pH of Weak Acid and Weak Base Solutions

Weak-acid and weak-base pH calculations underlie pharmaceutical formulation, where a drug's ionization state at physiological pH (set by its $\text{p}K_a$ relative to pH 7.4) governs membrane permeability and bioavailability — the same $K_a=x^2/(C-x)$ machinery, applied to compounds like ibuprofen ($\text{p}K_a\approx4.9$) or morphine's conjugate acid, predicts what fraction is neutral (membrane-permeant) versus charged at a given pH. In environmental and industrial water chemistry, natural organic acids (humic and fulvic acids), carbonic acid equilibria in rainwater, and preservative acids like benzoic or sorbic acid in food products are all analyzed with this exact ICE-table-plus-quadratic approach, often at concentrations dilute enough that the 5% rule genuinely fails and the full quadratic (or even the water-inclusive charge balance) is required for regulatory-grade accuracy. Analytical instrumentation — pH meters calibrated against NIST-traceable buffers — routinely reports pH to two decimal places, and getting the underlying equilibrium math right (quadratic versus approximation) is exactly the difference between a value that matches instrumental measurement and one that is systematically off by a tenth of a unit, indistinguishable from instrument miscalibration unless the calculation itself is scrutinized. This same equilibrium logic is the direct on-ramp to Buffers and the Henderson-Hasselbalch Equation, where a mixture of HA and A⁻ is present simultaneously, and to Salt Solutions and Hydrolysis, where the conjugate base or acid of a salt reacts with water using precisely this $K_b$ or $K_a$ formalism inherited from the parent acid or base.

:::faq[How does formal concentration C differ from equilibrium concentration [HA]?] Formal concentration $C$ is the total amount of acid put into solution per liter, regardless of how much has since reacted — it's what you calculate from mass and volume when preparing the solution. Equilibrium concentration $[\text{HA}]$ is what remains un-ionized once the system reaches equilibrium, equal to $C-x$. Conflating the two is a common setup error: always let $C$ be the fixed constant from preparation, and let $x$ (the unknown) represent how much has reacted.

Ask the tutor: How do I decide, without first solving anything, whether a given weak acid problem will need the quadratic — just from looking at Ka and C?

Consider a weak monoprotic acid HA dissolved in water at formal concentration $C$ (mol/L), meaning $C$ moles of acid were weighed out and diluted to one liter, whether or not any of it has since ionized. The acid establishes the equilibrium

Before any reaction, essentially all the acid sits as HA; only water itself contributes a background $[\text{H}^+] = [\text{OH}^-] = 10^{-7}$ M, negligible next to what a millimolar-scale acid will produce. As the system relaxes to equilibrium, some quantity $x$ of HA ionizes, producing $x$ mol/L of H$^+$ and $x$ mol/L of A$^-$, while $(C-x)$ mol/L of HA remains intact. This is the same ICE-table logic used for any single-reaction equilibrium, but here the payoff — pH — is a logarithm of the answer, which means small errors in $x$ get compressed, but systematic errors (like using the wrong approximation) can still shift the reported pH by several tenths of a unit, a large error by analytical standards. The physical picture becomes richer once you plot the fraction ionized, $\alpha = x/C$, against concentration. For a fixed $K_a$, diluting the solution increases $\alpha$ — this is Ostwald's dilution law in disguise. A 1.0 M acetic acid solution is only about 0.42% ionized; a $1.0\times10^{-4}$ M solution is about 12% ionized. The absolute amount of H$^+$ still falls with dilution (since $x = \alpha C$ and $C$ is shrinking faster than $\alpha$ grows, up to a point), but the relative extent of reaction rises — a manifestation of Le Châtelier's principle: diluting a solution favors the side with more dissolved particles, here the ionized side. At extreme dilution, this trend cannot continue forever, because the solution's $[\text{H}^+]$ can never fall below what pure water itself supplies. This is the second essential physical idea: water is always a competing equilibrium,

For any acid concentration high enough to swamp this background (say $C \gtrsim 10^{-4}$ M for a moderately weak acid), water's contribution is safely ignored. But as $C$ falls toward $10^{-7}$–$10^{-8}$ M, the two equilibria compete on equal footing, and only a full charge-balance treatment — combining mass-action, mass balance, and charge balance simultaneously — gives the correct pH.

