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The acid dissociation constant quantifies the chemical equilibrium for loss of one proton from a chemical acid. In water, write the actual proton transfer as $\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{A}}^{-}$. Its full dimensionless equilibrium constant isThe usual aqueous acid dissociation constant uses the solvent convention with water thermodynamic activity approximated by one:Here hydrated hydrogen ions are understood; ${\mathrm{H}}^{+}$ is shorthand, not an isolated bare proton in the chemical solution. All thermodynamic activities and $K_a$ are dimensionless quantities, using consistent standard states. If water thermodynamic activity differs appreciably from one, retain it in the full equilibrium constant instead of silently mixing conventions.

With activity coefficients ${\gamma }_i$ and reference concentration $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, put $a_i={\gamma }_i[i]/c^{\circ }$ for each solute. ThenThe last expression is the dilute ideal concentration approximation, not the definition in a nonideal chemical solution. Omitting $c^{\circ }$ instead produces a quotient with units of concentration; its numerical value depends on the chosen units. Do not take a logarithm of that dimensional quotient. Conditional concentration quotients can change with ionic strength even at fixed temperature, whereas the thermodynamic acid dissociation constant retains the activity coefficient corrections. The acid dissociation constant entry distinguishes these conventions.

At the same temperature, solvent and standard states, a larger acid dissociation constant, equivalently a smaller pKa, indicates a stronger tendency to donate a proton. It does not specify an analytical concentration, a pH, or a reaction rate. Composition and the weak acid mass and charge balances are needed to predict pH; chemical kinetics addresses how quickly the chemical equilibrium is approached.

A conjugate base is what remains after a chemical acid donates one proton. In the conjugate acid-base pair $\mathrm{HA}/{\mathrm{A}}^{-}$, $\mathrm{HA}$ is the conjugate acid of ${\mathrm{A}}^{-}$; accepting a proton reverses the conversion. For example, acetic acid $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ loses a proton to give its conjugate base $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}$. Partners differ by one proton and one unit of electric charge; they need not always be neutral and negative. Hydronium/water and water/hydroxide are also conjugate acid-base pairs.

At a prescribed pH, the ideal dilute acid dissociation constant determines the partition of a monoprotic acid inventory between its chemical species:These dimensionless fractions sum to one. They are the weak acid dissociation fraction and its complement when the only acid-bearing chemical species are $\mathrm{HA}$ and ${\mathrm{A}}^{-}$. Equal fractions occur at $\mathrm{pH}=\mathrm{p}{K}_a$ in this ideal approximation. For nonideal chemical solutions, equality of thermodynamic activities need not mean equality of concentrations.

For stipulated pKa $5$, the conjugate base fractions at pH $4$, $5$ and $6$ are $1/11$, $1/2$ and $10/11$. Here an external reservoir sets pH; its ions must supply the additional electric charge balance. The diagram omits those ions and water populations. It does not represent three pure $\mathrm{HA}$ samples at identical analytical concentration with freely chosen pH. In the latter experiment, the weak acid mass and charge balances determine pH as well as the partition.

Consider a neutral monoprotic acid $\mathrm{HA}$ dissolved in water, with no added salts or other acids. Assume dilute ideal thermodynamic activities, fixed temperature, water thermodynamic activity one, and negligible self-ionization of water. Write $C_T$ for its analytical concentration, $c=C_T/c^{\circ }$ with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and $\alpha$ for its weak acid dissociation fraction. Conservation of acid units and electric charge givesHere ${\mathrm{A}}^{-}$ is the conjugate base, and ${\mathrm{H}}^{+}$ is conventional shorthand for hydrated hydrogen ions, represented by hydronium in this model. Substitution into the dimensionless acid dissociation constant givesSolving this quadratic equation and rejecting its negative root of a polynomial gives an exact result within the stated model:The second form avoids subtracting nearly equal numbers. Unlike the approximation $\alpha \approx \sqrt{K_a/c}$, it retains depletion of neutral weak acid and always satisfies $0<\alpha <1$ for positive $K_a,c$. The broader weak acid mass and charge balances restore self-ionization of water when that omitted contribution matters.

