酸与碱

AP 化学 · 第 8 主题

训练

讲义词汇表

8.1

酸碱导论

大纲

Learning Objective

Essential Knowledge

8.1.A
Calculate the values of $\mathrm{pH}$ and $\mathrm{pOH}$, based on $K_w$ and the concentration of all species present in a neutral solution of water.

8.1.A.1 The concentrations of hydronium ion and hydroxide ion are often reported as $\mathrm{pH}$ and $\mathrm{pOH}$, respectively.
- Equation: $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$
- Equation: $\mathrm{pOH} = -\log[\mathrm{OH^-}]$
- The terms "hydrogen ion" and "hydronium ion" and the symbols $\mathrm{H^+}(aq)$ and $\mathrm{H_3O^+}(aq)$ are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and $\mathrm{H_3O^+}(aq)$ are preferred, but $\mathrm{H^+}(aq)$ is also accepted on the AP Exam.
8.1.A.2 Water autoionizes with an equilibrium constant $K_w$.
- Equation: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ at 25°C
8.1.A.3 In pure water, $\mathrm{pH} = \mathrm{pOH}$ is called a neutral solution. At 25°C, $\mathrm{p}K_w = 14.0$ and thus $\mathrm{pH} = \mathrm{pOH} = 7.0$.
- Equation: $\mathrm{p}K_w = 14 = \mathrm{pH} + \mathrm{pOH}$ at 25°C
8.1.A.4 The value of $K_w$ is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.

来源:美国大学理事会 AP 课程与考试说明

按布朗斯特-劳里(Brønsted–Lowry)定义,一个酸(acid)是一个质子($\text{H}^+$)给体而一个碱(base)是一个质子受体。当一个酸给出一个质子时它变成它的共轭碱(conjugate base);一个碱获得一个质子变成它的共轭酸(conjugate acid)。水是两性(amphoteric)的——它能作为任一个。

一个强酸完全解离;一个弱酸只部分解离

pH test strips: acid–base strength and concentration set the pH you measure

词汇表训练

英文	中文	拼音
acid	酸	suān
base	碱	jiǎn
conjugate base	共轭碱	gòng è jiǎn
conjugate acid	共轭酸	gòng è suān
weak acid	弱酸	ruò suān

练习卷

8.1

酸碱导论

大纲

Learning Objective

Essential Knowledge

8.1.A
Calculate the values of $\mathrm{pH}$ and $\mathrm{pOH}$, based on $K_w$ and the concentration of all species present in a neutral solution of water.

8.1.A.1 The concentrations of hydronium ion and hydroxide ion are often reported as $\mathrm{pH}$ and $\mathrm{pOH}$, respectively.
- Equation: $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$
- Equation: $\mathrm{pOH} = -\log[\mathrm{OH^-}]$
- The terms "hydrogen ion" and "hydronium ion" and the symbols $\mathrm{H^+}(aq)$ and $\mathrm{H_3O^+}(aq)$ are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and $\mathrm{H_3O^+}(aq)$ are preferred, but $\mathrm{H^+}(aq)$ is also accepted on the AP Exam.
8.1.A.2 Water autoionizes with an equilibrium constant $K_w$.
- Equation: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ at 25°C
8.1.A.3 In pure water, $\mathrm{pH} = \mathrm{pOH}$ is called a neutral solution. At 25°C, $\mathrm{p}K_w = 14.0$ and thus $\mathrm{pH} = \mathrm{pOH} = 7.0$.
- Equation: $\mathrm{p}K_w = 14 = \mathrm{pH} + \mathrm{pOH}$ at 25°C
8.1.A.4 The value of $K_w$ is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.

来源:美国大学理事会 AP 课程与考试说明

即使纯水也导一点点电流,因为水分子彼此反应,这个反应叫做自偶电离(autoionization):

$$2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-.$$

一个水分子把一个质子传给另一个,生成一个水合氢离子(hydronium ion)($\text{H}_3\text{O}^+$,我们松散地写作 $\text{H}^+$ 的那个离子)和一个氢氧根离子。这个平衡有它自己的常数,叫水的离子积(ion-product constant of water):

$$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0\times10^{-14}$$

在 25 °C。定义 $\text{pOH}=-\log[\text{OH}^-]$,对 $K_w$ 表达式两边取 $-\log$,就得到本主题其余部分所依赖的规则:

$$\text{pH} + \text{pOH} = \text{p}K_w = 14.0$$

(同样在 25 °C)。知道 pH 或 pOH 其中之一,用 14 减去就得到另一个。

在纯水里两种离子以相等的数目生成,所以 $[\text{H}_3\text{O}^+]=[\text{OH}^-]$ 而 $\text{pH}=\text{pOH}=7.0$。这就是中性(neutral)的意思。酸性意味着 $[\text{H}_3\text{O}^+]>[\text{OH}^-]$(pH 低于 7);碱性则相反。

