Edexcel化学 pH Ka 缓冲溶液 滴定曲线 指示剂
Edexcel A Level 化学的 Topic 12 叫 Acid-base Equilibria，是整套 9CH0 里计算最密的一块。它把 pH、Ka、Kw、缓冲溶液和滴定曲线绑在一起，题目常常一题连三问，第一问的 pH 算错，后面的缓冲比和指示剂选择就一起丢分。这篇文章按考纲顺序拆开讲，每个知识点给出可以直接套用的式子和一个算到答案的例子。
Topic 12 Acid-base Equilibria is the most calculation-heavy part of the Edexcel 9CH0 course. It ties pH, Ka, Kw, buffers and titration curves into chains of questions, where one wrong hydrogen ion concentration costs marks in every later part. This article works through the topic in specification order, with the formula and a worked number for each point.
Topic 12 在 9CH0 的卷面分工与分值
9CH0 的三张卷子分工很清楚，Paper 1 考 Topics 1 到 5 与 8、10 到 15，占 30%，时长 1 小时 45 分钟，满分 90。Paper 2 走有机路线，考 Topics 2、3、5、6、7、9 与 16 到 19，同样 30%。Paper 3 叫 General and Practical Principles in Chemistry，2 小时 30 分钟、120 分、占 40%，范围是 Topics 1 到 19。Topic 12 的内容落在 Paper 1 与 Paper 3，Paper 2 不会单独出酸碱题。Core Practical 9 也挂在 Topic 12 下，Paper 3 的实验题可以直接问它的步骤和数据处理。
Paper 1 covers Topics 1 to 5 and 8, 10 to 15 at 30 percent, lasting 1 hour 45 minutes for 90 marks. Paper 2 takes the organic route through Topics 2, 3, 5, 6, 7, 9 and 16 to 19 at the same weight. Paper 3 is the synoptic paper at 2 hours 30 minutes and 120 marks for 40 percent, so Topic 12 appears in Paper 1 and Paper 3 while Paper 2 never carries a standalone acid-base question.
Brønsted-Lowry 定义与共轭酸碱对
Edexcel 用的是 Brønsted-Lowry 口径：酸是质子给体，碱是质子受体，判断标准只看这一步有没有 H+ 转移。HCl 把质子交给水，水在这里当碱；NH3 从水里拿走质子，水在这里当酸，所以水既能当酸又能当碱。共轭对只差一个 H+，HA 给出质子后剩 A-，A- 就是 HA 的共轭碱，反过来 HA 是 A- 的共轭酸。写共轭对时先找变化前后只差一个质子的两个物种，电荷要跟着一起变。很多学生以为共轭对就是「反应物和生成物」，把 Na+ 和 Cl- 也配成一对，阅卷时这类答案一分不给。
A Brønsted-Lowry acid donates a proton and a base accepts one, so classification depends only on whether H+ moves in that step. Water accepts a proton from HCl and donates one to NH3, which makes it amphoteric. A conjugate pair differs by exactly one proton, so HA and A- form a pair and the charges change together.
强度与浓度是两件事
强酸弱酸说的是电离程度，浓溶液稀溶液说的是每升里有多少摩尔，两件事不能混。0.10 mol dm-3 的 HCl 完全电离，[H+] 等于 0.10，pH 是 1.00。同样 0.10 mol dm-3 的 ethanoic acid 只有约 1.3% 电离，[H+] 约 1.3 x 10-3，pH 约 2.88。把强酸稀释十倍，pH 上升 1.00；把弱酸稀释十倍，pH 只上升约 0.50，因为稀释会促进电离，补回一部分 H+。所以一瓶很稀的 HCl 可以比一瓶很浓的 ethanoic acid 更酸，pH 只反映 [H+]，不反映瓶子上写的浓度。常见的错误是把「弱酸」当成「稀酸」，答题时拿浓度直接当 [H+]，整道计算从这里就错了。
Strength describes how far an acid dissociates while concentration describes how many moles sit in a cubic decimetre. Hydrochloric acid at 0.10 mol dm-3 is fully dissociated and gives pH 1.00, whereas ethanoic acid at the same concentration is only about 1.3 percent dissociated and lands near pH 2.88. A dilute strong acid can therefore be more acidic than a concentrated weak acid, because pH reports hydrogen ion concentration and not the label on the bottle.
