Quick Estimation Techniques快速估算技巧

AP Science Practice 5 · Mathematical Routines~30 min

Learning goals

  • Multiply and divide quickly with scientific notation and rounding
  • Remember a few common logarithms and estimate pH, [H⁺] and the order of magnitude of K
  • Use estimates to eliminate wrong multiple-choice options and to check that answers are reasonable

Before you start

  • Weak acid and base equilibria (U8.3)
  • Free energy and equilibrium (U9.5)

Why estimate when calculators are allowed?

Calculators are allowed on both sections, but the four options of a multiple-choice question often differ by a factor of 10 or a sign, or one is a pH and another a pKa. A 15-second estimate often picks the answer and saves calculator time; estimating after a calculation also catches slips like a mistyped key.

Technique 1: rounding and scientific notation

Round to easy numbers and handle the powers of 10 separately:

  • 0.0249 mol × 18.02 g/mol ≈ 0.025 × 18 = 0.45 g;
  • (6.0 × 10⁻³)(4.0 × 10⁻²) ÷ (8.0 × 10⁻⁴): coefficients 6 × 4 ÷ 8 = 3, exponents −3 − 2 + 4 = −1, so the result is 3 × 10⁻¹.

Technique 2: a few logarithms to remember

x2357
log x0.300.480.700.85

From these: log 4 = 0.60, log 6 = 0.78, log 8 = 0.90, log 1.5 = 0.18.

  • pH from [H⁺]: when [H⁺] = a × 10⁻ⁿ, pH = n − log a. For [H⁺] = 2 × 10⁻⁴ M, pH = 4 − 0.30 = 3.70.
  • [H⁺] from pH: split −pH into a positive decimal plus a whole number. For pH 3.30: 10−3.30 = 100.70 × 10⁻⁴ ≈ 5 × 10⁻⁴ M.
  • Bracket first: if [H⁺] is between 10⁻⁴ and 10⁻³, the pH must be between 3 and 4.

Technique 3: make the exponent even before a square root

√(4.0 × 10⁻⁷) is awkward, so rewrite it as √(40 × 10⁻⁸): √40 ≈ 6.3 and √(10⁻⁸) = 10⁻⁴, giving about 6.3 × 10⁻⁴. The weak-acid shortcut [H⁺] ≈ √(Ka·c) uses this step all the time.

Technique 4: the order of magnitude of K

At 25 °C, 2.303RT ≈ 5.7 kJ/mol, so

  • every 5.7 kJ/mol more negative in ΔG° makes K 10 times larger: log K ≈ −ΔG° ÷ 5.7 (ΔG° in kJ/mol);
  • every 0.0592/n V more positive in E° makes K 10 times larger: log K = nE° ÷ 0.0592 (U9.8).

For example, ΔG° = −34 kJ/mol gives 34 ÷ 5.7 ≈ 6, so K ≈ 10⁶.

Technique 5: use the options

First cross out options that are impossible: a pH above 7 for an acid, a negative concentration, a yield over 100%, a positive ΔH for an exothermic reaction. Then compare the orders of magnitude of what’s left; usually only one fits.

Pick a question type and choose answers by estimating, without a calculator:

Estimation practice: pick a question type and choose the answer by estimating, without a calculator

Example 1: estimating the pH of a strong acid

Estimate the pH of 0.015 M HNO₃.

Step-by-step solution

  1. Write [H⁺]. HNO₃ is a strong acid and ionizes completely:

    [HX+]=0.015=1.5×10−2 M[\ce{H+}] = 0.015 = 1.5 \times 10^{-2}\ \text{M}

  2. Bracket it. 1.5 × 10⁻² is between 10⁻² and 10⁻¹, so the pH is between 1 and 2, and closer to 2.

  3. Use the logarithms. log 1.5 = log 3 − log 2 ≈ 0.48 − 0.30 = 0.18:

    pH=2−log⁡1.5≈2−0.18=1.82\begin{aligned} \text{pH} &= 2 - \log 1.5 \approx 2 - 0.18 \\ &= \boxed{1.82} \end{aligned}

  4. Compare. A calculator gives 1.82, so the estimate is all you need.

Example 2: eliminating options by estimating

A 0.20 M solution of a weak acid HA has Ka = 1.8 × 10⁻⁵. Its pH is closest to: (A) 1.36　(B) 2.72　(C) 4.74　(D) 5.44

Step-by-step solution

  1. Estimate [H⁺]. For a weak acid, [H⁺] ≈ √(Ka·c):

    Ka⋅c=1.8×10−5×0.20=3.6×10−6[HX+]≈3.6×10−6=1.9×10−3\begin{aligned} K_a \cdot c &= 1.8 \times 10^{-5} \times 0.20 \\ &= 3.6 \times 10^{-6} \\ [\ce{H+}] &\approx \sqrt{3.6 \times 10^{-6}} \\ &= 1.9 \times 10^{-3} \end{aligned}

    The exponent −6 is already even, and √3.6 ≈ 1.9.

  2. Estimate the pH. [H⁺] is between 10⁻³ and 10⁻², so the pH is between 2 and 3; log 1.9 ≈ log 2 = 0.30:

    pH≈3−0.30=2.7\text{pH} \approx 3 - 0.30 = 2.7

    The answer is (B).

