Edexcel化学 酸碱滴定 pH计算 缓冲溶液 Ka
Edexcel A Level Chemistry 的官方 spec（9CH0，2015 年 9 月起施教，2017 年首次考试）把酸碱平衡放在 Topic 12 Acid-base equilibria，紧接 Equilibrium II 之后，Paper 1 和 Paper 3 都会直接考。这一章的计算看似只有几个公式，失分却集中在三处：强弱酸碱的判断、浓度与稀释的换算、缓冲溶液里共轭酸碱比例的取值。很多学生以为 pH 题就是套 -log，一遇到弱酸仍按完全电离处理，一道五分题只剩公式分。滴定曲线的部分更麻烦，曲线形状、半等当点、指示剂选择常被压在同一个大题的六分里。下面按 Bronsted-Lowry 定义、强酸计算、弱酸 Ka、水的离子积、滴定曲线、半等当点与指示剂、缓冲溶液、盐水解这八条线逐段拆开，每一段都配一个可以直接照算的数字例子。
Topic 12 of the Edexcel A Level Chemistry specification (9CH0, first teaching September 2015, first assessment 2017) covers acid-base equilibria, and it is examined in both Paper 1 and Paper 3. The calculations look short, but the marks disappear in three predictable places: deciding whether an acid is strong or weak, converting between concentration and dilution, and picking the right conjugate ratio in a buffer. Many students treat every acid as if it dissociated completely, which turns a five-mark calculation into a single mark for the formula; titration curves are harder still, because the shape, the half-equivalence point and the choice of indicator often share one six-mark question. The sections below follow the specification order, from Bronsted-Lowry definitions through to salt hydrolysis, and each one carries a worked number you can check.
Bronsted-Lowry 定义与共轭酸碱对 / Bronsted-Lowry Pairs
Edexcel 在 Topic 12 里只使用 Bronsted-Lowry 定义：酸是质子给体，碱是质子受体，反应前后恰好相差一个 H+ 的两种微粒互为共轭酸碱对。HCl 失去一个 H+ 变成 Cl-，所以 HCl 与 Cl- 是一对；NH3 得到一个 H+ 变成 NH4+，所以 NH3 与 NH4+ 是另一对。判断是否共轭只看两点：相差的 H+ 是不是一个，电荷数是不是差一。水的特别之处在于两边都能当：H2O 与 H3O+ 构成一对，H2O 与 OH- 构成另一对，这种既能给出又能接受质子的性质称为两性。写弱酸的电离方程式时要把平衡箭头写对，因为 Ka 表达式的分子分母直接来自这个方程式。常见的错误是把水的浓度写进 Ka 的分母，纯液体与固体的浓度不出现在平衡常数里。另一个失分点是方程式没有配平电荷，后面的浓度关系就全部对不上。
Edexcel works with the Bronsted-Lowry definition throughout Topic 12: an acid donates a proton, a base accepts one, and two species that differ by a single H+ are called a conjugate pair. Hydrogen chloride loses a proton to become a chloride ion and ammonia gains one to become the ammonium ion, so both reactions hand you a pair to identify; water sits on both sides of the definition because it can donate a proton to form H3O+ or accept one to form OH-, which is why the specification calls it amphiprotic. Many students write the concentration of water into the denominator of the Ka expression, but pure liquids and solids never appear in an equilibrium constant, and an unbalanced charge in the equation wrecks every concentration ratio that follows.
