CH12 Unit 6: pH Calculations, Buffers & Titrations — pH 计算、缓冲液和滴定 完整解答¶
状态: ✅ 已完成
创建日期: 2026-02-18
最后更新: 2026-02-18
6.1 Calculations — 计算¶
Q1¶
EN (Original): If the pH of a solution decreases by 1, by how much does the acidity of a solution increase?
中文翻译： 如果溶液的 pH 降低 1，酸度增加多少？
Answer 答案:
EN: Acidity increases by a factor of 10 (pH is logarithmic; each unit = 10× change in [H⁺]).
中文： 酸度增加 10 倍。pH 是对数标度，每变化 1 个单位 = [H⁺] 变化 10 倍。
✅ Answer Key 答案: 10 times — 一致 ✓
Q2¶
EN (Original): If the pH of a solution increases by 3, by how much does the acidity of a solution decrease?
中文翻译： 如果 pH 增加 3，酸度减少多少？
Answer 答案:
EN: Acidity decreases by a factor of 10³ = 1000.
中文： 酸度减少 1000 倍（10³）。
✅ Answer Key 答案: 10³ = 1000 times — 一致 ✓
Q3¶
EN (Original): A 5000. mL sample of HCl has a pH of 2.500, how much water needs to be added to raise the pH to 4.200?
中文翻译： 5000 mL 的 HCl 样品 pH 为 2.500，需要加多少水才能将 pH 升至 4.200？
Answer 答案:
EN:
Initial: [H⁺]ᵢ = 10⁻²·⁵⁰⁰ = 3.162 × 10⁻³ M
Final: [H⁺]f = 10⁻⁴·²⁰⁰ = 6.310 × 10⁻⁵ M
Moles H⁺ stay constant (dilution only):
$$[\text{H}^+]_i \times V_i = [\text{H}^+]_f \times V_f$$
$$V_f = \frac{3.162 \times 10^{-3} \times 5000.}{6.310 \times 10^{-5}} = 250{,}600\ \text{mL}$$
$$\text{Water added} = 250{,}600 - 5000 = 245{,}600 \approx \boxed{2.46 \times 10^5\ \text{mL} = 246\ \text{L}}$$
中文： 稀释不改变 H⁺ 总摩尔数。Vf = 250,600 mL，需加水 = 250,600 - 5000 ≈ 2.46 × 10⁵ mL (246 L)。
✅ Answer Key 答案: 2.46 × 10⁵ mL or 246 L — 一致 ✓
Q4¶
EN (Original): The pH of a solution of Ca(OH)₂ is 8.57. Find the [Ca(OH)₂].
中文翻译： Ca(OH)₂ 溶液的 pH 为 8.57，求 [Ca(OH)₂]。注意它产生 2 个 OH⁻。
Answer 答案:
EN:
$$\text{pOH} = 14.00 - 8.57 = 5.43$$
$$[\text{OH}^-] = 10^{-5.43} = 3.72 \times 10^{-6}\ \text{M}$$
Since Ca(OH)₂ → Ca²⁺ + 2 OH⁻:
$$[\text{Ca(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{3.72 \times 10^{-6}}{2} = \boxed{1.9 \times 10^{-6}\ \text{M}}$$
中文： pOH = 5.43，[OH⁻] = 3.72 × 10⁻⁶ M。Ca(OH)₂ 提供 2 个 OH⁻：[Ca(OH)₂] = [OH⁻]/2 = 1.9 × 10⁻⁶ M。
✅ Answer Key 答案: 1.9 × 10⁻⁶ M — 一致 ✓
Q5¶
EN (Original): How small must the percent ionization be for the "x is small" assumption to be valid?
中文翻译： 百分电离度必须多小才能使"x 很小"的假设成立？
Answer 答案:
EN: The percent ionization must be less than 5% for the assumption to be valid.
中文： 百分电离度必须小于 5%。
Q6¶
EN (Original): Find the pH of a 1.75 M solution of HF.
中文翻译： 求 1.75 M HF 溶液的 pH。
Answer 答案:
EN: Ka(HF) = 3.5 × 10⁻⁴
| HF | H₃O⁺ | F⁻ | |
|---|---|---|---|
| I | 1.75 | 0 | 0 |
| C | −x | +x | +x |
| E | 1.75−x | x | x |
$$K_a = \frac{x^2}{1.75 - x} \approx \frac{x^2}{1.75}$$
$$x^2 = 3.5 \times 10^{-4} \times 1.75 = 6.125 \times 10^{-4}$$
$$x = 0.02475\ \text{M} \quad \text{Check: } \frac{0.02475}{1.75} = 1.4\% < 5\%\ ✓$$
$$\text{pH} = -\log(0.02475) = \boxed{1.606}$$
中文： ICE 表计算，假设 x ≪ 1.75（验证：1.4% < 5% ✓）。pH = 1.606。
✅ Answer Key 答案: pH = 1.606 — 一致 ✓
Q7¶
EN (Original): Find the % ionization of a 1.75 M solution of HF.
