Equilibria (A2)
A-Level Chemistry Topic 25 7:56 English narration · English + 中文 subtitles burned in
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Your blood is held at pH seven point four.
你的血液被稳定保持在 pH 7.4。
Move it a few tenths either way and you are in hospital.
只要向任一方向偏离零点几，你就得进医院。
Yet all day, acid pours in — from your food, and from your own muscles when you exercise.
可是酸整天不停地产生——来自你吃的食物，也来自你运动时的肌肉。
Something removes it almost as fast as it arrives.
有一种东西几乎在酸到来的同时就把它清除掉，把 pH 稳稳地固定住。
That something is a buffer, and by the end of this lesson you will write the equations that make it work.
那种东西就是缓冲溶液。 学完这节课，你就能写出让它起作用的方程式。
This is the A2 half of equilibrium.
这是平衡这一部分的 A2 内容。
We will calculate pH for strong and weak acids, build a buffer and prove it works, then meet salts that hardly dissolve, and finish with a solute shared between two liquids.
我们会计算强酸和弱酸的 pH， 配出一份缓冲溶液并证明它确实有效，然后认识几乎不溶的盐， 最后讲一个溶质在两种液体之间的分配。 让我们开始吧。
You met conjugate pairs at AS.
共轭酸碱对你在 AS 阶段已经见过。
Here is the A2 version. A conjugate acid-base pair is two species that differ by one proton.
这里是 A2 的说法，一句话： 共轭酸碱对就是相差一个质子的两种微粒。
Take the proton away from an acid and you have its conjugate base. Give a proton to a base and you have its conjugate acid.
把质子从酸上拿走，剩下的就是它的共轭碱； 给碱一个质子，就得到它的共轭酸。
The exam favourite is a species that does both. The hydrogen phosphate ion can lose a proton, or gain one, so it owns both at once.
考试最爱考的是既能给、又能拿的微粒： 磷酸氢根离子既可以失去一个质子，也可以得到一个质子， 所以它同时拥有自己的共轭碱和共轭酸。
Four definitions carry every calculation here.
这个专题的每一道计算都建立在四个定义上。
pH is minus the log of the hydrogen ion concentration, and it runs backwards: the concentration is ten to the power minus pH.
第一，pH 等于氢离子浓度的负对数—— 反过来也成立，所以浓度等于十的负 pH 次方。
For a weak acid, the acid dissociation constant puts the two ions on top and the un-split acid underneath.
第二，对弱酸来说， 酸解离常数是两种离子在上面，未解离的酸在下面。
Its minus log is the pKa value — the smaller the pKa, the stronger the acid.
第三，它的负对数就是 pKa 值： pKa 越小，酸性越强。
And water itself splits a little, giving the ionic product of water.
第四，水本身也会少量解离，由此得到水的离子积， 等于一乘十的负十四次方。
Three kinds of question, three routes.
三类题目，三条不同的路线。
A strong acid is fully ionised, so the hydrogen ion concentration is the concentration of the acid; take minus the log and stop.
强酸完全电离，所以氢离子浓度就等于酸的浓度， 取负对数就做完了。
A strong alkali hands you the hydroxide concentration, so go through the ionic product of water first.
强碱给你的是氢氧根浓度，所以要先通过水的离子积换算成氢离子浓度。
A weak acid is only partly ionised, so take the square root of the acid constant times the concentration.
弱酸只部分电离，所以要用酸常数乘以浓度再开平方根。
Choose the wrong route and the question is gone.
选错路线，整道题就没了。
A real exam question.
一道真题。
The hydrogenchromate ion is a weak acid with a pKa value of six point four nine.
铬酸氢根离子是弱酸，pKa 值为六点四九。
Find the pH at zero point zero two five zero moles per cubic decimetre.
求浓度为零点零二五零摩尔每立方分米的溶液的 pH。
First, turn the pKa back into the constant: ten to the power minus six point four nine is three point two four times ten to the minus seven.