Worked example. 一个溶液有 $[\text{H}_3\text{O}^+]=2.0\times10^{-3}\ \text{M}$。求 $[\text{OH}^-]$。因为 $K_w$ 在水里始终成立,

$$[\text{OH}^-]=\frac{K_w}{[\text{H}_3\text{O}^+]}=\frac{1.0\times10^{-14}}{2.0\times10^{-3}}=5.0\times10^{-12}\ \text{M}.$$

氢氧根浓度远小于水合氢离子浓度,证实这个溶液是酸性的。

$K_w$ 是温度依赖的:自偶电离是吸热的,所以加热水使它正向移动并升高 $K_w$。在 50 °C,$K_w>1.0\times10^{-14}$,所以中性水有 $\text{pH}=\text{pOH}<7.0$。它仍然是中性的,因为离子浓度仍然相等——中性意味着离子相等,而不是 pH 恰好为 7。

pH strips: water autoionizes so pure water is pH 7; acids and bases shift H+ and OH-

词汇表训练

英文	中文	拼音
autoionization	自偶电离	zì ǒu diàn lí
hydronium ion	水合氢离子	shuǐ hé qīng lí zi
ion-product constant of water	水的离子积	shuǐ de lí zi jī
neutral	中性	zhōng xìng
pH	pH值	pH zhí

练习卷

8.2

强酸与强碱的 pH 与 pOH

大纲

Learning Objective

Essential Knowledge

8.2.A
Calculate $\mathrm{pH}$ and $\mathrm{pOH}$ based on concentrations of all species in a solution of a strong acid or a strong base.

8.2.A.1 Molecules of a strong acid (e.g., $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HClO_4}$, $\mathrm{H_2SO_4}$, and $\mathrm{HNO_3}$) will completely ionize in aqueous solution to produce hydronium ions and the conjugate base of the acid. As such, the concentration of $\mathrm{H_3O^+}$ in a strong acid solution is equal to the initial concentration of the strong acid, and thus the $\mathrm{pH}$ of the strong acid solution is easily calculated.
8.2.A.2 When dissolved in solution, strong bases (e.g., group I and II hydroxides) completely dissociate to produce hydroxide ions. As such, the concentration of $\mathrm{OH^-}$ in a strong base solution is equal to the initial concentration of a group I hydroxide and double the initial concentration of a group II hydroxide, and thus the $\mathrm{pOH}$ (and $\mathrm{pH}$) of the strong base solution is easily calculated.

来源:美国大学理事会 AP 课程与考试说明

pH值(pH)标度测量酸度:$\text{pH}=-\log[\text{H}^+]$,在 25 °C 有 $\text{pH}+\text{pOH}=14$。一个强酸(strong acid)或强碱(strong base)完全解离,所以它的离子浓度等于它的浓度——直接读 pH。更低的 pH 意味着更酸。

pH 标度:pH 是氢离子浓度的负对数

Worked example. 求 $0.010\ \text{M}$ HCl 的 pH。因为 HCl 是一个强酸它完全解离,所以 $[\text{H}^+]=0.010\ \text{M}$ 而

$$\text{pH}=-\log(0.010)=2.0.$$

对于一个强碱像 $0.010\ \text{M}$ NaOH,$\text{pOH}=2.0$,所以 $\text{pH}=14-2.0=12.0$。

探索

Move along the pH scale

pH measures hydrogen-ion concentration on a log scale: each unit is a tenfold change. Strong acids sit low, strong bases high, 7 is neutral.

词汇表训练

英文	中文	拼音
strong acid	强酸	qiáng suān
strong base	强碱	qiáng jiǎn

练习卷

8.3

弱酸与弱碱的平衡

大纲

Learning Objective

Essential Knowledge

8.3.A
Explain the relationship among $\mathrm{pH}$, $\mathrm{pOH}$, and concentrations of all species in a solution of a monoprotic weak acid or weak base.