中文翻译： 求 1.75 M HF 溶液的百分电离度。
Answer 答案:
$$\%\ \text{ionization} = \frac{x}{[\text{HF}]_0} \times 100 = \frac{0.02475}{1.75} \times 100 = \boxed{1.4\%}$$
✅ Answer Key 答案: 1.4% — 一致 ✓
Q8¶
EN (Original): Find the pH of a 1.85 M solution of HSO₄⁻. Note: Ka is large enough that you can NOT assume x is much smaller than 1.85!
中文翻译： 求 1.85 M HSO₄⁻ 溶液的 pH。注意：Ka 较大，不能假设 x ≪ 1.85！
Answer 答案:
EN: Ka(HSO₄⁻) = 1.2 × 10⁻²
| HSO₄⁻ | H₃O⁺ | SO₄²⁻ | |
|---|---|---|---|
| I | 1.85 | 0 | 0 |
| C | −x | +x | +x |
| E | 1.85−x | x | x |
$$K_a = \frac{x^2}{1.85 - x} = 1.2 \times 10^{-2}$$
Cannot assume x is small → use quadratic:
$$x^2 + 0.012x - 0.0222 = 0$$
$$x = \frac{-0.012 + \sqrt{(0.012)^2 + 4(0.0222)}}{2} = \frac{-0.012 + \sqrt{0.0889}}{2} = \frac{-0.012 + 0.2982}{2} = 0.1431$$
$$\text{pH} = -\log(0.1431) = \boxed{0.844}$$
中文： Ka 较大，必须用二次方程。x² + 0.012x − 0.0222 = 0，解得 x = 0.1431。pH = 0.844。
✅ Answer Key 答案: pH = 0.844 — 一致 ✓
Q9¶
EN (Original): Find the % ionization of a 1.85 M solution of HSO₄⁻.
中文翻译： 求百分电离度。
Answer 答案:
$$\%\ \text{ionization} = \frac{0.1431}{1.85} \times 100 = \boxed{7.73\%}$$
✅ Answer Key 答案: 7.73% — 一致 ✓
Q10¶
EN (Original): A 0.045 M solution of hypobromous acid (HBrO) has a pH of 5.02. Calculate Ka.
中文翻译： 0.045 M 次溴酸 (HBrO) 溶液的 pH 为 5.02。计算 Ka。
Answer 答案:
EN:
$$[\text{H}_3\text{O}^+] = 10^{-5.02} = 9.55 \times 10^{-6}\ \text{M} = x$$
| HBrO | H₃O⁺ | BrO⁻ | |
|---|---|---|---|
| I | 0.045 | 0 | 0 |
| C | −x | +x | +x |
| E | 0.045−x ≈ 0.045 | x | x |
$$K_a = \frac{x^2}{0.045} = \frac{(9.55 \times 10^{-6})^2}{0.045} = \frac{9.12 \times 10^{-11}}{0.045} = \boxed{2.0 \times 10^{-9}}$$
中文： 从 pH 求得 x = 9.55 × 10⁻⁶ M，Ka = x²/0.045 = 2.0 × 10⁻⁹。
✅ Answer Key 答案: Ka = 2.0 × 10⁻⁹ — 一致 ✓
Q11¶
EN (Original): A 750. mL sample of nitrous acid (HNO₂) has a pH of 2.35. How many grams of HNO₂ were initially added?
中文翻译： 750 mL 的亚硝酸 (HNO₂) 样品 pH 为 2.35。最初加入了多少克 HNO₂？
Answer 答案:
EN: Ka(HNO₂) = 4.6 × 10⁻⁴
$$x = [\text{H}_3\text{O}^+] = 10^{-2.35} = 4.467 \times 10^{-3}\ \text{M}$$
| HNO₂ | H₃O⁺ | NO₂⁻ | |
|---|---|---|---|
| I | C | 0 | 0 |
| C | −x | +x | +x |
| E | C−x | x | x |
$$K_a = \frac{x^2}{C - x} \Rightarrow C - x = \frac{(4.467 \times 10^{-3})^2}{4.6 \times 10^{-4}} = \frac{1.995 \times 10^{-5}}{4.6 \times 10^{-4}} = 0.04338$$
$$C = 0.04338 + 0.004467 = 0.04784\ \text{M}$$
$$\text{Moles} = 0.04784 \times 0.750 = 0.03588\ \text{mol}$$
$$M_{\text{HNO}_2} = 1.01 + 14.01 + 2(16.00) = 47.02\ \text{g/mol}$$
$$\text{Mass} = 0.03588 \times 47.02 = \boxed{1.7\ \text{g}}$$
中文： 从 pH 求得 x，反求初始浓度 C = 0.0478 M。在 750 mL 中：0.0359 mol × 47.02 g/mol = 1.7 g。
✅ Answer Key 答案: 1.7 g — 一致 ✓
Q12¶
EN (Original): What is the value of Kb for CN⁻?
