Ionic Equilibrium and pH

Drop a spoon of vinegar into water and something subtler than simple mixing happens: acetic acid molecules do not all fall apart into ions at once. A fraction ionise, a fraction stay whole, and the two sides settle into a dynamic equilibrium — molecules breaking into ions exactly as fast as ions recombine. This tug-of-war, called ionic equilibrium, is what chemists tame with a single number, pH, to say precisely how acidic or basic a solution is. Behind that one number sits a family of constants — KaK_a, KbK_b, KwK_w — that quantify how far the tug-of-war leans, and a scale that turns a huge range of concentrations into a friendly 0-to-14 number line. This article builds pH from first principles: three definitions of acid and base, the ionisation constants that measure strength, water's own quiet self-ionisation, and Ostwald's law linking concentration to how ionised a weak electrolyte actually is.

The core insight

A weak acid or base sits in dynamic equilibrium with its ions, and the equilibrium constant (KaK_a or KbK_b) that measures how far that balance leans, together with water's constant self-ionisation (Kw=[H+][OH−]=10−14K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}), fixes the pH via pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+].

The core idea

Three definitions, in increasing order of generality. The Arrhenius definition says an acid produces H+\text{H}^+ (as H3O+\text{H}_3\text{O}^+) in water and a base produces OH−\text{OH}^- in water — simple, but limited to aqueous solutions. The Brønsted–Lowry definition widens this: an acid is a proton (H+\text{H}^+) donor and a base is a proton acceptor, working in any solvent. Every such reaction produces a conjugate acid–base pair — when HCl\text{HCl} donates H+\text{H}^+ it becomes its conjugate base Cl−\text{Cl}^-, and the water that accepts the proton becomes its conjugate acid H3O+\text{H}_3\text{O}^+. The Lewis definition is broader still: an acid is an electron-pair acceptor and a base is an electron-pair donor, which covers species like BF3\text{BF}_3 that have no proton to give at all.

Ionisation constants measure strength. For a weak monoprotic acid HA\text{HA} ionising in water, the reaction and its equilibrium constant are

HA+H2O⇌H3O++A−,Ka=[H3O+][A−][HA]\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-,\qquad K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}

where: KaK_a is the acid ionisation constant, and water — a pure liquid held at effectively constant concentration — is dropped from the expression as usual for KcK_c. A weak base BOH\text{BOH} ionises analogously, BOH⇌B++OH−\text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^-, with Kb=[B+][OH−]/[BOH]K_b = [\text{B}^+][\text{OH}^-]/[\text{BOH}]. A larger KaK_a or KbK_b means more of the substance has ionised at equilibrium — a stronger weak acid or base.

Water's own equilibrium sets a floor. Even pure water self-ionises very slightly, 2H2O⇌H3O++OH−2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-, giving the ionic product

Kw=[H+][OH−]=1.0×10−14 at 298 KK_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}\ \text{at}\ 298\ \text{K}

where: in pure water the two ion concentrations are equal, so [H+]=[OH−]=1.0×10−7 mol L−1[\text{H}^+] = [\text{OH}^-] = 1.0\times10^{-7}\ \text{mol L}^{-1}. This constant applies to every aqueous solution, acidic, neutral or basic — it is what lets you compute [OH−][\text{OH}^-] from [H+][\text{H}^+] or vice versa.

pH turns concentration into a friendly scale.

pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+]

Defining pOH=−log⁡10[OH−]\text{pOH} = -\log_{10}[\text{OH}^-] the same way, and taking −log⁡10-\log_{10} of the KwK_w expression, gives the companion relation pH+pOH=14\text{pH} + \text{pOH} = 14 at 298 K. Neutral solutions sit at pH=7\text{pH} = 7; acidic solutions have pH<7\text{pH} < 7; basic solutions have pH>7\text{pH} > 7. The scale is logarithmic, so each unit step is a tenfold change in [H+][\text{H}^+].

