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Química Geral · Acid–base and solubility equilibria

Precipitação seletiva e íons complexos

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O que você vai aprender

  • Calculate the minimum ion concentration or pH at which a salt or hydroxide begins to precipitate, and the conditions for complete precipitation.
  • Predict which ion precipitates first from a mixture and how completely it is removed before the next solid starts to form.
  • Write formation-constant expressions and calculate free metal-ion concentrations in solutions with excess ligand.
  • Combine Ksp and Kf to predict and calculate how far a complexing agent dissolves a precipitate.
  • Explain the behaviour of amphoteric hydroxides and use precipitation and complexation to interpret qualitative-analysis tests.
  • Draw a ligand donating a lone pair to a metal cation as a Lewis acid–base reaction.

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When does a precipitate start to form?

A solid starts to precipitate at the moment the ion product Q, written exactly like Ksp, reaches Ksp. Below Ksp the solution is unsaturated; above it the excess ions come out as solid. To find the minimum concentration of an added ion, set Q = Ksp and solve with the other ion's concentration fixed. With [Cl⁻] = 0.050 M, AgCl (Ksp = 1.8 × 10⁻¹⁰) starts to form when [Ag⁺] = 1.8 × 10⁻¹⁰ ÷ 0.050 = 3.6 × 10⁻⁹ M. Exponents follow the formula. For Mg(OH)₂ (Ksp = 5.6 × 10⁻¹²) with [Mg²⁺] = 0.10 M, [OH⁻]² = 5.6 × 10⁻¹¹, so [OH⁻] = 7.5 × 10⁻⁶ M, pOH = 5.13 and pH = 14.00 − 5.13 = 8.87; for an M(OH)₃ hydroxide take a cube root instead. The concentration of an ion already in solution enters Ksp as it is, with no factor of 2 or 3; those factors belong only to a salt dissolving in pure water. When two solutions are mixed, first dilute every ion to the total volume (10.0 mL made up to 100.0 mL is a tenfold dilution), then compare Q with Ksp. Precipitation is usually called complete when the ion left in solution has fallen to about 1.0 × 10⁻⁵ M; putting that small concentration into the same Ksp equation gives the reagent concentration or pH needed.

Which ion precipitates first?

When a reagent is added slowly to a mixture, the solid that needs the smallest concentration of reagent forms first. Calculate the onset concentration for each candidate solid. Compare Ksp values directly only when the salts have the same formula type and the ions start at the same concentration. For a solution 0.010 M in both I⁻ and Cl⁻, AgI (Ksp = 8.3 × 10⁻¹⁷) starts at [Ag⁺] = 8.3 × 10⁻¹⁵ M and AgCl (Ksp = 1.8 × 10⁻¹⁰) at 1.8 × 10⁻⁸ M, so AgI forms first. When AgCl just begins, the iodide left is 8.3 × 10⁻¹⁷ ÷ 1.8 × 10⁻⁸ = 4.6 × 10⁻⁹ M, a tiny fraction of 0.010 M, so the separation is clean. When formula types differ, a smaller Ksp does not mean an earlier precipitate: Ag₂CrO₄ (Ksp = 1.1 × 10⁻¹²) with 0.010 M CrO₄²⁻ needs [Ag⁺] = √(1.1 × 10⁻¹² ÷ 0.010) = 1.0 × 10⁻⁵ M, far more silver than AgCl needs. The Mohr titration of chloride uses exactly this: red Ag₂CrO₄ appears only after nearly all the chloride has precipitated. Metal ions can be separated by pH in the same way, because each hydroxide starts and completes its precipitation at its own pH: Fe(OH)₃ comes down in acidic solution, Mg(OH)₂ only above about pH 9.