中文翻译： CN⁻ 的 Kb 值是多少？
Answer 答案:
$$K_b(\text{CN}^-) = \frac{K_w}{K_a(\text{HCN})} = \frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}} = \boxed{2.0 \times 10^{-5}}$$
✅ Answer Key 答案: 2.0×10⁻⁵ — 一致 ✓
Q13¶
EN (Original): Find the pH of a 0.75 M solution of NaF.
中文翻译： 求 0.75 M NaF 溶液的 pH。
Answer 答案:
EN: NaF → Na⁺ + F⁻. F⁻ is the conjugate base of weak acid HF → hydrolyzes:
F⁻ + H₂O ⇌ HF + OH⁻
$$K_b(\text{F}^-) = \frac{K_w}{K_a(\text{HF})} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} = 2.86 \times 10^{-11}$$
| F⁻ | HF | OH⁻ | |
|---|---|---|---|
| I | 0.75 | 0 | 0 |
| C | −x | +x | +x |
| E | 0.75−x | x | x |
$$x^2 = 2.86 \times 10^{-11} \times 0.75 = 2.14 \times 10^{-11}$$
$$x = 4.63 \times 10^{-6}\ \text{M} = [\text{OH}^-]$$
$$\text{pOH} = -\log(4.63 \times 10^{-6}) = 5.33$$
$$\text{pH} = 14.00 - 5.33 = \boxed{8.67}$$
中文： F⁻ 水解为碱性。Kb = 2.86 × 10⁻¹¹，x = 4.63 × 10⁻⁶ M，pOH = 5.33，pH = 8.67。
✅ Answer Key 答案: pH = 8.67 — 一致 ✓
Q14¶
EN (Original): A 2.0 M solution of a weak base B⁻ has a pH of 10.67. Determine Kb for B⁻.
中文翻译： 2.0 M 弱碱 B⁻ 溶液的 pH 为 10.67。求 Kb。
Answer 答案:
$$\text{pOH} = 14.00 - 10.67 = 3.33$$
$$[\text{OH}^-] = 10^{-3.33} = 4.68 \times 10^{-4}\ \text{M} = x$$
$$K_b = \frac{x^2}{C - x} = \frac{(4.68 \times 10^{-4})^2}{2.0 - 4.68 \times 10^{-4}} = \frac{2.19 \times 10^{-7}}{2.0} = \boxed{1.1 \times 10^{-7}}$$
✅ Answer Key 答案: Kb = 1.1×10⁻⁷ — 一致 ✓
Q15¶
EN (Original): Identify the base B⁻ from question 14.
中文翻译： 识别 Q14 中的碱 B⁻。
Answer 答案:
EN: Find Ka of conjugate acid HB:
$$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-7}} = 9.1 \times 10^{-8}$$
From data table: Ka = 9.1 × 10⁻⁸ corresponds to H₂S (hydrogen sulphide).
Therefore B⁻ = HS⁻ (hydrogen sulphide ion / bisulphide ion).
中文： Ka(HB) = 9.1 × 10⁻⁸ → 对应 H₂S。所以 B⁻ = HS⁻（硫氢根离子）。
✅ Answer Key 答案: B⁻ is HS⁻ — 一致 ✓
Q16¶
EN (Original): Ethylamine (C₂H₅NH₂) has Kb = 5.60 × 10⁻⁴. pH = 12.215. Find initial [C₂H₅NH₂].
中文翻译： 乙胺的 Kb = 5.60 × 10⁻⁴，溶液 pH = 12.215。求初始浓度。
Answer 答案:
$$\text{pOH} = 14.000 - 12.215 = 1.785$$
$$[\text{OH}^-] = 10^{-1.785} = 1.641 \times 10^{-2}\ \text{M} = x$$
$$K_b = \frac{x^2}{C - x} \Rightarrow C - x = \frac{(1.641 \times 10^{-2})^2}{5.60 \times 10^{-4}} = \frac{2.693 \times 10^{-4}}{5.60 \times 10^{-4}} = 0.4809$$
$$C = 0.4809 + 0.01641 = \boxed{0.497\ \text{M}}$$
中文： 从 pH 求得 [OH⁻] = 0.01641 M，反求浓度 C = 0.4809 + 0.01641 = 0.497 M。
✅ Answer Key 答案: 0.497 M — 一致 ✓
Q17¶
EN (Original): What will happen to both the pH and the pOH of pure water if you cool it below 25°C? Explain.
中文翻译： 将纯水冷却到 25°C 以下，pH 和 pOH 会怎样变化？解释原因。
Answer 答案:
EN: Both pH and pOH will increase (rise above 7.00).
The dissociation of water is endothermic. Cooling (removing heat) shifts equilibrium to the left, producing fewer H₃O⁺ and OH⁻ ions → Kw decreases below 1.0 × 10⁻¹⁴.
