[{"data":1,"prerenderedAt":204},["ShallowReactive",2],{"topic-solubility-and-complex-ion-equilibria":3},{"topic":4,"prev":189,"next":196},{"id":5,"slug":6,"title":7,"shortTitle":8,"lesson":5,"category":9,"description":10,"metaDescription":11,"objectives":12,"conceptSections":34,"workedExample":101,"oerSources":110,"selfStudyQuestions":112,"relatedTopicSlugs":132,"relatedElements":135,"relatedReferences":140,"keyEquations":145,"howTo":158,"conceptProbe":166,"practiceTeaser":170,"gatedContent":178,"seoKeywords":181,"lastAlignmentAudit":187,"objectiveCount":188},24,"solubility-and-complex-ion-equilibria","Solubility and Complex-Ion Equilibria","Solubility Equilibria","equilibrium-acids","Study solubility product constants (Ksp), molar solubility, common ion effect, and selective precipitation.","Learn solubility equilibria: Ksp calculations, molar solubility, common ion effect, and selective precipitation. Step-by-step worked examples and practice.",[13,16,19,22,25,28,31],{"id":14,"text":15},"24.1","Write Ksp expressions for sparingly soluble salts",{"id":17,"text":18},"24.2","Calculate molar solubility from Ksp",{"id":20,"text":21},"24.3","Calculate Ksp from experimental solubility data",{"id":23,"text":24},"24.4","Predict whether precipitation will occur by comparing Q to Ksp",{"id":26,"text":27},"24.5","Apply the common-ion effect to calculate how solubility changes in the presence of a common ion",{"id":29,"text":30},"24.6","Apply selective precipitation to predict precipitation order and calculate residual ion concentration (fractional precipitation)",{"id":32,"text":33},"24.7","Write Kf expressions and calculate solubility from combined Ksp·Kf equilibria",[35,43,53,58,66,75,84,93,97],{"heading":36,"content":37,"relatedObjectives":38,"deepDive":39},"Writing Ksp Expressions","\u003Cp>The \u003Cstrong>solubility product constant (K\u003Csub>sp\u003C/sub>)\u003C/strong> describes the equilibrium between an ionic solid and its dissolved ions. For the general dissolution A\u003Csub>a\u003C/sub>B\u003Csub>b\u003C/sub>(s) → aA\u003Csup>n+\u003C/sup>(aq) + bB\u003Csup>m−\u003C/sup>(aq):\u003C/p>\u003Cp>\u003Cstrong>K\u003Csub>sp\u003C/sub> = [A\u003Csup>n+\u003C/sup>]\u003Csup>a\u003C/sup>[B\u003Csup>m−\u003C/sup>]\u003Csup>b\u003C/sup>\u003C/strong>\u003C/p>\u003Cp>The solid does not appear in the expression because its activity is defined as 1. Each ion concentration is raised to the power of its coefficient in the balanced equation.\u003C/p>\u003Cp>Examples: AgCl → Ag\u003Csup>+\u003C/sup> + Cl\u003Csup>−\u003C/sup> gives K\u003Csub>sp\u003C/sub> = [Ag\u003Csup>+\u003C/sup>][Cl\u003Csup>−\u003C/sup>]. Ca\u003Csub>3\u003C/sub>(PO\u003Csub>4\u003C/sub>)\u003Csub>2\u003C/sub> → 3 Ca\u003Csup>2+\u003C/sup> + 2 PO\u003Csub>4\u003C/sub>\u003Csup>3−\u003C/sup> gives K\u003Csub>sp\u003C/sub> = [Ca\u003Csup>2+\u003C/sup>]\u003Csup>3\u003C/sup>[PO\u003Csub>4\u003C/sub>\u003Csup>3−\u003C/sup>]\u003Csup>2\u003C/sup>. The exponents dramatically affect the relationship between K\u003Csub>sp\u003C/sub> and solubility, so you cannot compare the solubility of salts with different formulas by K\u003Csub>sp\u003C/sub> alone.