOH 的体积 M1·A1·A1

At the equivalence point, $n(\text{NaOH}) = n(\text{HCl}) = 0.0120\ \text{mol}$ (1:1 ratio).等当点时，$n(\text{NaOH}) = n(\text{HCl}) = 0.0120\ \text{mol}$（1:1 比例）。 $$ V(\text{NaOH}) = \frac{n}{C} = \frac{0.0120\ \text{mol}}{0.200\ \text{mol/L}} = 0.0600\ \text{L} = 60.0\ \text{mL.} $$

(c) Complete ionic equation完整离子方程式 M1·A1

Both HCl and NaOH are strong electrolytes, fully dissociated. The spectator ions $\text{Na}^+$ and $\text{Cl}^-$ cancel:HCl 和 NaOH 均为强电解质，完全解离。旁观离子 $\text{Na}^+$ 和 $\text{Cl}^-$ 抵消： $$ \text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\ell). $$ This net ionic equation is universal for all strong acid/strong base neutralizations.该净离子方程式适用于所有强酸/强碱的中和反应。

Net ionic equation strips away spectator ions to reveal the essential chemistry.净离子方程式去除旁观离子，揭示本质化学。 The molecular equation $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ is correct but misleading: it implies HCl and NaOH exist as molecules in solution, which they do not. The complete ionic equation separates all strong electrolytes into ions, then the net ionic equation cancels species that appear identically on both sides. For strong acid/strong base reactions the net ionic equation is always $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$, regardless of the specific acid and base used. The mole ratio method (step b) is the core stoichiometry tool for titration calculations.分子方程式 $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ 虽然正确，但会误导人以为 HCl 和 NaOH 以分子形式存在于溶液中，实际并非如此。完整离子方程式将所有强电解质拆写为离子，再消去两侧相同的物种，得到净离子方程式。强酸与强碱的净离子方程式始终为 $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$，与具体选用何种酸碱无关。步骤 (b) 的物质的量比方法是滴定计算的核心工具。

Q5MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §5 Titration滴定 · Chem 30 D2 [8 marks][8 分]

$25.00\ \text{mL}$ HCl titrated with $0.150\ \text{mol/L}$ NaOH; endpoint at $31.50\ \text{mL}$. (a) Moles NaOH. (b) $[\text{HCl}]$. (c) pH at equivalence point and indicator class. (d) One source of error and its effect.$25.00\ \text{mL}$ 盐酸用 $0.150\ \text{mol/L}$ NaOH 滴定，加入 $31.50\ \text{mL}$ 后到达终点。(a) NaOH 的物质的量。(b) $[\text{HCl}]$。(c) 等当点 pH 及指示剂类别。(d) 一个误差来源及其影响。

Answer:答案： (a) $n(\text{NaOH}) = 4.73 \times 10^{-3}\ \text{mol}$ · (b) $[\text{HCl}] = 0.189\ \text{mol/L}$ · (c) pH $= 7.00$; neutral-range indicator (e.g. bromothymol blue)pH $= 7.00$；中性范围指示剂（如溴百里酚蓝） · (d) see below见下

(a) Moles of NaOH used所用 NaOH 的物质的量 M1·A1

$$ n(\text{NaOH}) = 0.150\ \text{mol/L} \times 0.03150\ \text{L} = 4.725 \times 10^{-3}\ \text{mol} \approx 4.73 \times 10^{-3}\ \text{mol.} $$

(b) Concentration of HClHCl 的浓度 M1·A1

At the equivalence point, $n(\text{HCl}) = n(\text{NaOH}) = 4.725 \times 10^{-3}\ \text{mol}$.等当点时，$n(\text{HCl}) = n(\text{NaOH}) = 4.725 \times 10^{-3}\ \text{mol}$。 $$ [\text{HCl}] = \frac{4.725 \times 10^{-3}\ \text{mol}}{0.02500\ \text{L}} = 0.189\ \text{mol/L.} $$

(c) pH at equivalence point and indicator等当点 pH 及指示剂 A1·A1

The reaction produces NaCl (a neutral salt) and water. Neither $\text{Na}^+$ nor $\text{Cl}^-$ hydrolyzes, so the solution is neutral: pH $= 7.00$ at 25 °C. An appropriate indicator must change color near pH 7; bromothymol blue (transition range pH 6.0-7.6) or phenolphthalein (8.2-10.0) are both commonly accepted, though bromothymol blue is optimal for this endpoint.反应生成 NaCl（中性盐）和水。$\text{Na}^+$ 和 $\text{Cl}^-$ 均不水解，因此溶液为中性：25 °C 时 pH $= 7.00$。合适的指示剂应在 pH 7 附近变色；溴百里酚蓝（变色范围 pH 6.0-7.6）或酚酞（8.2-10.0）均被普遍接受，但溴百里酚蓝对此终点最为理想。

(d) Source of error and effect on calculated $[\text{HCl}]$误差来源及对计算所得 $[\text{HCl}]$ 的影响 A1·A1