pH 与氢离子浓度的双向换算
pH 的定义是 -log10[H+]，反过来 [H+] = 10^-pH，这两个方向都要练到不用想。Edexcel 的答案习惯把 pH 写到两位小数，[H+] 一般给两位或三位有效数字。强一元酸完全电离，[H+] 直接等于酸的浓度，0.010 mol dm-3 的 HNO3 算出 pH 2.00。硫酸是二元酸，按题目要求可以近似当作两步都完全电离，[H+] 取浓度的两倍，这个近似要在答案里写明是假设。很多学生把 10^-pH 按成 -log，从 2.88 算回 [H+] 时差了一个数量级，这一步最容易在第二问连锁失分。
The definition pH = -log[H+] runs in both directions, so [H+] = 10^-pH must be equally automatic. Edexcel answers normally quote pH to two decimal places and hydrogen ion concentration to two or three significant figures. A strong monobasic acid dissociates completely, so nitric acid at 0.010 mol dm-3 gives pH 2.00, while sulfuric acid is treated as doubly dissociated only when the assumption is stated.
Kw 与强碱溶液的 pH 路线
水的离子积 Kw = [H+][OH-]，298 K 时取 1.00 x 10^-14 mol2 dm-6。碱性溶液的题意是先给 [OH-]，再用 [H+] = Kw / [OH-] 换回氢离子浓度。0.050 mol dm-3 的 NaOH 完全电离，[OH-] = 0.050，[H+] = 2.0 x 10^-13，pH 12.70。把这瓶碱稀释十倍，[OH-] 变成 5.0 x 10-3，pH 就从 12.70 降到 11.70。也可以先算 pOH 再取 14 减，两条路结果一样，写清楚用了哪条就行。水的电离是吸热的，温度升高 Kw 变大，中性溶液的 pH 会低于 7，题目给了 298 K 以外的温度时不要默认 14 这个和。常见的错误是把 pOH 直接当 pH 交上去，尤其在 0.10 mol dm-3 强碱这种一眼看出 13 的题里。
Kw equals the product of hydrogen and hydroxide ion concentrations and takes the value 1.00 x 10^-14 mol2 dm-6 at 298 K. For a base the route runs from hydroxide concentration to hydrogen ion concentration through Kw divided by [OH-], so sodium hydroxide at 0.050 mol dm-3 gives pH 12.70. Because the ionisation of water is endothermic, Kw rises with temperature and a neutral solution then sits below pH 7.
Ka 与 pKa 的写法与单位
弱酸 HA 在水里建立平衡 HA <=> H+ + A-，Ka = [H+][A-] / [HA]，单位是 mol dm-3。Ka 越小酸越弱，为了书写方便再定义 pKa = -log10 Ka，pKa 越小酸性越强。ethanoic acid 在 298 K 的 Ka 约 1.74 x 10^-5 mol dm-3，pKa 约 4.76，这两个数在 Core Practical 9 里会直接用到。Ka 只随温度变，加水稀释不会改变 Ka，只会改变 [H+] 和 pH，这一点在稀释题里常被写错。写 Ka 表达式时不要把水的浓度写进去，水是溶剂，活度按 1 处理。
A weak acid sets up the equilibrium HA <=> H+ + A-, and Ka is the product of the two ion concentrations divided by the concentration of intact acid, with units of mol dm-3. A smaller Ka or a larger pKa means a weaker acid, and ethanoic acid gives 1.74 x 10^-5 mol dm-3 and pKa 4.76 at 298 K. Ka changes only with temperature, so dilution alters pH without touching Ka.
弱酸 pH 的两个假设与平方根近似
把 Ka 表达式整理后得到 [H+] = sqrt(Ka x [HA])，这一步依赖两个假设：[H+] 与 [A-] 相等，平衡时的 [HA] 近似等于初始浓度。第一个假设来自电离的化学计量比，第二个假设要求电离度足够小，通常电离度低于 5% 时误差可以接受。0.10 mol dm-3 ethanoic acid 代进去，[H+] = sqrt(1.74 x 10^-5 x 0.10) = 1.32 x 10^-3，pH 2.88，与实测的 2.87 到 2.88 一致。答案里把两个假设写出来本身就有分，只写公式不给假设通常扣一分。很多学生以为这个平方根公式对所有弱酸都成立，遇到浓度很低或 Ka 较大的酸它就会偏差，此时要解二次方程。水自身电离出的 H+ 在极稀溶液里也不能忽略，浓度低到 10-6 mol dm-3 量级时第一个假设先失效。
Rearranging Ka gives the square root expression for hydrogen ion concentration, and it stands only on two assumptions: that [H+] equals [A-] and that the equilibrium concentration of HA stays close to the starting value. Ethanoic acid at 0.10 mol dm-3 gives 1.32 x 10-3 mol dm-3 and pH 2.88. Both assumptions must be stated in the answer, and very dilute solutions or acids with a larger Ka need the quadratic route instead.