8.3.A.1 Weak acids react with water to produce hydronium ions. However, only a small percentage of molecules of a weak acid will ionize in this way. Thus, the concentration of $\mathrm{H_3O^+}$ is much less than the initial concentration of the molecular acid, and the vast majority of the acid molecules remain un-ionized.
8.3.A.2 A solution of a weak acid involves equilibrium between an un-ionized acid and its conjugate base. The equilibrium constant for this reaction is $K_a$, often reported as $\mathrm{p}K_a$. The pH of a weak acid solution can be determined from the initial acid concentration and the $\mathrm{p}K_a$.
- Equation: $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$
- Equation: $\mathrm{p}K_a = -\log K_a$
8.3.A.3 Weak bases react with water to produce hydroxide ions in solution. However, ordinarily just a small percentage of the molecules of a weak base in solution will ionize in this way. Thus, the concentration of $\mathrm{OH^-}$ in the solution does not equal the initial concentration of the base, and the vast majority of the base molecules remain un-ionized.
8.3.A.4 A solution of a weak base involves equilibrium between an un-ionized base and its conjugate acid. The equilibrium constant for this reaction is $K_b$, often reported as $\mathrm{p}K_b$. The $\mathrm{pH}$ of a weak base solution can be determined from the initial base concentration and the $\mathrm{p}K_b$.
- Equation: $K_b = \dfrac{[\mathrm{OH^-}][\mathrm{HB^+}]}{[\mathrm{B}]}$
- Equation: $\mathrm{p}K_b = -\log K_b$
8.3.A.5 The percent ionization of a weak acid (or base) can be calculated from its $\mathrm{p}K_a$ ($\mathrm{p}K_b$) and the initial concentration of the acid (base). The percent ionization can also be calculated from the initial concentration of the acid (base) and the equilibrium concentration of any of the species in the equilibrium expression.
8.3.A.6 For any conjugate acid-base pair, the acid ionization constant and base ionization constant are related by $K_w$:
- Equation: $K_w = K_a \times K_b$
- Equation: $\mathrm{p}K_w = \mathrm{p}K_a + \mathrm{p}K_b$

来源:美国大学理事会 AP 课程与考试说明

一个弱酸(weak acid)只部分解离,由一个平衡常数 $K_a$ 描述(更大的 $K_a$ = 更强的弱酸);一个弱碱有 $K_b$。因为解离很小,用一张 ICE 表和 $K_a$ 表达式求 $[\text{H}^+]$,常常用小-$x$ 近似。

Worked example. 求 $0.10\ \text{M}$ 乙酸的 pH,$K_a=1.8\times10^{-5}$。令 $x=[\text{H}^+]$ 在平衡处;ICE 表给出 $K_a=\dfrac{x^2}{0.10-x}\approx\dfrac{x^2}{0.10}$(小-$x$ 近似),所以

$$x=\sqrt{K_a\times0.10}=\sqrt{1.8\times10^{-6}}=1.3\times10^{-3}\ \text{M}\;\Rightarrow\;\text{pH}=-\log(1.3\times10^{-3})=2.9.$$

弱酸远不如一个相同浓度的强酸酸(它会是 pH $1.0$)。

练习卷

8.4

酸碱反应与缓冲溶液

大纲

Learning Objective

Essential Knowledge

8.4.A
Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.