Example: Overshooting the endpoint (adding excess NaOH past the color change). This makes the recorded volume of NaOH larger than the true equivalence volume, so $n(\text{NaOH})$ is overestimated, and the calculated $[\text{HCl}]$ is higher than the true value.示例：超过终点（在变色后继续加入过量 NaOH）。这使记录的 NaOH 体积大于真实等当点体积，从而高估 $n(\text{NaOH})$，导致计算所得 $[\text{HCl}]$ 高于真实值。
Alternative accepted error: Rinsing the burette with distilled water instead of NaOH solution before filling. This dilutes the NaOH, reduces its effective concentration, requires more volume to reach the endpoint, and again overestimates $[\text{HCl}]$.另一可接受的误差：向酸式滴定管中加液前用蒸馏水而非 NaOH 溶液润洗。这会稀释 NaOH，降低其有效浓度，需要更多体积才能到达终点，同样导致高估 $[\text{HCl}]$。

Every titration error question has three parts: what went wrong, how it changed the volume, and therefore how $C$ was affected.每道滴定误差题都有三个要素：出了什么问题、体积如何变化、$C$ 因此如何受影响。 On AB Chem 30 and BC Chemistry 12 exams, error analysis is a guaranteed multi-mark question. Structure your answer as: (1) name the error, (2) state whether the recorded titrant volume is too high or too low, and (3) conclude whether the calculated concentration is too high, too low, or unaffected. Errors that affect the analyte volume (e.g. misreading the initial burette reading) change the result in the opposite direction from errors affecting the titrant volume.在阿省化学 30 和卑诗化学 12 的考试中，误差分析是必考的多分题。建议按以下结构作答：(1) 指出误差，(2) 说明记录的滴定剂体积偏高还是偏低，(3) 得出计算浓度偏高、偏低还是不受影响的结论。影响待测液体积的误差（如初读数误差）与影响滴定剂体积的误差方向相反。

Section B · Worked SolutionsB 部分 · 详细解答

Q6EASY易 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 pH / pOH calculationspH / pOH 计算 · HS-PS1-6 [8 marks][8 分]

(a) $[\text{OH}^-] = 5.0 \times 10^{-3}\ \text{mol/L}$: find pOH, pH, classify. (b) pH $= 4.50$: find $[\text{H}^+]$ and $[\text{OH}^-]$. (c) $[\text{H}^+] = 2.5 \times 10^{-11}\ \text{mol/L}$: find pH and pOH, classify. All at 25 °C.(a) $[\text{OH}^-] = 5.0 \times 10^{-3}\ \text{mol/L}$：求 pOH、pH，判断酸碱性。(b) pH $= 4.50$：求 $[\text{H}^+]$ 和 $[\text{OH}^-]$。(c) $[\text{H}^+] = 2.5 \times 10^{-11}\ \text{mol/L}$：求 pH 和 pOH，判断酸碱性。均在 25 °C 下。

Answer:答案： (a) pOH $= 2.30$; pH $= 11.70$; basicpOH $= 2.30$；pH $= 11.70$；碱性 · (b) $[\text{H}^+] = 3.16 \times 10^{-5}\ \text{mol/L}$; $[\text{OH}^-] = 3.16 \times 10^{-10}\ \text{mol/L}$ · (c) pH $= 10.60$; pOH $= 3.40$; basicpH $= 10.60$；pOH $= 3.40$；碱性

(a) From $[\text{OH}^-]$ to pOH to pH从 $[\text{OH}^-]$ 求 pOH 再求 pH M1·A1·A1

$$ \text{pOH} = -\log[\text{OH}^-] = -\log(5.0 \times 10^{-3}) = 3 - \log 5.0 = 3 - 0.699 = 2.30. $$ $$ \text{pH} = 14.00 - \text{pOH} = 14.00 - 2.30 = 11.70. $$ Since pH $> 7$, the solution is basic.因为 pH $> 7$，该溶液为碱性。

(b) From pH to $[\text{H}^+]$ and $[\text{OH}^-]$从 pH 求 $[\text{H}^+]$ 和 $[\text{OH}^-]$ M1·A1

$$ [\text{H}^+] = 10^{-\text{pH}} = 10^{-4.50} = 10^{-5} \times 10^{0.50} = 10^{-5} \times 3.162 = 3.16 \times 10^{-5}\ \text{mol/L.} $$ $$ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.0 \times 10^{-14}}{3.16 \times 10^{-5}} = 3.16 \times 10^{-10}\ \text{mol/L.} $$

(c) From $[\text{H}^+]$ to pH and pOH从 $[\text{H}^+]$ 求 pH 和 pOH M1·A1·A1

$$ \text{pH} = -\log(2.5 \times 10^{-11}) = 11 - \log 2.5 = 11 - 0.398 = 10.60. $$ $$ \text{pOH} = 14.00 - 10.60 = 3.40. $$ pH $> 7$, so the solution is basic.pH $> 7$，该溶液为碱性。

