Logarithms in Chemistry: Mastering pH and pOH
Quick Answer
What are logarithms used for in chemistry? Logarithms compress the enormous range of hydrogen ion concentrations (spanning 14 orders of magnitude) into the simple 0–14 pH scale. The formulas are straightforward: pH = −log[H⁺] and pOH = −log[OH⁻], with pH + pOH = 14 at 25°C. If you need help with pH calculations, buffers, or acid-base equilibrium, Finish My Math Class handles the chemistry and the math—with guaranteed results.
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If you’ve ever wondered why chemists use pH instead of just stating hydrogen ion concentration, the answer is logarithms. Concentrations of H⁺ ions can range from 10 M (extremely acidic) to 10⁻¹⁴ M (extremely basic)—a span of 15 orders of magnitude. Logarithms compress this enormous range into a manageable scale.
This guide covers everything you need to master logarithms in chemistry: the pH and pOH formulas, calculator techniques, buffer calculations with Henderson-Hasselbalch, ICE tables for weak acids, sig-fig rules, and the common mistakes that cost students points.
Why Chemistry Uses Logarithms
Hydrogen ion concentration [H⁺] spans up to 14 orders of magnitude in aqueous solutions. Comparing numbers like 0.00000001 M to 0.1 M is cumbersome. Logarithms solve this by converting multiplicative relationships into additive ones.
The Power of the pH Scale
The pH scale compresses extreme concentration differences into simple, comparable numbers:
- A solution with [H⁺] = 10⁻³ M has pH = 3
- A solution with [H⁺] = 10⁻⁷ M has pH = 7
- A solution with [H⁺] = 10⁻¹¹ M has pH = 11
What a pH Difference Really Means
Because pH is logarithmic, each unit represents a 10× change in [H⁺]:
- pH 3 is 10× more acidic than pH 4
- pH 3 is 100× more acidic than pH 5
- pH 3 is 1000× more acidic than pH 6
This is why small pH changes can have dramatic effects in biological and chemical systems.
The Algebra Connection
pH math is essentially “undoing” powers of 10. If you’re comfortable with exponents and basic algebra, logarithms won’t be difficult—they’re just the inverse operation.
pH and pOH Basics (with Calculator Workflow)
The fundamental definitions you need to memorize:
pH = −log[H⁺]
pOH = −log[OH⁻]
pH + pOH = 14 (at 25°C, where Kw = 1.0 × 10⁻¹⁴)
Calculator Workflow
The four basic conversions you’ll perform repeatedly:
| Given | Find | Formula | Example |
|---|---|---|---|
| [H⁺] | pH | pH = −log[H⁺] | [H⁺] = 2.5×10⁻⁵ M → pH = 4.60 |
| pH | [H⁺] | [H⁺] = 10⁻ᵖᴴ | pH = 3.40 → [H⁺] = 3.98×10⁻⁴ M |
| [OH⁻] | pH | pOH = −log[OH⁻], then pH = 14 − pOH | [OH⁻] = 1.0×10⁻³ M → pOH = 3.00 → pH = 11.00 |
| pH | [OH⁻] | pOH = 14 − pH, then [OH⁻] = 10⁻ᵖᴼᴴ | pH = 9.50 → pOH = 4.50 → [OH⁻] = 3.16×10⁻⁵ M |
Calculator Keystrokes
TI-84 / TI-83
To find pH from [H⁺]: Enter concentration, press LOG, then multiply by −1 (or press (−) first)
To find [H⁺] from pH: Press 2ND LOG (which gives 10^x), then enter the negative pH value
Casio fx Series
To find pH from [H⁺]: Enter concentration, press LOG, negate the result
To find [H⁺] from pH: Press SHIFT LOG (10^x), enter the negative pH value
Worked Examples: Strong Acids and Bases
Strong acids and bases dissociate completely, making pH calculations straightforward.
Strong Acid Example
Problem: Find the pH of 0.0350 M HCl.
Solution: HCl is a strong acid—it dissociates completely.