8.4.A.1 When a strong acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: $\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{H_2O}(l)$. The pH of the resulting solution may be determined from the concentration of excess reagent.
8.4.A.2 When a weak acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: $\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftarrows \mathrm{A^-}(aq)\ \mathrm{H_2O}(l)$. If the weak acid is in excess, then a buffer solution is formed, and the $\mathrm{pH}$ can be determined from the Henderson-Hasselbalch (H–H) equation (see 8.9.A.1). If the strong base is in excess, then the $\mathrm{pH}$ can be determined from the moles of excess hydroxide ion and the total volume of solution. If they are equimolar, then the (slightly basic) $\mathrm{pH}$ can be determined from the equilibrium represented by the equation: $\mathrm{A^-}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{HA}(aq) + \mathrm{OH^-}(aq)$.
8.4.A.3 When a weak base and a strong acid are mixed, they will react quantitatively in a reaction represented by the equation: $\mathrm{B}(aq) + \mathrm{H_3O^+}(aq) \rightleftarrows \mathrm{HB^+}(aq) + \mathrm{H_2O}(l)$. If the weak base is in excess, then a buffer solution is formed, and the $\mathrm{pH}$ can be determined from the H–H equation. If the strong acid is in excess, then the $\mathrm{pH}$ can be determined from the moles of excess hydronium ion and the total volume of solution. If they are equimolar, then the (slightly acidic) $\mathrm{pH}$ can be determined from the equilibrium represented by the equation: $\mathrm{HB^+}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{B}(aq) + \mathrm{H_3O^+}(aq)$.
8.4.A.4 When a weak acid and a weak base are mixed, they will react to an equilibrium state whose reaction may be represented by the equation: $\mathrm{HA}(aq) + \mathrm{B}(aq) \rightleftarrows \mathrm{A^-}(aq) + \mathrm{HB^+}(aq)$.

来源:美国大学理事会 AP 课程与考试说明

一个缓冲溶液(buffer)抵抗 pH 变化。它含一个弱酸和它的共轭碱(或一个弱碱和它的共轭酸)以相当的量。加入的酸被共轭碱中和,而加入的碱被弱酸中和,所以 pH 几乎不移动。

词汇表训练

英文	中文	拼音
buffer	缓冲溶液	huǎn chōng róng yè

练习卷

8.5

酸碱滴定

大纲

Learning Objective

Essential Knowledge

8.5.A
Explain results from the titration of a mono- or polyprotic acid or base solution, in relation to the properties of the solution and its components.

8.5.A.1 An acid-base reaction can be carried out under controlled conditions in a titration. A titration curve, plotting $\mathrm{pH}$ against the volume of titrant added, is useful for summarizing results from a titration.
8.5.A.2 At the equivalence point for titrations of monoprotic acids or bases, the number of moles of titrant added is equal to the number of moles of analyte originally present. This relationship can be used to obtain the concentration of the analyte. This is the case for titrations of strong acids/bases and weak acids/bases.
8.5.A.3 For titrations of weak acids/bases, it is useful to consider the point halfway to the equivalence point, that is, the half-equivalence point. At this point, there are equal concentrations of each species in the conjugate acid-base pair, for example, for a weak acid $[\mathrm{HA}] = [\mathrm{A^-}]$. Because $\mathrm{pH} = \mathrm{p}K_a$ when the conjugate acid and base have equal concentrations, the $\mathrm{p}K_a$ can be determined from the $\mathrm{pH}$ at the half-equivalence point in a titration.
8.5.A.4 At the equivalence point, pH is determined by the major species in solution. Strong acid and strong base titrations result in neutral pH at the equivalence point. However, in titrations of weak acids (weak bases), the conjugate base of the weak acid (conjugate acid of the weak base) is present at the equivalence point and can undergo proton-transfer reactions with the surrounding water, producing basic (acidic) solutions.
8.5.A.5 For polyprotic acids, titration curves can be used to determine the number of acidic protons. In doing so, the major species present at any point along the curve can be identified, along with the $\mathrm{p}K_a$ associated with each proton in a weak polyprotic acid.
- Exclusion statement: Computation of the concentration of each species present in the titration curve for polyprotic acids will not be assessed on the AP Exam. Such computations for titration of monoprotic acids are within the scope of the course (see 8.4.A.2 and 8.4.A.3), as is qualitative reasoning regarding what species are present in large versus small concentrations at any point in a titration of a polyprotic acid.

来源:美国大学理事会 AP 课程与考试说明

酸碱滴定曲线

一条滴定曲线(titration curve)在加入碱(或酸)时画出 pH。关键点:等当点(equivalence point)(酸的摩尔 = 碱的摩尔——一个陡峭的跳跃),和半等当点(half-equivalence point),那里 $\text{pH}=\text{p}K_a$(一半的弱酸被转化,所以一个缓冲溶液在它的中心)。

一种酸碱指示剂(acid-base indicator)本身是一种弱酸,它的质子化形式和去质子化形式有不同的颜色,所以它的颜色随 pH 而响应。你要挑选一种颜色变化发生在滴定等当点 pH 附近的指示剂,这样颜色恰好在反应完成时翻转。

一条滴定曲线在等当点附近有一个陡峭的跳跃

探索

Titrate to the equivalence point

A titration curve rises gently, then sharply at the equivalence point where moles of acid and base match. The steep jump pinpoints that volume.

