Drawing Questions: Particle Diagrams, Energy Diagrams, Titration Curves画图题:粒子图、能量图、滴定曲线

AP Science Practice 3 · Representing Data and Phenomena~35 min

Learning goals

  • Draw particle diagrams with correct ratios, conservation, a key, and water molecules oriented correctly around ions
  • Draw energy diagrams showing the relative levels of reactants and products, peaks, Ea and ΔH arrows, catalysts and multistep reactions
  • Sketch titration curves through the starting point, half-equivalence point, equivalence volume and pH, and the end behavior

Before you start

  • Particle-diagram questions (4.2.1)
  • Reaction energy profile (U5.6)
  • Acid-base titrations (U8.5)

How drawing questions are scored

The scoring guidelines for a drawing list the features it must show, such as “products lower in energy than reactants” or “equivalence at 25 mL with pH above 7.” Neatness doesn’t matter; every listed feature must be clearly visible, with labels where needed. Here are checklists for the three most common kinds of drawing.

Particle diagrams

  • Include a key saying what each circle represents.
  • Use ratios from formulas and equations: a CaCl₂ solution has twice as many Cl⁻ as Ca²⁺, and atoms are conserved in a reaction.
  • Draw strong electrolytes as ions, weak electrolytes mostly as molecules, and insoluble compounds as solid at the bottom.
  • Around ions, the O end of each water molecule faces cations and the H ends face anions.
  • States: gas particles far apart, liquid particles touching but disordered, solid particles in an ordered arrangement.

Energy diagrams

  • Reaction progress on the x-axis, energy on the y-axis.
  • Products lower than reactants means exothermic (ΔH < 0); higher means endothermic.
  • Peaks are transition states; the forward Ea runs from the reactants up to the peak; the ΔH arrow points from reactants to products.
  • With a catalyst: lower peak(s), same reactant and product energies (same ΔH).
  • Multistep reactions: one peak per step, with intermediates in the valleys between; the step with the largest activation energy is slowest.

Titration curves

Calculate a few key points first, then join them in a smooth curve (for a weak acid titrated with a strong base):

  1. Start: pH from the weak acid’s Ka.
  2. Buffer region: a gentle rise before equivalence; pH = pKa at half-equivalence.
  3. Equivalence point: volume from the stoichiometry; the solution contains the weak base A⁻, so the pH is above 7, with a steep jump nearby.
  4. After equivalence: the pH levels off toward the titrant’s pH.

Strong acid with strong base: a very low starting pH, no buffer region, pH = 7 at equivalence. Weak base with strong acid: the curve falls from high pH, and the equivalence pH is below 7.

Look at each student’s sketch and tick the real problems:

Find the errors in a drawing: look at the student’s sketch, tick the real problems, then check

Task: sketch the titration curve for 25.0 mL of 0.10 M weak acid HA (pKa = 4.7) titrated with 0.10 M NaOH. Here is a student’s sketch.

Volume of NaOH added (mL)pH071402550equivalence pH = 7

Example 1: four points before sketching a weak acid titration

20.0 mL of 0.10 M monoprotic weak acid HA (Ka = 1.0 × 10⁻⁵) is titrated with 0.10 M NaOH. Calculate the four key points needed to sketch the curve.

Step-by-step solution
  1. Start (0 mL). A weak acid solution, so [H⁺] ≈ √(Ka·c):

    [HX+]≈1.0×10−5×0.10=1.0×10−3pH=3.00\begin{aligned} [\ce{H+}] &\approx \sqrt{1.0 \times 10^{-5} \times 0.10} \\ &= 1.0 \times 10^{-3} \\ \text{pH} &= 3.00 \end{aligned}
  2. Equivalence volume and half-equivalence. NaOH and HA have the same concentration and react 1 : 1, so equivalence is at 20.0 mL and half-equivalence at 10.0 mL, where:

    pH=pKa=−log⁡(1.0×10−5)=5.00\text{pH} = \text{p}K_a = -\log(1.0 \times 10^{-5}) = 5.00
  3. pH at equivalence. Only A⁻ remains: 2.0 mmol in 40.0 mL, so [A⁻] = 0.050 M; Kb = Kw/Ka = 1.0 × 10⁻⁹:

    [OHX−]≈1.0×10−9×0.050=7.1×10−6pH=14.00−5.15=8.85\begin{aligned} [\ce{OH-}] &\approx \sqrt{1.0 \times 10^{-9} \times 0.050} = 7.1 \times 10^{-6} \\ \text{pH} &= 14.00 - 5.15 = 8.85 \end{aligned}
  4. Excess NaOH (30.0 mL). 1.0 mmol of excess OH⁻ in a total of 50.0 mL:

    [OHX−]=1.0 mmol50.0 mL=0.020pH=14.00−1.70=12.30\begin{aligned} [\ce{OH-}] &= \frac{1.0\ \text{mmol}}{50.0\ \text{mL}} = 0.020 \\ \text{pH} &= 14.00 - 1.70 = 12.30 \end{aligned}
  5. Join them. Start at pH 3.00, rise gently through (10.0 mL, 5.00), climb steeply near 20.0 mL through pH 8.85, then level off above pH 12. Label the half-equivalence and equivalence points on the sketch.