\u003C/p>",[],[40],{"label":41,"body":42},"Why this matters: patients drink barium because of Ksp","\u003Cp>Barium ions are toxic, yet hospitals hand patients a barium “milkshake” before an X-ray of the digestive tract. The safety margin is a K\u003Csub>sp\u003C/sub>: barium sulfate’s is about 1.1 × 10\u003Csup>−10\u003C/sup>, so the suspension releases only a vanishing trace of free Ba\u003Csup>2+\u003C/sup>, far below harm, while the insoluble solid coats the gut and shows up brilliantly on the image. The same class of number decides which lead compounds can pigment paint, which fluoride salts can go in toothpaste, and which forms of a drug survive the stomach. A K\u003Csub>sp\u003C/sub> is not an abstraction; it is often the line between medicine and poison.\u003C/p>",{"heading":44,"content":45,"relatedObjectives":46,"deepDive":49},"Calculating Molar Solubility from Ksp","\u003Cp>\u003Cstrong>Molar solubility (s)\u003C/strong> is the number of moles of a sparingly soluble salt that dissolve per liter to form a saturated solution. To calculate it from K\u003Csub>sp\u003C/sub>:\u003C/p>\u003Col>\u003Cli>Write the dissolution equation and the K\u003Csub>sp\u003C/sub> expression.\u003C/li>\u003Cli>Let s = molar solubility. Express each ion concentration in terms of s using the stoichiometric coefficients (e.g., for PbI\u003Csub>2\u003C/sub>: [Pb\u003Csup>2+\u003C/sup>] = s, [I\u003Csup>−\u003C/sup>] = 2s).\u003C/li>\u003Cli>Substitute into K\u003Csub>sp\u003C/sub> and solve for s.\u003C/li>\u003C/ol>\u003Cp>For a 1:1 salt like AgCl: K\u003Csub>sp\u003C/sub> = s\u003Csup>2\u003C/sup>, so s = √K\u003Csub>sp\u003C/sub>. For a 1:2 salt like PbI\u003Csub>2\u003C/sub>: K\u003Csub>sp\u003C/sub> = (s)(2s)\u003Csup>2\u003C/sup> = 4s\u003Csup>3\u003C/sup>, so s = (K\u003Csub>sp\u003C/sub>/4)\u003Csup>1/3\u003C/sup>. The algebra becomes more complex for salts with higher coefficients, but the approach is the same.\u003C/p>\u003Cp>Molar solubility can be converted to grams per liter by multiplying by the molar mass, which is often more practical for laboratory work.\u003C/p>",[47,48],180,182,[50],{"label":51,"body":52},"Common mistake: ranking solubility by comparing Ksp directly","\u003Cp>Smaller K\u003Csub>sp\u003C/sub> means less soluble, right? Only within the same stoichiometry. Compare AgCl (K\u003Csub>sp\u003C/sub> = 1.8 × 10\u003Csup>−10\u003C/sup>) with Ag\u003Csub>2\u003C/sub>CrO\u003Csub>4\u003C/sub> (K\u003Csub>sp\u003C/sub> = 1.1 × 10\u003Csup>−12\u003C/sup>): the chromate’s constant is a hundred times smaller, yet its molar solubility is \u003Cem>higher\u003C/em>: s = 1.3 × 10\u003Csup>−5\u003C/sup> M for AgCl (from s\u003Csup>2\u003C/sup> = K\u003Csub>sp\u003C/sub>) versus 6.5 × 10\u003Csup>−5\u003C/sup> M for Ag\u003Csub>2\u003C/sub>CrO\u003Csub>4\u003C/sub> (from 4s\u003Csup>3\u003C/sup> = K\u003Csub>sp\u003C/sub>). The exponents change the game: a 1:2 salt’s K\u003Csub>sp\u003C/sub> is a cubic in s, not a square.\u003C/p>\u003Cp>Rule: compare K\u003Csub>sp\u003C/sub> values directly only between salts of matching formulas; otherwise, convert both to molar solubility first.\u003C/p>",{"heading":54,"content":55,"relatedObjectives":56},"Determining Ksp from Solubility Data","\u003Cp>The reverse calculation — finding K\u003Csub>sp\u003C/sub> from experimentally measured solubility — follows the same logic in reverse:\u003C/p>\u003Col>\u003Cli>Convert the measured solubility to molar solubility (mol/L) if given in g/L.\u003C/li>\u003Cli>Use stoichiometric ratios to find the equilibrium concentration of each ion.\u003C/li>\u003Cli>Substitute the ion concentrations into the K\u003Csub>sp\u003C/sub> expression and evaluate.