The four interconversions: $[\text{H}^+] \leftrightarrow \text{pH} \leftrightarrow \text{pOH} \leftrightarrow [\text{OH}^-]$.四种相互换算：$[\text{H}^+] \leftrightarrow \text{pH} \leftrightarrow \text{pOH} \leftrightarrow [\text{OH}^-]$。 At 25 °C the four quantities are linked by two relations: $\text{pH} = -\log[\text{H}^+]$ and $\text{pH} + \text{pOH} = 14.00$. Together with $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ you can find any one from any other. Common log tip: $-\log(a \times 10^{-n}) = n - \log a$. Memorise $\log 2 \approx 0.301$, $\log 3 \approx 0.477$, $\log 5 \approx 0.699$, $\log 2.5 \approx 0.398$ to do these without a calculator in MCQ sections.25 °C 时，四个量由两个关系联系：$\text{pH} = -\log[\text{H}^+]$ 和 $\text{pH} + \text{pOH} = 14.00$。结合 $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$，可由任意一个量求得其他量。常用对数技巧：$-\log(a \times 10^{-n}) = n - \log a$。记住 $\log 2 \approx 0.301$、$\log 3 \approx 0.477$、$\log 5 \approx 0.699$、$\log 2.5 \approx 0.398$，可在选择题部分不用计算器完成计算。

Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 + §5 Neutralization and titration中和与滴定 · SCH3U E2.7 [8 marks][8 分]

$20.00\ \text{mL}$ of $0.250\ \text{mol/L}$ NaOH titrated with $0.100\ \text{mol/L}$ HCl. (a) Moles NaOH. (b) Volume HCl to equivalence. (c) pH at equivalence. (d) After $40.00\ \text{mL}$ HCl added: what is in solution, pH above/below/equal to 7?用 $0.100\ \text{mol/L}$ HCl 滴定 $20.00\ \text{mL}$ 的 $0.250\ \text{mol/L}$ NaOH。(a) NaOH 的物质的量。(b) 到达等当点所需 HCl 体积。(c) 等当点 pH。(d) 加入 $40.00\ \text{mL}$ HCl 后：溶液中有什么，pH 高于/低于/等于 7？

Answer:答案： (a) $5.00 \times 10^{-3}\ \text{mol}$ · (b) $50.0\ \text{mL}$ · (c) pH $= 7.00$pH $= 7.00$ · (d) excess NaOH + NaCl; pH $> 7$过量 NaOH + NaCl；pH $> 7$

(a) Moles of NaOH in flask锥形瓶中 NaOH 的物质的量 A1

$$ n(\text{NaOH}) = 0.250\ \text{mol/L} \times 0.02000\ \text{L} = 5.00 \times 10^{-3}\ \text{mol.} $$

(b) Volume of HCl to reach equivalence point到达等当点所需 HCl 体积 M1·A1·A1

At equivalence, $n(\text{HCl}) = n(\text{NaOH}) = 5.00 \times 10^{-3}\ \text{mol}$.等当点时，$n(\text{HCl}) = n(\text{NaOH}) = 5.00 \times 10^{-3}\ \text{mol}$。 $$ V(\text{HCl}) = \frac{5.00 \times 10^{-3}\ \text{mol}}{0.100\ \text{mol/L}} = 0.0500\ \text{L} = 50.0\ \text{mL.} $$

(c) pH at the equivalence point等当点 pH M1·A1

Strong acid + strong base produces NaCl (neutral salt). Neither $\text{Na}^+$ nor $\text{Cl}^-$ undergoes hydrolysis, so the solution is neutral: pH $= 7.00$ at 25 °C.强酸 + 强碱生成 NaCl（中性盐），$\text{Na}^+$ 和 $\text{Cl}^-$ 均不水解，溶液为中性：25 °C 时 pH $= 7.00$。

(d) State of solution after $40.00\ \text{mL}$ HCl added加入 $40.00\ \text{mL}$ HCl 后的溶液状态 M1·A1

$n(\text{HCl added}) = 0.100 \times 0.04000 = 4.00 \times 10^{-3}\ \text{mol}$. Since $n(\text{NaOH}) = 5.00 \times 10^{-3}\ \text{mol} > n(\text{HCl}) = 4.00 \times 10^{-3}\ \text{mol}$, NaOH is in excess.$n(\text{HCl加入}) = 0.100 \times 0.04000 = 4.00 \times 10^{-3}\ \text{mol}$。因为 $n(\text{NaOH}) = 5.00 \times 10^{-3}\ \text{mol} > n(\text{HCl}) = 4.00 \times 10^{-3}\ \text{mol}$，NaOH 过量。 $$ n(\text{excess NaOH}) = 5.00 \times 10^{-3} - 4.00 \times 10^{-3} = 1.00 \times 10^{-3}\ \text{mol.} $$ $$ [\text{OH}^-] = \frac{1.00 \times 10^{-3}\ \text{mol}}{(20.00 + 40.00) \times 10^{-3}\ \text{L}} = \frac{1.00 \times 10^{-3}}{0.06000} = 0.01667\ \text{mol/L.} $$ $$ \text{pOH} = -\log(0.01667) \approx 1.78; \quad \text{pH} = 14.00 - 1.78 = 12.22. $$ The solution contains excess NaOH and NaCl. pH $> 7$ (basic).溶液中含有过量 NaOH 和 NaCl，pH $> 7$（碱性）。

