Logarithms in Chemistry: Mastering pH and pOH
TL;DR: pH and pOH are base-10 logs of [H+] and [OH⁻]. With the right algebra, you can handle acid–base, buffers, and equilibrium. If you’d rather have experts do it fast (A/B Guarantee, private), we handle the chemistry and the math.
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Table of Contents
- Why Chemistry Uses Logarithms
- pH & pOH Basics (with Calculator Workflow)
- Worked Examples: Strong Acids/Bases
- Buffers: Henderson–Hasselbalch + Targeting a pH
- Weak Acid ICE Table — Full Quadratic Solution
- Rounding & Sig-Fig Rules for pH (Mini-Guide)
- Platform Notes: ALEKS • WebAssign • MyLab
- Common Log Mistakes (and Fast Fixes)
- Practice Set (with Answers)
- How FMMC Helps
- FAQs: Logs in Chemistry
- Next Reads (Internal Links)
1) Why Chemistry Uses Logarithms
[H+] spans up to 14 orders of magnitude. Logs compress extremes into a readable scale (0–14), so comparisons are easy. A change of 1.0 pH unit = 10× change in [H+].Brush Up the Math, Ace the Chem
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2) pH & pOH Basics (with Calculator Workflow)
Definitions: pH = −log[H+], pOH = −log[OH⁻], and at 25°C: pH + pOH = 14 (since Kw=1.0×10⁻¹⁴).
Calculator Workflow
- Given [H+] → pH = −log([H+])
- Given pH → [H+] = 10−pH
- Given [OH⁻] → pOH = −log([OH⁻]) → pH = 14 − pOH (25°C)
- Given pH → pOH = 14 − pH → [OH⁻] = 10−pOH
Quick Examples
Given [H+] = 2.5×10⁻⁵ M → pH = −log(2.5×10⁻⁵) ≈ 4.60
Given pH = 3.40 → [H+] = 10−3.40 ≈ 3.98×10⁻⁴ M
Given [OH⁻] = 1.0×10⁻³ M → pOH = 3.00 → pH = 11.00
TI-84:
2ND LOG is 10^x. For pH from 3.40: 2ND LOG (-) 3.40 gives [H+] = 10−3.40. For pH from [H+], use LOG then negate.Casio fx:
SHIFT LOG is 10^x. Use LOG for base-10 logs; mind the negative sign when computing pH.
Sig-fig rule: Digits after the decimal in pH match sig figs in [H+].
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3) Worked Examples: Strong Acids/Bases
Strong Acid
0.0350 M HCl (fully dissociates): [H+] = 0.0350 M → pH = −log(0.0350) = 1.456 (report 1.456 or 1.46 per instructions).
Strong Base
0.0120 M NaOH (fully dissociates): [OH⁻] = 0.0120 M → pOH = 1.920 → pH = 14 − 1.920 = 12.080 (≈ 12.08).
Mixtures/titrations: After mixing volumes, compute new molarities before pH. For titrations, track limiting reagent, then use excess strong acid/base to get [H+] or [OH⁻].
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4) Buffers: Henderson–Hasselbalch + Targeting a pH
H–H equation: pH = pKa + log([A⁻]/[HA]). Works well when both acid/base forms are present in decent amounts.
Example — Acetate Buffer
pKa=4.76, [A⁻]=0.50 M, [HA]=0.20 M → pH = 4.76 + log(2.5) ≈ 5.16.
Target a pH
Desired pH 5.00, pKa=4.76 → log([A⁻]/[HA])=0.24 → ratio=[A⁻]/[HA]=100.24≈1.74.
When to skip H–H: Very dilute solutions or when activities matter → use ICE with Ka directly.
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5) Weak Acid ICE Table — Full Quadratic Solution
HA (C=0.0100 M), Ka=1.8×10⁻⁵. Find pH without small-x approximation.
- HA ⇌ H+ + A⁻ | I: C, 0, 0 C: −x, +x, +x E: C−x, x, x
- Ka = x²/(C−x) → x² + Kax − KaC = 0.
- Numbers: x² + (1.8×10⁻⁵)x − 1.8×10⁻⁷=0.
- Quadratic (positive root): x ≈ 4.1538×10⁻⁴ M → pH ≈ 3.381 (≈ 3.38).
Small-x check: √(KaC)=√(1.8×10⁻⁷)=4.243×10⁻⁴ M (pH≈3.372). Close, but you showed the exact path—great for strict graders.