Example 2: a catalyzed two-step energy diagram

An exothermic reaction takes place in two steps, and the first step is rate-determining. On one set of axes, draw its energy diagram with and without a catalyst. Before drawing, list the features the diagram must show.

Step-by-step solution
  1. Start and end. Draw the products lower than the reactants (exothermic), with a downward arrow labeled ΔH.
  2. Peaks and valley. Two steps mean two peaks, with the intermediate in the valley between them.
  3. Which peak is higher. The first step is rate-determining, so its activation energy (reactants up to the first peak) is larger than that of the second step (intermediate up to the second peak).
  4. The catalyzed curve. Both peaks are lower; the start and end coincide with the original curve, so ΔH is unchanged.
  5. Labels. Label reactants, products, intermediate, Ea and ΔH, and distinguish the two curves (for example, a dashed line for “catalyzed”).
Practice: when drawing these, is each choice right?
  1. Drawing the products lower in energy after adding a catalyst
  2. Placing the equivalence-point pH above 7 for a weak acid titrated with a strong base
  3. Drawing twice as many Cl⁻ as Ca²⁺ in a CaCl₂ solution
  4. Drawing the products lower than the reactants for an endothermic reaction
  5. Drawing two peaks for a two-step reaction, with the intermediate in the valley between
  6. Starting a strong acid–strong base titration curve near pH 7
  7. Drawing liquid water molecules in a neat lattice
  8. Drawing the Ea arrow from the reactant energy up to the peak

Common mistakes

Mistake 1: “A catalyst makes a reaction release more heat.” A catalyst only lowers the activation energy; the reactant and product energies, and so ΔH, are unchanged. The two curves must start and end at the same levels.

Mistake 2: “The pH at equivalence is always 7.” Only for a strong acid with a strong base. It’s above 7 for a weak acid titrated with a strong base and below 7 for a weak base titrated with a strong acid.

Mistake 3: “A particle diagram only has to look roughly right.” Particle ratios, conservation of atoms and whether ions are separated are all specific scoring requirements. Count before you draw, and include a key.

Quick check

NH₃ (a weak base) is titrated with a strong acid. Where should the equivalence-point pH be?

On the energy diagram of an exothermic reaction, how do the products compare with the reactants?

The half-equivalence point of a weak acid titration curve is at pH 4.2. What is the pKa of HA?

On one energy diagram, how does the catalyzed curve compare with the uncatalyzed one?

In boxes of the same size, 0.10 M Na₂SO₄ and 0.10 M NaCl are drawn. How do the numbers of Na⁺ compare?

How should a gas be drawn in a particle diagram?

AP-style practice

AP-style free response

25.0 mL of 0.200 M NH₃ (Kb = 1.8 × 10⁻⁵) is titrated with 0.200 M HCl.

(a) Calculate the volume of HCl needed to reach the equivalence point.

(b) Calculate the pH of the solution before any HCl is added.

(c) Calculate the pH after 12.5 mL of HCl has been added.

(d) Calculate the pH at the equivalence point.

(e) Describe the shape of the titration curve (which points you would label on a sketch), and explain whether methyl red (range 4.4–6.2) or phenolphthalein (range 8.3–10.0) is the better indicator.

Model answer and scoring

(a) n(NH₃) = 0.0250 L × 0.200 M = 5.00 × 10⁻³ mol, which needs an equal amount of HCl: V = 5.00 × 10⁻³ ÷ 0.200 = 0.0250 L = 25.0 mL.

(b) [OH⁻] ≈ √(Kb·c) = √(1.8 × 10⁻⁵ × 0.200) = 1.9 × 10⁻³ M, so pOH = 2.72 and pH = 11.28.

(c) 12.5 mL is the half-equivalence point, where [NH₃] = [NH₄⁺], so pOH = pKb = 4.74 and pH = 14.00 − 4.74 = 9.26.

(d) Only NH₄⁺ remains: 5.00 mmol in 50.0 mL, so its concentration is 0.100 M; Ka = Kw/Kb = 5.6 × 10⁻¹⁰, so [H⁺] ≈ √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ M and pH = 5.13.

(e) The curve starts at pH 11.28, falls gently through the buffer region and the half-equivalence point (12.5 mL, 9.26), drops steeply near 25.0 mL through the equivalence pH of 5.13, and levels off toward the pH of the HCl solution (about 1). A sketch should label the start, the half-equivalence point and the equivalence point. Methyl red is better, because its range includes 5.13; phenolphthalein would change color well before equivalence.

Scoring: (a) 25.0 mL; (b) pH 11.28; (c) pH 9.26; (d) pH 5.13; (e) curve shape with key points, and methyl red with a reason.

Summary

  • Drawings are scored by features: show every feature clearly, with labels where needed.
  • Particle diagrams: key, ratios, conservation, strong vs weak electrolytes and precipitates, water orientation, arrangement for each state.
  • Energy diagrams: relative levels of reactants and products, peaks, Ea and ΔH arrows, and catalysts that don’t move the start or end. Titration curves: calculate the start, half-equivalence, equivalence and excess points first.

AP tip: Before sketching a titration curve or energy diagram, calculate the key values on scratch paper; after sketching, label those points on the drawing (for example, “half-equivalence: pH = pKa”) so the reader sees at a glance which features you’ve shown.

Sources: Kb of NH₃ from OpenStax Chemistry 2e Appendix I; indicator ranges from the Wikipedia article “pH indicator.”

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