\u003C/li>\u003C/ol>\u003Cp>For example, if the solubility of CaF\u003Csub>2\u003C/sub> is measured as 0.017 g/L, convert to molar solubility: s = 0.017 / 78.08 = 2.2 × 10\u003Csup>−4\u003C/sup> M. The ions are [Ca\u003Csup>2+\u003C/sup>] = s and [F\u003Csup>−\u003C/sup>] = 2s, so K\u003Csub>sp\u003C/sub> = (s)(2s)\u003Csup>2\u003C/sup> = 4s\u003Csup>3\u003C/sup> = 4(2.2 × 10\u003Csup>−4\u003C/sup>)\u003Csup>3\u003C/sup> = 4.3 × 10\u003Csup>−11\u003C/sup>.\u003C/p>\u003Cp>This approach assumes the only source of ions is the dissolving solid. If other electrolytes are present, their contributions must be accounted for separately.\u003C/p>",[57],181,{"heading":59,"content":60,"relatedObjectives":61,"deepDive":62},"Predicting Precipitation with Q vs. Ksp","\u003Cp>To predict whether a precipitate forms when two solutions are mixed, calculate the \u003Cstrong>ion product Q\u003C/strong> — the same mathematical form as K\u003Csub>sp\u003C/sub>, but using the \u003Cem>initial\u003C/em> ion concentrations after mixing (before any reaction):\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Q > K\u003Csub>sp\u003C/sub>\u003C/strong> — the solution is supersaturated and a precipitate \u003Cem>will\u003C/em> form, driving ion concentrations down until Q = K\u003Csub>sp\u003C/sub>.\u003C/li>\u003Cli>\u003Cstrong>Q = K\u003Csub>sp\u003C/sub>\u003C/strong> — the solution is exactly saturated; no precipitate forms, but the system is at equilibrium.\u003C/li>\u003Cli>\u003Cstrong>Q < K\u003Csub>sp\u003C/sub>\u003C/strong> — the solution is unsaturated; no precipitate forms and more solute could still dissolve.\u003C/li>\u003C/ul>\u003Cp>A critical step is accounting for \u003Cstrong>dilution\u003C/strong>: when equal volumes are mixed, each ion’s concentration is halved. Always compute the post-mixing concentrations before evaluating Q.\u003C/p>",[],[63],{"label":64,"body":65},"Why this matters: caves are Q-versus-Ksp in slow motion","\u003Cp>Every stalactite is a supersaturation event running for millennia. Groundwater seeping through limestone carries dissolved Ca\u003Csup>2+\u003C/sup> and bicarbonate; when a droplet reaches the cave ceiling, its dissolved CO\u003Csub>2\u003C/sub> escapes into the cave air, the carbonate concentration rises, and Q for calcium carbonate climbs past K\u003Csub>sp\u003C/sub>: a microscopic ring of calcite precipitates. Drop after drop, the ring becomes an icicle of stone.\u003C/p>\u003Cp>Your kettle performs the same computation faster: heating drives off CO\u003Csub>2\u003C/sub> and concentrates the ions, Q overtakes K\u003Csub>sp\u003C/sub>, and limescale precipitates onto the heating element. Same inequality, different timescales.\u003C/p>",{"heading":67,"content":68,"relatedObjectives":69,"deepDive":71},"The Common Ion Effect on Solubility","\u003Cp>The \u003Cstrong>common ion effect\u003C/strong> reduces the solubility of a sparingly soluble salt when one of its ions is already present in solution from another source. This is a direct application of Le Châtelier’s principle: the added common ion shifts the dissolution equilibrium to the left, favoring the solid.\u003C/p>\u003Cp>For example, AgCl has K\u003Csub>sp\u003C/sub> = 1.8 × 10\u003Csup>−10\u003C/sup>. In pure water, s = √(1.8 × 10\u003Csup>−10\u003C/sup>) = 1.3 × 10\u003Csup>−5\u003C/sup> M. In 0.10 M NaCl, the Cl\u003Csup>−\u003C/sup> is already 0.10 M, so K\u003Csub>sp\u003C/sub> = [Ag\u003Csup>+\u003C/sup>](0.10), giving [Ag\u003Csup>+\u003C/sup>] = 1.8 × 10\u003Csup>−9\u003C/sup> M — about 7,500 times less soluble than in pure water.