Acids, Bases and pH酸、碱与 pH

Two frameworks, one direction of travel.两种框架，一个发展方向。

• Arrhenius acid阿伦尼乌斯酸 — a substance that produces $\text{H}^+$ (or $\text{H}_3\text{O}^+$) in aqueous solution. Example: HCl, HNO₃, CH₃COOH. AB Chemistry 20 GO2: "define Arrhenius (modified) acids as substances that produce H₃O⁺(aq) in aqueous solutions and recognize that the definition is limited."— 在水溶液中产生 $\text{H}^+$（或 $\text{H}_3\text{O}^+$）的物质。例如：HCl、HNO₃、CH₃COOH。AB Chemistry 20 GO2："将阿伦尼乌斯（修正）酸定义为在水溶液中产生 H₃O⁺(aq) 的物质，并认识到该定义的局限性。"
• Arrhenius base阿伦尼乌斯碱 — a substance that produces $\text{OH}^-$ in aqueous solution. Example: NaOH, KOH, Ca(OH)₂. SCH3U E3.5: "explain the Arrhenius theory of acids and bases."— 在水溶液中产生 $\text{OH}^-$ 的物质。例如：NaOH、KOH、Ca(OH)₂。SCH3U E3.5："解释阿伦尼乌斯酸碱理论。"
• Brønsted-Lowry acid布朗斯特-劳里酸 — a proton donor (donates $\text{H}^+$). Broader than Arrhenius: works in non-aqueous media too. SCH4U E3.6: "explain the Brønsted-Lowry theory of acids and bases."— 质子供体（提供 $\text{H}^+$）。比阿伦尼乌斯更广泛：也适用于非水体系。SCH4U E3.6："解释布朗斯特-劳里酸碱理论。"
• Brønsted-Lowry base布朗斯特-劳里碱 — a proton acceptor (accepts $\text{H}^+$). Every Brønsted-Lowry acid produces a conjugate base after donating; every base produces a conjugate acid after accepting.— 质子受体（接受 $\text{H}^+$）。每个布朗斯特-劳里酸在提供质子后产生共轭碱；每个碱在接受质子后产生共轭酸。

Conjugate pair:共轭对：

$$ \underbrace{\text{HA}}_{\text{acid}} + \underbrace{\text{B}}_{\text{base}} \rightleftharpoons \underbrace{\text{A}^-}_{\text{conj. base}} + \underbrace{\text{BH}^+}_{\text{conj. acid}} $$ AB Chemistry 30 GO1: "describe Brønsted-Lowry acids as proton donors and bases as proton acceptors; write Brønsted-Lowry equations; identify conjugate pairs."AB Chemistry 30 GO1："将布朗斯特-劳里酸描述为质子供体，碱为质子受体；书写布朗斯特-劳里方程；识别共轭对。"

Strong = complete dissociation; weak = partial dissociation.强酸碱 = 完全解离；弱酸碱 = 部分解离。

• Strong acid强酸 — essentially 100% dissociated in aqueous solution. The six common ones to memorise: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄. Example: $\text{HCl}(aq) \to \text{H}^+(aq) + \text{Cl}^-(aq)$ (single arrow, irreversible).— 在水溶液中基本上 100% 解离。六种常见强酸需背熟：HCl、HBr、HI、HNO₃、H₂SO₄、HClO₄。例如：$\text{HCl}(aq) \to \text{H}^+(aq) + \text{Cl}^-(aq)$（单箭头，不可逆）。
• Weak acid弱酸 — only partially dissociates; a dynamic equilibrium exists between the molecular and ionic forms. Example: $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)$ (double arrow).— 只部分解离；分子形式与离子形式之间存在动态平衡。例如：$\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)$（双箭头）。
• Strong base强碱 — completely dissociates to give OH⁻. Group 1 and heavier Group 2 hydroxides: NaOH, KOH, LiOH, Ca(OH)₂, Ba(OH)₂.— 完全解离产生 OH⁻。第 1 族和较重的第 2 族氢氧化物：NaOH、KOH、LiOH、Ca(OH)₂、Ba(OH)₂。
• Weak base弱碱 — partially reacts with water to produce OH⁻. Example: $\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$.— 与水部分反应产生 OH⁻。例如：$\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$。

SCH3U E3.6: "explain the difference between strong and weak acids, and between strong and weak bases, in terms of degree of ionization." BC Chemistry 12 Big Idea: "Acid or base strength depends on the degree of ion dissociation." AB Chemistry 20 GO2: "differentiate, qualitatively, between strong and weak acids and bases on the basis of ionization and dissociation."SCH3U E3.6："从电离程度角度解释强酸与弱酸、强碱与弱碱的区别。"BC Chemistry 12 大概念："酸碱强度取决于离子解离程度。"AB Chemistry 20 GO2："定性区分强弱酸碱，依据是电离和解离程度。"

pH measures H⁺ concentration on a log scale; pH + pOH = 14 at 25 °C.pH 在对数标度上衡量 H⁺ 浓度；25 °C 时 pH + pOH = 14。