  3. Where the wrong options come from. (C) 4.74 is the pKa, which equals the pH only in a buffer with [HA] = [A⁻]; (D) 5.44 is −log(3.6 × 10⁻⁶), forgetting the square root; (A) 1.36 is half of 2.72, applying the square root to the pH instead.

  4. Conclusion. Once you know the pH is between 2 and 3, only (B) is left, with no quadratic needed.

Practice: without a calculator, in which range is the pH of each solution at 25 °C?
  1. 0.050 M HCl
  2. 1.0 × 10⁻³ M HNO₃
  3. 0.10 M CH₃COOH (Ka = 1.8 × 10⁻⁵)
  4. 0.010 M NaOH
  5. A buffer with [HF] = [F⁻] (pKa of HF = 3.19)
  6. 0.10 M NH₄Cl (Ka of NH₄⁺ = 5.6 × 10⁻¹⁰)
  7. 0.10 M NH₃ (Kb = 1.8 × 10⁻⁵)
  8. 0.30 M HCl

Common mistakes

Mistake 1: “With a calculator allowed, there’s no need to estimate.” Estimating is for speed and for catching errors: an order-of-magnitude estimate answers many multiple-choice questions outright, and a quick estimate after a calculation catches a mistyped key or a wrong exponent.

Mistake 2: “log and ln are close enough to swap.” ln x = 2.303 log x, so swapping them is off by more than a factor of two; in an exponent, that puts K off by several orders of magnitude. pH uses log; ΔG° = −RT ln K uses ln.

Mistake 3: “If an estimate is a bit off from the exact value, it can’t be used to choose an answer.” Multiple-choice options are usually far apart (often by a factor of 10 or a sign), so one decimal place is plenty. Only FRQs need the full, exact calculation written out.

Quick check

[H⁺] = 3.0 × 10⁻⁵ M. About what is the pH?

A solution at 25 °C has pH 9.70. About what is [OH⁻]?

At 25 °C a reaction has ΔG° = −34 kJ/mol. What is the order of magnitude of K?

About what is √(9.0 × 10⁻⁹)?

About what volume does 0.25 mol of an ideal gas occupy at STP?

A cell reaction has n = 2 and E° = +0.118 V at 25 °C. What is the order of magnitude of K?

AP-style practice

AP-style free response

Without a calculator, estimate each value and show brief working.

(a) The pH of 0.0020 M Ba(OH)₂ at 25 °C.

(b) ΔG° (in kJ/mol) for a reaction with K = 2.0 × 10⁻⁷ at 25 °C.

(c) A student reports pH 5.0 for 0.10 M HA, a weak acid with Ka = 1.0 × 10⁻⁵. Without a detailed calculation, explain why this answer is unreasonable and give a reasonable pH.

(d) The volume at STP of the CO₂ formed when 2.50 g of CaCO₃ (M ≈ 100 g/mol) decomposes completely into CaO and CO₂.

(e) The value of (6.0 × 10⁻³)(4.0 × 10⁻²) ÷ (8.0 × 10⁻⁴).

Model answer and scoring

(a) [OH⁻] = 2 × 0.0020 = 4 × 10⁻³ M, pOH = 3 − log 4 ≈ 3 − 0.60 = 2.40, so pH ≈ 14.00 − 2.40 = 11.60.

(b) log K = log 2.0 − 7 ≈ −6.7, so ΔG° ≈ −5.7 × log K = −5.7 × (−6.7) ≈ +38 kJ/mol (K < 1 and ΔG° is positive, which makes sense).

(c) [H⁺] ≈ √(Ka·c) = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ M, so pH ≈ 3.0. A pH of 5.0 equals the pKa, which is true only in a buffer with [HA] = [A⁻]; it could also come from a slip such as forgetting the square root.

(d) n(CO₂) = n(CaCO₃) ≈ 2.50 ÷ 100 = 0.025 mol, so V ≈ 0.025 × 22.4 ≈ 0.56 L.

(e) Coefficients 6 × 4 ÷ 8 = 3 and exponents −3 − 2 + 4 = −1, so the result is 3 × 10⁻¹ = 0.30.

Scoring: (a) [OH⁻] and pH; (b) log K and the sign and size of ΔG°; (c) a reasonable pH of about 3, with where 5.0 came from; (d) moles and volume; (e) 0.30.

Summary

  • Rounding and scientific notation: handle coefficients and powers of 10 separately.
  • Remember log 2 = 0.30, log 3 = 0.48, log 5 = 0.70, log 7 = 0.85; pH = n − log a; bracket first, then refine.
  • Make the exponent even before a square root; at 25 °C, every −5.7 kJ/mol in ΔG° or +0.0592/n V in E° makes K 10 times larger.

AP tip: On multiple choice, estimate first, then look at the options; if only one has the right order of magnitude, there’s no need for an exact calculation. On FRQs you still write out the full calculation (4.3.2) and use the estimate only as a check.

Sources: Ka and Kb values from OpenStax Chemistry 2e Appendices H and I.

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