\u003C/p>\u003Cp>This effect is exploited in analytical chemistry: adding excess precipitating reagent ensures nearly complete removal of the target ion from solution.\u003C/p>",[70],185,[72],{"label":73,"body":74},"Purifying salt with more of its own ion","\u003Cp>A classic preparative trick makes the common ion effect tangible: bubble HCl gas through a saturated NaCl solution and solid salt crystallizes out. Nothing was removed and no temperature changed; the added chloride simply raised [Cl\u003Csup>−\u003C/sup>], pushing the ion product past K\u003Csub>sp\u003C/sub> and forcing the dissolution equilibrium backwards. The crystals that form are purer than the starting salt, because the impurities are still far from their own saturation limits and stay dissolved.\u003C/p>\u003Cp>The same lever works everywhere in this topic: any dissolved salt can be squeezed out of solution by flooding the system with either of its own ions.\u003C/p>",{"heading":76,"content":77,"relatedObjectives":78,"deepDive":80},"Selective Precipitation","\u003Cp>\u003Cstrong>Selective precipitation\u003C/strong> separates ions from a mixture by exploiting differences in their precipitation thresholds. As a precipitating agent is added gradually, the first salt to reach Q = K\u003Csub>sp\u003C/sub> precipitates first; that order depends on K\u003Csub>sp\u003C/sub>, salt stoichiometry, and the dissolved-ion concentrations.\u003C/p>\u003Cp>The strategy: for each ion that could form a precipitate, calculate the concentration of precipitating agent needed to just begin precipitation (set Q = K\u003Csub>sp\u003C/sub> and solve). The ion requiring the \u003Cem>lowest\u003C/em> precipitant concentration precipitates first. Comparing K\u003Csub>sp\u003C/sub> values alone is safe only for salts with the same stoichiometric form when the relevant ion concentrations are equal.\u003C/p>\u003Cp>For example, a solution containing comparable concentrations of Ag\u003Csup>+\u003C/sup> and Cu\u003Csup>2+\u003C/sup> can be separated by slowly adding Cl\u003Csup>−\u003C/sup>. AgCl (K\u003Csub>sp\u003C/sub> = 1.8 × 10\u003Csup>−10\u003C/sup>) reaches its precipitation threshold while CuCl\u003Csub>2\u003C/sub> remains soluble, so Ag\u003Csup>+\u003C/sup> can be removed while Cu\u003Csup>2+\u003C/sup> remains in solution.\u003C/p>\u003Cp>Selective precipitation is widely used in qualitative analysis schemes to identify unknown cations by systematically precipitating groups of ions with specific reagents.\u003C/p>",[79],184,[81],{"label":82,"body":83},"Why this matters: separating metals one Ksp at a time","\u003Cp>Selective precipitation is a workhorse of both classical analysis and modern cleanup. The traditional qualitative-analysis scheme identifies the metals in an unknown mixture by adding precipitating agents in a strict sequence and collecting each group as its precipitation threshold is crossed. Wastewater plants run the industrial version: dosing sulfide or hydroxide in controlled stages drops toxic metals like cadmium and lead out of solution as filterable solids while leaving other ions dissolved.