Always track moles, not volumes, to determine what is in excess.判断过量物质时，始终追踪物质的量，而非体积。 A common error is to assume that since 40 mL $<$ 50 mL equivalence volume, there must be equal amounts of acid and base remaining. Instead, calculate moles: $n(\text{NaOH}) - n(\text{HCl added}) = 1.00 \times 10^{-3}\ \text{mol}$ NaOH in excess. The pH calculation then uses the total volume (60 mL) as the denominator, not just the excess volume. This dilution step is frequently missed and costs marks on titration problems.一个常见错误是认为只要 40 mL $<$ 50 mL（等当点体积），溶液就含有等量的酸和碱。正确做法是计算物质的量：$n(\text{NaOH}) - n(\text{HCl加入}) = 1.00 \times 10^{-3}\ \text{mol}$ NaOH 过量。计算 pH 时，需用总体积（60 mL）作为分母，而非仅用过量体积。这个稀释步骤经常被遗漏，在滴定题中容易失分。

Q8HARD难Honors荣誉级 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Buffers缓冲溶液 · Chemistry 12 [9 marks][9 分]

Buffer: $0.10\ \text{mol/L}\ \text{CH}_3\text{COOH}$ and $0.15\ \text{mol/L}\ \text{CH}_3\text{COONa}$; $K_a = 1.8 \times 10^{-5}$ ($\text{p}K_a = 4.74$). (a) pH by Henderson-Hasselbalch. (b) New pH after $0.010\ \text{mol}$ HCl added to $1.00\ \text{L}$. (c) Why buffer resists pH change (Le Chatelier).缓冲液：$0.10\ \text{mol/L}\ \text{CH}_3\text{COOH}$ 和 $0.15\ \text{mol/L}\ \text{CH}_3\text{COONa}$；$K_a = 1.8 \times 10^{-5}$（$\text{p}K_a = 4.74$）。(a) 用 Henderson-Hasselbalch 计算 pH。(b) 向 $1.00\ \text{L}$ 缓冲液中加入 $0.010\ \text{mol}$ HCl 后的新 pH。(c) 用勒夏特列原理解释缓冲液抗 pH 变化的原因。

Answer:答案： (a) pH $= 4.92$pH $= 4.92$ · (b) pH $= 4.84$pH $= 4.84$ · (c) see below见下

(a) Henderson-Hasselbalch equationHenderson-Hasselbalch 方程 M1·A1·A1

$$ \text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = 4.74 + \log\frac{0.15}{0.10} = 4.74 + \log(1.5) = 4.74 + 0.176 = 4.92. $$

(b) New pH after adding $0.010\ \text{mol}$ HCl to $1.00\ \text{L}$向 $1.00\ \text{L}$ 缓冲液中加入 $0.010\ \text{mol}$ HCl 后的新 pH M1·A1·M1·A1

The added $\text{H}^+$ reacts with the conjugate base $\text{CH}_3\text{COO}^-$:加入的 $\text{H}^+$ 与共轭碱 $\text{CH}_3\text{COO}^-$ 反应： $$ \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{CH}_3\text{COOH}(\text{aq}). $$ In $1.00\ \text{L}$: $n(\text{CH}_3\text{COO}^-)_\text{initial} = 0.15\ \text{mol}$; $n(\text{CH}_3\text{COOH})_\text{initial} = 0.10\ \text{mol}$.在 $1.00\ \text{L}$ 中：$n(\text{CH}_3\text{COO}^-)_{\text{初}} = 0.15\ \text{mol}$；$n(\text{CH}_3\text{COOH})_{\text{初}} = 0.10\ \text{mol}$。 $$ n(\text{CH}_3\text{COO}^-)_\text{new} = 0.15 - 0.010 = 0.140\ \text{mol;} \quad n(\text{CH}_3\text{COOH})_\text{new} = 0.10 + 0.010 = 0.110\ \text{mol.} $$ $$ \text{pH} = 4.74 + \log\frac{0.140}{0.110} = 4.74 + \log(1.2727) = 4.74 + 0.105 = 4.84. $$ The pH dropped only from 4.92 to 4.84 (a change of 0.08 units), demonstrating effective buffering.pH 仅从 4.92 降至 4.84（变化 0.08 个单位），体现了有效的缓冲作用。

(c) Le Chatelier explanation勒夏特列原理解释 M1·A1

Adding HCl introduces $\text{H}^+$ ions, which disturb the equilibrium $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$. By Le Chatelier's principle, the system shifts to the left to consume the added $\text{H}^+$: $\text{CH}_3\text{COO}^-$ reacts with $\text{H}^+$ to reform $\text{CH}_3\text{COOH}$. This consumes most of the added protons, preventing a large drop in pH.加入 HCl 引入 $\text{H}^+$ 离子，扰动平衡 $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$。根据勒夏特列原理，体系向左移动以消耗加入的 $\text{H}^+$：$\text{CH}_3\text{COO}^-$ 与 $\text{H}^+$ 反应重新生成 $\text{CH}_3\text{COOH}$。这消耗了大部分加入的质子，阻止了 pH 的大幅下降。

