Solve acid-base equilibrium problems.
A weak acid only partly dissociates, and how far it goes is set by its Ka and its concentration. Writing x for the hydrogen ion concentration produced, the equilibrium gives x squared divided by (C - x) equal to Ka, and solving that quadratic exactly avoids the usual shortcut of assuming x is negligible against C — an assumption that fails for dilute or fairly strong weak acids and can throw the pH off by several tenths. Degree of ionisation follows as x/C, and it rises as the solution is diluted, which is Ostwald's dilution law. For a base the same algebra applies to hydroxide, and pH is recovered as 14 minus pOH.
Weak acid or base equilibrium
Ka = 10^-pKa; solve x^2 + Ka x - Ka C = 0 giving x = (-Ka + sqrt(Ka^2 + 4 Ka C)) / 2; pH = -log10(x) for an acid, or 14 + log10(x) for a base; ionisation = x / C x 100
Concentrations here are ideal-solution approximations that ignore activity coefficients and ionic strength, so they diverge from measurement above roughly 0.1 M. Strong acids and bases are corrosive: consult the safety data sheet and use appropriate protection before preparing any solution.
Because pH = (pKa - log C) / 2 assumes dissociation is negligible. Once ionisation passes about 5% — dilute solutions, or acids with pKa below 3 — that assumption breaks down and the shortcut overstates the pH.
Ostwald's dilution law: diluting shifts the equilibrium towards more particles to keep Ka constant. The absolute hydrogen ion concentration still falls, so pH rises even though a larger fraction of the acid has dissociated.
No. It treats a single monoprotic equilibrium. For a buffer containing both the acid and its conjugate base, use the Henderson-Hasselbalch relation instead; for diprotic acids the second dissociation needs its own step.