• pH definition:pH 定义：

$$ \text{pH} = -\log[\text{H}^+] \qquad \Leftrightarrow \qquad [\text{H}^+] = 10^{-\text{pH}} $$

• pOH definition:pOH 定义：

$$ \text{pOH} = -\log[\text{OH}^-] \qquad \Leftrightarrow \qquad [\text{OH}^-] = 10^{-\text{pOH}} $$

• Water equilibrium at 25 °C:25 °C 时水的平衡：

$$ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad \Rightarrow \quad \text{pH} + \text{pOH} = 14 $$

• Neutral solution: pH $= 7$; acidic: pH $< 7$; basic: pH $> 7$ (at 25 °C).中性溶液：pH $= 7$；酸性：pH $< 7$；碱性：pH $> 7$（25 °C）。
• Each unit change in pH = tenfold change in $[\text{H}^+]$. pH 3 is 10× more acidic than pH 4.pH 每变化一个单位 = $[\text{H}^+]$ 变化十倍。pH 3 比 pH 4 酸性强 10 倍。

AB Chemistry 20 GO2: "calculate [H₃O⁺] and [OH⁻] and the pH and pOH … based on logarithmic expressions, i.e., pH = −log[H₃O⁺] and pOH = −log[OH⁻]." SCH4U E3.5: "use the ionization constant of water (Kw) to calculate pH, pOH, [H₃O⁺], and [OH⁻]."AB Chemistry 20 GO2："基于对数表达式计算 [H₃O⁺] 和 [OH⁻] 以及 pH 和 pOH，即 pH = −log[H₃O⁺] 和 pOH = −log[OH⁻]。"SCH4U E3.5："用水的电离常数（Kw）计算 pH、pOH、[H₃O⁺] 和 [OH⁻]。"

Acid + base → salt + water; H⁺ meets OH⁻.酸 + 碱 → 盐 + 水；H⁺ 与 OH⁻ 结合。

• Neutralization reaction中和反应 — a reaction between an acid and a base that produces a salt and water. The net ionic equation for strong acid + strong base is always:— 酸与碱之间的反应，产生盐和水。强酸 + 强碱的净离子方程式始终为：

$$ \text{H}^+(aq) + \text{OH}^-(aq) \to \text{H}_2\text{O}(l) $$

• Molecular equation example:分子方程式示例：

$$ \text{HCl}(aq) + \text{NaOH}(aq) \to \text{NaCl}(aq) + \text{H}_2\text{O}(l) $$

• The salt is the ionic compound formed from the cation of the base and the anion of the acid (here: Na⁺ and Cl⁻ form NaCl).盐是由碱的阳离子与酸的阴离子形成的离子化合物（此处：Na⁺ 和 Cl⁻ 形成 NaCl）。
• For stoichiometry: moles of acid $\times$ (H⁺ per formula unit) = moles of base $\times$ (OH⁻ per formula unit) at the equivalence point.化学计量：在终点处，酸的摩尔数 $\times$（每式量的 H⁺ 数）= 碱的摩尔数 $\times$（每式量的 OH⁻ 数）。

AB Chemistry 20 GO2: "define neutralization as a reaction between hydronium and hydroxide ions." SCH3U Strand C: C2.1 lists "neutralization" in the assessed terminology list; E2.5 "write balanced net ionic equations to represent neutralization reactions."AB Chemistry 20 GO2："将中和定义为水合氢离子与氢氧根离子之间的反应。"SCH3U C 单元：C2.1 把"中和"列入评估术语列表；E2.5"书写平衡的净离子方程式以表示中和反应。"

Titration measures unknown concentration; the equivalence point is where moles match.滴定测定未知浓度；终点处摩尔数匹配。

• Titration滴定 — a technique where a solution of known concentration (the titrant) is added to a solution of unknown concentration (the analyte) until the reaction is complete.— 将已知浓度溶液（滴定液）加入未知浓度溶液（被分析物）中，直到反应完全的技术。
• Equivalence point终点 — the point at which stoichiometrically equivalent amounts of acid and base have been mixed ($n_{\text{H}^+} = n_{\text{OH}^-}$). For strong acid/strong base, the equivalence point is at pH 7. For weak acid/strong base, the equivalence point is at pH $> 7$ (conjugate base is present).— 酸和碱的化学计量等量混合的点（$n_{\text{H}^+} = n_{\text{OH}^-}$）。对于强酸/强碱，终点在 pH 7。对于弱酸/强碱，终点在 pH $> 7$（共轭碱存在）。
• Indicator end point指示剂终点 — the colour change of the indicator. Choose an indicator whose $\text{p}K_a$ is close to the equivalence point pH. AB Chemistry 20 distinguishes end point (indicator change) from equivalence point (stoichiometric point).— 指示剂的颜色变化。选择 $\text{p}K_a$ 接近终点 pH 的指示剂。AB Chemistry 20 区分指示剂终点（颜色变化）与化学计量终点（等量点）。
• Key formula:关键公式：

