Acid-Base Calculations: Question Types and Solution Strategies | 酸碱计算题型与解法

📚 Acid-Base Calculations: Question Types and Solution Strategies | 酸碱计算题型与解法

In IB Chemistry HL, acid-base equilibria are a major assessment area. Students are expected to calculate pH, pOH, [H⁺], [OH⁻], K_a, K_b and buffer compositions, and to interpret titration curves. This guide breaks down every common question type and gives a step-by-step solution strategy.

在IB化学HL中，酸碱平衡是重点考查模块。学生需要计算pH、pOH、[H⁺]、[OH⁻]、K_a、K_b，以及缓冲溶液的组成，并解读滴定曲线。本文梳理常见题型，并给出分步解题策略。

1. Overview of Question Types | 考点概览与题型分类

Acid-base calculations can be grouped into strong acid/base, weak acid/base, buffer solutions, salt hydrolysis, mixtures after reaction, and titration curves. Each type requires a different starting equation and set of assumptions.

酸碱计算可分为强酸/强碱、弱酸/弱碱、缓冲溶液、盐水解、反应后混合物以及滴定曲线等类型。不同题型需要选择不同的起始方程和近似条件。

Before attempting any question, first identify whether the species is a strong electrolyte or a weak electrolyte. Strong acids and bases dissociate completely, while weak acids and bases establish an equilibrium with their ions.

做题前首先要判断物质是强电解质还是弱电解质。强酸和强碱完全解离，而弱酸和弱碱则与它们生成的离子之间建立平衡。

ines whether you use pH = -log₁₀[H⁺], an ICE table, or the Henderson-Hasselbalch equation.

同时还必须判断最终溶液中是缓冲体系、过量强酸/强碱，还是仅含盐类。这决定了你该使用 pH = -log₁₀[H⁺]、ICE表格，还是Henderson-Hasselbalch方程。

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2. Strong Acids and Strong Bases | 强酸强碱计算

ydrogen ions is equal to the acid concentration because dissociation is complete: [H⁺] = C_a. Therefore pH = -log₁₀ C_a.

对于一元强酸，由于完全解离，氢离子浓度等于酸的浓度：即 [H⁺] = C_a。因此 pH = -log₁₀ C_a。

base, first calculate the hydroxide ion concentration, then pOH, and finally pH: [OH⁻] = C_b, pOH = -log₁₀ [OH⁻], pH = 14 - pOH at 25 °C.

对于强碱，先计算氢氧根离子浓度，再算pOH，最后求pH：即 [OH⁻] = C_b，pOH = -log₁₀ [OH⁻]，在25 °C时 pH = 14 - pOH。

[H⁺] = C_a    pH = -log₁₀ [H⁺]

[OH⁻] = C_b    pOH = -log₁₀ [OH⁻]

. Do not forget that the volume increases during dilution, so [H⁺] decreases and pH increases.

强酸稀释时，用 C₁V₁ = C₂V₂ 求出新浓度。不要忘记稀释后体积增大，因此 [H⁺] 下降，pH 升高。

When a strong acid and a strong base are mixed, calculate the moles of H⁺ and OH⁻ separately. The smaller amount is fully neutralised, and the excess ions determine the final pH.

当强酸与强碱混合时，分别计算 H⁺ 和 OH⁻ 的物质的量。较少量完全中和，剩余的离子种类决定最终pH。

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3. Weak Acids and Weak Bases | 弱酸弱碱与平衡常数

A weak acid HA dissociates partially according to HA ⇌ H⁺ + A⁻. The equilibrium expression is K_a = [H⁺][A⁻] / [HA].

弱酸 HA 部分解离：HA ⇌ H⁺ + A⁻。其平衡表达式为 K_a = [H⁺][A⁻] / [HA]。

Set up an ICE table: initial concentration C_a, change -x, +x, +x. At equilibrium, [H⁺] = x, [A⁻] = x, and [HA] = C_a - x.

用ICE表格：初始浓度为 C_a，变化量分别为 -x、+x、+x。平衡时，[H⁺] = x，[A⁻] = x，[HA] = C_a - x。

If C_a is much larger than x, the approximation K_a ≈ x² / C_a gives x = √K_aC_a. Always check the 5% rule: x / C_a × 100% < 5%.

如果 C_a 远大于 x，可使用近似式 K_a ≈ x² / C_a，得到 x = √K_aC_a。务必验证5%规则：x / C_a × 100% < 5%。

If the approximation fails, solve the quadratic equation x² + K_ax - K_aC_a = 0 and take the positive root.

如果近似不成立，需要解二次方程 x² + K_ax - K_aC_a = 0，并取正根。

Weak bases are solved in the same way using K_b = x² / (C_b - x). Once x = [OH⁻] is found, calculate pOH = -log₁₀ x and pH = 14 - pOH at 25 °C.

弱碱用同样的方法处理，K_b = x² / (C_b - x)。求出 x = [OH⁻] 后，pOH = -log₁₀ x，再在25 °C下用 pH = 14 - pOH 转换。

For a conjugate acid-base pair, K_a × K_b = K_w .0 × 10⁻¹⁴ at 25 °C. This relationship is essential when solving salt hydrolysis problems.

对于共轭酸碱对，在25 °C下有 K_a × K_b = K_w = 1.0 × 10⁻¹⁴。这个关系式是解决盐水解问题的关键。

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4. Water Autoionisation and K_w | 水的自电离与K_w

Water undergoes autoionisation: H₂O ⇌ H⁺ + OH⁻. The ion product is K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.

水会发生自电离：H₂O ⇌ H⁺ + OH⁻。离子积为 K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴（25 °C）。

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

Taking negative logarithms gives pK_wPublished by TutorHao | IB Chemistry Revision Series | aleveler.com

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