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6) Rounding & Sig-Fig Rules for pH (Mini-Guide)
| Given | Rule of Thumb | Example |
|---|---|---|
| [H+] with n sig figs | pH has n decimals | [H+]=2.5×10⁻⁵ (2 SF) → pH=4.60 |
| [OH⁻] at 25°C | pOH→pH (carry extra digits, round at end) | [OH⁻]=1.00×10⁻³ (3 SF) → pOH=3.000 → pH=11.000 |
| Intermediate logs | Don’t round mid-calc | Keep 4–5 extra digits in log; round final pH |
| T ≠ 25°C | Check Kw | Use course-specified Kw to relate pH/pOH |
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7) Platform Notes: ALEKS • WebAssign • MyLab
- ALEKS: Often strict on decimal places; follow the on-screen instruction (e.g., 2 decimals). Use exact keystrokes for scientific notation.
- WebAssign: Tends to want more sig-figs; beware of rounding inside logs. Carry extra digits. See our WebAssign Help.
- MyLab Chemistry: Watch for temperature assumptions (pH+pOH=14 default) unless a different T is stated. See MyLab Chemistry Help.
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8) Common Log Mistakes (and Fast Fixes)
- Using ln instead of log: pH work uses base-10 logs. Fix: Use
logunless told otherwise. - Dropping the minus: pH = −log[H+]. Fix: Keep the sign; typical pH should be positive.
- Early rounding: Rounding inside logs compounds error. Fix: Round at the end.
- Forgetting units/temperature: pH+pOH=14 holds at 25°C. Fix: Confirm T or use proper Kw.
- Wrong species: Using [H+] when you actually have strong base (or vice versa). Fix: Convert via neutrality.
- “Why is my pH negative?” Extremely strong acids can produce pH < 0 when [H+] > 1 M. It’s rare but valid.
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9) Practice Set (with Answers)
- pH from [H+]: [H+] = 7.9×10⁻⁴ M → pH = ?
- [H+] from pH: pH = 2.35 → [H+] = ?
- pH from [OH⁻] at 25°C: [OH⁻] = 3.2×10⁻⁶ M → pH = ?
- Buffer (H–H): pKa=4.74, [A⁻]=0.18 M, [HA]=0.12 M → pH = ?
- Weak acid ICE (approx): HA 0.050 M, Ka=6.5×10⁻⁶ → pH ≈ ?
- Dilution trap: 25.0 mL of 0.100 M HCl diluted to 250.0 mL. New pH?
- Mixture trap: Mix 50.0 mL of 0.0400 M HCl with 100.0 mL of water, then with 50.0 mL of 0.0200 M NaOH. Final pH?
Show Answers
- pH = −log(7.9×10⁻⁴) ≈ 3.10 (2 SF → 2 decimals)
- [H+] = 10−2.35 ≈ 4.47×10⁻³ M
- pOH = −log(3.2×10⁻⁶)=5.49 → pH = 14 − 5.49 = 8.51
- pH = 4.74 + log(0.18/0.12) = 4.74 + log(1.5) ≈ 4.74 + 0.176 = 4.92
- x ≈ √(Ka·C) = √(6.5×10⁻⁶ × 0.050) = √(3.25×10⁻⁷) ≈ 5.70×10⁻⁴ M → pH ≈ 3.24
- New M = 0.100×(25.0/250.0)=0.0100 M → pH = −log(0.0100) = 2.000
- Moles HCl = 0.0500 L×0.0400 M = 0.00200 mol. Add 0.100 L water (no reaction), then add NaOH: moles NaOH = 0.0500 L×0.0200 M = 0.00100 mol. Excess HCl = 0.00200−0.00100=0.00100 mol in total volume 0.200 L → [H+] = 0.00500 M → pH = 2.301 (≈ 2.30)
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10) How FMMC Helps
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11) FAQs: Logs in Chemistry
Why does chemistry use logarithms for pH?
Because [H+] spans many orders of magnitude. Logs compress that range into a simple, comparable scale.
What’s the difference between log and ln here?
pH uses base-10 log. Natural log (ln) shows up in kinetics/thermo; use whichever the formula specifies.
Is pH + pOH always 14?
At 25°C, yes. At other temperatures, Kw changes slightly; your course may still assume 14 unless stated.
Can pH be negative? What’s the highest pH?
Yes, extremely strong acids can give pH < 0 when [H+] > 1 M. At the other extreme, very strong bases can push pH > 14 in idealized calculations.
How many decimals should pH have?
Decimals in pH = significant figures in [H+] (mantissa rule). Follow your platform’s instructions.
Can you help with ALEKS/WebAssign pH problems?
Yes—rounding/formatting included. See Complete ALEKS Topics Fast, WebAssign Help, and MyLab Chemistry Help.
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