Complex ions and formation constants

A metal cation is a Lewis acid. Ligands such as NH₃, OH⁻, CN⁻, S₂O₃²⁻ and SCN⁻ each donate a lone pair to it, forming a complex ion whose charge is the metal charge plus the ligand charges. The formation constant is written like any equilibrium constant: for Cu²⁺ + 4 NH₃ ⇌ [Cu(NH₃)₄]²⁺, Kf = [Cu(NH₃)₄²⁺]/([Cu²⁺][NH₃]⁴). Ligands add one at a time, and the overall constant is the product of the stepwise constants (for silver, Kf = K₁ × K₂), so log Kf is the sum of the stepwise log K values. Formation constants are large ([Ag(NH₃)₂]⁺ 1.7 × 10⁷, [Cu(NH₃)₄]²⁺ 5.0 × 10¹²), so with excess ligand nearly all the metal is complexed. Worked example: 0.010 M Cu²⁺ in 1.0 M NH₃. Complexation uses 4 × 0.010 = 0.040 M NH₃, leaving 0.96 M, so [Cu²⁺] = 0.010 ÷ (5.0 × 10¹² × 0.96⁴) = 2.4 × 10⁻¹⁵ M. Subtract the bound ligand and raise [L] to the power n. Because n differs between complexes, a larger Kf does not always mean less free metal: at a fixed free ligand concentration the ratio [complex]/[free metal ion] equals Kf × [L]ⁿ.

Dissolving a precipitate with a ligand

A ligand lowers the free metal-ion concentration, so Q falls below Ksp and more solid dissolves. Adding the two equilibria gives one equation whose constant is the product of theirs: AgCl(s) + 2 NH₃ ⇌ [Ag(NH₃)₂]⁺ + Cl⁻, K = Ksp × Kf = 1.8 × 10⁻¹⁰ × 1.7 × 10⁷ = 3.1 × 10⁻³. If s mol/L dissolves in ligand of starting concentration C, then K = s²/(C − 2s)², so s/(C − 2s) = √K. The same steps run backwards: a measured solubility gives K, and K ÷ Ksp gives Kf. AgI has a far smaller Ksp (8.3 × 10⁻¹⁷), so K = 1.4 × 10⁻⁹ and even concentrated ammonia leaves it almost untouched, while thiosulfate (Kf = 2.9 × 10¹³) gives K greater than 1 with AgBr, which is how photographic fixer works. To stop a salt precipitating, keep the free metal ion below Ksp ÷ [anion] and use Kf to find the free ligand needed. Removing the ligand reverses everything: acid turns NH₃ into NH₄⁺, which has no lone pair, so AgCl precipitates again when a silver–ammonia solution containing chloride is acidified.

Amphoteric hydroxides

Some hydroxides, notably Al(OH)₃, Zn(OH)₂, Cr(OH)₃, Pb(OH)₂ and Sn(OH)₂, dissolve both in acid and in excess strong base. With acid they act as bases: Zn(OH)₂(s) + 2 H₃O⁺ → Zn²⁺ + 4 H₂O. With excess OH⁻ the metal ion accepts more hydroxide lone pairs and forms a soluble hydroxo complex: Zn(OH)₂(s) + 2 OH⁻ ⇌ [Zn(OH)₄]²⁻ and Al(OH)₃(s) + OH⁻ ⇌ [Al(OH)₄]⁻. The constant for dissolving in base is again Ksp × Kf, and the zincate concentration rises with [OH⁻]², so the solubility of an amphoteric hydroxide is lowest at intermediate pH and rises on both sides. Mg(OH)₂, Ca(OH)₂, Fe(OH)₃ and Ni(OH)₂ are not amphoteric: excess NaOH leaves them as solids. That difference separates ions, because excess NaOH keeps aluminium and zinc in solution while Fe(OH)₃ or Mg(OH)₂ is filtered off. Aqueous ammonia behaves differently: it supplies too little OH⁻ to form hydroxo complexes, but Zn²⁺, Cu²⁺, Ni²⁺ and Ag⁺ form ammine complexes, so Zn(OH)₂ dissolves in excess NH₃ while Al(OH)₃ does not.

