The acid dissociation constant (Ka) is the equilibrium constant for a weak acid ionizing in water, Ka = [H3O+][A−]/[HA]. A bigger Ka means a stronger weak acid that ionizes more. In AP Chem (Topic 8.3), Ka plus the initial acid concentration lets you calculate the pH of a weak acid solution.
The acid dissociation constant, Ka, is just an equilibrium constant with a job title. When a weak acid HA sits in water, a small fraction of its molecules react with water to form hydronium (H3O+) and the conjugate base (A−). Most molecules stay un-ionized. That partial ionization sets up an equilibrium, and Ka measures where that equilibrium sits, with Ka = [H3O+][A−]/[HA] at equilibrium.
The size of Ka tells you acid strength among weak acids. A larger Ka means the equilibrium lies further toward the products, so more H3O+ forms and the solution is more acidic at the same concentration. Because Ka values for weak acids are tiny (often 10⁻⁵ or smaller), chemists usually report pKa = −log(Ka) instead. Per the CED (8.3.A.2), you can find the pH of any weak acid solution from just two pieces of information, the initial acid concentration and the Ka (or pKa). That's the core calculation of Topic 8.3.
Ka lives in Unit 8 (Acids and Bases), Topic 8.3: Weak Acid and Base Equilibria, and it directly supports learning objective 8.3.A, which asks you to explain the relationship among pH, pOH, and the concentrations of all species in a weak acid solution. The essential knowledge here (8.3.A.1 and 8.3.A.2) is blunt about the big idea. Weak acids only partially ionize, so [H3O+] is much smaller than the initial acid concentration, and Ka is the tool that quantifies exactly how much smaller. Ka is also the bridge concept of Unit 8. It connects equilibrium math from Unit 7 to acid-base chemistry, it pairs with Kb through Kw, and it reappears in titrations and buffers. If you can set up an ICE table with Ka, half of Unit 8 unlocks.
Keep studying AP® Chemistry Unit 8
Kb, the base dissociation constant (Unit 8)
Kb is the mirror image of Ka, the equilibrium constant for a weak base grabbing a proton from water. For a conjugate acid-base pair, Ka × Kb = Kw, so knowing one constant always gives you the other. A strong-ish weak acid has a weak conjugate base, and the math proves it.
Equilibrium constants and ICE tables (Unit 7)
Ka isn't a new kind of math, it's Unit 7 equilibrium applied to acids. Every weak acid pH calculation is an ICE table where you solve for x = [H3O+], usually with the shortcut [HA]eq ≈ [HA]initial because so little acid actually ionizes.
Percent ionization (Unit 8)
Percent ionization is the fraction of acid molecules that actually dissociated, and Ka controls it. Here's the counterintuitive part the exam loves. Diluting a weak acid keeps Ka constant but increases percent ionization, because the equilibrium shifts toward the side with more dissolved particles.
Titration curves and pKa (Unit 8)
At the half-equivalence point of a weak acid titration, exactly half the acid has been converted to its conjugate base, so [HA] = [A−] and the pH equals the pKa. Reading pKa off a titration curve is one of the fastest ways the exam tests whether you really understand what Ka means.
Ka shows up constantly in multiple choice, usually in one of four moves. First, pure calculation, where you get an initial concentration and a Ka and find pH (or get pH and find Ka). Second, the approximation question, which tests whether you know that for a small Ka, [HA]eq ≈ [HA]initial, so [H3O+] ≈ √(Ka·[HA]initial). That square root means quadrupling the acid concentration only doubles [H3O+], a relationship MCQs test directly. Third, dilution and percent ionization, where diluting a weak acid raises percent ionization even though Ka never changes. Fourth, titration curves, where the pH at the half-equivalence point equals pKa, so a curve passing through pH 4.75 at half the equivalence volume tells you Ka = 10⁻⁴·⁷⁵. On free-response questions, weak acid equilibrium setups are a Section II staple. You write the ionization equation, write the Ka expression with correct equilibrium concentrations, and justify any approximation you make.
Ka and pKa carry the same information but run in opposite directions, and that flip burns people. Since pKa = −log(Ka), a LARGER Ka means a stronger acid, but a SMALLER pKa means a stronger acid. An acid with pKa = 3 is stronger than one with pKa = 5, even though 5 is the bigger number. When you're ranking acid strength on an MCQ, check which constant you're given before you decide which way 'bigger' points.
Ka is the equilibrium constant for a weak acid ionizing in water, written as Ka = [H3O+][A−]/[HA].
A larger Ka means a stronger weak acid, but a larger pKa means a weaker acid, because pKa = −log(Ka).
Because weak acids only partially ionize, [H3O+] is much less than the initial acid concentration, which justifies the approximation [HA]eq ≈ [HA]initial when Ka is small.
With that approximation, [H3O+] ≈ √(Ka × [HA]initial), so quadrupling the initial concentration only doubles the hydronium concentration.
Diluting a weak acid increases its percent ionization even though Ka stays constant at constant temperature.
For a conjugate acid-base pair, Ka × Kb = Kw, and on a titration curve the pH at the half-equivalence point equals the pKa.
Ka is the equilibrium constant for a weak acid HA reacting with water to form H3O+ and its conjugate base A−, so Ka = [H3O+][A−]/[HA]. In Topic 8.3, you use Ka with the initial acid concentration to calculate the pH of a weak acid solution.
Stronger. A larger Ka means the ionization equilibrium lies further toward H3O+ and A−, so more of the acid dissociates. Just watch the flip with pKa, where smaller pKa means stronger acid.
No. Ka is constant at a given temperature no matter the concentration. What changes is percent ionization, which actually increases on dilution. A common MCQ gives you a 1.0 M acid diluted to 0.1 M and asks what happens, and the answer is that percent ionization goes up while Ka stays the same.
Ka describes a weak acid donating a proton to water, Kb describes a weak base accepting one, and Kw (1.0 × 10⁻¹⁴ at 25°C) is the autoionization constant of water itself. They're linked by Ka × Kb = Kw for any conjugate acid-base pair.
Read the pH at the half-equivalence point, which is half the volume needed to reach the equivalence point. There, [HA] = [A−], so pH = pKa and Ka = 10^(−pH). For example, if pH = 4.75 at 12.5 mL when the equivalence point is at 25.0 mL, then Ka = 10⁻⁴·⁷⁵.
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