Core Practical 9 用半中和法测 Ka
Core Practical 9 的题目就叫 Determine Ka for a weak acid，做法是半中和。先用 25.0 cm3 的 0.10 mol dm-3 ethanoic acid 滴定到 phenolphthalein 刚好变粉，记下到达终点的 NaOH 体积；再往同一只锥形瓶里加入同样 25.0 cm3 的 ethanoic acid。这时被中和的酸正好是一半，[HA] 与 [A-] 相等，Ka = [H+]，pH 数值上等于 pKa。Pearson 的教师手册给的典型读数是 pH 4.77，反算 [H+] 约 1.7 x 10^-5 mol dm-3。pH 计要先拿已知 pH 的缓冲液校准，两次读数之间用去离子水冲洗电极，残留离子会把结果拉偏。这一步最容易丢分的地方不是计算，而是没写清「加入等体积酸后体系处于半中和点」这个前提。
Core Practical 9 asks students to determine Ka for a weak acid by half-neutralising it. A 25.0 cm3 portion of 0.10 mol dm-3 ethanoic acid is titrated against sodium hydroxide to the phenolphthalein end point, then a second identical portion of acid is added so that [HA] equals [A-] and pH equals pKa. The Pearson teacher sheet quotes pH 4.77, which gives [H+] of about 1.7 x 10-5 mol dm-3.
缓冲溶液里发生的两个反应
缓冲溶液能抵抗外加少量酸或碱带来的 pH 变化，酸性缓冲由弱酸和它的共轭碱组成，最常用的是 ethanoic acid 加 sodium ethanoate。外加 H+ 时，共轭碱 A- 把它接走生成 HA；外加 OH- 时，弱酸 HA 把它中和掉生成 A- 和水。两个反应都只消耗少量组分，[HA] 与 [A-] 的比值变化很小，pH 因此基本不动。缓冲能力有上限，加的酸或碱多到把其中一组分耗尽，pH 就会突然跳变。人体血液里的 carbonic acid 与 hydrogencarbonate 体系靠同一套原理把 pH 稳在 7.35 到 7.45。
A buffer resists pH change because it contains a weak acid together with its conjugate base. Added acid is removed by the conjugate base and added hydroxide is removed by the weak acid, so the ratio between the two components barely moves. Buffer capacity is finite, and once one component is exhausted the pH jumps sharply.
缓冲 pH 计算可以直接用摩尔数
缓冲溶液的氢离子浓度用 [H+] = Ka x [HA]/[A-]，取负对数就是 pH = pKa + log([A-]/[HA])。因为两个组分处在同一体积里，体积在比值中约掉，直接用摩尔数代入即可。0.10 mol dm-3 ethanoic acid 与 0.10 mol dm-3 sodium ethanoate 等体积混合，比值为 1，pH 等于 pKa，也就是 4.76。把酸和盐的比例改成 2 比 1，pH = 4.76 + log(0.5) = 4.46，酸多了一份，pH 低了 0.30。常见的错误是把盐和酸的浓度都乘上体积再相除，多算一步还容易把单位弄混。答案要求两位小数时，log 项要算到第三位再进位。配液时把两者的摩尔数算准比量体积更可靠，0.050 mol 的酸配 0.025 mol 的盐，比值就是 2，pH 比 pKa 高 0.30。
Buffer pH follows pH = pKa + log([A-]/[HA]), and because both components share the same volume the mole ratio can be used directly. Equal amounts of ethanoic acid and sodium ethanoate give pH 4.76, while a two to one acid to salt ratio lowers it to 4.46. The logarithm term should be carried to a third decimal place before the final answer is rounded to two.