In pure water, [H⁺] = [OH⁻] always (neutral). Since both decrease, pH = −log[H⁺] increases and pOH = −log[OH⁻] also increases. The water is still neutral (not basic) because [H⁺] = [OH⁻]; the neutral point simply shifts above 7.00.
中文： pH 和 pOH 都升高（超过 7.00）。水的解离是吸热反应，降温使平衡左移，[H⁺] 和 [OH⁻] 同时减小，Kw 变小。水仍然是中性的（[H⁺] = [OH⁻]），只是中性点不再是 7.00。
Q18¶
EN (Original): The [H⁺] of pure water at 30°C is 1.4×10⁻⁷. a) pH? b) [OH⁻]? c) Acidic, basic or neutral? d) Calculate pKw.
中文翻译： 30°C 纯水中 [H⁺] = 1.4×10⁻⁷。求 pH、[OH⁻]、酸碱性、pKw。
Answer 答案:
a) pH = −log(1.4 × 10⁻⁷) = 6.85
b) Pure water: [OH⁻] = [H⁺] = 1.4 × 10⁻⁷ M
c) Neutral. Even though pH < 7, the solution is neutral because [H⁺] = [OH⁻]. At 30°C, the neutral pH is 6.85, not 7.00. The pH = 7.00 neutral point applies only at 25°C.
d) Kw = [H⁺][OH⁻] = (1.4 × 10⁻⁷)² = 1.96 × 10⁻¹⁴
$$\text{p}K_w = -\log(1.96 \times 10^{-14}) = \boxed{13.70}$$
中文： a) pH = 6.85 b) [OH⁻] = 1.4 × 10⁻⁷ M c) 中性（[H⁺] = [OH⁻]，30°C 中性点不是 7） d) pKw = 13.70。
✅ Answer Key 答案: d) pKw = 13.70 — 一致 ✓
Q19¶
EN (Original): Which compound causes 70% of the acid in acid rain?
中文翻译： 哪种化合物造成酸雨中 70% 的酸？
Answer 答案:
EN: SO₂ (sulphur dioxide), which forms H₂SO₃ and ultimately H₂SO₄ (sulphuric acid) in the atmosphere.
中文： SO₂（二氧化硫），在大气中形成 H₂SO₃ 和 H₂SO₄（硫酸）。
Q20¶
EN (Original): What substance is most commonly used to reverse the effect of acid rain?
中文翻译： 最常用什么物质来逆转酸雨的影响？
Answer 答案:
EN: CaCO₃ (calcium carbonate / limestone). It's a base that neutralizes the acid: CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂.
中文： CaCO₃（碳酸钙/石灰石）。作为碱中和酸。
Q21¶
EN (Original): Fill in: Metal oxide and non-metal oxide — acid or base anhydride?
中文翻译： 金属氧化物和非金属氧化物分别是酸酐还是碱酐？
Answer 答案:
| Anhydride 酸碱酐 | Acid/Base 酸/碱 |
|---|---|
| Metal oxide (ionic) 金属氧化物 | Base 碱 |
| Non-metal oxide 非金属氧化物 | Acid 酸 |
6.2 More about Acids and Bases — 酸碱深入¶
6.2 Q1¶
EN (Original): Define buffer.
中文翻译： 定义缓冲液。
Answer 答案:
EN: A buffer is a solution that resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in significant amounts.
中文： 缓冲液 是在加入少量酸或碱时能抵抗 pH 变化的溶液。由弱酸及其共轭碱（或弱碱及其共轭酸）以足量组成。
6.2 Q2¶
EN (Original): What type of acid and base can we NOT use to prepare a buffer?
中文翻译： 我们不能用哪种类型的酸碱来制备缓冲液？
Answer 答案:
EN: We cannot use a strong acid and a strong base — they fully ionize and completely neutralize each other, leaving no conjugate pair to buffer pH changes.
中文： 不能使用强酸和强碱 — 它们完全电离并互相中和，没有共轭对来缓冲 pH 变化。
6.2 Q3¶
EN (Original): To create an equimolar buffer using 1.0 mol of NaHS in 1.0 L of water, what acid would you need? How many moles?
中文翻译： 用 1.0 mol NaHS 在 1.0 L 水中制备等摩尔缓冲液，需要什么酸？多少摩尔？
Answer 答案:
EN: NaHS → Na⁺ + HS⁻. HS⁻ is amphiprotic. To make an equimolar buffer of HS⁻/H₂S, add a strong acid (e.g., HCl) to convert half of HS⁻ to H₂S:
HS⁻ + H⁺ → H₂S
Need 0.50 mol HCl to convert 0.50 mol HS⁻ → 0.50 mol H₂S. Result: 0.50 mol HS⁻ + 0.50 mol H₂S (equimolar).