强酸的 pH 计算与稀释倍数 / Strong Acids and Dilution
强酸在水中完全电离，0.0500 mol dm^-3 的 HCl 溶液里 H+ 浓度就是 0.0500 mol dm^-3，pH 等于 -log(0.0500)，算出来是 1.30。这个关系可以反着用：pH = 1.30 的强酸溶液浓度就是 0.0500 mol dm^-3。H2SO4 需要小心，第一步电离完全、第二步只部分电离，所以 0.0500 mol dm^-3 的硫酸 pH 略大于 1.30，spec 只要求写清两步方程式并说明第二步不完全。稀释的算法是把体积换算成倍数：稀释 10 倍，强酸 pH 上升 1；稀释 100 倍，pH 上升 2。反过来看，pH 从 3 变成 5 说明稀释了 100 倍，不是 2 倍。这类题常与摩尔计算连在一起：取 25.00 cm³ 的 0.100 mol dm^-3 HCl 稀释到 250 cm³，新浓度是 0.0100 mol dm^-3，pH 就是 2.00。很多学生以为稀释题要重新数一遍摩尔数，其实物质的量不变，变的只有体积。
A strong acid dissociates completely, so 0.0500 mol dm^-3 hydrochloric acid has a hydrogen ion concentration of 0.0500 mol dm^-3 and a pH of 1.30, and the reverse reading works just as well. Sulfuric acid needs care because only the first ionisation is complete, which is why 0.0500 mol dm^-3 sulfuric acid sits slightly above pH 1.30; dilution is a matter of multiplying by tens, so a tenfold dilution lifts the pH of a strong acid by 1 and a hundredfold dilution lifts it by 2, which means a pH change from 3 to 5 describes a hundredfold dilution rather than a doubling. A typical question hides the moles inside the volumes, such as 25.00 cm³ of 0.100 mol dm^-3 hydrochloric acid diluted to 250 cm³, giving 0.0100 mol dm^-3 and a pH of 2.00.
弱酸的 Ka 近似计算与电离度 / Weak Acids and Ka
弱酸只部分电离，HA、H+、A- 在溶液中达成平衡，Ka 等于 [H+][A-] 除以 [HA]，单位是 mol dm^-3。Edexcel 的算法建立在两条假设上：忽略水自身电离贡献的 H+，并认为平衡时的 [HA] 约等于初始浓度。于是 [H+] 等于 Ka 乘浓度的平方根，pH 再取负对数。以 0.100 mol dm^-3 的乙酸为例，Ka = 1.75 x 10^-5 mol dm^-3，[H+] = sqrt(1.75 x 10^-5 x 0.100) = 1.32 x 10^-3 mol dm^-3，pH = 2.88。电离度等于电离出的浓度除以初始浓度，上例为 1.32%，远小于 5%，两条假设因此站得住。把乙酸稀释到 0.0100 mol dm^-3，[H+] 降到 4.18 x 10^-4，pH 升到 3.38，只升了 0.5，而强酸稀释 10 倍会升 1，差别来自稀释推动电离平衡右移。错得最多的是把弱酸当强酸算，直接套 -log(0.100) 得 1.00，与真实值 2.88 差了接近两个数量级。
A weak acid only partly ionises, so the equilibrium expression Ka = [H+][A-] / [HA] carries units of mol dm^-3 and the calculation relies on two assumptions: the hydrogen ions from water are ignored, and the equilibrium concentration of HA is taken as the starting concentration. With those in place, [H+] is the square root of Ka multiplied by the concentration, so 0.100 mol dm^-3 ethanoic acid with Ka = 1.75 x 10^-5 mol dm^-3 gives [H+] = 1.32 x 10^-3 mol dm^-3 and a pH of 2.88, a degree of ionisation of 1.32 per cent that sits far below the five per cent the approximation needs. Diluting that acid to 0.0100 mol dm^-3 lowers [H+] only to 4.18 x 10^-4 and lifts the pH to 3.38, a rise of 0.5 against the full unit a strong acid would show, because dilution pushes the ionisation equilibrium to the right; treating the weak acid as strong gives 1.00 and misses the true value by nearly two orders of magnitude.