Figure
A number line from pH 0 to pH 14 shaded from red at the acidic end through neutral at 7 to blue at the basic end
The pH scale — a factor of ten in H+ per unit step 0 7 14 acidic, pH less than 7 neutral basic, pH greater than 7 pH = -log10[H+], each step is a tenfold change in [H+]

Strength constants and dilution. pKa\text{p}K_a and pKb\text{p}K_b are defined the same way as pH, pKa=−log⁡10Ka\text{p}K_a = -\log_{10}K_a and pKb=−log⁡10Kb\text{p}K_b = -\log_{10}K_b — a smaller pKa\text{p}K_a means a stronger acid. For a conjugate acid–base pair, Ka×Kb=KwK_a \times K_b = K_w, so a weak base's conjugate acid can never be a strong acid, since the product of the two constants is a small, fixed number. Finally, the degree of ionisation α\alpha (the fraction of a weak electrolyte that ionises, 0<α<10 < \alpha < 1) obeys Ostwald's dilution law: for a weak acid of initial concentration cc, Ka=cα2/(1−α)≈cα2K_a = c\alpha^2/(1-\alpha) \approx c\alpha^2 when α≪1\alpha \ll 1, so α≈Ka/c\alpha \approx \sqrt{K_a/c} — diluting a weak electrolyte increases its degree of ionisation.

Derivation AFrom concentration and Ka to pH, via Ostwald's law

The goal is to turn a weak acid's initial concentration cc and its ionisation constant KaK_a into a numerical pH, using only an ICE table, one algebraic approximation, and a single logarithm — the standard method for any weak-acid pH problem on this syllabus.

Step 1. Set up the ICE table for HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-: initial concentrations are c,0,0c, 0, 0; the change is −x,+x,+x-x, +x, +x; equilibrium concentrations are c−x,x,xc-x, x, x, where x=[H+]x = [\text{H}^+] at equilibrium.

Step 2. Substitute into the equilibrium expression, Ka=x2c−xK_a = \dfrac{x^2}{c-x}. Because weak acids typically ionise only slightly (α≪1\alpha \ll 1, i.e. x≪cx \ll c), approximate the denominator as c−x≈cc - x \approx c, giving the simplified relation Ka≈x2cK_a \approx \dfrac{x^2}{c} — exactly Ostwald's dilution law in the form Ka≈cα2K_a \approx c\alpha^2 with α=x/c\alpha = x/c.

Step 3. Solve for xx: x=Kacx = \sqrt{K_a c}. This xx is [H+][\text{H}^+] at equilibrium, so take −log⁡10-\log_{10} of it to get pH=−log⁡10Kac=12(pKa−log⁡10c)\text{pH} = -\log_{10}\sqrt{K_a c} = \tfrac{1}{2}(\text{p}K_a - \log_{10}c).

Step 4. Always check the approximation afterwards by computing α=x/c\alpha = x/c. If α\alpha comes out below about 0.050.05–0.10.1, the simplification was valid; if it is larger, the quadratic x2+Kax−Kac=0x^2 + K_a x - K_a c = 0 must be solved exactly instead of approximated.

Worked example

Problem: Calculate the pH of a 0.01 mol L−10.01\ \text{mol L}^{-1} solution of acetic acid (CH3COOH\text{CH}_3\text{COOH}), given Ka=1.8×10−5K_a = 1.8\times10^{-5}.

Step 1. Write the equilibrium and ICE table: CH3COOH⇌H++CH3COO−\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-; initial c=0.01c = 0.01, change −x,+x,+x-x, +x, +x, equilibrium 0.01−x,x,x0.01-x, x, x.

Step 2. Apply the approximation from the derivation: Ka≈x2/cK_a \approx x^2/c, so x2=Kac=(1.8×10−5)(0.01)=1.8×10−7x^2 = K_a c = (1.8\times10^{-5})(0.01) = 1.8\times10^{-7}.

Step 3. Take the square root: x=1.8×10−7=4.24×10−4 mol L−1=[H+]x = \sqrt{1.8\times10^{-7}} = 4.24\times10^{-4}\ \text{mol L}^{-1} = [\text{H}^+].

