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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.

An analytical concentration records the total conserved inventory of a specified component per volume of chemical solution, summed over its relevant chemical species. Here $C_T=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$; it is not generally the chemical equilibrium concentration $[\mathrm{HA}]$ of the undissociated weak acid.

Stipulate pKa $5$ and $K_w=10^{-14}$; these are model inputs, not measurements of an identified substance. At pH $5$ the pair contribution is $0.575646\dots C$. For analytical concentrations $C=0.010$ and $0.100\mathrm{mol}{\mathrm{L}}^{-1}$, respectively, it isThe free hydrogen ion and hydroxide contribution is the same at this pH, ${\beta }_{\mathrm{water}}=(\ln 10)(10^{-5}+10^{-9})\mathrm{mol}{\mathrm{L}}^{-1}=0.0000230282\dots \mathrm{mol}{\mathrm{L}}^{-1}$ per pH. The full buffer capacities are therefore approximately $0.00577949$ and $0.0575877\mathrm{mol}{\mathrm{L}}^{-1}$ per pH. Identical pH fixes the free hydrogen ion thermodynamic activity; it does not fix the inventory of weak acid and conjugate base available for neutralization.

At one unit of pH on either side of pKa, $r=10$ or $0.1$, giving ${\beta }_{\mathrm{pair}}/{\beta }_{\mathrm{pair},\max }=40/121\approx 0.331$. The partners still form a buffer solution, but its local resistance is only about one third of the maximum for that analytical concentration. At fixed pH, the pair contribution scales exactly with $C$; the full buffer capacity contains the additional $C$-independent water contribution. Consequently, extreme dilution does not justify dropping self-ionization of water.

A proposed modern measurement extends the weak acid dissociation measurement by estimating the local slope of an acid-base titration. Transfer a measured sample with a pipette into a stirred vessel; clamp a combination glass electrode/reference electrode clear of the stirring magnet and place a temperature probe nearby. A calibrated burette delivers a chemical base of independently standardized concentration $c_T$, such as sodium hydroxide. Record the initial sample volume $V_0$, cumulative delivered volume $v$, stable pH, and temperature after each small addition. Use a separate fresh portion for an acidward series. These are proposed measurement conditions, not reported experimental results.

For the instrument calibration, use fresh standard buffer solutions bracketing the measured pH range, with values appropriate to the actual temperature. The Hanna manual specifies separate rinse and calibration portions, gentle stirring, and approximately $4\mathrm{cm}$ immersion of the probe tip. It calls for stable readings; the sensing bulb and reference electrode junction must remain immersed. Rinse between standards and samples and avoid carrying rinse water into the measured portion.^[ref]

The readout pairs each added amount of substance with the measured pH. With negligible added volume, neighboring readings give the finite difference estimateassigned to the interval's midpoint. High buffer capacity appears as a small change of pH for a measured reagent dose. Choose increments small enough to resolve changing buffer capacity, but large enough that the change of pH exceeds measurement noise. Repeat with a fresh portion and different increment sizes; remeasure a standard afterward to check drift. A water blank can reveal contamination, but its pH must not be subtracted from the sample pH. These controls are proposed tests of the measurement and model assumptions.

If additions appreciably increase volume, even the analytical concentration changes: $C(v)=C_0{V}_0/(V_0+v)$. Merely dividing the added moles by the current volume does not restore the constant-volume derivative. One alternative is explicitly to report a changing-volume convention. For $c(v)=c_{T}v/(V_0+v)$, defineThis definition includes the reagent's changing concentration and differs from the constant-volume buffer capacity defined above.^[ref]

To compare with the constant-volume theory, fit the changing-volume chemical equilibrium model instead of treating the observed slope as $\beta$. In the same ideal monoprotic acid system, define $S_0=[{\mathrm{A}}^{-}{]}_0+[{\mathrm{OH}}^{-}{]}_0-[{\mathrm{H}}^{+}{]}_0$. For baseward additions, the electric charge balance at each reading isUse a minus sign before $c_{T}v$ for the corresponding acidward series. From the model parameters, evaluate the constant-volume buffer capacity at the chosen analytical concentration and pH. The fit also tests whether one acid dissociation constant describes the observations.

