Logarithms & Exponentials — Maths for Chemistry

Logarithms compress, exponentials explode

One pair of inverse functions runs through pH, kinetics and equilibrium

A logarithm answers “what power do I raise the base to?” Because chemistry deals with quantities spanning many orders of magnitude — from molar acid to 10⁻¹⁴ M — logs let us write them as small, friendly numbers.

lo g_{b} (x) = y ⟺ b^{y} = x ln x \equiv lo g_{e} x, e \approx 2.718

The reverse of a log is the antilog: if $lo g x = y$ then $x = 1 0^{y}$ . That single step turns a pH back into a concentration. The three log laws then turn hard arithmetic into easy arithmetic:

Product → sum

lo g (A B) = lo g A + lo g B

e.g. log(2×500)=log2+log500=3

Quotient → difference

lo g \frac{A}{B} = lo g A - lo g B

the basis of Henderson–Hasselbalch

Power → multiply

lo g (A^{n}) = n lo g A

pulls exponents down to ground level

Everywhere in chemistry

pH = −log[H⁺]; pKa = −log Ka; the Henderson–Hasselbalch equation carries a log ratio; the Nernst equation a log term; absorbance A = log(I₀/I); and the Arrhenius plot uses ln k. The exponential

e^{- k t}

is the solution to every first-order rate law. Open the pH explorer to feel the log scale, then the Arrhenius linearizer to see logs straighten a curve.

Common pitfalls

log and ln are different bases. pH, pKa and absorbance use base-10 $lo g$ ; kinetics and the Boltzmann factor use base-e $ln$ . Convert with $ln x = 2.303 lo g x$ , never interchange them blindly.
Significant figures live after the decimal point. In $pH = 3.35$ only the .35 counts; the “3” just fixes the power of ten.
Logs only add for a product. $lo g (A + B) \neq = lo g A + lo g B$ . The sum rule is $lo g (A B) = lo g A + lo g B$ — a multiplication inside, addition outside.
The Arrhenius slope is negative. $ln k = ln A - \frac{E _{a}}{R} \frac{1}{T}$ , so the slope is $- E_{a} / R$ . Forgetting the sign gives a negative activation energy.

pH explorer

Slide [H⁺] across fourteen orders of magnitude and watch a single log compress it onto the 0–14 scale.

Slide the hydrogen-ion concentration and watch pH respond

024678101214

[H⁺] = 10^p, set the power pp = -3.0

pH3.0

pOH11.0

[H⁺]1.0e-3 M

[OH⁻]1.0e-11 M

This solution is acidic. Notice pH = −log[H⁺]: every single pH unit is a ten-fold change in [H⁺]. A log scale lets one number span fourteen orders of magnitude — from 1 M acid to 10⁻¹⁴ M.

pH = - lo g_{10} [H^{+}] pH + pOH = 14

Worked example 1pH from a hydrogen-ion concentration

A solution has $[H^{+}] = 4.5 \times 1 0^{- 4} M$ . Calculate its pH.

Worked example 2Concentration from a pH (the antilog)

Human blood is held at $pH = 7.40$ . What is $[H^{+}]$ ?

Worked example 3A strong base — go through pOH

Find the pH of $0.025 M$ NaOH at 25 °C.

Arrhenius linearizer

Taking ln of k = A e^(−Ea/RT) turns an exponential curve into a straight line whose slope hands you the activation energy.

Taking ln of the Arrhenius equation turns a curve into a straight line

ln k = ln A - \frac{E _{a}}{R} (\frac{1}{T})

Eₐ (activation energy)50 kJ/mol

A = 10^(power)10^13

slope = −Eₐ/R-6014 K

intercept = ln A29.9

k at 298 K1.7e+4

A steeper (more negative) slope means a larger activation energy — the rate is more temperature-sensitive. Reading Eₐ off this slope is one of the most common log-based calculations in all of chemistry.

Worked example 4Activation energy from two rate constants

A reaction has $k_{1} = 1.0 \times 1 0^{- 3} s^{- 1}$ at 300 K and $k_{2} = 8.0 \times 1 0^{- 3} s^{- 1}$ at 330 K. Find the activation energy $E_{a}$ .

Worked example 5Reading Eₐ straight off the slope

An Arrhenius plot of $ln k$ against $1/ T$ has a slope of $- 6.0 \times 1 0^{3} K$ . What is the activation energy?

Worked examples

Logs at work across equilibrium, buffers and spectroscopy.

Worked example 6pKa from Ka

Acetic acid has $K_{a} = 1.8 \times 1 0^{- 5}$ . Find its $p K_{a}$ .

Worked example 7Buffer pH with Henderson–Hasselbalch

An acetate buffer contains $[C H_{3} CO O^{-}] = 0.20 M$ and $[C H_{3} COOH] = 0.10 M$ ( $p K_{a} = 4.74$ ). Find the pH.

Worked example 8Absorbance from transmittance (Beer–Lambert)

A coloured solution transmits 25 % of the incident light ( $T = 0.25$ ) in a 1.00 cm cell. Find its absorbance, and the concentration if $ε = 8.0 \times 1 0^{3} M^{- 1} c m^{- 1}$ .

ChallengeA buffer and its pH change on dilution and addition

A buffer is 0.10 M in $N H_{4}^{+}$ and 0.10 M in $N H_{3}$ ( $p K_{a} (N H_{4}^{+}) = 9.25$ ). (a) Find the pH. (b) 0.010 mol of strong acid is added to 1.0 L of this buffer — find the new pH.

Check yourself

Six quick questions on logs, pH, pKa and the Arrhenius plot.

Question 1 of 6 · Score 0

A solution has [H⁺] = 1 × 10⁻³ M. What is its pH?

Choose an answer.