\u003C/p>\u003Cp>The arithmetic in this section, comparing which Q = K\u003Csub>sp\u003C/sub> threshold is crossed first as the reagent concentration rises, is exactly the design calculation behind both.\u003C/p>",{"heading":85,"content":86,"relatedObjectives":87,"deepDive":89},"Factors Affecting Solubility","\u003Cp>Beyond the common ion effect, several other factors influence the solubility of ionic compounds in water:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>pH effects:\u003C/strong> salts containing the anion of a weak acid (e.g., CaCO\u003Csub>3\u003C/sub>, Mg(OH)\u003Csub>2\u003C/sub>) are more soluble in acidic solution. The added H\u003Csup>+\u003C/sup> reacts with the anion (CO\u003Csub>3\u003C/sub>\u003Csup>2−\u003C/sup> + H\u003Csup>+\u003C/sup> → HCO\u003Csub>3\u003C/sub>\u003Csup>−\u003C/sup>), removing it from the equilibrium and shifting dissolution to the right.\u003C/li>\u003Cli>\u003Cstrong>Complex ion formation:\u003C/strong> metal ions that form stable complexes with ligands (e.g., Ag\u003Csup>+\u003C/sup> with NH\u003Csub>3\u003C/sub>) become more soluble because the free metal-ion concentration decreases, driving more solid to dissolve.\u003C/li>\u003Cli>\u003Cstrong>Temperature:\u003C/strong> for most ionic solids, solubility increases with temperature, though a few salts (e.g., Ce\u003Csub>2\u003C/sub>(SO\u003Csub>4\u003C/sub>)\u003Csub>3\u003C/sub>) show the opposite trend.\u003C/li>\u003C/ul>\u003Cp>When solving solubility problems, always check whether pH or complexation effects apply before using the simple K\u003Csub>sp\u003C/sub> expression.\u003C/p>",[88],183,[90],{"label":91,"body":92},"Why this matters: cavities are a pH-solubility problem","\u003Cp>Tooth enamel is hydroxyapatite, a calcium phosphate salt whose anions belong to weak acids, which makes it exactly the kind of solid this section says dissolves in acid. Mouth bacteria ferment sugar into acids, the local pH drops, and the equilibrium tips toward dissolution: that is a cavity forming, ion by ion.\u003C/p>\u003Cp>Fluoride toothpaste intervenes thermodynamically: fluoride exchanges into the mineral to form fluorapatite, which resists acid attack to a meaningfully lower pH than the original enamel. Dentistry’s cheapest, most effective intervention is a solubility equilibrium nudged in the tooth’s favor.\u003C/p>",{"heading":94,"relatedObjectives":95,"content":96},"Complex Ions and the Formation Constant (Kf)",[],"\u003Cp>A \u003Cstrong>complex ion\u003C/strong> is a central metal cation bonded to one or more molecules or ions called \u003Cstrong>ligands\u003C/strong> (e.g., NH\u003Csub>3\u003C/sub>, CN\u003Csup>−\u003C/sup>, S\u003Csub>2\u003C/sub>O\u003Csub>3\u003C/sub>\u003Csup>2−\u003C/sup>). Complex-ion formation is itself an equilibrium, described by the \u003Cstrong>formation constant K\u003Csub>f\u003C/sub>\u003C/strong>. For the general formation M\u003Csup>n+\u003C/sup>(aq) + j L(aq) ⇌ ML\u003Csub>j\u003C/sub>\u003Csup>z+\u003C/sup>(aq):\u003C/p>\u003Cp>K\u003Csub>f\u003C/sub> = [ML\u003Csub>j\u003C/sub>\u003Csup>z+\u003C/sup>] / ([M\u003Csup>n+\u003C/sup>][L]\u003Csup>j\u003C/sup>)\u003C/p>\u003Cp>For example, for Ag\u003Csup>+\u003C/sup> + 2 NH\u003Csub>3\u003C/sub> ⇌ Ag(NH\u003Csub>3\u003C/sub>)\u003Csub>2\u003C/sub>\u003Csup>+\u003C/sup>: K\u003Csub>f\u003C/sub> = [Ag(NH\u003Csub>3\u003C/sub>)\u003Csub>2\u003C/sub>\u003Csup>+\u003C/sup>] / ([Ag\u003Csup>+\u003C/sup>][NH\u003Csub>3\u003C/sub>]\u003Csup>2\u003C/sup>). Typical K\u003Csub>f\u003C/sub> values are very large (10\u003Csup>7\u003C/sup> to 10\u003Csup>25\u003C/sup> or more), meaning complex formation is strongly product-favored. The reverse (dissociation) reaction has the reciprocal constant: \u003Cstrong>K\u003Csub>d\u003C/sub> = 1/K\u003Csub>f\u003C/sub>\u003C/strong>.