第一步，把 pKa 换回常数： 十的负六点四九次方等于三点二四乘十的负七次方。
Multiply by the concentration and take the square root — the hydrogen ion concentration is eight point nine nine times ten to the minus five.
第二步，把它乘以浓度再开平方根—— 氢离子浓度是八点九九乘十的负五次方。
Minus the log gives a pH of four point zero five.
再取负对数，得到 pH 等于四点零五。
Check: a strong acid this dilute would sit at pH one point six, so a higher value is right.
最后检查：同样稀的强酸 pH 只有一点六，所以弱酸的数值高得多，正好合理。
A buffer solution resists a change in pH when a small amount of acid or alkali is added.
缓冲溶液能在加入少量酸或碱时抵抗 pH 的变化。
You make one from a weak acid together with its own conjugate base — ethanoic acid plus sodium ethanoate.
它由一种弱酸和它自己的共轭碱配成—— 例如乙酸加乙酸钠。 这种混合物同时储备了两个搭档，而这份储备就是全部秘密。
That store of both partners is the secret. Add acid, and the ethanoate ion mops up the extra hydrogen ions.
加入酸时，乙酸根离子把多出来的氢离子清除掉；加入碱时，乙酸把氢氧根离子清除掉。
Add alkali, and the ethanoic acid mops up the hydroxide ions. Your blood does the same job with the hydrogencarbonate ion.
你的血液用碳酸氢根离子做同样的事，正是它把血液稳定在 pH 7.4。
Now a real exam trap.
现在来看一个真实的考试陷阱。
A student mixes propanoic acid with sodium chloride and calls it a buffer. Test it.
一位同学把丙酸和氯化钠混在一起，说这就是缓冲溶液。
Add alkali, and the acid itself reacts with the hydroxide ions, so the pH holds — against alkali it really does buffer.
来检验一下。 加入碱时，酸本身会与氢氧根反应，pH 确实保持住了——对碱它真的能缓冲。
Add acid instead, and nothing removes the extra hydrogen ions, because the conjugate base of propanoic acid is not present.
但换成加入酸，就没有任何东西能清除多出来的氢离子，因为丙酸的共轭碱并不存在。 氯化钠是另一种完全不同的酸的盐，在这里毫无用处。
A buffer needs both partners — the acid and its own conjugate base.
缓冲溶液需要两个搭档都在场——弱酸和它自己的共轭碱。
Now a buffer pH.
现在来算一个缓冲溶液的 pH。
Propanoic acid sits at zero point one zero zero moles per cubic decimetre, sodium propanoate at zero point one five zero, and the acid constant is one point three five times ten to the minus five.
缓冲液中丙酸的浓度是零点一零零摩尔每立方分米， 丙酸钠是零点一五零，酸常数是一点三五乘十的负五次方。
The hydrogen ion concentration is the acid constant times the acid over the salt.
规则很短： 氢离子浓度等于酸常数乘以酸与盐的浓度之比。
So multiply by zero point one zero zero over zero point one five zero: nine times ten to the minus six.
所以乘以零点一零零除以零点一五零， 得到九乘十的负六次方。
Minus the log gives a pH of five point zero five.
取负对数，得到 pH 等于五点零五。
Check: there is more salt than acid, so the pH must sit above the pKa value, four point eight seven.
检查一下：盐比酸多，所以 pH 必须高于四点八七的 pKa 值——确实如此。
Some salts barely dissolve, and for those we use the solubility product.
有些盐几乎不溶，对这类盐我们使用溶度积。
Write the dissolving equilibrium, then multiply the ion concentrations in the saturated solution, each raised to the power of its number in the formula.
先写出溶解平衡， 再把饱和溶液中各离子的浓度相乘，每个浓度按它在化学式中的个数取幂。
For silver chloride that is silver ion times chloride ion.
对氯化银来说，就是银离子浓度乘以氯离子浓度；对氟化钙来说有两个氟离子， 所以氟离子那一项要平方。
For calcium fluoride there are two fluoride ions, so that term is squared.
单位本身也有一分：数一数浓度项的个数。
And the units carry a mark of their own: count the terms — two give moles squared, three give moles cubed.
