Acids, Bases and pH: Proton Transfer and the Logarithmic Scale

From Arrhenius acids to Brønsted–Lowry proton transfer

Arrhenius defined an acid as a compound that produces hydrogen ions in water and a base as one that produces hydroxide ions. The definition works for hydrochloric acid and sodium hydroxide, but it cannot explain why ammonia, NH₃, which contains no hydroxide, makes water basic.

The Brønsted–Lowry definition fixes this by focusing on the proton. An acid is a proton donor and a base is a proton acceptor, so every acid–base reaction is a proton transfer from one species to another. Ammonia is a base because it accepts a proton from water: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq).

A bare proton does not exist free in water. It bonds to a water molecule to form the hydronium ion, H₃O⁺. Chemists often write H⁺(aq) as shorthand, and the two notations mean the same thing in pH calculations.

Finding conjugate acid–base pairs

When an acid donates a proton, what remains is its conjugate base; when a base accepts a proton, it becomes its conjugate acid. The two members of a conjugate pair differ by exactly one H⁺. In HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq), the pairs are HF/F⁻ and H₃O⁺/H₂O.

Water appears as a base in the HF reaction and as an acid in the ammonia reaction. A species that can either donate or accept a proton is called amphiprotic. Hydrogen carbonate, HCO₃⁻, is another example: it can lose a proton to become CO₃²⁻ or gain one to become H₂CO₃.

• Write the equation and find the species that loses an H⁺ on going from left to right. That is the acid.
• Find the species that gains an H⁺. That is the base.
• Pair each reactant with the product that differs from it by one H⁺, and check that the charges differ by one.

Strong and weak describe extent, not concentration

A strong acid ionizes essentially completely in water, so every acid molecule dissolved gives H₃O⁺ and its conjugate base. The common strong acids are HCl, HBr, HI, HNO₃, HClO₄ and H₂SO₄ (for its first proton). The common strong bases are the hydroxides of the group 1 metals and of calcium, strontium and barium. A weak acid ionizes only partly, and an equilibrium between the molecule and its ions is established.

Strength and concentration are separate ideas. A dilute solution of a strong acid can have fewer hydronium ions than a concentrated solution of a weak one. At equal concentration, though, the difference is large. A 0.10 mol/L HCl solution has [H₃O⁺] = 0.10 mol/L and pH 1.00. Acetic acid, with an acid ionization constant Ka = 1.8 × 10⁻⁵ (OpenStax Appendix H), is only about 1 percent ionized at 0.10 mol/L, and its pH is about 2.9. Weak-acid equilibrium calculations are a separate topic; this guide only uses the comparison.

The strength of an acid and its conjugate base are linked. The conjugate bases of strong acids, such as Cl⁻ and NO₃⁻, have negligible tendency to accept a proton in water.

Water's ion product, pH and pOH

Even pure water contains a few ions, because water molecules transfer protons to one another: 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). The equilibrium constant for this autoionization is Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. In pure water the two concentrations are equal, 1.0 × 10⁻⁷ mol/L each. In any aqueous solution at 25 °C, if you know one concentration you can find the other by dividing Kw by it.

Because these concentrations span many powers of ten, chemists use logarithms: pH = −log[H₃O⁺] and pOH = −log[OH⁻]. Taking the negative logarithm of the Kw expression gives pH + pOH = 14.00 at 25 °C. A change of one pH unit is a tenfold change in hydronium concentration, so a solution at pH 3 has 100 times the [H₃O⁺] of one at pH 5.

Neutral means [H₃O⁺] = [OH⁻], which is pH 7.00 only at 25 °C. Kw increases with temperature; OpenStax gives about 5.6 × 10⁻¹³ at 100 °C, where neutral water has a pH near 6.1. For logarithms, the number of decimal places in a pH equals the number of significant figures in the concentration.

Calculating pH for strong acids and strong bases

Worked example 1. Find the pH of 4.6 × 10⁻³ mol/L nitric acid. HNO₃ is strong, so [H₃O⁺] = 4.6 × 10⁻³ mol/L. pH = −log(4.6 × 10⁻³) = 2.34. The concentration has two significant figures, so the pH has two decimal places. Then [OH⁻] = (1.0 × 10⁻¹⁴)/(4.6 × 10⁻³) = 2.2 × 10⁻¹² mol/L, tiny as expected in an acidic solution.

Worked example 2. 0.350 g of potassium hydroxide is dissolved to make 750.0 mL of solution. Using IUPAC standard atomic weights rounded to two decimals (K 39.10, O 16.00, H 1.01), M(KOH) = 56.11 g/mol. n = 0.350 g / 56.11 g/mol = 6.24 × 10⁻³ mol, and [OH⁻] = 6.24 × 10⁻³ mol / 0.7500 L = 8.32 × 10⁻³ mol/L. pOH = 2.080, and pH = 14.00 − 2.080 = 11.92, limited to two decimal places by the 14.00.

Worked example 3. Find the pH of 3.60 × 10⁻³ mol/L strontium hydroxide. Sr(OH)₂ releases two hydroxide ions per formula unit, so [OH⁻] = 2 × 3.60 × 10⁻³ = 7.20 × 10⁻³ mol/L. pOH = 2.143 and pH = 14.00 − 2.143 = 11.86. Forgetting the factor of 2 would give a pH about 0.30 too low.

Worked example 4. A solution has pH 4.62. Reverse the logarithm: [H₃O⁺] = 10 raised to the power −4.62 = 2.4 × 10⁻⁵ mol/L. The two decimal places in the pH allow two significant figures in the concentration. The pOH is 14.00 − 4.62 = 9.38, so [OH⁻] = 4.2 × 10⁻¹⁰ mol/L.

Logarithm and stoichiometry mistakes in pH work

Most pH errors come from the logarithm or from the formula of the acid or base. A quick sanity check catches many of them: acidic solutions at 25 °C have pH below 7, basic ones above 7, and each factor of ten in concentration moves the pH by one unit.

Be careful with very dilute strong acids. A 1.0 × 10⁻⁸ mol/L HCl solution does not have pH 8, which would be basic. At such low concentrations, the hydronium ions from water's own ionization are not negligible, and a full treatment gives a pH of about 6.98, just below neutral.

• Dropping the minus sign: −log of a number below 1 is positive. A negative pH from an ordinary dilute solution signals a sign error.
• Forgetting stoichiometry: Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂ each release two OH⁻ per formula unit.
• Using pH when the base gives pOH: find pOH from [OH⁻] first, then convert.
• Treating a weak acid as strong: [H₃O⁺] equals the acid concentration only for strong acids.
• Applying pH + pOH = 14.00 at temperatures far from 25 °C.