词汇表训练

英文	中文	拼音
titration curve	滴定曲线	dī dìng qū xiàn
equivalence point	等当点	děng dāng diǎn
acid-base indicator	酸碱指示剂	suān jiǎn zhǐ shì jì

练习卷

8.6

酸碱的分子结构

大纲

Learning Objective

Essential Knowledge

8.6.A
Explain the relationship between the strength of an acid or base and the structure of the molecule or ion.

8.6.A.1 The protons on a molecule that will participate in acid-base reactions, and the relative strength of these protons, can be inferred from the molecular structure.
- i. Strong acids (such as $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HClO_4}$, $\mathrm{H_2SO_4}$, and $\mathrm{HNO_3}$) have very weak conjugate bases that are stabilized by electronegativity, inductive effects, resonance, or some combination thereof.
- ii. Carboxylic acids are one common class of weak acid.
- iii. Strong bases (such as group I and II hydroxides) have very weak conjugate acids.
- iv. Common weak bases include nitrogenous bases such as ammonia as well as carboxylate ions.
- v. Electronegative elements tend to stabilize the conjugate base relative to the conjugate acid, and so increase acid strength.

来源:美国大学理事会 AP 课程与考试说明

强度有结构的根源。一个酸更强,当它的共轭碱更稳定时——例如,更多电负性原子或共振散布负电荷稳定它。对于含氧酸,中心原子上更多氧原子意味着一个更强的酸。

练习卷

8.7

pH 与 pKa

大纲

Learning Objective

Essential Knowledge

8.7.A
Explain the relationship between the predominant form of a weak acid or base in solution at a given $\mathrm{pH}$ and the $\mathrm{p}K_a$ of the conjugate acid or the $\mathrm{p}K_b$ of the conjugate base.

8.7.A.1 The protonation state of an acid or base (i.e., the relative concentrations of $\mathrm{HA}$ and $\mathrm{A^-}$) can be predicted by comparing the $\mathrm{pH}$ of a solution to the $\mathrm{p}K_a$ of the acid in that solution. When solution $\mathrm{pH} < \mathrm{acid}\ \mathrm{p}K_a$, the acid form has a higher concentration than the base form. When solution $\mathrm{pH} > \mathrm{acid}\ \mathrm{p}K_a$, the base form has a higher concentration than the acid form.
8.7.A.2 Acid-base indicators are substances that exhibit different properties (such as color) in their protonated versus deprotonated state, making that property respond to the $\mathrm{pH}$ of a solution.
8.7.A.3 To ensure accurate results in a titration experiment, acid-base indicators should be selected that have a $\mathrm{p}K_a$ close to the $\mathrm{pH}$ at the equivalence point.

来源:美国大学理事会 AP 课程与考试说明

$\text{p}K_a=-\log K_a$:一个更小的 $\text{p}K_a$ 意味着一个更强的酸。把 pH 与 $\text{p}K_a$ 比较告诉你主导的形式:低于 $\text{p}K_a$ 酸形式主导;高于它,共轭碱形式主导。

练习卷

8.8

缓冲溶液的性质

大纲

Learning Objective	Essential Knowledge
8.8.A Explain the relationship between the ability of a buffer to stabilize $\mathrm{pH}$ and the reactions that occur when an acid or a base is added to a buffered solution.	8.8.A.1 A buffer solution contains a large concentration of both members in a conjugate acid-base pair. The conjugate acid reacts with added base and the conjugate base reacts with added acid. These reactions are responsible for the ability of a buffer to stabilize $\mathrm{pH}$.

来源:美国大学理事会 AP 课程与考试说明

缓冲溶液如何抵抗 pH 变化

一个缓冲溶液在弱酸和共轭碱浓度相似(pH 接近 $\text{p}K_a$)时工作最好。选择一个 $\text{p}K_a$ 接近目标 pH 的缓冲溶液。稀释一个缓冲溶液几乎不改变它的 pH,因为酸对碱的比保持相同。

一个缓冲溶液通过吸收加入的酸或碱来抵抗 pH 变化

练习卷

8.9

亨德森-哈塞尔巴尔赫方程

大纲

Learning Objective

Essential Knowledge

8.9.A
Identify the $\mathrm{pH}$ of a buffer solution based on the identity and concentrations of the conjugate acid-base pair used to create the buffer.