水的离子积 Kw 与温度影响 / Kw and Temperature
Kw 等于 [H+] 与 [OH-] 的乘积，298 K 时取 1.00 x 10^-14，所以中性水的 pH 是 7.00。在给定温度下，任何水溶液中这两个浓度的乘积都等于 Kw，酸的加入只会改变两者的比例。水的电离是吸热过程，温度升高 Kw 变大，中性水的 pH 降到 7 以下，但溶液仍然是中性，因为 [H+] 与 [OH-] 依旧相等。考试里看到 50 摄氏度的水，不要条件反射写 7.00，先看题目给的 Kw 数值。pOH 与 pH 之和等于 pKw，298 K 时是 14.00，求碱溶液 pH 时先算 pOH 再相减最省时间。例如 0.0100 mol dm^-3 的 NaOH，pOH = 2.00，pH = 14.00 - 2.00 = 12.00。学生最常把碱溶液的 pH 直接写成 pOH，或者忘记 Kw 随温度变化这一步。
The ionic product of water, Kw, is the product of the hydrogen and hydroxide ion concentrations and equals 1.00 x 10^-14 at 298 K, which sets the pH of neutral water at 7.00. Adding an acid shifts the ratio between those two concentrations without changing their product, and because the ionisation of water is endothermic, a higher temperature raises Kw and pushes the pH of neutral water below 7.00 while the solution stays neutral; that is why a question quoting water at 50 degrees should be read with the supplied Kw rather than a reflex 7.00. For alkaline solutions the quickest route runs through pOH, since pH plus pOH equals pKw, so 0.0100 mol dm^-3 sodium hydroxide gives pOH 2.00 and pH 12.00.
滴定曲线的形状与等当点 / Titration Curves
滴定曲线的形状取决于酸的强弱，四种酸碱组合的曲线都要能默画出来。强酸对强碱，起始 pH 低，等当点 pH 为 7.00，等当点前后 pH 从 3 跳到 11，竖直段跨度约 8 个单位。弱酸对强碱（乙酸对 NaOH）起始 pH 偏高、上升缓慢，中途有一段平台，等当点 pH 大于 7，典型值在 8 到 9 之间。等当点偏碱的原因是此时溶液里主要是乙酸钠，乙酸根水解产生 OH-。强酸对弱碱（盐酸对氨水）等当点 pH 小于 7，因为生成的 NH4+ 水解显酸性。画图题的给分点通常落在起始 pH、等当点 pH、等当点体积与竖直段位置这四处。阅卷时看到弱酸弱碱组合配上 pH = 7.00 的等当点就直接判错，这类答案几乎每一轮都会出现。
The shape of a titration curve follows the strength of the acid, and all four combinations have to be reproducible from memory. A strong acid against a strong base starts low and reaches an equivalence point at pH 7.00, with the pH jumping from about 3 to about 11 across a vertical section roughly eight units wide; a weak acid against a strong base starts higher and climbs slowly, showing a flat buffer region and an equivalence point above pH 7, usually between 8 and 9, because the solution there is mainly sodium ethanoate and the ethanoate ion hydrolyses to give hydroxide ions. Reverse the pairing and the equivalence point falls below 7.00, since ammonium ions hydrolyse to make the solution acidic, and the marks in a sketch question are awarded for four things: starting pH, equivalence pH, equivalence volume and the width of the vertical section.
半等当点与指示剂选择 / Half-equivalence and Indicators
弱酸滴定曲线在加入一半碱量处有一个特殊点，此时 [HA] 与 [A-] 相等，代入 Ka 表达式得 [H+] = Ka，所以 pH = pKa，这个点叫半等当点。乙酸的 pKa 为 4.76，半等当点 pH 就是 4.76，题目即使不给 Ka 数值也能靠它反推。指示剂的选择标准是变色范围必须落在滴定曲线的竖直段里：强酸强碱滴定可选甲基橙（3.1 到 4.4）或酚酞（8.3 到 10.0）。弱酸强碱滴定只能用酚酞，强酸弱碱滴定只能用甲基橙，选错指示剂即使滴定操作无误也会得到偏斜的终点体积。常见的错误是以为指示剂要恰好在等当点变色，其实只要在竖直段内变色就行，因为半滴碱就足以让 pH 跨越整个竖直段。指示剂本身是弱酸或弱碱，它的 Ka 决定变色范围，用量必须少，否则指示剂会消耗滴定液并让终点提前。滴定时读数记到 0.05 cm³，平行滴定至少两次落在 0.10 cm³ 以内才算符合要求。
Halfway to the equivalence point of a weak acid titration, the concentrations of HA and A- are equal, so substituting into the Ka expression leaves [H+] equal to Ka and the pH equal to pKa; for ethanoic acid that pins the half-equivalence point at 4.76, which lets a question hide the value of Ka and still expect you to find it. An indicator is chosen on one criterion only, that its colour change lies inside the vertical section of the curve: methyl orange (3.1 to 4.4) or phenolphthalein (8.3 to 10.0) for a strong acid with a strong base, phenolphthalein alone for a weak acid with a strong base, and methyl orange alone for a strong acid with a weak base. Many students believe the indicator has to change colour exactly at the equivalence point, when in practice half a drop of titrant moves the pH across the whole vertical section; the indicator is itself a weak acid or base, so it has to be used sparingly, and burette readings are recorded to 0.05 cm³ with titres agreeing within 0.10 cm³.