Qualitative-analysis logic

Qualitative analysis separates the cations of a mixture by adding group reagents in a fixed order, each chosen to precipitate one group while the rest stay dissolved. Dilute HCl comes first, because AgCl, PbCl₂ and Hg₂Cl₂ are the only common insoluble chlorides. H₂S in acidic solution then precipitates only the least soluble sulfides, such as CuS, because acid keeps the sulfide concentration very low. Making the solution basic precipitates the remaining sulfides and hydroxides, such as ZnS, FeS and Al(OH)₃. Carbonate then removes Ca²⁺, Sr²⁺ and Ba²⁺, and Na⁺ and K⁺ are found by flame tests. Within a group, solubility and complexation tell the ions apart. In the chloride group, hot water dissolves PbCl₂, and chromate then gives yellow PbCrO₄; aqueous NH₃ dissolves AgCl as [Ag(NH₃)₂]⁺, and acidifying with HNO₃ brings white AgCl back; Hg₂Cl₂ turns grey-black with NH₃. Confirmatory colours come from complexes: blood-red FeNCS²⁺ when thiocyanate meets Fe³⁺, deep-blue [Cu(NH₃)₄]²⁺ when Cu²⁺ meets excess ammonia. Read every test both ways: a test that would have given a colour or a residue and did not rules that ion out.

What to carry forward

Precipitation starts when Q reaches Ksp: solve Ksp for the missing concentration with the full exponent and the diluted concentrations. In a mixture, the solid that needs the least reagent forms first; compare onset concentrations, not raw Ksp values, when formula types differ. Ligands donate lone pairs to metal cations, and large Kf values leave very little free metal ion when the ligand is in excess. A ligand dissolves a precipitate with K = Ksp × Kf; amphoteric hydroxides dissolve in excess OH⁻ as hydroxo complexes; qualitative analysis chains these ideas into selective precipitations and confirmatory colour tests.

Problemas de prática (40)

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Desenhe o mecanismo 2

Complex ions and Kf

  • Thiocyanate binds iron(III): the blood-red testDraw thiocyanate donating a nitrogen lone pair to hydrated Fe³⁺, displacing a water ligand, to form the blood-red FeNCS²⁺ ion used to confirm iron(III), and predict how a competing ligand affects the colour.

Amphoteric hydroxides and qualitative analysis

  • Hydroxide ions bond to Zn²⁺ to form zinc hydroxideDraw two hydroxide ions donating lone pairs to Zn²⁺ to form Zn(OH)₂, and predict why the precipitate dissolves again in excess strong base.

Preveja o produto 5

Amphoteric hydroxides and qualitative analysis

  • Acidifying a silver–ammonia solutionPredict how removing a ligand by protonation makes a dissolved precipitate reappear.
  • Zinc hydroxide in excess NaOHPredict the balanced equation for an amphoteric hydroxide dissolving in excess strong base.

Selective precipitation

  • Mg²⁺ or Ca²⁺ first?Predict which of two hydroxides of the same formula type precipitates first by comparing their Ksp values.
  • First silver halide to appearPredict the first precipitate when a common reagent is added to equal concentrations of ions that form 1:1 salts.

Dissolving precipitates with ligands

  • Which white solid dissolves in ammonia?Predict which precipitate dissolves in aqueous ammonia from whether its cation forms an ammine complex.

Classificar 4

Amphoteric hydroxides and qualitative analysis

  • Spot the amphoteric hydroxideClassify a metal hydroxide as amphoteric from its dissolution in both excess acid and excess strong base.

Dissolving precipitates with ligands

  • Constant for dissolving Cu(OH)₂ in ammoniaIdentify the equilibrium constant of a dissolution-by-complexation reaction as the product Ksp × Kf.

Complex ions and Kf

  • Formation constant of [Cu(NH₃)₄]²⁺Write the formation-constant expression for a complex ion with the correct ligand exponent.

Starting precipitation

  • The instant AgCl appearsIdentify the condition Q = Ksp that marks the onset of precipitation.

Comparar 3

Complex ions and Kf

  • Rank complexation at fixed ammoniaRank metal ions by the ratio of complexed to free metal at a fixed ligand concentration, accounting for the number of ligands.

Dissolving precipitates with ligands

  • Which solution dissolves the most AgBr?Compare the solubility of a precipitate in several solutions by computing Ksp × Kf for each ligand that binds the cation.

Selective precipitation

  • Order of silver precipitatesRank several precipitates by their onset reagent concentration when both Ksp and ion concentrations differ.