四种滴定组合的等当点位置
滴定曲线的形状由酸和碱的强弱决定，等当点的 pH 也跟着变。强酸加强碱，等当点在 pH 7，曲线中段有一段几乎垂直的跳跃。弱酸加强碱，等当点偏碱，ethanoic acid 与 NaOH 大约在 8.7，因为生成的 ethanoate 会水解。强酸加弱碱，等当点偏酸，通常在 5 附近，NH4+ 水解给出酸性溶液。弱酸加弱碱，两端都没有明显的垂直段，等当点很难读，这类组合一般不选指示剂滴定。曲线的垂直段越陡，指示剂颜色变化的 pH 范围越容易落进去，读数误差也越小。弱酸的曲线在加入一半碱量处有一段平缓区，此时的 pH 约等于 pKa，与 Core Practical 9 的半中和法对应。
The shape of a titration curve depends on the strength of both partners. Strong acid with strong base gives an equivalence point at pH 7 with a nearly vertical section, weak acid with strong base shifts it to about 8.7, and strong acid with weak base lands near pH 5. A weak acid with a weak base produces no sharp jump, so no indicator can locate the end point reliably.
指示剂选择的 pKa 判据
选指示剂的判据只有一条：它的 pKa 要落在滴定曲线垂直段的中点附近，实际做法是让变色范围覆盖等当点上下各一个 pH 单位。phenolphthalein 的变色范围约 8.3 到 10.0，适合弱酸加强碱；methyl orange 约 3.1 到 4.4，适合强酸加弱碱。强酸加强碱两种都能用，因为垂直段跨过 pH 3 到 11。指示剂本身是弱酸或弱碱，加几滴不会改变体系浓度，加多了才会影响终点。很多学生以为变色范围越宽越好，用了 universal indicator，颜色一路渐变，终点反而读不出来。
An indicator works when its pKa sits inside the steep section of the curve, which in practice means its colour change range covers the equivalence point plus or minus one pH unit. Phenolphthalein changes between about 8.3 and 10.0 and suits weak acid titrations, while methyl orange changes between about 3.1 and 4.4 and suits strong acid with weak base. Either one works for a strong acid and strong base pair because the vertical section spans pH 3 to 11.
中和焓在强弱酸上的差别
强酸与强碱的中和焓几乎相同，约 -57 kJ mol-1，因为溶液里真正发生的反应只有 H+ + OH- -> H2O，其余离子都是旁观离子。换成弱酸，数值会明显变小，部分能量被用来让弱酸继续电离，电离本身吸热。Edexcel 的实验题常把这条规律做成比较题，给你两组温度变化，让你解释为什么弱酸的绝对值更小。测量时热量散失、溶液比热容按 4.18 J g-1 K-1 处理、密度按 1.00 g cm-3 处理，这三点写进答案才算完整。阅卷时错得最多的是把中和焓写成正值，放热反应的焓变必须带负号。热量用 q = mcΔT 计算，m 取溶液总质量，温差取最高温度减起始温度，再除以生成的水的摩尔数才得到中和焓。
Strong acids all give nearly the same enthalpy of neutralisation, about -57 kJ mol-1, because the only reaction in solution is H+ + OH- -> H2O. Weak acids release less heat since some energy is consumed in pushing the remaining acid molecules to ionise. Heat loss and the standard assumptions of 4.18 J g-1 K-1 for specific heat capacity and 1.00 g cm-3 for density belong in the answer.
复习顺序与刷题节奏
Topic 12 的复习顺序建议按计算链走：先把 pH 与 [H+] 的换算练到 30 秒一题，再练 Kw 与强碱，然后 Ka 与弱酸平方根近似，最后接缓冲与滴定曲线。缓冲题的错误大多来自前面 pH 换算出错，前面不稳，后面刷再多也只是重复同一个错误。每做完一套卷，把算错的题按「假设没写」「单位漏了」「有效数字不对」三类归档，一周后回头只看错题本。Core Practical 9 的步骤和两个假设要能默写，Paper 3 的实验题常从这里出两到三分。
A sensible order moves from pH and hydrogen ion conversions to Kw and strong bases, then to Ka and the square root approximation, and only then to buffers and curves. Most buffer errors trace back to an earlier pH slip, so speed on the first step matters more than volume of practice. Sorting wrong answers into missing assumptions, dropped units and rounding faults turns a marked paper into a short revision list.
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