Logarithms — The Rules Students Misapply Most
Logarithms are not an optional maths topic for a chemistry student. pH, pOH, pKa, the Nernst equation, first-order kinetics, the Arrhenius equation and half-life problems are all built on them. Most marks lost in these chapters are not chemistry mistakes at all — they are log mistakes. This guide sets out the rules, shows the four calculations you will actually perform, and lists the specific misuses that examiners see every year.
What a logarithm is
A logarithm is an exponent — nothing more. "log 1000 = 3" says "10 raised to the power 3 gives 1000". Two bases matter in practice: base 10, written simply log, and base e ≈ 2.71828, written ln (the natural log). The number inside must be strictly positive: log 0 and log(−5) do not exist as real numbers.
The five rules
2. Quotient: log(m ÷ n) = log m − log n
3. Power: log(mn) = n log m
4. Change of base: logb x = log x / log b (any consistent base on the right)
5. Basics: loga a = 1, loga 1 = 0, and aloga x = x
Rules 1 and 2 are what make logs useful: they turn multiplication into addition and division into subtraction. Rule 3 is what lets you pull an exponent down and solve for it — the reason logs appear whenever an unknown sits in a power.
A relation used constantly in physical chemistry links the two bases:
That is where the factor 2.303 in the first-order rate equation and the Arrhenius equation comes from. It is not a chemistry constant; it is a change of base.
Worked example 1 — pH from a hydrogen-ion concentration
Find the pH of a solution with [H⁺] = 2.5 × 10⁻⁴ mol/L.
pH = −log[H⁺] = −log(2.5 × 10⁻⁴)
Apply the product rule: log(2.5 × 10⁻⁴) = log 2.5 + log 10⁻⁴
log 2.5 = 0.3979 and log 10⁻⁴ = −4
So log[H⁺] = 0.3979 − 4 = −3.6021
pH = −(−3.6021) = 3.60
Sanity check: the concentration lies between 10⁻⁴ (pH 4) and 10⁻³ (pH 3), so the pH must be between 3 and 4 ✓. Doing that check first catches sign errors instantly.
Worked example 2 — going backwards, pH to concentration
A solution has pH 4.70. Find [H⁺].
[H⁺] = 10−4.70. Split the exponent into a whole part and a decimal part:
10−4.70 = 100.30 × 10−5
100.30 = 1.995
[H⁺] = 2.0 × 10⁻⁵ mol/L
Splitting the exponent this way is the antilog step. It is also how log tables work — see the section on characteristic and mantissa below.
Worked example 3 — first-order kinetics
A first-order reaction has k = 1.0 × 10⁻³ s⁻¹. How long until it is 90% complete?
If 90% has reacted, 10% remains, so a/(a − x) = 100/10 = 10, and log 10 = 1.
t = (2.303 / 1.0 × 10⁻³) × 1 = 2303 s
In minutes: 2303 ÷ 60 = 38.4 min
Because log 10 = 1, the time for 90% completion is always 2.303/k — and, by the same logic, the time for 99% completion (log 100 = 2) is exactly twice that. Recognising the log of a clean power of ten saves a full calculation.
Worked example 4 — an unknown stuck in the exponent
Solve 3x = 20.
Take log of both sides: x log 3 = log 20 (power rule)
x = log 20 / log 3 = 1.3010 / 0.4771 = 2.727
Check: 3² = 9 and 3³ = 27, so x must sit between 2 and 3, closer to 3 ✓
The same manipulation gives the change-of-base rule. For instance log2 50 = log 50 / log 2 = 1.6990 / 0.3010 = 5.644, which is consistent with 2⁵ = 32 and 2⁶ = 64 ✓
Characteristic and mantissa — for log tables
If you are using four-figure log tables rather than a calculator, every logarithm is split into two parts: the characteristic (the whole-number part, decided by the power of ten) and the mantissa (the decimal part, read from the table and always positive).
| Number | Scientific form | Characteristic | Mantissa | log |
|---|---|---|---|---|
| 250 | 2.5 × 10² | 2 | 0.3979 | 2.3979 |
| 2.5 | 2.5 × 10⁰ | 0 | 0.3979 | 0.3979 |
| 0.0025 | 2.5 × 10⁻³ | −3 | 0.3979 | −2.6021 |
The mantissa is the same for all three, because they differ only by powers of ten. For the last row, the old notation writes the negative characteristic with a bar: 3̄.3979, meaning −3 + 0.3979 = −2.6021. Both forms are correct; the bar form keeps the mantissa positive so it can be looked up directly. State which form you are using so an examiner can follow.
The mistakes that cost marks
- log(m + n) ≠ log m + log n. This is the number one error. Test it: log(2 + 3) = log 5 = 0.6990, but log 2 + log 3 = 0.3010 + 0.4771 = 0.7781. Different numbers. There is no rule for the log of a sum.
- (log m) / (log n) ≠ log(m/n). log 10 / log 2 = 1 / 0.3010 = 3.322, whereas log(10/2) = log 5 = 0.6990. The first is a change of base, the second is the quotient rule — completely different operations.
- (log m)² ≠ 2 log m. The power rule applies to log(m²), not to squaring the log itself.
- Mixing ln and log. Writing k = A e−Ea/RT as "log k = log A − Ea/RT" drops the 2.303. Correct forms: ln k = ln A − Ea/RT, or log k = log A − Ea/(2.303 RT).
- Forgetting the minus sign in pH. pH = −log[H⁺]. Since [H⁺] is small, its log is negative and the pH comes out positive. If you get a negative pH for a dilute solution, you dropped the sign.
- Taking the log of a quantity with units. Only a pure number can go inside a log. In practice this means the concentration is divided by the standard concentration (1 mol/L) first, which is why the number is used as it stands — but never write "log(0.1 mol/L)" in a derivation.
- Rounding the mantissa too early. In pH work, the digits before the decimal point come from the power of ten and carry no significant figures. A concentration known to 2 significant figures gives a pH quoted to 2 decimal places — 2.5 × 10⁻⁴ gives pH 3.60, not pH 3.6021.
- Taking the log of a negative number or zero. Undefined. If your working produces one, the algebra above it is wrong.
Where logs appear in the chemistry syllabus
| Topic | The log relation used |
|---|---|
| Ionic equilibrium | pH = −log[H⁺], pOH = −log[OH⁻], pKa = −log Ka |
| Buffers | Henderson–Hasselbalch: pH = pKa + log([salt]/[acid]) |
| Chemical kinetics | First order: k = (2.303/t) log[a/(a − x)] |
| Temperature dependence | Arrhenius: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂) |
| Electrochemistry | Nernst at 298 K: E = E° − (0.0591/n) log Q |
| Class 12 maths | Logarithmic differentiation, integrals giving ln|x| |
Six chapters, one skill. For the exact syllabus and question pattern of any exam, read its current official notification rather than relying on second-hand summaries.
Check your log and antilog steps. The suite's scientific calculator handles log, ln, 10x and ex, and the dedicated pH and Henderson–Hasselbalch tools apply these relations for you when you want to confirm a full chemistry answer.
Open the ABC Chemistry Calculator Suite →A weak log foundation quietly damages five chemistry chapters at once, so it is worth fixing early. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — abcchemistry.in.