\u003C/p>\u003Cp>\u003Cstrong>Free-metal concentration in excess ligand:\u003C/strong> when K\u003Csub>f\u003C/sub> is large and the ligand is in excess, treat the complexation as complete first (stoichiometry step: each mole of metal consumes j moles of ligand), then let the complex dissociate back by a tiny x. The equilibrium free-metal concentration follows from [M\u003Csup>n+\u003C/sup>] = [ML\u003Csub>j\u003C/sub>] / (K\u003Csub>f\u003C/sub>[L]\u003Csup>j\u003C/sup>) — typically vanishingly small (10\u003Csup>−15\u003C/sup> M or less).\u003C/p>\u003Cp>\u003Cstrong>Complex-ion formation increases solubility.\u003C/strong> When a sparingly soluble salt MX dissolves in the presence of a complexing ligand, the two equilibria couple, and their constants multiply:\u003C/p>\u003Cul>\u003Cli>MX(s) ⇌ M\u003Csup>+\u003C/sup> + X\u003Csup>−\u003C/sup>  (K\u003Csub>sp\u003C/sub>)\u003C/li>\u003Cli>M\u003Csup>+\u003C/sup> + j L ⇌ ML\u003Csub>j\u003C/sub>  (K\u003Csub>f\u003C/sub>)\u003C/li>\u003Cli>\u003Cstrong>Net: MX(s) + j L ⇌ ML\u003Csub>j\u003C/sub> + X\u003Csup>−\u003C/sup>  (K\u003Csub>net\u003C/sub> = K\u003Csub>sp\u003C/sub> × K\u003Csub>f\u003C/sub>)\u003C/strong>\u003C/li>\u003C/ul>\u003Cp>Even when K\u003Csub>sp\u003C/sub> is tiny, a huge K\u003Csub>f\u003C/sub> can make K\u003Csub>net\u003C/sub> large enough for substantial dissolution. The classic application is photographic fixing: AgBr (K\u003Csub>sp\u003C/sub> = 5.0 × 10\u003Csup>−13\u003C/sup>) dissolves in sodium thiosulfate solution because Ag(S\u003Csub>2\u003C/sub>O\u003Csub>3\u003C/sub>)\u003Csub>2\u003C/sub>\u003Csup>3−\u003C/sup> has K\u003Csub>f\u003C/sub> = 4.7 × 10\u003Csup>13\u003C/sup>, giving K\u003Csub>net\u003C/sub> ≈ 24. Similarly, AgCl dissolves in ammonia via Ag(NH\u003Csub>3\u003C/sub>)\u003Csub>2\u003C/sub>\u003Csup>+\u003C/sup>, and Al(OH)\u003Csub>3\u003C/sub> dissolves in strong base via Al(OH)\u003Csub>4\u003C/sub>\u003Csup>−\u003C/sup>.\u003C/p>\u003Cp>For a 1:1 salt dissolving via the net reaction, if s mol/L dissolves then [ML\u003Csub>j\u003C/sub>] = [X\u003Csup>−\u003C/sup>] = s and K\u003Csub>net\u003C/sub> = s\u003Csup>2\u003C/sup>/[L]\u003Csup>j\u003C/sup>, so \u003Cstrong>s = √(K\u003Csub>net\u003C/sub> × [L]\u003Csup>j\u003C/sup>)\u003C/strong> when the ligand stays approximately fixed; if dissolution consumes a significant fraction of the ligand, account for the 1:j stoichiometry in an ICE table (each mole dissolved consumes j moles of ligand) and solve the resulting quadratic.\u003C/p>\u003Cp>Watch the charges when writing K\u003Csub>f\u003C/sub> expressions: the complex's net charge is the metal charge plus the total ligand charge (e.g., Cu\u003Csup>2+\u003C/sup> + 4 CN\u003Csup>−\u003C/sup> gives Cu(CN)\u003Csub>4\u003C/sub>\u003Csup>2−\u003C/sup>, net −2).\u003C/p>",{"heading":98,"content":99,"relatedObjectives":100},"Solubility Equilibria Decision Framework and Common Mistakes","\u003Cp>Follow this workflow for solubility equilibrium problems:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Write the dissolution equation and the K\u003Csub>sp\u003C/sub> expression\u003C/strong>. Each ion concentration is raised to its stoichiometric coefficient.