中文： 加入0.50 mol HCl（强酸）将一半 HS⁻ 转化为 H₂S，得到 0.50 mol HS⁻ + 0.50 mol H₂S 等摩尔缓冲液。
6.2 Q4¶
EN (Original): Write an equilibrium equation for the above buffer. Acid on the left. No spectator ions.
中文翻译： 写出上述缓冲液的平衡方程，酸在左侧，不含旁观离子。
Answer 答案:
$$\text{H}_2\text{S}(aq) \rightleftharpoons \text{H}^+(aq) + \text{HS}^-(aq)$$
6.2 Q5¶
EN (Original): What happens if a small amount of acid is added? Direction of shift and effect on pH?
中文翻译： 加入少量酸后平衡向哪个方向移动？pH 如何变化？
Answer 答案:
EN: Equilibrium shifts left (←). Added H⁺ is consumed by HS⁻: HS⁻ + H⁺ → H₂S. pH decreases slightly but much less than without a buffer.
中文： 平衡左移。加入的 H⁺ 被 HS⁻ 消耗：HS⁻ + H⁺ → H₂S。pH 略微降低，但远小于无缓冲液时的变化。
6.2 Q6¶
EN (Original): What happens if a small amount of base is added?
中文翻译： 加入少量碱后呢？
Answer 答案:
EN: Equilibrium shifts right (→). Added OH⁻ reacts with H₂S: H₂S + OH⁻ → HS⁻ + H₂O. pH increases slightly but is buffered.
中文： 平衡右移。OH⁻ 与 H₂S 反应：H₂S + OH⁻ → HS⁻ + H₂O。pH 略微升高，但被缓冲。
6.2 Q7¶
EN (Original): Does ammonia act as an acid or base in water? Write the formula for the compound it forms.
中文翻译： 氨在水中是酸还是碱？写出它形成的化合物。
Answer 答案:
EN: Ammonia acts as a base in water:
$$\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$$
It forms NH₄OH (ammonium hydroxide).
中文： 氨在水中是碱，形成 NH₄OH（氢氧化铵）。
6.2 Q8¶
EN (Original): What is the pOH and pH of a 0.250 M solution of NH₄OH at 25°C?
中文翻译： 25°C 时 0.250 M NH₄OH 溶液的 pOH 和 pH。
Answer 答案:
EN:
$$K_b(\text{NH}_3) = \frac{K_w}{K_a(\text{NH}_4^+)} = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-10}} = 1.79 \times 10^{-5}$$
| NH₄OH | NH₄⁺ | OH⁻ | |
|---|---|---|---|
| I | 0.250 | 0 | 0 |
| C | −x | +x | +x |
| E | 0.250−x | x | x |
$$x^2 = 1.79 \times 10^{-5} \times 0.250 = 4.46 \times 10^{-6}$$
$$x = 2.11 \times 10^{-3}\ \text{M}\ [\text{OH}^-] \quad (0.85\% < 5\%\ ✓)$$
$$\text{pOH} = -\log(2.11 \times 10^{-3}) = \boxed{2.67}$$
$$\text{pH} = 14.00 - 2.67 = \boxed{11.33}$$
中文： Kb = 1.79 × 10⁻⁵，x = 2.11 × 10⁻³ M。pOH = 2.67，pH = 11.33。
✅ Answer Key 答案: pOH = 2.67, pH = 11.33 — 一致 ✓
6.2 Q9¶
EN (Original): What three-letter symbol do we use as a short form for an indicator?
中文翻译： 指示剂的三字母缩写是什么？
Answer 答案:
EN: HIn (the protonated/acid form of the indicator). The color change occurs between HIn and In⁻.
中文： HIn（指示剂的酸式，即质子化形式）。颜色变化发生在 HIn 和 In⁻ 之间。
6.2 Q10¶
EN (Original): What is the Ka for the second ionization of the indicator Thymol blue?
中文翻译： 指示剂百里酚蓝第二次电离的 Ka 是多少？
Answer 答案:
EN: Thymol blue has two transitions. The second transition range is pH 8.0 – 9.6 (yellow to blue).
At the midpoint: pH = pKa
$$\text{pH}_{\text{mid}} = \frac{8.0 + 9.6}{2} = 8.8$$
$$K_a = 10^{-8.8} = \boxed{2 \times 10^{-9}}$$
中文： 百里酚蓝第二变色范围 pH 8.0–9.6，中点 pH = 8.8。Ka = 10⁻⁸·⁸ ≈ 2 × 10⁻⁹。
✅ Answer Key 答案: Ka = 2 × 10⁻⁹ — 一致 ✓
6.2 Q11¶
EN (Original): What is the Kb for the indicator phenolphthalein?
