Ion concentration is the molarity of a specific ion dissolved in solution, written in brackets like [Cl⁻] or [H⁺]. In AP Chem, ion concentrations plug directly into Ksp expressions, common-ion-effect predictions, pH calculations, and the Henderson-Hasselbalch equation.
Ion concentration is the amount of a particular ion present per liter of solution, expressed in molarity (mol/L) and written with square brackets, like [Pb²⁺] = 0.015 M. The brackets matter. When you see [A⁻], the exam is asking about the equilibrium molarity of that specific ion, not the amount of salt you dissolved.
Here's the part that trips people up. Ion concentration is tied to stoichiometry. When PbCl₂ dissolves, every formula unit releases one Pb²⁺ and two Cl⁻, so in a saturated solution [Cl⁻] is exactly twice [Pb²⁺]. That 1:2 ratio is why a saturated PbCl₂ solution has [Pb²⁺] = 0.015 M but [Cl⁻] = 0.030 M. Once you have the ion concentrations, you can build any equilibrium expression the exam throws at you, from Ksp to Ka to the buffer ratio in Henderson-Hasselbalch.
Ion concentration is the working currency of Units 7 and 8. In Topic 7.12, LO 7.12.A asks you to find a salt's solubility or Ksp based on the concentration of a common ion already in solution. The common-ion effect (EK 7.12.A.1) is really a statement about ion concentrations: if [Cl⁻] is already high, less PbCl₂ can dissolve, which you can predict with Le Châtelier or calculate from Ksp. In Topic 8.9, LO 8.9.A has you identify a buffer's pH from the concentrations of the conjugate acid-base pair. The Henderson-Hasselbalch equation's log term is literally a ratio of ion (and molecule) concentrations, [A⁻]/[HA], and EK 8.9.A.1 explains that buffers work because adding small amounts of acid or base barely changes that ratio. If you can track ion concentrations through stoichiometry and equilibrium, both topics open up.
Keep studying AP Chemistry Unit 8
Molar Solubility (Unit 7)
Molar solubility tells you how many moles of salt dissolve per liter; ion concentrations come from multiplying that by the stoichiometric coefficients. For PbCl₂, solubility s gives [Pb²⁺] = s and [Cl⁻] = 2s. Mixing these up is the most common Ksp error on the exam.
Common Ion Effect (Unit 7)
The common ion effect is what happens when an ion concentration starts above zero. Dissolving AgCl into a solution that already contains Cl⁻ shifts the dissolution equilibrium left, so less salt dissolves. The starting [Cl⁻] goes straight into your Ksp math.
Henderson-Hasselbalch Equation (Unit 8)
pH = pKa + log([A⁻]/[HA]) is built entirely from concentrations. The log term is the ratio of conjugate base concentration to weak acid concentration, and when [A⁻] = [HA] the ratio is 1, the log is 0, and pH = pKa. That's the half-equivalence point.
Molarity (Unit 3 foundation)
Ion concentration is just molarity applied to one species instead of the whole solute. The mol/L skills you built early in the course (dilutions, mole conversions) are the same moves you make inside every equilibrium problem in Units 7 and 8.
Multiple-choice questions hand you ion concentrations and ask you to do something with them. Classic stems give you [Pb²⁺] and [Cl⁻] in a saturated solution and ask for Ksp, ask which Ksp expression matches a dissolution equation like CaCO₃, or give you [H⁺] = 1 × 10⁻⁵ M and ask for pH (it's 5, since pH = -log[H⁺]). Buffer questions ask what the log term in Henderson-Hasselbalch represents, and the answer is the concentration ratio of conjugate base to weak acid. On free-response, acid-base FRQs like the 2025 long FRQ on ascorbic acid expect you to move between concentrations, equilibrium expressions, and pH, often justifying your answer with the [A⁻]/[HA] ratio or a Le Châtelier argument about a common ion. The skill being tested is always the same. Translate the chemistry into bracket quantities, respect the stoichiometry, and plug into the right equilibrium expression.
Molar solubility is how many moles of the whole salt dissolve per liter. Ion concentration is the molarity of one specific ion afterward. They're only equal when the stoichiometry is 1:1. For PbCl₂ with molar solubility 0.015 M, [Pb²⁺] = 0.015 M but [Cl⁻] = 0.030 M because each formula unit releases two chloride ions. Plugging solubility in where an ion concentration belongs (or forgetting to square the 2s term in Ksp = (s)(2s)²) is a guaranteed point loss.
Ion concentration is the molarity of a specific ion in solution, written in square brackets like [Cl⁻] = 0.030 M.
Ion concentrations follow the dissolution stoichiometry, so a salt like PbCl₂ produces twice as much Cl⁻ as Pb²⁺.
The common ion effect (Topic 7.12) means a salt dissolves less in a solution that already contains one of its ions, which you can predict with Le Châtelier or calculate from Ksp.
In the Henderson-Hasselbalch equation (Topic 8.9), buffer pH depends on pKa plus the log of the concentration ratio [A⁻]/[HA].
Buffers resist pH change because adding small amounts of acid or base doesn't significantly change the [A⁻]/[HA] ratio.
pH itself is an ion concentration in disguise, since pH = -log[H⁺].
It's the molarity of a specific ion in solution, written in brackets like [H⁺] or [Pb²⁺]. It's the quantity you plug into Ksp expressions, the Henderson-Hasselbalch equation, and pH calculations in Units 7 and 8.
No. Molar solubility counts moles of the entire salt that dissolve per liter, while ion concentration counts one ion. For PbCl₂ with solubility 0.015 M, [Pb²⁺] = 0.015 M but [Cl⁻] = 0.030 M because of the 1:2 stoichiometry.
No. Ksp is a constant at a given temperature. Adding a common ion lowers the salt's solubility, but the equilibrium ion concentrations adjust so their product still equals Ksp. That's exactly what EK 7.12.A.1 means by the common-ion effect.
Take the negative log of the hydrogen ion concentration: pH = -log[H⁺]. So a solution with [H⁺] = 1 × 10⁻⁵ M has a pH of 5. For buffers, use Henderson-Hasselbalch with the [A⁻]/[HA] ratio instead.
It's the log of the concentration ratio of conjugate base to weak acid, log([A⁻]/[HA]). When the two concentrations are equal, the log term is 0 and pH equals pKa, which is what happens at the half-equivalence point of a titration.