8.9.A.1 The $\mathrm{pH}$ of the buffer is related to the $\mathrm{p}K_a$ of the acid and the concentration ratio of the conjugate acid-base pair. This relation is a consequence of the equilibrium expression associated with the dissociation of a weak acid, and is described by the Henderson-Hasselbalch equation. Adding small amounts of acid or base to a buffered solution does not significantly change the ratio of $[\mathrm{A^-}]/[\mathrm{HA}]$ and thus does not significantly change the solution $\mathrm{pH}$. The change in $\mathrm{pH}$ on addition of acid or base to a buffered solution is therefore much less than it would have been in the absence of the buffer.
- Equation: $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$
- Exclusion statement: Computation of the change in pH resulting from the addition of an acid or a base to a buffer will not be assessed on the AP Exam.
- Exclusion statement: Derivation of the Henderson-Hasselbalch equation will not be assessed on the AP Exam.

来源:美国大学理事会 AP 课程与考试说明

对于一个缓冲溶液,pH 由酸对碱的比得出:

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

当浓度相等时,对数是零而 $\text{pH}=\text{p}K_a$。用它来设计一个缓冲溶液或快速求它的 pH。

Worked example. 一个缓冲溶液持有 $0.20\ \text{M}$ 乙酸($\text{p}K_a=4.74$)和 $0.30\ \text{M}$ 乙酸盐。它的 pH 是

$$\text{pH}=4.74+\log\frac{0.30}{0.20}=4.74+\log(1.5)=4.74+0.18=4.92.$$

比酸稍多一点的共轭碱把 pH 推到 $\text{p}K_a$ 稍上,正如方程所预测。

练习卷

8.10

缓冲容量

大纲

Learning Objective

Essential Knowledge

8.10.A
Explain the relationship between the buffer capacity of a solution and the relative concentrations of the conjugate acid and conjugate base components of the solution.

8.10.A.1 Increasing the concentration of the buffer components (while keeping the ratio of these concentrations constant) keeps the $\mathrm{pH}$ of the buffer the same but increases the capacity of the buffer to neutralize added acid or base.
8.10.A.2 When a buffer has more conjugate acid than base, it has a greater buffer capacity for addition of added base than acid. When a buffer has more conjugate base than acid, it has a greater buffer capacity for addition of added acid than base.

来源:美国大学理事会 AP 课程与考试说明

缓冲容量(buffer capacity)是一个缓冲溶液在它的 pH 急剧变化之前能吸收多少酸或碱。它在组分浓且以大致相等的量时最大。一旦一个组分被用尽,缓冲溶液失效。

词汇表训练

英文	中文	拼音
Buffer capacity	缓冲容量	huǎn chōng róng liàng

练习卷

8.11

pH 与溶解度

大纲

Learning Objective	Essential Knowledge
8.11.A Identify the qualitative effect of changes in pH on the solubility of a salt.	8.11.A.1 The solubility of a salt is pH sensitive when one of the constituent ions is a weak acid, a weak base, or the hydroxide ion. These effects can be understood qualitatively using Le Châtelier's principle. Exclusion statement: Computations of solubility as a function of pH will not be assessed on the AP Exam.

来源:美国大学理事会 AP 课程与考试说明

一个带一个碱性阴离子的盐的溶解度在酸性溶液里上升:加入的 $\text{H}^+$ 与阴离子反应,把它从溶解度平衡里移除并拉更多固体溶解(勒沙特列应用于 $K_{sp}$)。所以 pH 能控制一个离子固体是否溶解。

练习卷

8.11

考试技巧

$\text{pH}=-\log[\text{H}^+]$ 是对数的——每一个单位是 $[\text{H}^+]$ 的一个十倍变化。
一个强酸/碱完全解离(所以 $[\text{H}^+]$ = 浓度);一个弱的只部分,需要 $K_a$。
一个缓冲溶液含一个弱酸和它的共轭碱在一起;它通过吸收加入的酸或碱来抵抗 pH 变化。
在一条滴定曲线上等当点是陡峭的跳跃;半等当点有 $\text{pH}=\text{p}K_a$。
一个更小的 $\text{p}K_a$ 意味着一个更强的酸。

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