平衡

When a Brønsted acid loses an $\text{H}^+$, what is left is its conjugate base 共轭碱. When a base gains an $\text{H}^+$, it becomes its conjugate acid 共轭酸. The two species that differ by just one $\text{H}^+$ form a conjugate acid–base pair 共轭酸碱对.

For example, in $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$, the pair is $\text{CH}_3\text{COOH}$ (acid) and $\text{CH}_3\text{COO}^-$ (its conjugate base).

• pH measures acidity: $\text{pH} = -\log[\text{H}^+]$, so $[\text{H}^+] = 10^{-\text{pH}}$.
• the acid dissociation constant 酸解离常数 of a weak acid $\text{HA}$ is $K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$, and $\text{p}K_a = -\log K_a$. A larger $K_a$ (smaller $\text{p}K_a$) means a stronger acid.
• the ionic product of water 水的离子积 is $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $298\ \text{K}$.

Calculating pH

• strong acid 强酸: fully ionised, so $[\text{H}^+]$ equals the acid concentration; then take $-\log$.
• strong alkali 强碱: find $[\text{OH}^-]$ from the concentration, then use $[\text{H}^+] = K_w / [\text{OH}^-]$.
• weak acid 弱酸: only partly ionised, so use $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$.

Worked example. Find the pH of $0.050\ \text{mol dm}^{-3}$ hydrochloric acid (a strong acid).

It is fully ionised, so $[\text{H}^+] = 0.050\ \text{mol dm}^{-3}$, giving $\text{pH} = -\log(0.050) = 1.30$.

Worked example. Find the pH of $0.10\ \text{mol dm}^{-3}$ sodium hydroxide (a strong base). ($K_w = 1.0 \times 10^{-14}$.)

$[\text{OH}^-] = 0.10$, so $[\text{H}^+] = K_w/[\text{OH}^-] = (1.0 \times 10^{-14})/0.10 = 1.0 \times 10^{-13}$, giving $\text{pH} = 13.0$.

Worked example. Find the pH of $0.10\ \text{mol dm}^{-3}$ ethanoic acid (a weak acid). ($K_a = 1.8 \times 10^{-5}$.)

A buffer solution 缓冲溶液 resists a change in pH when a small amount of acid or alkali is added. You make one from a weak acid and its conjugate base (for example ethanoic acid and sodium ethanoate).

It works because the mixture holds a store of both partners:

• added $\text{H}^+$ is removed by the conjugate base: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$.
• added $\text{OH}^-$ is removed by the weak acid: $\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$.

To find the pH, put the concentrations of the acid and its salt into the $K_a$ expression. Buffers are important in living things — for example, $\text{HCO}_3^-$ keeps the pH of blood close to $7.4$.

Worked example (buffer pH). A buffer contains $0.10\ \text{mol dm}^{-3}$ ethanoic acid and $0.20\ \text{mol dm}^{-3}$ sodium ethanoate. $K_a = 1.8 \times 10^{-5}\ \text{mol dm}^{-3}$. Find its pH.

The salt is fully ionised, so $[\text{CH}_3\text{COO}^-] = 0.20$; the weak acid hardly ionises, so $[\text{CH}_3\text{COOH}] = 0.10$. Rearranging the $K_a$ expression:

Do not use $\sqrt{K_a[\text{HA}]}$ here: that formula assumes $[\text{H}^+] = [\text{A}^-]$, which is not true once the salt is added.

For a salt that barely dissolves, the solubility product 溶度积 ($K_{\text{sp}}$) is the product of the ion concentrations in a saturated solution, each raised to the power of its number in the formula:

You can find $K_{\text{sp}}$ from the solubility, or the solubility from $K_{\text{sp}}$.

Worked example. The solubility of calcium fluoride, $\text{CaF}_2$, is $2.0 \times 10^{-4}\ \text{mol dm}^{-3}$. Find $K_{\text{sp}}$.

Each $\text{CaF}_2$ that dissolves gives one $\text{Ca}^{2+}$ and two $\text{F}^-$, so $[\text{Ca}^{2+}] = 2.0 \times 10^{-4}$ and $[\text{F}^-] = 4.0 \times 10^{-4}$:

The common ion effect

The common ion effect 同离子效应 is the way a salt becomes less soluble in a solution that already contains one of its ions. The extra ion pushes the dissolving equilibrium back (Le Chatelier), so less salt dissolves. You can calculate the new solubility using $K_{\text{sp}}$ and the concentration of the common ion.

When a solute is shaken with two solvents that do not mix, it spreads between them. The partition coefficient 分配系数 ($K_{\text{pc}}$) is the ratio of its concentrations in the two layers (at constant temperature):

This works when the solute 溶质 is in the same physical state in both solvents. The value depends on the polarity 极性 of the solute and of each solvent 溶剂: a non-polar solute dissolves more in the non-polar solvent, while a polar solute prefers the polar solvent.

Common mistakes.

• Using $\sqrt{K_a[\text{HA}]}$ for a buffer. Use $[\text{H}^+] = K_a[\text{acid}]/[\text{salt}]$, because the salt supplies most of the $\text{A}^-$.
• In $K_{\text{sp}}$, doubling the concentration of the ion that appears twice but forgetting to square it, or squaring it but forgetting to double it. For $\text{CaF}_2$ with solubility $s$, $K_{\text{sp}} = s(2s)^2 = 4s^3$.
• Saying a buffer keeps the pH constant. It keeps the pH almost constant when small amounts of acid or alkali are added; a large amount uses up the acid or its conjugate base.
• Saying a weak acid has a high pH. A concentrated weak acid can have a lower pH than a dilute strong acid; "weak" means only partly ionised.
• Writing the conjugate base of $\text{H}_2\text{O}$ as $\text{H}_3\text{O}^+$. A conjugate pair differs by one $\text{H}^+$: the conjugate base of water is $\text{OH}^-$ and its conjugate acid is $\text{H}_3\text{O}^+$.

• $\text{pH} = -\log[\text{H}^+]$; for a strong acid $[\text{H}^+]$ = concentration, for a weak acid use $[\text{H}^+] = \sqrt{K_a[\text{HA}]}$.
• For a base, get $[\text{H}^+]$ from $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$.
• For a buffer use $[\text{H}^+] = K_a \times [\text{acid}]/[\text{salt}]$ and explain how it removes added $\text{H}^+$/$\text{OH}^-$.
• Write the $K_{sp}$ expression with the correct powers; the number of decimal places in a pH equals the significant figures of $[\text{H}^+]$.