For an original worked example, stipulate $K_a=10^{-5}$ without identifying a real substance. At $C_T=10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$, the weak acid dissociation fraction is approximately $0.0311267$, or $3.11267\%$, and $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=3.11267\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$. A hundredfold dilution to $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ raises the fraction to $0.270156$, or $27.0156\%$, but lowers $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to $2.70156\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}$. A completely ionized strong acid at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ would have $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\approx 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$: less than the more concentrated weak acid in the first calculation, despite its greater converted fraction.

The directions of both changes follow without relying on the plot. Put $x=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=c\alpha$. Rearranging the chemical equilibrium relation givesThus smaller $c$ forces smaller $x$, which in turn makes $\alpha$ larger. The derivative proves that total hydronium concentration and converted fraction move in opposite directions during dilution. In this model, adding solvent also makes additional acid units ionize at fixed total acid amount of substance, even though their ions are spread over more volume. That distinction explains why “more ionized” alone does not predict greater electrical conductivity.

Do not extend the no-background model to unlimited dilution: its prediction $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\to 0$ omits water's own ions. The weak acid mass and charge balances include that background. Added salts, other acids or bases, and nonideal activity coefficients also change the relation between analytical concentration, converted fraction and hydronium concentration.

For a monoprotic acid with analytical concentration $C_T=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$, define the dimensionless weak acid dissociation fractionThis counts acid units in the conjugate base form. Identifying it with $[{\mathrm{H}}_3{\mathrm{O}}^{+}]/C_T$ additionally requires negligible self-ionization of water and no other sources of electric charge. In the ideal dilute chemical solution model, with $h=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }$, the acid dissociation constant gives $\alpha =K_a/(K_a+h)$. Thus lowering hydronium thermodynamic activity favors the conjugate base.

The diagram uses stipulated dimensionless acid dissociation constant $K_a=10^{-5}$ and self-ionization of water ion product $K_w=10^{-14}$ at fixed temperature; these are model inputs, not measurements of a named substance. It solves the weak acid mass and charge balances. At $C_T=0.010\mathrm{mol}{\mathrm{L}}^{-1}$, the calculated weak acid dissociation fraction is about $3.11\%$ and pH about $3.51$; at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$, they are about $27.0\%$ and $4.57$. More acid units are dissociated as a fraction of the total, yet fewer hydronium ions occur per volume.

At extreme dilution, self-ionization of water makes $h$ approach $\sqrt{K_w}$, so pH approaches $7$ for the chosen model. The limiting weak acid dissociation fraction is $K_a/(K_a+\sqrt{K_w})\approx 0.9901$, rather than exactly one. The calculation assumes pure added water, a neutral monoprotic acid, and no competing chemical reactions; actual contamination becomes increasingly important as $C_T$ falls.

Consider only neutral $\mathrm{HA}$ added to pure water, at fixed temperature, with ideal dilute thermodynamic activities, water thermodynamic activity approximated by one, and no added salts or other chemical species. Normalize every concentration by $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$:Conservation of acid units and zero net electric charge, respectively, requireThe second equation balances positive hydronium against negative conjugate base and hydroxide. It would need all additional ions if salts or a chemical base were added. Together with acid dissociation constant $K_a=hx/y$ and the self-ionization of water relation $K_w=ho$, these giveThis is exact within the stated ideal model, including depletion and self-ionization of water. Multiplication by $h(K_a+h)$ yields the cubic equationFor positive $K_a,K_w,c$, the unmultiplied left-hand side increases strictly from negative to positive values as $h$ increases from zero to infinity; its derivative is $1+K_w/h^2+K_{a}c/(K_a+h{)}^2>0$. There is therefore exactly one positive root of a polynomial. A bisection method can find it without choosing an unphysical root of a polynomial.

If $K_w/h^2\ll 1$, neglect hydroxide, so $h\approx x$ and the quadratic equation retaining acid depletion givesThe last form avoids subtracting nearly equal numbers. If also $x/c\ll 1$, neglect depletion to obtainCheck both conditions after computing: $x/c$ measures the fraction lost from the initial undissociated weak acid inventory, while $K_w/h^2$ measures the omitted hydroxide contribution to electric charge balance. A $5\%$ depletion criterion is a chosen tolerance, not a universal boundary. Keeping depletion does not cure the neglect of self-ionization of water. These weak acid mass and charge balances also explain why the square root estimate cannot be extended to arbitrarily dilute chemical solutions.