Limitations include incomplete mixing, slow glass electrode response, uncertain reagent concentration, and changes in temperature or ionic strength. Dissolved atmospheric carbon dioxide introduces an additional chemical acid system; comparable exposure and freshly prepared samples help distinguish it from the intended monoprotic acid response. Additional reacting chemical species and nonideal activity coefficients invalidate the simple formula. Especially when pH increments are tiny, measurement noise can dominate the inferred derivative; reporting the dose, pH interval, volume convention, and repeatability is part of reporting a meaningful buffer capacity.

Assume an ideal dilute aqueous chemical solution with just one monoprotic acid pair $\mathrm{HA}/{\mathrm{A}}^{-}$, self-ionization of water, and other ions that balance electric charge without undergoing further chemical reactions. Hold the analytical concentration $C=[\mathrm{HA}]+[{\mathrm{A}}^{-}]$ constant. Keep the acid dissociation constant $K_a$ and the self-ionization of water constant $K_w$ dimensionless, consistently with pH:Thus $H$ denotes normalized hydrogen ion concentration, not a dimensional concentration. The chemical equilibrium and conserved analytical concentration giveAdding chemical base converts weak acid to conjugate base, increases free hydroxide, and decreases free hydrogen ion concentration. Balancing electric charge, including the added reagent's accompanying ions, gives the signed dose relative to any initial chemical equilibrium at $p_0$:Differentiate with respect to pH, using $\mathrm{d}H/\mathrm{d}p=-(\ln 10)H$. The result isThe first term is the monoprotic acid pair's contribution, ${\beta }_{\mathrm{pair}}$; the remaining terms are the free hydrogen ion and hydroxide contributions, ${\beta }_{\mathrm{water}}$. Each has units $\mathrm{mol}{\mathrm{L}}^{-1}$ per pH. Equivalently, with dimensional $H_c=[{\mathrm{H}}^{+}]$, $K_{a,c}=c^{\circ }{K}_a$ and $K_{w,c}=(c^{\circ }{)}^2{K}_w$, the expression is $(\ln 10)(C{K}_{a,c}{H}_c/(K_{a,c}+H_c{)}^2+H_c+K_{w,c}/H_c)$. Mixing the dimensional and dimensionless conventions would give incorrect units.

For $r=K_a/H=10^{\mathrm{pH}-\mathrm{p}{K}_a}$,because $(1+r{)}^2-4r=(1-r{)}^2\ge 0$. Equality occurs at $r=1$: equal weak acid and conjugate base concentrations, or pH equal to pKa. This is the exact maximum of the pair contribution in this model. The full buffer capacity need not have its maximum there: free hydrogen ions and hydroxide make it grow toward the acidic and alkaline extremes. When these contributions are small, the central peak lies close to pKa.

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.

Dilution lowers a solute's concentration by adding solvent. For conserved amount of substance, molar concentration satisfiesBoth $V_1$ and $V_2$ are chemical solution volumes; $V_2$ is the final total, not the added solvent.^[ref]

The proof is simply the definition $n=cV$ applied before and after. Its useful assumptions are full transfer of the selected portion, no loss or gain of the counted solute, and complete mixing. An ideal chemical solution approximation is not required for this inventory equation. If chemical equilibrium redistributes dissolved chemical species, use a conserved analytical concentration or chemical element inventory rather than assuming each individual species is conserved. This conservation statement does not by itself preserve the pH: a chemical equilibrium may change as the chemical solution is diluted.

In an ideal aqueous chemical solution, set $h=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=10^{-\mathrm{pH}}$, with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and approximate activity coefficients by one. Put $K_1=10^{-2.35}$ and $K_2=10^{-9.78}$ for the stipulated dimensionless acid dissociation constants. The Henderson-Hasselbalch equation gives the adjacent-state ratiosWith analytical concentration $C_T=[G^{+}]+[G^0]+[G^{-}]$, normalization gives the three fractions and mean electric charge in units of the elementary charge:All three fractions are nonnegative and sum to one. At pH $7$, the model gives about $99.832\%$ zwitterion, $0.00224\%$ cation and $0.1657\%$ anion; $\bar{z}\approx -0.00163$. Predominantly zwitterionic therefore need not mean exactly at the isoelectric point.

Setting $\bar{z}=0$ requires $h^2=K_1{K}_2$. Taking the logarithm yieldsAt this model isoelectric point, ${\alpha }_{+}={\alpha }_{-}\approx 0.0001927$ and ${\alpha }_0\approx 0.9996146$: about $0.0193\%$ of each charged form remains. At either pKa, the relevant adjacent fractions are equal, but the third fraction is not mathematically zero. This three-state model neglects the minor uncharged form ${\mathrm{H}}_2\mathrm{N}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COOH}$; macroscopic pKa values alone do not determine its fraction separately from the zwitterion.