The Henderson-Hasselbalch equation works because the ratio $[\text{A}^-]/[\text{HA}]$ changes very little when small amounts of strong acid or base are added.Henderson-Hasselbalch 方程之所以有效，是因为加入少量强酸或强碱时，比值 $[\text{A}^-]/[\text{HA}]$ 变化很小。 When $[\text{A}^-] = [\text{HA}]$, pH $=$ p$K_a$ and the buffer is at its most effective point. The buffer capacity is greatest within one pH unit of p$K_a$ (here pH 3.74 to 5.74). Outside this range the ratio becomes very large or very small, so small additions produce large pH swings. The calculation in (b) shows that adding 0.010 mol HCl to unbuffered water would drop the pH to 2.00, while the buffer held it at 4.84 -- a difference of nearly 3 pH units.当 $[\text{A}^-] = [\text{HA}]$ 时，pH $=$ p$K_a$，此时缓冲能力最强。缓冲能力在 p$K_a$ 两侧各 1 个 pH 单位内最大（此处为 pH 3.74 到 5.74）。超出此范围，比值变得过大或过小，少量加入即引起较大 pH 变化。(b) 的计算说明：若向无缓冲的纯水加入 0.010 mol HCl，pH 将降至 2.00，而缓冲液将其保持在 4.84，相差近 3 个 pH 单位。

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]

Weak acid HA, $K_a = 2.0 \times 10^{-5}$ at 25 °C. (a) Ionization equilibrium and $K_a$ expression. (b) $K_b$ for $\text{A}^-$ and p$K_b$. (c) What larger $K_a$ means; is $\text{A}^-$ stronger or weaker base than conjugate base of acid with $K_a = 1.0 \times 10^{-3}$?弱酸 HA，25 °C 时 $K_a = 2.0 \times 10^{-5}$。(a) 电离平衡方程式及 $K_a$ 表达式。(b) $\text{A}^-$ 的 $K_b$ 和 p$K_b$。(c) 较大 $K_a$ 意味着什么；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？

Answer:答案： (a) see below见下 · (b) $K_b = 5.0 \times 10^{-10}$; p$K_b = 9.30$p$K_b = 9.30$ · (c) $\text{A}^-$ is a stronger base$\text{A}^-$ 是更强的碱

(a) Ionization equilibrium and $K_a$ expression电离平衡方程式及 $K_a$ 表达式 A1·A1

$$ \text{HA}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{A}^-(\text{aq}) $$ $$ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} $$ Alternatively written with $\text{H}_3\text{O}^+$: $\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$ (both are accepted).也可写为含 $\text{H}_3\text{O}^+$ 的形式：$\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$（两种形式均可接受）。

(b) $K_b$ for $\text{A}^-$ and p$K_b$$\text{A}^-$ 的 $K_b$ 和 p$K_b$ M1·A1·A1

Using the conjugate acid-base relationship $K_a \times K_b = K_w$:利用共轭酸碱关系 $K_a \times K_b = K_w$： $$ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} = 5.0 \times 10^{-10}. $$ $$ \text{p}K_b = -\log(5.0 \times 10^{-10}) = 10 - \log 5.0 = 10 - 0.699 = 9.30. $$

(c) Interpretation of $K_a$ and comparison of base strength$K_a$ 的含义及碱性强弱比较 M1·A1·A1

A larger $K_a$ means the acid is a stronger weak acid: a greater proportion of molecules ionize, producing more $\text{H}^+$ at equilibrium. The acid with $K_a = 1.0 \times 10^{-3}$ is stronger than HA ($K_a = 2.0 \times 10^{-5}$).较大的 $K_a$ 意味着该酸是更强的弱酸：在平衡时，更大比例的分子发生电离，产生更多 $\text{H}^+$。$K_a = 1.0 \times 10^{-3}$ 的酸比 HA（$K_a = 2.0 \times 10^{-5}$）更强。

For the conjugate base of the stronger acid: $K_b' = K_w / (1.0 \times 10^{-3}) = 1.0 \times 10^{-11}$. Compare:对于更强酸的共轭碱：$K_b' = K_w / (1.0 \times 10^{-3}) = 1.0 \times 10^{-11}$。比较： $$ K_b(\text{A}^-) = 5.0 \times 10^{-10} \gg K_b' = 1.0 \times 10^{-11}. $$ Therefore $\text{A}^-$ is a stronger base than the conjugate base of the acid with $K_a = 1.0 \times 10^{-3}$.因此 $\text{A}^-$ 是比 $K_a = 1.0 \times 10^{-3}$ 的酸的共轭碱更强的碱。

Inverse relationship: stronger acid $\Rightarrow$ weaker conjugate base; weaker acid $\Rightarrow$ stronger conjugate base.反向关系：酸越强，共轭碱越弱；酸越弱，共轭碱越强。 The equation $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at 25 °C encodes this inverse relationship. As $K_a$ increases, $K_b$ decreases proportionally. This is why the conjugate base of a strong acid (like $\text{Cl}^-$ from HCl) is so weak it does not measurably hydrolyze water, while the conjugate base of a very weak acid (like $\text{F}^-$ from HF, $K_a = 6.8 \times 10^{-4}$) is a detectably basic species. On AP exams this relationship is often tested by asking whether a salt solution is acidic, basic, or neutral at equilibrium.25 °C 时 $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ 体现了这种反向关系。$K_a$ 增大时，$K_b$ 等比例减小。这就是为什么强酸的共轭碱（如 HCl 的 $\text{Cl}^-$）弱到几乎不水解，而非常弱的酸的共轭碱（如 HF 的 $\text{F}^-$，$K_a = 6.8 \times 10^{-4}$）是可测量到的碱性物种。AP 考试常通过提问盐溶液在平衡时是酸性、碱性还是中性来考查这一关系。

