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Logarithms for pH, pKa and Kw

pH is a logarithm, and a logarithm is a power of ten. Everything about pH that looks odd, from the decimal places to why pH 2 is ten times pH 3, follows from that one fact.

What a log is

A logarithm to base ten answers one question: ten to what power gives this number? log 100 is 2, because 10² is 100; log 0.001 is −3, because 10⁻³ is 0.001; log 0.0500 is −1.30, because 10⁻¹·³⁰ is 0.0500. Hydrogen ion concentrations in solution run from about 1 mol dm⁻³ down to 10⁻¹⁴ mol dm⁻³, fourteen powers of ten, and a scale that wide is only readable as the power. pH is that power with the sign changed: pH = −log[H⁺].

The minus sign is there so that the numbers come out positive. A concentration below 1 mol dm⁻³ has a negative log, and acids are almost always below 1 mol dm⁻³, so −log[H⁺] is positive for nearly every solution you meet. It also means the scale runs backwards: more H⁺ is a smaller pH.

The reverse move is the antilog. If pH = −log[H⁺] then [H⁺] = 10⁻ᵖᴴ, and on a calculator that is the 10ˣ key, usually SHIFT then log. Press it, then type the pH with its minus sign in front, and close the bracket if the calculator opens one. Typing −10ˣ or forgetting the minus gives a concentration of ten million for a pH of 7, which should look wrong.

Why pH 2 is ten times pH 3

A solution of pH 2 has [H⁺] = 10⁻² = 0.01 mol dm⁻³. A solution of pH 3 has [H⁺] = 10⁻³ = 0.001 mol dm⁻³. One unit of pH is one power of ten in concentration, so the first has ten times the hydrogen ion concentration of the second, and pH 1 has a hundred times. Diluting an acid tenfold raises its pH by one; a thousandfold dilution raises it by three.

The same logic sets the number of figures. A concentration quoted to three significant figures, 0.0500 mol dm⁻³, fixes the pH to two decimal places, 1.30, because the digit before the point only records the power of ten and the digits after it carry the precision. Give pH to two decimal places for a three-figure concentration, and when going back from pH to [H⁺] give the concentration to as many significant figures as the pH had decimal places.

pH of a strong acid, and back again

Calculate the pH of 0.0500 mol dm⁻³ hydrochloric acid, then show that the pH you get returns the concentration.

Show the working

Hydrochloric acid is fully dissociated, so [H⁺] = 0.0500 mol dm⁻³.

pH = −log(0.0500) = −(−1.301) = 1.30, to two decimal places because the concentration has three significant figures.

Back again: [H⁺] = 10⁻¹·³⁰ = 0.0501 mol dm⁻³, which rounds to the 0.0500 started with. The small difference is the rounding of the pH, so a pH is carried to at least two decimal places before it is used in a further step.

pKa and Kw: the same move on a constant

Ka for a weak acid is a small number in standard form, 1.74 × 10⁻⁵ mol dm⁻³ for ethanoic acid, and pKa is its negative log, 4.76. The p prefix always means −log of whatever follows. A smaller pKa is a stronger acid, for the same reason that a smaller pH is a higher [H⁺]: the number has the sign reversed. Going back, Ka = 10⁻ᵖᴷᵃ, the same antilog key.

Kw is 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K, so pKw = 14.00, and taking logs of [H⁺][OH⁻] = Kw gives pH + pOH = 14.00. A log turns a product into a sum, and it means the pH of a strong base can be found two ways: divide Kw by [OH⁻] and take the negative log, or take −log[OH⁻] for pOH and subtract from 14.

pKa from Ka, and the pH of a weak acid

Ethanoic acid has Ka = 1.74 × 10⁻⁵ mol dm⁻³. Find its pKa, then the pH of a 0.100 mol dm⁻³ solution.

Show the working

pKa = −log(1.74 × 10⁻⁵) = −(−4.759) = 4.76. On the calculator the standard-form number goes in with the EXP or ×10ˣ key, which keeps it one number. Typed as × 10 ^ −5 it is two operations, and inside a division, 0.100 ÷ 1.74 × 10 ^ −5 is worked as (0.100 ÷ 1.74) × 10⁻⁵, not 0.100 ÷ (1.74 × 10⁻⁵).

For the weak acid, [H⁺] = [A⁻] and [HA] stays close to 0.100, so Ka = [H⁺]² ÷ 0.100 and [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³.

pH = −log(1.32 × 10⁻³) = 2.88. Compare 1.00 for hydrochloric acid at the same concentration: two pH units is a hydrogen ion concentration nearly a hundred times smaller, which is what partial dissociation of about one molecule in a hundred means.

The pH of a strong base goes through Kw. For 0.0500 mol dm⁻³ sodium hydroxide, [OH⁻] = 0.0500 mol dm⁻³, [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³ and pH = 12.70. A mark scheme accepts either route and rejects a pH given to one decimal place when the data had three figures.

TRY IT: Four logs and antilogs

(a) The pH of 0.0250 mol dm⁻³ nitric acid.

(b) [H⁺] in a solution of pH 4.20.

(d) The pH of 0.200 mol dm⁻³ potassium hydroxide at 298 K.

Check your answer

(a) pH = −log(0.0250) = 1.60.

(b) [H⁺] = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵ mol dm⁻³, three figures for two decimal places.

(d) [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.200 = 5.00 × 10⁻¹⁴ mol dm⁻³, so pH = 13.30. Or pOH = −log(0.200) = 0.70 and pH = 14.00 − 0.70 = 13.30.

The lessons that use this: Brønsted–Lowry acids and bases, pH and strong acids · Weak acids, Ka and pKa; Kw and strong bases · Buffer solutions and buffer calculations. Every skill the boards number, and where each is taught: maths skills.