- [H⁺] = 0.0350 M (same as HCl concentration)
- pH = −log(0.0350)
- pH = 1.456 (or 1.46, depending on sig-fig requirements)
Strong Base Example
Problem: Find the pH of 0.0120 M NaOH at 25°C.
Solution: NaOH is a strong base—it dissociates completely.
- [OH⁻] = 0.0120 M
- pOH = −log(0.0120) = 1.921
- pH = 14 − 1.921 = 12.079 (or 12.08)
Dilution Example
Problem: 25.0 mL of 0.100 M HCl is diluted to 250.0 mL. What is the new pH?
Solution:
- M₁V₁ = M₂V₂
- (0.100 M)(25.0 mL) = (M₂)(250.0 mL)
- M₂ = 0.0100 M
- pH = −log(0.0100) = 2.000
Mixture and Titration Tip
When mixing acids and bases, always calculate the new total volume and track moles of each species. Find the excess strong acid or base, then calculate pH from the excess concentration.
Buffers: Henderson-Hasselbalch Equation
Buffers resist pH changes when small amounts of acid or base are added. The Henderson-Hasselbalch equation relates buffer pH to the ratio of conjugate base to weak acid:
pH = pKₐ + log([A⁻]/[HA])
Finding pH of a Buffer
Problem: An acetate buffer contains 0.50 M sodium acetate (A⁻) and 0.20 M acetic acid (HA). The pKₐ of acetic acid is 4.76. Find the pH.
Solution:
- pH = pKₐ + log([A⁻]/[HA])
- pH = 4.76 + log(0.50/0.20)
- pH = 4.76 + log(2.5)
- pH = 4.76 + 0.40
- pH = 5.16
Targeting a Specific pH
Problem: You need a buffer with pH = 5.00 using the acetic acid/acetate system (pKₐ = 4.76). What ratio of [A⁻]/[HA] do you need?
Solution:
- 5.00 = 4.76 + log([A⁻]/[HA])
- log([A⁻]/[HA]) = 0.24
- [A⁻]/[HA] = 10⁰·²⁴
- [A⁻]/[HA] = 1.74
You need about 1.74 times as much acetate as acetic acid.
When to Skip Henderson-Hasselbalch
The H-H equation works well when both [A⁻] and [HA] are present in reasonable amounts (typically within a factor of 10 of each other). For very dilute solutions or when one species dominates heavily, use the full ICE table approach with Kₐ.
Weak Acid ICE Table — Full Quadratic Solution
For weak acids that don’t fully dissociate, you need an ICE (Initial-Change-Equilibrium) table to find [H⁺].
The Setup
For a weak acid HA with initial concentration C and acid dissociation constant Kₐ:
HA ⇌ H⁺ + A⁻
I: C 0 0
C: −x +x +x
E: C−x x x
The equilibrium expression: Kₐ = x²/(C − x)
Worked Example (Full Quadratic)
Problem: Find the pH of 0.0100 M acetic acid (Kₐ = 1.8 × 10⁻⁵) without using the small-x approximation.
Solution:
- Kₐ = x²/(C − x) → 1.8 × 10⁻⁵ = x²/(0.0100 − x)
- Rearranging: x² + (1.8 × 10⁻⁵)x − (1.8 × 10⁻⁷) = 0
- Using the quadratic formula: x = [−1.8×10⁻⁵ + √((1.8×10⁻⁵)² + 4(1.8×10⁻⁷))]/2
- x = [−1.8×10⁻⁵ + √(7.2×10⁻⁷)]/2
- x = [−1.8×10⁻⁵ + 8.49×10⁻⁴]/2
- x = 4.15 × 10⁻⁴ M = [H⁺]
- pH = −log(4.15 × 10⁻⁴) = 3.38
The Small-x Approximation
If x is much smaller than C (typically x < 5% of C), you can simplify:
Kₐ ≈ x²/C → x ≈ √(Kₐ × C)
For our example: x ≈ √(1.8×10⁻⁵ × 0.0100) = √(1.8×10⁻⁷) = 4.24×10⁻⁴ M → pH ≈ 3.37
The approximation gives pH = 3.37; the exact answer is pH = 3.38. Close enough for most purposes, but show the exact method if your instructor requires it.