中文翻译： 酚酞指示剂的 Kb 是多少？
Answer 答案:
EN: Phenolphthalein range: pH 8.2 – 10.0
$$\text{pH}_{\text{mid}} = \frac{8.2 + 10.0}{2} = 9.1$$
$$K_a = 10^{-9.1} = 7.94 \times 10^{-10}$$
$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{7.94 \times 10^{-10}} = \boxed{1 \times 10^{-5}}$$
中文： 酚酞变色范围 pH 8.2–10.0，中点 pH = 9.1。Ka = 10⁻⁹·¹。Kb = Kw/Ka = 1 × 10⁻⁵。
✅ Answer Key 答案: Kb = 1 × 10⁻⁵ — 一致 ✓
6.3 Titrations — 滴定¶
6.3 Q1¶
EN (Original): Provide two definitions for the word titration.
中文翻译： 给"滴定"下两个定义。
Answer 答案:
EN:
- An analytical procedure where a solution of known concentration (titrant) is gradually added to a solution of unknown concentration (analyte) until the reaction is complete.
- A volumetric method used to determine the concentration of an unknown solution by measuring the volume of a standard solution required to react completely with it.
中文：
- 将已知浓度的溶液（滴定剂）逐步加入未知浓度的溶液（被分析物）直至反应完成的分析方法。
- 通过测量标准溶液完全反应所需体积来确定未知溶液浓度的容量分析法。
6.3 Q2¶
EN (Original): What is the point at which a titration results in a color change called?
中文翻译： 滴定中指示剂变色的那个点叫什么？
Answer 答案:
EN: The endpoint (终点). This is different from the equivalence point (the theoretical point where moles of acid = moles of base).
中文： 终点（endpoint）。它不同于等当量点（equivalence point，即酸碱摩尔数相等的理论点）。
6.3 Q3¶
EN (Original): Complete and balance: __ H₃PO₄(aq) + __ Sr(OH)₂(aq) →
中文翻译： 完成并配平方程。
Answer 答案:
$$2\ \text{H}_3\text{PO}_4(aq) + 3\ \text{Sr(OH)}_2(aq) \rightarrow \text{Sr}_3(\text{PO}_4)_2(s) + 6\ \text{H}_2\text{O}(l)$$
验证： H: 6+6=12→12 ✓; P: 2→2 ✓; Sr: 3→3 ✓; O: 8+6=14→8+6=14 ✓
6.3 Q4¶
EN (Original): Write a balanced complete ionic equation for Q3.
中文翻译： 写出 Q3 的完全离子方程式。
Answer 答案:
H₃PO₄ is a weak acid (stays molecular). Sr(OH)₂ is a strong base (fully ionizes). Sr₃(PO₄)₂ is a precipitate (solid).
$$2\ \text{H}_3\text{PO}_4(aq) + 3\ \text{Sr}^{2+}(aq) + 6\ \text{OH}^-(aq) \rightarrow \text{Sr}_3(\text{PO}_4)_2(s) + 6\ \text{H}_2\text{O}(l)$$
6.3 Q5¶
EN (Original): Write a balanced net ionic equation for Q3 and Q4.
中文翻译： 写出净离子方程式。
Answer 答案:
No spectator ions to remove (all ions participate in the reaction):
$$2\ \text{H}_3\text{PO}_4(aq) + 3\ \text{Sr}^{2+}(aq) + 6\ \text{OH}^-(aq) \rightarrow \text{Sr}_3(\text{PO}_4)_2(s) + 6\ \text{H}_2\text{O}(l)$$
中文： 没有旁观离子可以删除，净离子方程与完全离子方程相同。
6.3 Q6¶
EN (Original): Strong acid + strong base → equivalence point pH?
中文翻译： 强酸 + 强碱的等当量点 pH 是多少？
Answer 答案:
pH = 7 (neutral). Neither the conjugate acid nor the conjugate base hydrolyzes.
6.3 Q7¶
EN (Original): Weak acid + strong base → equivalence point pH above or below 7?
Answer 答案:
Above 7 (basic). At equivalence, the conjugate base of the weak acid remains → hydrolyzes to produce OH⁻.
中文： 高于 7（碱性）。等当量点时，弱酸的共轭碱水解产生 OH⁻。
6.3 Q8¶
EN (Original): Strong acid + weak base → equivalence point pH above or below 7?
Answer 答案:
Below 7 (acidic). At equivalence, the conjugate acid of the weak base remains → hydrolyzes to produce H⁺.
中文： 低于 7（酸性）。等当量点时，弱碱的共轭酸水解产生 H⁺。
6.3 Q9¶
EN (Original): Titration of 20.00 mL HBr with 0.700 M KOH. Find [HBr].
| Burette Reading (mL) | Trial 1 | Trial 2 | Trial 3 |
|---|---|---|---|
| Initial | 1.07 | 1.02 | 2.12 |
| Final | 33.42 | 30.18 | 31.32 |
中文翻译： 用 0.700 M KOH 滴定 20.00 mL HBr。求 [HBr]。
Answer 答案:
EN — Step 1: Calculate volumes added: - Trial 1: 33.42 − 1.07 = 32.35 mL - Trial 2: 30.18 − 1.02 = 29.16 mL - Trial 3: 31.32 − 2.12 = 29.20 mL
Trial 1 (32.35) is an outlier — discard it.