For a weak monoprotic acid HA with formal concentration $C$, dissociation constant $K_a$, letting $x = [\text{H}^+]$ at equilibrium:

where:

• $K_a$ — acid dissociation constant (dimensionless, referenced to standard state 1 M), known from tables or from $\text{p}K_a$
• $x$ — equilibrium $[\text{H}^+]$, equal to $[\text{A}^-]$ by 1:1 stoichiometry (mol/L)
• $C$ — formal (label) concentration of the acid before any dissociation (mol/L)

Reading: this is exact only under the approximation that water's autoionization contributes negligibly to $[\text{H}^+]$; it is the mass-action law applied directly to the ICE table. The exact quadratic form, obtained by clearing the denominator:

where:

• the negative root is discarded because $x=[\text{H}^+]$ must be positive
• this holds for any ratio of $x$ to $C$, so it is always safe to use, merely more laborious

Reading: whenever the simplification $C - x \approx C$ is not justified (dilute solutions, or acids with $K_a$ comparable to $C$), this quadratic is mandatory, not optional. The systematic approximation, valid when $x \ll C$:

where:

• valid when $x/C < 0.05$ (the "5% rule"), an empirical tolerance matched to the ~2–3 significant figures normally reported for pH

Reading: this shortcut converts a quadratic into a single square root; always check the 5% criterion after solving, and escalate to the quadratic if it fails. For a weak monoprotic base B with formal concentration $C_b$ and base constant $K_b$, the mirror relationship holds with $x = [\text{OH}^-]$:

where:

• $\text{pOH} = -\log x$, and then $\text{pH} = 14.00 - \text{pOH}$ at 25 °C, using $K_w = 1.0\times10^{-14}$
• $K_b$ relates to the conjugate acid's $K_a$ via $K_aK_b = K_w$ (the Ka–Kb relationship)

Reading: every weak-base problem is algebraically identical to a weak-acid problem; only the final pOH-to-pH conversion differs.

Example 1 (Foundational): 0.0500 M acetic acid. Acetic acid, CH₃COOH, has $K_a = 1.8\times10^{-5}$ (p$K_a = 4.74$). Find the pH of a 0.0500 M solution. Set up the ICE table with $x=[\text{H}^+]=[\text{CH}_3\text{COO}^-]$, $[\text{CH}_3\text{COOH}] = 0.0500-x$:

Try the small-x approximation first:

Check the 5% rule:

This is comfortably under 5%, so the approximation is valid. Report $x$ to appropriate significant figures — $K_a$ is given to 2 sig figs, so $x = 9.5\times10^{-4}$ M is reasonable, though carrying an extra digit through the log avoids rounding error:

Round to 2 decimal places (matching 2 sig figs in $K_a$, since pH decimals correspond to mantissa sig figs): pH = 3.02. Limiting-case check: if instead $C \to \infty$ conceptually (glacial acetic acid regime), $x/C \to 0$ and the approximation becomes exact in the limit — consistent with the derivation's Step 5. If $C$ were instead comparable to $K_a$ (say $C = 1.0\times10^{-4}$ M), the ratio $x/C$ would exceed 5% and the quadratic would be required — illustrated next.

Example 2 (Applied): pH of a very dilute benzoic acid solution near the breakdown of the shortcut. Benzoic acid, a food preservative, has $K_a = 6.3\times10^{-5}$. A quality-control chemist prepares a rinse solution at $C = 1.00\times10^{-3}$ M (an unusually dilute working concentration, chosen deliberately to test the approximation). Find the pH. First, try the shortcut:

Check:

This badly fails the 5% rule — the approximation would understate $[\text{HA}]$ depletion and is not trustworthy here. Fall back to the exact quadratic:

Apply the quadratic formula with $a=1$, $b=6.3\times10^{-5}$, $c=-6.3\times10^{-8}$:

Compare: the quadratic gives $x = 2.215\times10^{-4}$ M versus the naive $2.510\times10^{-4}$ M — an 11.7% overestimate from the flawed shortcut, which would have shifted the pH by

nearly a twentieth of a pH unit — small-looking but a real, systematic, avoidable error in an analytical report. Using the correct root:

pH = 3.65, versus the erroneous 3.60 from the uncorrected shortcut — a routine but consequential distinction for anyone reporting three significant figures. Limiting-case check: as $C \to 0$ further (say $10^{-6}$ M), water's own $10^{-7}$ M background stops being negligible relative to $x$, and even this quadratic (which already ignores water) would start to drift; that regime needs the full charge-balance treatment shown in Example 3.