$$ c_a V_a = c_b V_b \quad \text{(for 1:1 monoprotic acid-base)} $$ SCH3U E2.7: "determine the concentration of an acid or a base in a solution … using the acid–base titration technique." AB Chemistry 20 Unit D GO2: "draw and interpret titration curves, using data from titration experiments involving strong monoprotic acids and strong monoprotic bases; describe the function and choice of indicators in titrations; identify equivalence points."SCH3U E2.7："用酸碱滴定技术测定溶液中酸或碱的浓度。"AB Chemistry 20 D 单元 GO2："用强单质子酸-强单质子碱滴定实验数据绘制和解读滴定曲线；描述指示剂在滴定中的作用和选择；识别等量点。"

A buffer resists pH change when small amounts of acid or base are added.缓冲溶液在加入少量酸或碱时能抵抗 pH 变化。

• Composition组成 — a weak acid (HA) and its conjugate base (A⁻) in comparable amounts, or a weak base (B) and its conjugate acid (BH⁺).— 弱酸（HA）与其共轭碱（A⁻）等量混合，或弱碱（B）与其共轭酸（BH⁺）等量混合。
• How it works工作原理 — add H⁺: the base component (A⁻) absorbs it: $\text{A}^- + \text{H}^+ \to \text{HA}$. Add OH⁻: the acid component (HA) neutralizes it: $\text{HA} + \text{OH}^- \to \text{A}^- + \text{H}_2\text{O}$. In both cases, the ratio [A⁻]/[HA] barely changes, so pH barely changes.— 加入 H⁺：碱性组分（A⁻）吸收它：$\text{A}^- + \text{H}^+ \to \text{HA}$。加入 OH⁻：酸性组分（HA）中和它：$\text{HA} + \text{OH}^- \to \text{A}^- + \text{H}_2\text{O}$。两种情况下，[A⁻]/[HA] 比值几乎不变，故 pH 几乎不变。
• Henderson-Hasselbalch equation:亨德森-哈塞尔巴尔赫方程：

$$ \text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} $$

• Buffer capacity is greatest when $[\text{A}^-] = [\text{HA}]$, i.e., when pH $= \text{p}K_a$.当 $[\text{A}^-] = [\text{HA}]$ 时缓冲容量最大，即 pH $= \text{p}K_a$ 时。

SCH4U E3.8: "describe the chemical characteristics of buffer solutions." AB Chemistry 30 GO1: "define a buffer as relatively large amounts of a weak acid or base and its conjugate in equilibrium that maintain a relatively constant pH."SCH4U E3.8："描述缓冲溶液的化学特性。"AB Chemistry 30 GO1："将缓冲溶液定义为在平衡中含有较多弱酸或弱碱及其共轭物、能维持相对恒定 pH 的溶液。"

Ka and Kb quantify the strength of weak acids and bases; larger K = stronger.Ka 和 Kb 量化弱酸弱碱的强度；K 越大 = 越强。

• Acid ionization constant $K_a$:酸电离常数 $K_a$：

$$ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \qquad K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} $$

• Base ionization constant $K_b$:碱电离常数 $K_b$：

$$ \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \qquad K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} $$

• Relationship between Ka and Kb for a conjugate pair:共轭对的 Ka 与 Kb 关系：

$$ K_a \times K_b = K_w = 1.0 \times 10^{-14} \quad \text{(at 25 °C)} $$

• Square-root approximation平方根近似 (valid when $K_a \ll C_a$, i.e., <5% ionization): $[\text{H}^+] \approx \sqrt{K_a \cdot C_a}$, then pH $= -\log[\text{H}^+]$.（当 $K_a \ll C_a$ 时有效，即电离 <5%）：$[\text{H}^+] \approx \sqrt{K_a \cdot C_a}$，然后 pH $= -\log[\text{H}^+]$。

SCH4U E3.4: "identify common equilibrium constants, including Keq, Ksp, Kw, Ka, Kb" and E2.4: "solve problems … involving Ka, pH, pOH, Kb." AB Chemistry 30 GO2: "define Kw, Ka, Kb and use these to determine pH, pOH, [H₃O⁺] and [OH⁻]." BC Chemistry 12: quantitative Ka/Kb is explicit in the "quantitative relationships" elaboration.SCH4U E3.4："识别常见平衡常数，包括 Keq、Ksp、Kw、Ka、Kb"，以及 E2.4："求解涉及 Ka、pH、pOH、Kb 的问题。"AB Chemistry 30 GO2："定义 Kw、Ka、Kb 并用它们确定 pH、pOH、[H₃O⁺] 和 [OH⁻]。"BC Chemistry 12：Ka/Kb 的定量计算在"定量关系"细化中明确给出。