Section C · Worked SolutionsC 部分 · 详细解答

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3 + §4 pH after mixing混合后的 pH · Chem 30 D1 [8 marks][8 分]

$50.0\ \text{mL}$ of $0.200\ \text{mol/L}\ \text{H}_2\text{SO}_4$ (strong diprotic) mixed with $50.0\ \text{mL}$ of $0.150\ \text{mol/L}$ NaOH. (a) Total $n(\text{H}^+)$ and $n(\text{OH}^-)$. (b) Excess reagent and moles in excess. (c) $[\text{H}^+]$ and pH of final mixture. (d) One-sentence conclusion on acidity.$50.0\ \text{mL}$ 的 $0.200\ \text{mol/L}\ \text{H}_2\text{SO}_4$（强二元酸）与 $50.0\ \text{mL}$ 的 $0.150\ \text{mol/L}$ NaOH 混合。(a) $n(\text{H}^+)$ 总量和 $n(\text{OH}^-)$。(b) 过量试剂及过量物质的量。(c) 最终混合液的 $[\text{H}^+]$ 和 pH。(d) 一句话结论。

Answer:答案： (a) $n(\text{H}^+) = 0.0200\ \text{mol}$; $n(\text{OH}^-) = 7.50 \times 10^{-3}\ \text{mol}$ · (b) $\text{H}^+$ in excess; $1.25 \times 10^{-2}\ \text{mol}$$\text{H}^+$ 过量；$1.25 \times 10^{-2}\ \text{mol}$ · (c) $[\text{H}^+] = 0.125\ \text{mol/L}$; pH $= 0.90$pH $= 0.90$ · (d) The mixture is acidic.混合溶液为酸性。

(a) Total moles of $\text{H}^+$ and $\text{OH}^-$$\text{H}^+$ 和 $\text{OH}^-$ 的物质的量 A1·A1

$\text{H}_2\text{SO}_4$ is a strong diprotic acid; each mole yields 2 moles of $\text{H}^+$:$\text{H}_2\text{SO}_4$ 是强二元酸，每摩尔释放 2 摩尔 $\text{H}^+$： $$ n(\text{H}^+) = 2 \times 0.200\ \text{mol/L} \times 0.0500\ \text{L} = 2 \times 0.0100 = 0.0200\ \text{mol.} $$ $$ n(\text{OH}^-) = 0.150\ \text{mol/L} \times 0.0500\ \text{L} = 7.50 \times 10^{-3}\ \text{mol.} $$

(b) Excess reagent过量试剂 M1·A1

$n(\text{H}^+) = 0.0200\ \text{mol} > n(\text{OH}^-) = 7.50 \times 10^{-3}\ \text{mol}$, so $\text{H}^+$ is in excess:$n(\text{H}^+) = 0.0200\ \text{mol} > n(\text{OH}^-) = 7.50 \times 10^{-3}\ \text{mol}$，故 $\text{H}^+$ 过量： $$ n(\text{excess H}^+) = 0.0200 - 7.50 \times 10^{-3} = 1.25 \times 10^{-2}\ \text{mol.} $$

(c) $[\text{H}^+]$ and pH of the final mixture最终混合液的 $[\text{H}^+]$ 和 pH M1·A1·A1

$$ V_{\text{total}} = 50.0 + 50.0 = 100.0\ \text{mL} = 0.1000\ \text{L.} $$ $$ [\text{H}^+] = \frac{1.25 \times 10^{-2}\ \text{mol}}{0.1000\ \text{L}} = 0.125\ \text{mol/L.} $$ $$ \text{pH} = -\log(0.125) = -\log(1.25 \times 10^{-1}) = 1 - \log(1.25) = 1 - 0.097 = 0.90. $$

(d) One-sentence conclusion一句话结论 A1

The final mixture is acidic (pH $= 0.90 < 7$) because $\text{H}_2\text{SO}_4$ provided more $\text{H}^+$ ions than could be neutralized by the NaOH.最终混合溶液为酸性（pH $= 0.90 < 7$），因为 $\text{H}_2\text{SO}_4$ 提供的 $\text{H}^+$ 多于 NaOH 能中和的量。

For polyprotic acids, multiply moles by the number of ionizable protons before comparing with base.对于多元酸，须将物质的量乘以可电离质子数，再与碱比较。 The most common error here is to use $n(\text{H}_2\text{SO}_4) = 0.0100\ \text{mol}$ directly against $n(\text{NaOH}) = 7.50 \times 10^{-3}\ \text{mol}$ without doubling, which gives an incorrect excess of only $0.0100 - 7.50 \times 10^{-3} = 2.5 \times 10^{-3}\ \text{mol}$. The stoichiometric equation is $\text{H}_2\text{SO}_4 + 2\,\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$, confirming the 2:1 mole ratio. After dilution to 100 mL, the pH $= 0.90$ is below 1, which is physically achievable for concentrated acid mixtures.此处最常见的错误是直接用 $n(\text{H}_2\text{SO}_4) = 0.0100\ \text{mol}$ 与 $n(\text{NaOH}) = 7.50 \times 10^{-3}\ \text{mol}$ 比较而不乘以 2，得到错误的过量值 $0.0100 - 7.50 \times 10^{-3} = 2.5 \times 10^{-3}\ \text{mol}$。化学计量方程为 $\text{H}_2\text{SO}_4 + 2\,\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$，确认了 2:1 的摩尔比。稀释至 100 mL 后，pH $= 0.90$ 低于 1，这在浓酸混合物中是物理上可以实现的。

Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2 + §3 Weak acid pH approximation弱酸 pH 近似 · Chem 30 D3 [9 marks][9 分]

$0.100\ \text{mol/L}\ \text{CH}_3\text{COOH}$ ($K_a = 1.8 \times 10^{-5}$) at 25 °C. (a) ICE table. (b) $[\text{H}^+]$ and pH using approximation. (c) Verify approximation valid. (d) Compare to $0.100\ \text{mol/L}$ HCl.$0.100\ \text{mol/L}\ \text{CH}_3\text{COOH}$（$K_a = 1.8 \times 10^{-5}$）在 25 °C。(a) ICE 表格。(b) 用近似计算 $[\text{H}^+]$ 和 pH。(c) 验证近似有效。(d) 与 $0.100\ \text{mol/L}$ HCl 比较。

Answer:答案： (a) ICE belowICE 表见下 · (b) $[\text{H}^+] = 1.34 \times 10^{-3}\ \text{mol/L}$; pH $= 2.87$pH $= 2.87$ · (c) 1.34% ionization $< 5\%$; valid电离率 1.34% $< 5\%$；有效 · (d) HCl pH $= 1.00$; acetic acid pH $= 2.87$ (higher); weak acid partially ionizes, fewer $\text{H}^+$HCl pH $= 1.00$；乙酸 pH $= 2.87$（更高）；弱酸部分电离，$\text{H}^+$ 更少

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

The equilibrium: $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$平衡方程：$\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$

	$\text{CH}_3\text{COOH}$	$\text{H}^+$	$\text{CH}_3\text{COO}^-$
I (mol/L)初始	$0.100$	$0$	$0$
C (mol/L)变化	$-x$	$+x$	$+x$
E (mol/L)平衡	$0.100 - x$	$x$	$x$

(b) $[\text{H}^+]$ and pH using the approximation用近似计算 $[\text{H}^+]$ 和 pH M1·A1·A1

$$ K_a = \frac{x^2}{0.100 - x} \approx \frac{x^2}{0.100} = 1.8 \times 10^{-5}. $$ $$ x^2 = 1.8 \times 10^{-5} \times 0.100 = 1.8 \times 10^{-6}. $$ $$ x = [\text{H}^+] = \sqrt{1.8 \times 10^{-6}} = 1.342 \times 10^{-3}\ \text{mol/L.} $$ $$ \text{pH} = -\log(1.342 \times 10^{-3}) = 3 - \log(1.342) = 3 - 0.128 = 2.87. $$

(c) Verify the 5% approximation验证 5% 近似 M1·A1

$$ \%\ \text{ionization} = \frac{x}{C} \times 100\% = \frac{1.342 \times 10^{-3}}{0.100} \times 100\% = 1.34\%. $$ Since $1.34\% < 5\%$, the approximation $0.100 - x \approx 0.100$ is valid.因为 $1.34\% < 5\%$，近似 $0.100 - x \approx 0.100$ 有效。

(d) Comparison with $0.100\ \text{mol/L}$ HCl与 $0.100\ \text{mol/L}$ HCl 比较 M1·A1

$[\text{HCl}] = 0.100\ \text{mol/L}$ (strong acid, fully dissociated): pH $= -\log(0.100) = 1.00$. The acetic acid solution has pH $= 2.87$, which is higher (less acidic) because acetic acid is a weak acid that only partially ionizes, producing far fewer $\text{H}^+$ ions than HCl at the same concentration.$0.100\ \text{mol/L}$ HCl（强酸，完全解离）：pH $= -\log(0.100) = 1.00$。乙酸溶液的 pH $= 2.87$，更高（酸性更弱），因为乙酸是弱酸，只部分电离，在相同浓度下产生的 $\text{H}^+$ 离子远少于 HCl。

The 5% rule is a quick validity check: if percent ionization $> 5\%$, solve the quadratic exactly.5% 规则是快速有效性检验：若电离百分数 $> 5\%$，需精确求解二次方程。 The ICE table is the universal framework for all equilibrium problems. Steps: (1) write the balanced equation; (2) fill I/C/E rows with $x$ for the unknown change; (3) write the $K$ expression in terms of $x$; (4) apply the approximation if justified; (5) solve and verify. If the 5% check fails, substitute back and solve $K_a = x^2 / (C - x)$ as a quadratic: $x^2 + K_a x - K_a C = 0$, giving $x = (-K_a + \sqrt{K_a^2 + 4K_a C})/2$. For this problem the exact solution gives $x = 1.341 \times 10^{-3}$, confirming the approximation was essentially exact.ICE 表是所有平衡问题的通用框架。步骤：(1) 写出配平方程式；(2) 用 $x$ 填写 I/C/E 行；(3) 写出含 $x$ 的 $K$ 表达式；(4) 若合理则应用近似；(5) 求解并验证。若 5% 检验不通过，则将 $K_a = x^2 / (C - x)$ 作为二次方程精确求解：$x^2 + K_a x - K_a C = 0$，解为 $x = (-K_a + \sqrt{K_a^2 + 4K_a C})/2$。本题精确解为 $x = 1.341 \times 10^{-3}$，确认近似基本精确。