\u003C/li>\u003Cli>\u003Cstrong>For molar solubility from K\u003Csub>sp\u003C/sub>\u003C/strong>: let s = molar solubility. Express each ion concentration in terms of s using the stoichiometry (e.g., for Ca\u003Csub>3\u003C/sub>(PO\u003Csub>4\u003C/sub>)\u003Csub>2\u003C/sub>: [Ca\u003Csup>2+\u003C/sup>] = 3s, [PO\u003Csub>4\u003C/sub>\u003Csup>3−\u003C/sup>] = 2s).\u003C/li>\u003Cli>\u003Cstrong>For will-it-precipitate questions\u003C/strong>: calculate Q\u003Csub>sp\u003C/sub> from the actual ion concentrations after mixing. If Q\u003Csub>sp\u003C/sub> > K\u003Csub>sp\u003C/sub>, a precipitate forms.\u003C/li>\u003Cli>\u003Cstrong>For common-ion problems\u003C/strong>: the initial concentration of the common ion is not zero—include it in the ICE table.\u003C/li>\u003Cli>\u003Cstrong>For selective precipitation\u003C/strong>: compute the precipitant concentration at which each salt reaches Q = K\u003Csub>sp\u003C/sub>; the salt with the lowest threshold precipitates first. (Comparing K\u003Csub>sp\u003C/sub> values directly works only for salts of the same formula type whose target ions are present at comparable concentrations; each threshold depends on both K\u003Csub>sp\u003C/sub> and the ion’s existing concentration.)\u003C/li>\u003C/ol>\u003Cp>Common mistakes: forgetting to raise ion concentrations to their stoichiometric powers in the K\u003Csub>sp\u003C/sub> expression (e.g., writing [Ag\u003Csup>+\u003C/sup>] instead of [Ag\u003Csup>+\u003C/sup>]\u003Csup>2\u003C/sup> for Ag\u003Csub>2\u003C/sub>CrO\u003Csub>4\u003C/sub>), ignoring dilution when two solutions are mixed (volumes add, concentrations drop), assuming molar solubility equals K\u003Csub>sp\u003C/sub> directly without setting up the algebra, and neglecting the common-ion effect when a shared ion is already present in solution.\u003C/p>",[47,57,48,88,79,70],{"title":102,"problem":103,"steps":104,"answer":109},"Calculating Molar Solubility from Kₛₚ","The Kₛₚ of PbI₂ is 9.8 × 10⁻⁹. Calculate its molar solubility in pure water.",[105,106,107,108],"Write the dissolution equation: PbI₂(s) ⇌ Pb²⁺(aq) + 2 I⁻(aq).","Let s = molar solubility. Then [Pb²⁺] = s and [I⁻] = 2s.","Substitute into Kₛₚ: Kₛₚ = [Pb²⁺][I⁻]² = (s)(2s)² = 4s³.","Solve: 4s³ = 9.8 × 10⁻⁹, so s³ = 2.45 × 10⁻⁹, s = 1.35 × 10⁻³ M.","The molar solubility of PbI₂ in pure water is 1.35 × 10⁻³ M (about 0.62 g/L).",[111],"OpenStax Chemistry 2e, Ch 15.1: Precipitation and Dissolution (CC BY 4.0)",[113,115,117,120,122,124,126,128,130],{"question":114},"What is the solubility product constant (Ksp)?",{"question":116},"How do you write a Ksp expression for a slightly soluble ionic compound?",{"question":118,"hint":119},"What is molar solubility and how is it related to Ksp?","Set up an ICE table starting from the dissolution equilibrium.",{"question":121},"How do you calculate Ksp from experimental solubility data?",{"question":123},"How do you use Q versus Ksp to predict whether a precipitate will form?",{"question":125},"What is the common-ion effect and how does it influence solubility?",{"question":127},"What is selective precipitation?",{"question":129},"Why does adding a common ion decrease the solubility of a slightly soluble salt?",{"question":131},"What assumptions are typically made