讲解本题，通俗易懂，便于理解。---**Extracted Content:** **Question Stem:** 14. 甲醛法测定 NH₄⁺ 的反应原理为 4NH₄⁺ + 6HCHO = (CH₂)₆ N₄H⁺ + 3H⁺ + 6H₂O。 取含 NH₄Cl 的废 水浓缩至原体积的 1/10 后, 移取 20.00mL, 加入足量甲醛反应后, 用 0.01000mol·L⁻¹ 的 NaOH 标准溶液滴 定。滴定曲线如图 1, 含氮微粒的分布分数 δ 与 pH 关系如图 2[比如: δ[(CH₂)₆ N₄H⁺] = c[(CH₂)₆ N₄H⁺] / [c[(CH₂)₆ N₄H⁺] + c[(CH₂)₆ N₄]] 。下列说法正确的是 **Chemical Equation:** 4NH₄⁺ + 6HCHO = (CH₂)₆ N₄H⁺ + 3H⁺ + 6H₂O **Experiment Description:** Waste water containing NH₄Cl is concentrated to 1/10 of its original volume. 20.00mL is taken, sufficient formaldehyde is added and reacted. It is then titrated with 0.01000mol·L⁻¹ NaOH standard solution. **Distribution Fraction Formula:** δ[(CH₂)₆ N₄H⁺] = c[(CH₂)₆ N₄H⁺] / [c[(CH₂)₆ N₄H⁺] + c[(CH₂)₆ N₄]] **Figure 1 Description:** Type: Titration curve (Line chart). Title: 图1 X-axis: V(NaOH)/mL, scale from 0 to 30 in increments of 5. Y-axis: pH, scale from 2 to 12 in increments of 2. Curve: Shows pH change as NaOH volume increases. Starts around pH 2.5, increases gradually, then sharply around V=20mL, and levels off around pH 11. Labeled points: a (approx V=10mL, pH=3), b (V=20mL, pH approx 7.5), c (V=25mL, pH approx 11). Vertical dashed lines extend from points b and c to the x-axis at V=20 and V=25. **Figure 2 Description:** Type: Species distribution chart (Line chart). Title: 图2 X-axis: pH, scale from 0 to 12 in increments of 2. Y-axis: δ, scale from 0.0 to 1.0 in increments of 0.2. Curves: Solid line: Starts at δ=0 at low pH, increases to δ=1 at high pH. Represents δ[(CH₂)₆ N₄]. Dashed line: Starts at δ=1 at low pH, decreases to δ=0 at high pH. Represents δ[(CH₂)₆ N₄H⁺]. Intersection: The point where the solid and dashed lines meet, around pH 5.1-5.2, where δ = 0.5. Labeled point: (6.00, 0.88) on the solid line. **Options:** A. 废水中 NH₄⁺ 的含量为 20.00mg·L⁻¹ B. c点: c[(CH₂)₆ N₄H⁺] + c(H⁺) = c(OH⁻) C. a点: c[(CH₂)₆ N₄H⁺] > c(H⁺) > c(OH⁻) > c[(CH₂)₆ N₄] D. (CH₂)₆ N₄H⁺ ⇌ (CH₂)₆ N₄ + H⁺ 的平衡常数 K ≈ 7.3×10⁻⁶