Example 3 (Exam/competition level): pH of an extremely dilute weak acid where water's autoionization matters. A analytical chemist must report the pH of a $5.00\times10^{-7}$ M solution of a weak acid with $K_a = 1.0\times10^{-4}$ (chosen close to formic acid, illustrating a case where dilution is severe enough to invoke water explicitly). Estimate the pH rigorously. First, note that $C=5.00\times10^{-7}$ M is comparable in magnitude to $\sqrt{K_w}=1.0\times10^{-7}$ M, the natural scale of water's own ionization — a red flag that the simple $K_a=x^2/C$ approach (ignoring water) will be unreliable, and even the acid-only quadratic from Example 2 may be inadequate because it implicitly assumed $[\text{H}^+]$ comes only from the acid. Set up the full charge balance. Let $[\text{H}^+] = h$. Species present: HA, A⁻, H⁺, OH⁻. Charge balance:

Mass balance: $[\text{HA}]+[\text{A}^-] = C$. Water equilibrium: $[\text{OH}^-]=K_w/h$. Acid equilibrium:

Solve for $[\text{A}^-]$:

Substitute into the charge balance:

Because $K_a = 1.0\times10^{-4} \gg h$ (we expect $h$ on the order of $10^{-6}$–$10^{-7}$), the term $h+K_a \approx K_a$ to excellent approximation — since the acid is not "weak" in the sense of comparable-to-$h$ dissociation constant, the fraction simplifies:

This says: at such extreme dilution and with $K_a$ this much larger than $h$, essentially all the acid dissociates ($\alpha \to 1$), which makes physical sense — Ostwald's dilution law pushed to its limit. The charge balance reduces to

This is precisely the "strong acid at extreme dilution" formula (the acid, fully dissociated, behaves like a strong acid whose own concentration is small enough that water contributes measurably). Solve with $C=5.00\times10^{-7}$, $K_w=1.0\times10^{-14}$:

pH ≈ 6.28 — noticeably closer to neutral (7.00) than the naive weak-acid formula would ever predict, and importantly higher than the pH a "fully-dissociated strong acid at this concentration" estimate ignoring water would give ($-\log(5.00\times10^{-7})=6.30$, coincidentally close here only because dissociation is nearly complete). Had a student instead blindly applied $x\approx\sqrt{K_aC}$:

giving pH = 5.15 — wrong by more than a full pH unit, because that formula assumes $\alpha \ll 1$ when here $\alpha$ is close to 1, and it ignores water entirely. This example shows both failure modes at once and why extreme dilution demands the complete charge-balance derivation. Limiting-case check: setting $C\to 0$ entirely in the final quadratic recovers $h=\sqrt{K_w}=1.00\times10^{-7}$ M, pH = 7.00 — pure water, exactly as required.

The equilibrium framework used here — mass action, mass balance, systematic approximation — descends most directly from Wilhelm Ostwald, whose 1888 dilution law was the first quantitative statement that a weak electrolyte's dissociation fraction depends predictably on concentration through its dissociation constant. Ostwald measured conductivities of dilute acetic acid solutions at varying concentration and showed that $\Lambda/\Lambda_0$ (the fraction ionized, inferred from conductivity) obeyed exactly the relationship $K=\alpha^2C/(1-\alpha)$ — algebraically identical to the $K_a=x^2/(C-x)$ used throughout this article, with $x=\alpha C$. This was strong early evidence for the Arrhenius theory of electrolytic dissociation, published only a few years earlier, and it tied a measurable macroscopic property (conductivity) to a microscopic equilibrium constant for the first time.