when calculating molar solubility from Ksp?",[133,134],"equilibrium","chemical-reactions-in-solution",[136,137,138,139],"Ag","Ba","Pb","Ca",[141,142,143,144],"solubility-rules","solubility-product-constants","weak-acid-base-constants","formation-constants",[146,150,154],{"label":147,"equation":148,"note":149},"Solubility Product (General Form)","Ksp = [Aⁿ⁺]ᵃ[Bᵐ⁻]ᵇ","For the dissolution AₐBᵦ(s) → aAⁿ⁺(aq) + bBᵐ⁻(aq)",{"label":151,"equation":152,"note":153},"Molar Solubility of a 1:1 Salt","s = √Ksp","For salts like AgCl where Ksp = s²",{"label":155,"equation":156,"note":157},"Molar Solubility of a 1:2 Salt","s = ³√(Ksp / 4)","For salts like PbI₂ where Ksp = 4s³",{"title":159,"steps":160},"How to Predict Whether a Precipitate Will Form",[161,162,163,164,165],"Identify the potentially insoluble compound that could form when the two solutions are mixed.","Write the dissolution equation and the Ksp expression for that compound.","Calculate the ion concentrations immediately after mixing, accounting for dilution (total volume increases when solutions are combined).","Calculate the ion product Q using the same mathematical form as the Ksp expression but with post-mixing concentrations.","Compare Q to Ksp: if Q > Ksp, a precipitate forms; if Q \u003C Ksp, no precipitate forms; if Q = Ksp, the solution is exactly saturated.",{"question":167,"answer":168,"type":169},"Silver chloride (AgCl) has a Ksp of 1.8 x 10^-10. A student dissolves AgCl in pure water and then in 0.10 M NaCl. In which solution is the molar solubility of AgCl greater, and why does the common ion reduce solubility instead of increasing the total amount of dissolved ions?","The molar solubility is far greater in pure water. In 0.10 M NaCl, the Cl- already present shifts the dissolution equilibrium to the left (Le Chatelier's principle). The Ksp value is fixed at a given temperature, so if [Cl-] is already 0.10 M, then [Ag+] can only reach 1.8 x 10^-9 M. In pure water, both ions start at zero and reach 1.3 x 10^-5 M. The common ion does not change Ksp; it simply forces the other ion to a much lower equilibrium concentration.","conceptual",[171,174],{"id":172,"problem":173,"type":169},"pt-24-1","Two salts, AB and AB2, both have Ksp = 1.0 x 10^-12. Which one has the greater molar solubility? Explain your reasoning without calculating.",{"id":175,"problem":176,"type":177},"pt-24-2","The Ksp of BaSO4 is 1.1 x 10^-10. If 50.0 mL of 0.0020 M BaCl2 is mixed with 50.0 mL of 0.0040 M Na2SO4, will a precipitate form? Show your calculation of Q.","calculation",{"workedExampleCount":179,"hasWorksheets":180},10,true,[182,183,184,185,186],"Ksp","molar solubility","common ion effect","selective precipitation","solubility product","2026-07-17",7,{"id":190,"slug":191,"lesson":190,"title":192,"shortTitle":193,"description":194,"category":9,"objectiveCount":188,"problemCount":195},23,"buffers-and-titration-curves","Buffers and Titration Curves","Buffers","Understand buffer solutions: Henderson-Hasselbalch equation, buffer capacity, and buffer preparation.",46,{"id":197,"slug":198,"lesson":197,"title":199,"shortTitle":200,"description":201,"category":202,"objectiveCount":188,"problemCount":203},25,"entropy-and-free-energy","Entropy and Free Energy","Thermodynamics","Explore spontaneity, entropy, Gibbs free energy, and the second and third laws of thermodynamics.","thermodynamics-kinetics",51,1785108608189]