Calcular 13

Dissolving precipitates with ligands

  • Ammonia needed to keep AgCl dissolvedCombine Ksp and Kf to find the minimum free ligand concentration that prevents a precipitate from forming.
  • How much AgBr does fixer dissolve?Solve for the solubility of a precipitate in a complexing agent when K exceeds 1, accounting for the ligand consumed.
  • K for dissolving AgBr in thiosulfateCombine a solubility product and a formation constant into the equilibrium constant for dissolving a precipitate.

Starting precipitation

  • Chromate needed to start Ag₂CrO₄Calculate the minimum anion concentration that starts precipitation of a 2:1 salt when the cation concentration is fixed.
  • pH at which Fe(OH)₃ appearsUse a cube root to find the pH at which an M(OH)₃ hydroxide begins to precipitate.
  • pH at which Mg(OH)₂ appearsConvert the onset hydroxide concentration of a metal hydroxide into a pH.
  • Onset of PbI₂ after mixingFind the minimum concentration of a reagent solution that starts precipitation after both solutions are diluted on mixing.
  • Silver needed to start AgClCalculate the minimum cation concentration that starts precipitation of a 1:1 salt.

Complex ions and Kf

  • Free Ag⁺ in ammoniaCalculate the free metal-ion concentration in excess ligand after subtracting the ligand bound in the complex.
  • Overall Kf from stepwise constantsCombine stepwise formation constants into an overall formation constant and express it as a logarithm.
  • Charge of the silver–thiosulfate complexCalculate the charge of a complex ion from the metal-ion charge and the ligand charges.

Selective precipitation

  • Chloride left at the Mohr end pointCalculate how much of the first-precipitating ion remains when a second solid just begins to form.

Amphoteric hydroxides and qualitative analysis

  • Dissolved zinc at pH 13Calculate the solubility of an amphoteric hydroxide in strong base from Ksp × Kf and the hydroxide concentration.

Encontre o erro 2

Selective precipitation

  • Does the smaller Ksp precipitate first?Diagnose a precipitation-order prediction that compares Ksp values of salts with different formula types.

Starting precipitation

  • Find the slip in a PbI₂ onset calculationDiagnose a missing square root in a minimum-concentration calculation for a 1:2 salt.

Interprete as evidências 3

Selective precipitation

  • How clean is the Ba²⁺/Sr²⁺ separation?Interpret two Ksp values to judge how completely sulfate removes Ba²⁺ before SrSO₄ starts to form.

Amphoteric hydroxides and qualitative analysis

  • Reading the chloride-group testsInterpret a sequence of chloride-group observations to decide which of Ag⁺, Pb²⁺ and Hg₂²⁺ are present or absent.
  • Identify the cation from two testsInterpret hydroxide and ammonia tests to identify a cation from its amphoterism and ammine-complex formation.

Trabalhe de trás para frente 1

  • Kf from a measured solubilityWork backwards from a measured solubility in a ligand solution to the formation constant of the complex.

Planeje uma síntese 4

Amphoteric hydroxides and qualitative analysis

  • Separate Al³⁺ from Fe³⁺Plan a separation of two metal ions that uses the amphoterism of one hydroxide.

Starting precipitation

  • pH for complete copper removalChoose the pH at which a hydroxide precipitation becomes complete by using the remaining, not the initial, metal-ion concentration.

Selective precipitation

  • pH window to remove iron but keep magnesiumPlan a pH range that precipitates one hydroxide completely while a second hydroxide has not yet begun to form.
  • Separate Ba²⁺ from Mg²⁺Plan a selective precipitation that removes one cation and leaves another in solution.

Explique por quê 3

Complex ions and Kf

  • Pale-blue solid, then deep-blue solutionExplain how a weak-base ligand first precipitates a metal hydroxide and then dissolves it by complexation.
  • Why so little free Cu²⁺?Explain how a large formation constant and excess ligand make the free metal-ion concentration tiny.

Dissolving precipitates with ligands

  • Why AgCl dissolves in ammonia but AgI does notExplain why one precipitate dissolves in a complexing agent and another does not by comparing Ksp × Kf.

Travou em precipitação seletiva e íons complexos? Escaneie o problema.

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