Rounding and Sig-Fig Rules for pH
pH calculations have special sig-fig rules that differ from standard chemistry calculations.
The Mantissa Rule
In a logarithm, only the digits after the decimal point (the mantissa) carry significant figure information. The digits before the decimal just indicate the order of magnitude.
| Given Value | Sig Figs | pH Result | Decimals in pH |
|---|---|---|---|
| [H⁺] = 2.5 × 10⁻⁵ M | 2 | 4.60 | 2 |
| [H⁺] = 3.45 × 10⁻⁴ M | 3 | 3.462 | 3 |
| [OH⁻] = 1.00 × 10⁻³ M | 3 | 11.000 | 3 |
Practical Guidelines
- Don’t round intermediate values: Carry 4-5 extra digits through calculations; round only at the end
- Follow platform instructions: Some graders want 2 decimals, others want 3—check before submitting
- Temperature matters: pH + pOH = 14 only at 25°C. At other temperatures, use the specified Kw value
Common Log Mistakes (and How to Avoid Them)
These errors cost students points repeatedly. Learn to recognize and avoid them.
Using ln Instead of log
Mistake: Using natural logarithm (ln) when base-10 log is required.
Fix: pH always uses base-10 logarithms. Natural log (ln) appears in kinetics and thermodynamics—check which the formula requires.
Forgetting the Negative Sign
Mistake: Calculating pH = log[H⁺] instead of pH = −log[H⁺].
Fix: Remember that pH should typically be positive (between 0-14 for most solutions). If you get a negative pH for a dilute solution, you forgot the minus sign.
Rounding Too Early
Mistake: Rounding intermediate values, which compounds errors.
Fix: Keep 4-5 extra digits throughout your calculation. Round only at the final step.
Wrong Temperature Assumption
Mistake: Using pH + pOH = 14 when the temperature isn’t 25°C.
Fix: Check if the problem specifies a temperature. If T ≠ 25°C, use the given Kw value.
Confusing [H⁺] and [OH⁻]
Mistake: Using the wrong concentration (acid concentration for a base problem, or vice versa).
Fix: Identify whether you’re dealing with an acid (gives H⁺) or a base (gives OH⁻). Convert using pH + pOH = 14 when needed.
“My pH is Negative—Is That Wrong?”
Answer: Not necessarily! For very strong acids at concentrations above 1 M, pH can be negative. For example, 2 M HCl has [H⁺] = 2 M, so pH = −log(2) = −0.30. This is rare but valid.
Practice Problems (with Solutions)
Test your understanding with these problems covering all the major pH calculation types.
Problem Set
Problem 1: [H⁺] = 7.9 × 10⁻⁴ M. Find the pH.
Problem 2: pH = 2.35. Find [H⁺].
Problem 3: [OH⁻] = 3.2 × 10⁻⁶ M at 25°C. Find the pH.
Problem 4: An acetate buffer has pKₐ = 4.74, [A⁻] = 0.18 M, and [HA] = 0.12 M. Find the pH.
Problem 5: A weak acid HA has C = 0.050 M and Kₐ = 6.5 × 10⁻⁶. Find the approximate pH using the small-x approximation.
Problem 6: 25.0 mL of 0.100 M HCl is diluted to 250.0 mL. Find the new pH.
Problem 7: Mix 50.0 mL of 0.0400 M HCl with 100.0 mL of water, then add 50.0 mL of 0.0200 M NaOH. Find the final pH.
Problem 8: pH = 8.75. Find [OH⁻] at 25°C.
Problem 9: A buffer needs pH = 7.40 using the H₂PO₄⁻/HPO₄²⁻ system (pKₐ = 7.20). What ratio of [HPO₄²⁻]/[H₂PO₄⁻] is needed?
Problem 10: 0.0250 M HNO₃ (strong acid). Find the pH and pOH at 25°C.