缓冲溶液的 pH 与缓冲机理 / Buffers
缓冲溶液由弱酸与它的共轭碱盐组成，乙酸加乙酸钠是典型的酸性缓冲，氨水加氯化铵则是碱性缓冲。缓冲机理可以拆成两个方向：加入少量 H+ 时，H+ 与 A- 结合生成 HA；加入少量 OH- 时，OH- 与 HA 反应生成 A- 和水。两个方向都消耗掉外加的酸碱，所以 pH 变化很小。pH 用 [H+] = Ka 乘 [HA] 除以 [A-] 计算，整理后写成 pH = pKa + log([A-]/[HA])。当 [A-] 与 [HA] 相等时 pH 等于 pKa，缓冲能力最强。例题：0.100 mol dm^-3 的乙酸与 0.0500 mol dm^-3 的乙酸钠等体积混合，浓度比取 1 比 2，pH = 4.76 + log(0.0500/0.100) = 4.46。很多学生以为缓冲溶液能中和任意量的酸碱，其实它只对付少量外加酸碱，加得多了照样失效。
A buffer pairs a weak acid with the salt of its conjugate base, which is why ethanoic acid with sodium ethanoate buffers on the acidic side and ammonia with ammonium chloride buffers on the basic side. Added hydrogen ions are mopped up by A- to form HA, added hydroxide ions are consumed by HA to form A- and water, and either reaction leaves the pH almost unchanged; the arithmetic runs through [H+] = Ka x [HA] / [A-], which rearranges to pH = pKa + log([A-]/[HA]). Equal concentrations of A- and HA put the pH at pKa, the point of maximum buffer capacity, so equal volumes of 0.100 mol dm^-3 ethanoic acid and 0.0500 mol dm^-3 sodium ethanoate give pH = 4.76 + log(0.0500/0.100) = 4.46; many students assume a buffer neutralises any amount of acid or base, when it only copes with a small addition.
盐类水解与溶液酸碱性判断 / Salt Hydrolysis
盐类水解这一块考的是判断溶液酸碱性并给出理由。强酸强碱盐（NaCl、KNO3）不水解，溶液中性，pH 为 7.00。强碱弱酸盐（乙酸钠、碳酸钠）水解显碱性，乙酸根与水反应生成乙酸和 OH-，用 Kb = Kw 除以 Ka 来算。0.100 mol dm^-3 乙酸钠的 [OH-] 约为 7.6 x 10^-6 mol dm^-3，pOH = 5.12，pH = 8.88。强酸弱碱盐（NH4Cl）水解显酸性，NH4+ 给出质子生成 NH3 与 H3O+。弱酸弱碱盐要看 Ka 与 Kb 谁大，谁大就显谁的性质。很多学生以为水解是盐把水分解掉了，实际是离子与水之间的质子转移平衡，判断顺序固定为先看离子来源、再看是否水解、最后才动笔算 pH。
Salt hydrolysis questions ask you to decide whether a solution is acidic, basic or neutral and to justify the answer. Salts of a strong acid with a strong base such as sodium chloride or potassium nitrate do not hydrolyse and leave the pH at 7.00, while a salt of a weak acid with a strong base such as sodium ethanoate turns the solution basic because the ethanoate ion takes a proton from water to leave hydroxide ions, and Kb = Kw / Ka gives [OH-] = 7.6 x 10^-6 mol dm^-3 for a 0.100 mol dm^-3 solution, a pOH of 5.12 and a pH of 8.88. Salts of a strong acid with a weak base such as ammonium chloride go the other way, since the ammonium ion donates a proton to water, and for a salt of a weak acid with a weak base the larger of Ka and Kb decides the side; the reliable order of work is to trace each ion back to its parent acid or base, ask whether it hydrolyses, and only then calculate.
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