Average = (29.16 + 29.20) / 2 = 29.18 mL
Step 2: At equivalence (1:1 ratio):
$$[\text{HBr}] \times 20.00 = 0.700 \times 29.18$$
$$[\text{HBr}] = \frac{0.700 \times 29.18}{20.00} = \boxed{1.02\ \text{M}}$$
中文： Trial 1 为异常值，取 Trial 2 和 3 的平均值 29.18 mL。[HBr] = 0.700 × 29.18 / 20.00 = 1.02 M。
✅ Answer Key 答案: 1.02 M — 一致 ✓
6.3 Q10¶
EN (Original): 159 mL of 0.0085 M HI titrated with 125.0 mL NaOH. Find [NaOH].
中文翻译： 159 mL 的 0.0085 M HI 与 125.0 mL NaOH 完全反应。求 [NaOH]。
Answer 答案:
$$[\text{NaOH}] = \frac{[\text{HI}] \times V_{\text{HI}}}{V_{\text{NaOH}}} = \frac{0.0085 \times 159}{125.0} = \frac{1.3515}{125.0} = \boxed{0.011\ \text{M}}$$
✅ Answer Key 答案: 0.011 M — 一致 ✓
6.3 Q11¶
EN (Original): 300. mL of 0.758 M HBr + 650. mL of 0.362 M LiOH. Resulting pH?
中文翻译： 300 mL 的 0.758 M HBr 与 650 mL 的 0.362 M LiOH 混合，pH 是多少？
Answer 答案:
EN:
Moles HBr = 0.758 × 0.300 = 0.2274 mol Moles LiOH = 0.362 × 0.650 = 0.2353 mol
LiOH in excess: 0.2353 − 0.2274 = 0.0079 mol OH⁻ Total volume = 0.950 L
$$[\text{OH}^-] = \frac{0.0079}{0.950} = 8.32 \times 10^{-3}\ \text{M}$$
$$\text{pOH} = -\log(8.32 \times 10^{-3}) = 2.080$$
$$\text{pH} = 14.000 - 2.080 = \boxed{11.920}$$
中文： LiOH 过量 0.0079 mol，[OH⁻] = 8.32 × 10⁻³ M，pOH = 2.080，pH = 11.920。
✅ Answer Key 答案: pH = 11.920 — 一致 ✓
6.3 Q12¶
EN (Original): 500. mL of 0.00673 M H₂SO₄ + 400. mL of 0.00487 M KOH. Resulting pH and pOH?
中文翻译： 500 mL 的 0.00673 M H₂SO₄ 与 400 mL 的 0.00487 M KOH 混合。求 pH 和 pOH。
Answer 答案:
EN:
Moles H⁺ = 2 × 0.00673 × 0.500 = 0.00673 mol (H₂SO₄ is diprotic) Moles OH⁻ = 0.00487 × 0.400 = 0.001948 mol
H⁺ in excess: 0.00673 − 0.001948 = 0.004782 mol Total volume = 0.900 L
$$[\text{H}^+] = \frac{0.004782}{0.900} = 5.31 \times 10^{-3}\ \text{M}$$
$$\text{pH} = -\log(5.31 \times 10^{-3}) = 2.275$$
$$\text{pOH} = 14.000 - 2.275 = \boxed{11.725}$$
中文： H₂SO₄ 是双质子酸（× 2）。H⁺ 过量 0.00478 mol，[H⁺] = 5.31 × 10⁻³ M。pH = 2.275，pOH = 11.725。
✅ Answer Key 答案: pOH = 11.725 — 一致 ✓
6.3 Q13¶
EN (Original): Determine if each salt is acidic, basic or neutral.
中文翻译： 判断每种盐是酸性、碱性还是中性。
Answer 答案:
| Salt 盐 | Cation source 阳离子来源 | Anion source 阴离子来源 | Result 结果 |
|---|---|---|---|
| CaBr₂ | Ca²⁺ ← Ca(OH)₂ (strong base) | Br⁻ ← HBr (strong acid) | Neutral 中性 |
| NH₄Cl | NH₄⁺ ← NH₃ (weak base) | Cl⁻ ← HCl (strong acid) | Acidic 酸性 |
| KNO₂ | K⁺ ← KOH (strong base) | NO₂⁻ ← HNO₂ (weak acid) | Basic 碱性 |
6.3 Q14¶
EN (Original): For each acidic or basic salt in Q13, write a hydrolysis reaction.
中文翻译： 为 Q13 中的酸性或碱性盐写出水解反应。
Answer 答案:
NH₄Cl (acidic 酸性):
$$\text{NH}_4^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_3(aq) + \text{H}_3\text{O}^+(aq)$$
KNO₂ (basic 碱性):
$$\text{NO}_2^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HNO}_2(aq) + \text{OH}^-(aq)$$
6.3 Q15¶
EN (Original): Calculate the pH of a 0.0587 M solution of KHS.