Show All Solutions
Solution 1:
pH = −log(7.9 × 10⁻⁴) = 3.10 (2 sig figs → 2 decimals)
Solution 2:
[H⁺] = 10⁻²·³⁵ = 4.47 × 10⁻³ MSolution 3:
pOH = −log(3.2 × 10⁻⁶) = 5.49
pH = 14 − 5.49 = 8.51
Solution 4:
pH = pKₐ + log([A⁻]/[HA])
pH = 4.74 + log(0.18/0.12) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92
Solution 5:
x ≈ √(Kₐ × C) = √(6.5 × 10⁻⁶ × 0.050) = √(3.25 × 10⁻⁷) = 5.70 × 10⁻⁴ M
pH = −log(5.70 × 10⁻⁴) = 3.24
Solution 6:
New M = 0.100 × (25.0/250.0) = 0.0100 M
pH = −log(0.0100) = 2.000
Solution 7:
Moles HCl = 0.0500 L × 0.0400 M = 0.00200 mol
Adding water doesn’t change moles, just volume (now 150.0 mL)
Moles NaOH = 0.0500 L × 0.0200 M = 0.00100 mol
Excess HCl = 0.00200 − 0.00100 = 0.00100 mol in 200.0 mL total
[H⁺] = 0.00100/0.200 = 0.00500 MpH = −log(0.00500) = 2.30
Solution 8:
pOH = 14 − 8.75 = 5.25
[OH⁻] = 10⁻⁵·²⁵ = 5.62 × 10⁻⁶ MSolution 9:
7.40 = 7.20 + log([HPO₄²⁻]/[H₂PO₄⁻])
log(ratio) = 0.20
ratio = 10⁰·²⁰ = 1.58
Solution 10:
[H⁺] = 0.0250 M (strong acid, complete dissociation)pH = −log(0.0250) = 1.602
pOH = 14 − 1.602 = 12.398
Platform Notes: ALEKS and MyLab
Different homework platforms have different formatting requirements. Here’s what to watch for:
ALEKS Chemistry
- Often strict on decimal places—follow the on-screen instruction (e.g., “2 decimals”)
- Use exact keystrokes for scientific notation entry
- May require showing intermediate steps
- See our ALEKS Chemistry help for platform-specific guidance
MyLab Chemistry (Pearson Mastering)
- Watch for temperature assumptions—pH + pOH = 14 is the default unless another T is stated
- May require selecting units from dropdown menus
- Often wants more significant figures than you might expect
- See our MyLab Chemistry help
Platform-Specific Help
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Frequently Asked Questions
Why does chemistry use logarithms for pH?
Because [H⁺] spans many orders of magnitude (from 10 M to 10⁻¹⁴ M). Logarithms compress that enormous range into a simple, comparable scale where each unit represents a 10× change.
What’s the difference between log and ln?
pH uses base-10 logarithms (log). Natural logarithm (ln) uses base e and appears in kinetics and thermodynamics. Always check which the formula requires—using the wrong one will give incorrect answers.
Is pH + pOH always equal to 14?
Only at 25°C, where Kw = 1.0 × 10⁻¹⁴. At other temperatures, Kw changes (it increases with temperature), so pH + pOH will equal a different value. Your course may still assume 14 unless otherwise stated.
Can pH be negative or greater than 14?
Yes! Extremely strong acids at concentrations above 1 M can give pH < 0 (e.g., 2 M HCl has pH = −0.30). Similarly, very strong bases can push pH > 14 in idealized calculations. These are rare but valid.
How many decimal places should pH have?
The number of decimal places in pH should equal the number of significant figures in [H⁺]. This is called the mantissa rule. Always follow your platform’s specific instructions if given.
When should I use Henderson-Hasselbalch vs. an ICE table?
Use Henderson-Hasselbalch when both acid and conjugate base are present in reasonable amounts (within a factor of 10). Use an ICE table for weak acids alone, very dilute solutions, or when higher precision is needed.
Can you help with ALEKS or MyLab pH problems?
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