Step 4. Check the approximation: α=x/c=4.24×10−4/0.01=0.0424\alpha = x/c = 4.24\times10^{-4}/0.01 = 0.0424, about 4.2%4.2\% — comfortably small, so the approximation holds.

Step 5. Convert to pH: pH=−log⁡10(4.24×10−4)=3.37\text{pH} = -\log_{10}(4.24\times10^{-4}) = 3.37. The pH of the solution is approximately 3.37, acidic as expected for a weak acid.

Competitive stretch

Aniline (C6H5NH2\text{C}_6\text{H}_5\text{NH}_2) has Kb=4.3×10−10K_b = 4.3\times10^{-10}; its conjugate acid is the anilinium ion. Find KaK_a for anilinium using Ka×Kb=KwK_a \times K_b = K_w: Ka=Kw/Kb=(1.0×10−14)/(4.3×10−10)=2.3×10−5K_a = K_w/K_b = (1.0\times10^{-14})/(4.3\times10^{-10}) = 2.3\times10^{-5}. This is comparable to acetic acid's KaK_a, confirming that the conjugate acid of a weak base is itself only weak — never strong — because Ka×Kb=KwK_a \times K_b = K_w is a fixed small number.

Common traps

  • Treating Arrhenius, Brønsted–Lowry and Lewis as competing "correct" theories rather than successive generalisations — each widens the scope of the last without contradicting it.
  • Skipping the approximation check on Ka≈cα2K_a \approx c\alpha^2: it is only valid when α≪1\alpha \ll 1 (roughly α<0.05\alpha < 0.05–0.10.1); for larger α\alpha or larger KaK_a, solve the quadratic exactly.
  • Mixing up pKa\text{p}K_a with KaK_a — a smaller pKa\text{p}K_a means a stronger acid, which is the opposite of what many students first guess.
  • Assuming pH 7 always means neutral regardless of temperature — KwK_w itself is temperature-dependent, so pure water is exactly pH 7 only at 298 K.

Ask the tutor: want a graded set of pH and Ka/Kb problems, including ones where the quadratic approximation breaks down?

Why do we drop water from the Ka expression when water is a reactant?

Water is the solvent and is present in enormous excess compared to the small amount of acid dissolved in it — its concentration barely changes during the reaction, so it behaves as an effective constant. This is the same convention used for any KcK_c involving a pure liquid or solvent in large excess: its concentration is folded into the equilibrium constant itself rather than written explicitly. That is why Ka=[H3O+][A−]/[HA]K_a = [\text{H}_3\text{O}^+][\text{A}^-]/[\text{HA}] has no [H2O][\text{H}_2\text{O}] term, even though water clearly took part in the reaction.

If Ka times Kb always equals Kw, can a weak base ever have a strong conjugate acid?

No, and this is one of the most useful consequences of the relation. Since Kw=1.0×10−14K_w = 1.0\times10^{-14} is a small fixed number at a given temperature, if KbK_b for a base is small (a weak base), then Ka=Kw/KbK_a = K_w/K_b for its conjugate acid must also come out small — you cannot get a large KaK_a from dividing a tiny fixed number by another small number without the result staying modest. So the conjugate acid of any weak base is itself always weak; only the conjugate base of a strong acid can be essentially non-basic (like Cl−\text{Cl}^- from HCl\text{HCl}).

Why does pH plus pOH always equal 14, and does that ever change?

It comes directly from taking −log⁡10-\log_{10} of both sides of Kw=[H+][OH−]=1.0×10−14K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}: the log of a product splits into a sum of logs, giving −log⁡10[H+]+(−log⁡10[OH−])=−log⁡10(1.0×10−14)-\log_{10}[\text{H}^+] + (-\log_{10}[\text{OH}^-]) = -\log_{10}(1.0\times10^{-14}), which is exactly pH+pOH=14\text{pH} + \text{pOH} = 14. The number 14 is not a universal constant, though — it comes from the value of KwK_w at 298 K298\ \text{K}. At other temperatures KwK_w changes, so pH+pOH\text{pH} + \text{pOH} shifts away from 14 even though the underlying relation still holds.

Before this

Where this leads