中文翻译： 计算 0.0587 M KHS 溶液的 pH。
Answer 答案:
EN: KHS → K⁺ + HS⁻. HS⁻ is amphiprotic but acts mainly as a base here (Kb > Ka):
$$\text{HS}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{S}(aq) + \text{OH}^-(aq)$$
$$K_b = \frac{K_w}{K_a(\text{H}_2\text{S})} = \frac{1.0 \times 10^{-14}}{9.1 \times 10^{-8}} = 1.10 \times 10^{-7}$$
| HS⁻ | H₂S | OH⁻ | |
|---|---|---|---|
| I | 0.0587 | 0 | 0 |
| C | −x | +x | +x |
| E | 0.0587−x | x | x |
$$x^2 = 1.10 \times 10^{-7} \times 0.0587 = 6.45 \times 10^{-9}$$
$$x = 8.03 \times 10^{-5}\ \text{M}\ (0.14\% < 5\%\ ✓)$$
$$\text{pOH} = -\log(8.03 \times 10^{-5}) = 4.095$$
$$\text{pH} = 14.000 - 4.095 = \boxed{9.905}$$
中文： HS⁻ 水解为碱性。Kb = 1.10 × 10⁻⁷，x = 8.03 × 10⁻⁵ M，pOH = 4.095，pH = 9.905。
✅ Answer Key 答案: pH = 9.905 — 一致 ✓
术语表 | Terminology¶
| English 英文 | Chinese 中文 | Definition 定义 |
|---|---|---|
| Buffer | 缓冲液 | Solution that resists pH changes |
| Equivalence Point | 等当量点 | Moles acid = moles base |
| Endpoint | 终点 | Color change of indicator |
| Titrant | 滴定剂 | Solution of known concentration |
| Hydrolysis | 水解 | Ion reacting with water to produce H⁺ or OH⁻ |
| Acid Anhydride | 酸酐 | Non-metal oxide that forms acid in water |
| Base Anhydride | 碱酐 | Metal oxide that forms base in water |
| Indicator (HIn) | 指示剂 | Weak acid whose acid/base forms have different colors |
| Amphiprotic | 两性的 | Can donate or accept a proton |
答案核对总表 | Answer Key Verification Summary¶
| Question | Our Answer | Answer Key | Status |
|---|---|---|---|
| 6.1 Q1 | 10 times | 10 times | ✅ |
| 6.1 Q2 | 1000 times | 1000 times | ✅ |
| 6.1 Q3 | 2.46 × 10⁵ mL | 2.46 × 10⁵ mL | ✅ |
| 6.1 Q4 | 1.9 × 10⁻⁶ M | 1.9 × 10⁻⁶ M | ✅ |
| 6.1 Q6 | pH = 1.606 | pH = 1.606 | ✅ |
| 6.1 Q7 | 1.4% | 1.4% | ✅ |
| 6.1 Q8 | pH = 0.844 | pH = 0.844 | ✅ |
| 6.1 Q9 | 7.73% | 7.73% | ✅ |
| 6.1 Q10 | Ka = 2.0 × 10⁻⁹ | Ka = 2.0 × 10⁻⁹ | ✅ |
| 6.1 Q11 | 1.7 g | 1.7 g | ✅ |
| 6.1 Q12 | Kb = 2.0 × 10⁻⁵ | Kb = 2.0 × 10⁻⁵ | ✅ |
| 6.1 Q13 | pH = 8.67 | pH = 8.67 | ✅ |
| 6.1 Q14 | Kb = 1.1 × 10⁻⁷ | Kb = 1.1 × 10⁻⁷ | ✅ |
| 6.1 Q15 | HS⁻ | HS⁻ | ✅ |
| 6.1 Q16 | 0.497 M | 0.497 M | ✅ |
| 6.1 Q18d | pKw = 13.70 | pKw = 13.70 | ✅ |
| 6.2 Q8 | pOH=2.67, pH=11.33 | pOH=2.67, pH=11.33 | ✅ |
| 6.2 Q10 | Ka = 2 × 10⁻⁹ | Ka = 2 × 10⁻⁹ | ✅ |
| 6.2 Q11 | Kb = 1 × 10⁻⁵ | Kb = 1 × 10⁻⁵ | ✅ |
| 6.3 Q9 | 1.02 M | 1.02 M | ✅ |
| 6.3 Q10 | 0.011 M | 0.011 M | ✅ |
| 6.3 Q11 | pH = 11.920 | pH = 11.920 | ✅ |
| 6.3 Q12 | pOH = 11.725 | pOH = 11.725 | ✅ |
| 6.3 Q15 | pH = 9.905 | pH = 9.905 | ✅ |
Result: 24/24 answers verified — 100% match ✅