Q12HARD难Honors荣誉级 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 + §7 Buffer design and Ka/Kb integration缓冲溶液设计与 Ka/Kb 综合 · Chem 30 D4 [8 marks][8 分]

Lactic acid $K_a = 1.4 \times 10^{-4}$; buffer with $0.200\ \text{mol/L}$ lactic acid and $0.200\ \text{mol/L}$ sodium lactate. (a) p$K_a$ and pH of buffer. (b) $K_b$ for lactate ion. (c) Why buffer is effective and its effective pH range.乳酸 $K_a = 1.4 \times 10^{-4}$；用 $0.200\ \text{mol/L}$ 乳酸和 $0.200\ \text{mol/L}$ 乳酸钠配制缓冲液。(a) p$K_a$ 和缓冲液 pH。(b) 乳酸根离子 $K_b$。(c) 缓冲液为何有效及其有效 pH 范围。

Answer:答案： (a) p$K_a = 3.85$; pH $= 3.85$p$K_a = 3.85$；pH $= 3.85$ · (b) $K_b = 7.14 \times 10^{-11}$ · (c) see below见下

(a) p$K_a$ and pH of the bufferp$K_a$ 和缓冲液 pH M1·A1·A1

$$ \text{p}K_a = -\log(1.4 \times 10^{-4}) = 4 - \log(1.4) = 4 - 0.146 = 3.85. $$ Since $[\text{HA}] = [\text{A}^-] = 0.200\ \text{mol/L}$, the ratio $[\text{A}^-]/[\text{HA}] = 1$, so $\log(1) = 0$:因为 $[\text{HA}] = [\text{A}^-] = 0.200\ \text{mol/L}$，比值 $[\text{A}^-]/[\text{HA}] = 1$，故 $\log(1) = 0$： $$ \text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} = 3.85 + 0 = 3.85. $$

(b) $K_b$ for the lactate ion乳酸根离子的 $K_b$ M1·A1

$$ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.4 \times 10^{-4}} = 7.14 \times 10^{-11}. $$

(c) Why the buffer is effective and its pH range缓冲液为何有效及其 pH 范围 M1·A1·A1

The buffer is effective near pH 3.85 because it contains substantial amounts of both lactic acid (which neutralizes added base: $\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$) and lactate ion (which neutralizes added acid: $\text{A}^- + \text{H}^+ \rightarrow \text{HA}$). When equal concentrations of HA and A$^-$ are present, the buffer can handle equal additions of acid and base, and pH $=$ p$K_a$.该缓冲液在 pH 3.85 附近有效，因为它含有大量乳酸（中和加入的碱：$\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$）和乳酸根离子（中和加入的酸：$\text{A}^- + \text{H}^+ \rightarrow \text{HA}$）。当 HA 和 A$^-$ 浓度相等时，缓冲液对加酸和加碱的抵抗能力相同，此时 pH $=$ p$K_a$。

The effective buffering range is approximately p$K_a \pm 1$, i.e., pH $2.85$ to $4.85$. Outside this range, one component becomes so depleted relative to the other that the buffer capacity is essentially exhausted.有效缓冲范围约为 p$K_a \pm 1$，即 pH $2.85$ 到 $4.85$。超出此范围，一种组分相对另一种耗尽，缓冲容量基本消失。

When $[\text{HA}] = [\text{A}^-]$, pH $=$ p$K_a$ exactly: this is the maximum-capacity point of the buffer.当 $[\text{HA}] = [\text{A}^-]$ 时，pH $=$ p$K_a$ 恰好成立：这是缓冲液容量最大的点。 Lactic acid buffers are biologically important: blood lactate from anaerobic exercise requires pH regulation in muscle tissue (physiological pH 7.2-7.4, controlled by the bicarbonate system with p$K_a = 6.1$ and respiratory compensation). The lactic acid buffer itself (p$K_a = 3.85$) operates at a much more acidic pH and is used in food preservation and fermentation chemistry. The general design principle: choose a weak acid whose p$K_a$ is within 1 unit of the target pH for maximum buffer capacity.乳酸缓冲液在生物学上非常重要：无氧运动产生的血乳酸需要在肌肉组织中调节 pH（生理 pH 7.2-7.4，由碳酸氢盐体系控制，p$K_a = 6.1$，并有呼吸补偿）。乳酸缓冲液本身（p$K_a = 3.85$）在酸性更强的 pH 下工作，用于食品防腐和发酵化学。缓冲液设计的一般原则：选择 p$K_a$ 在目标 pH 1 个单位内的弱酸，以获得最大缓冲容量。