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.

Molar concentration is amount of substance of a named chemical species per final chemical solution volume: $c=n/V$. State whether the count describes dissolved species or a conserved analytical concentration.

Serial dilution repeatedly transfers a measured portion of one well-mixed chemical solution and dilutes that portion to a new final volume. With transferred volume $v_j$ and final volume $V_j$ at step $j$, conserved solute givesThis product follows by repeated substitution. Each step conserves the solute in the portion transferred into its receiving flask, not the entire inventory in the preceding flask. Uniform mixing makes the selected fraction representative.

Continue the original solution mass and volume bookkeeping example. A $10.00\mathrm{mL}$ portion of the $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$ stock contains $0.002000\mathrm{mol}$ of analytical sodium chloride, corresponding to $0.11688\mathrm{g}$ before rounding. After dilution to $100.0\mathrm{mL}$, its analytical concentration is $0.02000\mathrm{mol}{\mathrm{L}}^{-1}$. A new $5.000\mathrm{mL}$ portion of that mixed chemical solution contains $0.0001000\mathrm{mol}$; dilution to $50.00\mathrm{mL}$ gives $0.002000\mathrm{mol}{\mathrm{L}}^{-1}$. The two factors $10.00/100.0$ and $5.000/50.00$ multiply to $1/100$.

The remaining stock holds $0.05000-0.002000=0.04800\mathrm{mol}$; the remaining first dilution holds $0.002000-0.0001000=0.001900\mathrm{mol}$. Including the final flask restores the starting inventory:This is a direct check that making a dilute final sample does not destroy the solute. Most of it remains in earlier vessels.

Serial dilution makes a small final concentration accessible using measurable portions instead of a single extremely small transfer. Its limitation is that systematic errors propagate: if each transferred volume is $1\%$ below its intended value while final volumes are correct, two steps give $0.99^2=0.9801$ times the intended concentration, a $1.99\%$ deficit. Repeated readings alone do not remove that bias. The gain in manageable volume therefore depends on calibrated transfer and representative mixing.

The denominator of molar concentration is the final volume of the chemical solution, including the dissolved solute. It is not the volume of solvent initially poured in. A volumetric flask fixes the final total volume; weighing the solute supplies its mass, and a specified molar mass converts that mass to amount of substance.^[ref]

Consider an original illustrative account, not reported measurements. Suppose $2.922\mathrm{g}$ of pure dry sodium chloride, with stipulated molar mass $58.44\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, is completely dissolved and quantitatively transferred into a chemical solution whose final volume is $250.0\mathrm{mL}$. Assume no losses, no additional chemical reactions, and complete dissociation of an ionic compound. ThenHere analytical concentration counts the sodium chloride inventory, not intact dissolved $\mathrm{NaCl}$ molecules. Under these assumptions, the amount of substance of sodium cations and of chloride anions is separately $0.05000\mathrm{mol}$; each has molar concentration $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$. Their combined ionic amount of substance is $0.1000\mathrm{mol}$, but this does not double either the salt inventory or the mass. Each sodium cation still contains a sodium atomic nucleus, and each chloride anion a chlorine atomic nucleus. Counting chemical elements survives dissociation of an ionic compound.

The mass ledger is different from the volume ledger. If the final chemical solution had an illustrative net mass of $252.922\mathrm{g}$, the water contribution would be $252.922-2.922=250.000\mathrm{g}$, assuming nothing entered or escaped except those two components. The ratio $2.922/252.922\approx 0.011553$ expresses the solute share of total mass; the separate mass density would be $252.922/250.0=1.011688\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$ before rounding. These stipulated quantities are mutually consistent bookkeeping inputs, not a measured mass density for saltwater. Neither establishes that $250.0\mathrm{mL}$ of water was added.

More generally, with full transfer and conserved solute, $n=m/M$ and $c=m/(MV)$. No additive volume or ideal chemical solution assumption is needed when $V$ is the measured final volume. If only a fraction $f$ of the weighed solute reaches the flask, the actual result is $c=fm/(MV)$. This short calculation explains why an undissolved residue or an incomplete rinse creates a systematic deficit even when the final calibration mark is read correctly.