两项给出摩尔平方、分米的负六次方；三项给出摩尔立方、分米的负九次方。
One more real question, with a trap.
第二道真题，这一道藏着一个陷阱。
The solubility product of silver chromate is one point one two times ten to the minus twelve. Find the silver ion concentration in a saturated solution.
铬酸银的溶度积是一点一二乘十的负十二次方， 求饱和溶液中银离子的浓度。
Write the equation first: one formula unit gives two silver ions and one chromate ion.
先写方程：一个化学式单位给出两个银离子和一个铬酸根离子。
If the solubility is s, the silver is two s and the chromate is s.
所以如果溶解度是 s，银离子浓度就是 2s，铬酸根浓度是 s。
Substitute: two s, squared, times s, is four s cubed.
代入：2s 的平方乘以 s， 等于 4s 的立方。
Divide by four and take the cube root, and s is six point five four times ten to the minus five.
用溶度积除以四再开立方根，得到 s 等于六点五四乘十的负五次方。
But stop — the question asked for the silver ion concentration, and that is two s: one point three one times ten to the minus four.
但请停一下——题目问的是银离子浓度，也就是 2s：一点三一乘十的负四次方。
One last twist.
最后一个转折。
Silver chloride dissolving is an equilibrium, and from AS you know that an equilibrium opposes any change you make.
氯化银的溶解是一个平衡，而你在 AS 就已经知道， 平衡会抵抗你对它做的任何改变。
So drop it into a solution that already contains chloride ions: the extra chloride pushes that equilibrium back towards the solid, and less dissolves.
所以把氯化银放进本来就含有氯离子的溶液里， 多出来的氯离子就把这个平衡推回固体那一侧，溶解的盐就变少了。
This is the common ion effect.
这就是同离子效应。
In pure water, silver chloride reaches about one point three times ten to the minus five moles per cubic decimetre. In a zero point one zero chloride solution it reaches one point eight times ten to the minus nine — seven thousand times less soluble.
在纯水中，氯化银的溶解度约为一点三乘十的负五次方摩尔每立方分米； 在零点一零的氯离子溶液中，它降到一点八乘十的负九次方——大约小七千倍。
Finally, a different equilibrium.
最后，来看另一种平衡。
Shake a solute with two liquids that do not mix, and it spreads between them.
把溶质和两种互不相溶的液体一起振荡，它会在两者之间分布。
The partition coefficient is simply its concentration in one solvent divided by its concentration in the other, at a fixed temperature, and it only applies when the solute is in the same physical state in both.
分配系数就是它在一种溶剂中的浓度除以它在另一种溶剂中的浓度，温度保持不变； 而且只有当溶质在两种溶剂中处于同一物态时才适用。
What decides the value is polarity: like dissolves like.
决定这个数值的是极性： 相似相溶，所以非极性溶质会集中在非极性的那一层。
If eight tenths of the solute ends up in the ether and two tenths in the water, at equal volumes, the coefficient is four.
如果十分之八进入乙醚、十分之二留在水中，且体积相等，那么分配系数就是四。
Non-polar solute, non-polar layer — that is the whole rule.
非极性溶质进入非极性层——这就是全部规则。
Three marks students throw away.
三个学生常丢的分。
First, the units of the solubility product.
第一，溶度积的单位：单位从来不是自动得到的—— 数一数你自己写的表达式里有几个浓度项，每次都把单位写出来。
They are never automatic — count the concentration terms in your own expression and write the units down every time.
第二，缓冲溶液需要两个搭档都在场：一种弱酸和它自己的共轭碱； 别的酸的盐一点用也没有。
Second, a buffer needs both partners: a weak acid and its own conjugate base; the salt of some other acid does nothing.
第三，把两条酸的路线分清楚： 平方根只属于弱酸；对强酸来说，氢离子浓度就等于浓度本身。
Third, keep the two acid routes apart — the square root belongs to a weak acid only.
还有，pH 一律保留两位小数。
And give every pH to two decimal places.
把这些做对，这个专题就是你的了。