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L e c t u r e 2

L e c t u r e 2. Precipitation equilibrium. Associate prof . L.V. Vronska Associate prof . M.M. Mykhalkiv. Outline. Precipitation equilibrium it heterogeneous equilibrium Calculation of solubility and solubility product K sp Influence of chemical factors is on solubility of precipitate

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L e c t u r e 2

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  1. Lecture 2 Precipitation equilibrium Associate prof. L.V. Vronska Associate prof . M.M. Mykhalkiv

  2. Outline • Precipitation equilibrium it heterogeneous equilibrium • Calculation of solubility and solubility product Ksp • Influence of chemical factors is on solubility of precipitate • Completeness of precipitation and factors which influence on her • Conditions of dissolution of precipitation

  3. Precipitation equilibrium • precipitate An insoluble solid that forms when two or more soluble reagents are combined.

  4. The most common precipitation reaction is a metathesis reaction, in which two soluble ionic compounds exchange parts. Thus, the precipitation of PbCl2 is written as Pb2+(aq) + 2Cl–(aq) = PbCl2(s) • In the equilibrium treatment of precipitation, however, the reverse reaction describing the dissolution of the precipitate is more frequently encountered. PbCl2(s) = Pb2+(aq) + 2Cl–(aq)

  5. Precipitation equilibrium The equilibrium constant for this reaction is called the solubility product, Ksp, and is given as Ksp = [Pb2+][Cl–]2 = 1.7 ·10–5 • and for all electrolytesAmBn Ksp = [A]m[B]n • solubility product Ksp - the equilibrium constant for a reaction in which a solid dissociates into its ions

  6. Concentrational (real) constantsolubility product KRspfor all electrolytesAmBn (use, when we have real conditions (influence of ionic strength)) KRsp = [A]m[B]n Thermodynamicconstant solubility product KTspfor all electrolytesAmBn

  7. Thermodynamicconstant solubility product KTsp depends on: • Temperature • Pressure • Nature of solvent • Nature of precipitate

  8. Thermodynamicconstants solubility product KTsp are adduction in reference books

  9. Conditionalconstant solubility product KCspfor all electrolytesAmBn

  10. We use , when we have the following real conditions: • Temperature • Pressure • Influence of ionic strength • Influence of competitive reactions

  11. Calculation of solubility and solubility product Ksp • Solubility is a property of matter to give/ to form a solution with a certain solvent at certain conditions We determine Solubility as: • Coefficient of Solubility (ks) • Molar Solubility (S)

  12. Coefficient of Solubility (ks) • It is mass of matter which dissolves at this temperature in 100 g or 100 mL of solvent

  13. Coefficient of Solubility (ks)

  14. Molar Solubility (S) • It is a molar concentration of matter in the saturated solution АmВn mА + nВ KRsp = [A]m [B]n [A] = m[AmBn] = mS [B] = n[AmBn] = nS KRsp=(mS)m(nS)n = mmnnSm+n

  15. A rule of solubility product:in saturated solution above sediment product of ions concentrations is permanent at a stationarytemperature • In unsaturated solution [A]m[B]nKsp • In saturatedsolution [A]m[B]n=Ksp • In supersaturated solution [A]m[B]n>Ksp

  16. If ionic strength can be adopted even a zero and to neglect of competitive reactions, solubility of precipitate is expected on the size of KTsp (if μ→0 then f→1; α=1) • If to take into account ionic strength, but to neglect of competitive reactions, solubility is expected after the size of KRsp (if μ≠0 then f≠1; α=1) • If we cannot neglect by competitive reactions, then solubility is expected on the size of KCsp (if μ≠0 then f≠1; α ≠ 1)

  17. Influence of chemical factors is on solubility of precipitate • The Common-Ion effect • The activity effect • Acid-base reactions • Complexation reactions • Red-ox reactions

  18. Completeness of precipitation and factors which influence on her The precipitation is considered practically complete, if the concentration of the precipitate’s ions in solution above sediment does not exceed a 10-6 mol/L

  19. The Common-Ion Effect

  20. The common-ion effect. The solubility of at 25°C decreases markedly on addition of ions. Note that the calculated solubility is plotted on a logarithmic scale.

  21. The solubility of precipitate decreases in the presence of a solution that already contains one of its ions. This is known as the common ion effect.

  22. The activity effect • Clearly the equilibrium position for the reaction AgIO3(s)  Ag+(aq) + IO3–(aq) • depends on the composition of the solution. When the solubility product for AgIO3is calculated using the equilibrium concentrations of Ag+ and IO3– Ksp = [Ag+][IO3–] its apparent value increaseswhenan inert electrolyte such as KNO3is added.

  23. The true thermodynamic equilibrium constant, Ksp, for the solubility of AgIO3, therefore, is KTsp = a(Ag+) a(IO3–) KTsp=[Ag+][IO3-]f(Ag+)f(IO3–) KTsp= KRsp f(Ag+)f(IO3–) KRsp= KTsp/ f(Ag+)f(IO3–) • To accurately calculate the solubility of AgIO3, we must know the activity coefficients for Ag+ and IO3–.

  24. Mention !!! • First, as the ionic strength approaches zero, the activity coefficient approaches a value of one. Thus, in a solution where the ionic strength is zero, an ion’s activity and concentration are identical. We can take advantage of this fact to determine a reaction’s thermodynamic equilibrium constant. The equilibrium constant based on concentrations is measured for several increasingly smaller ionic strengths and the results extrapolated back to zero ionic strength to give the thermodynamic equilibrium constant. • Second, activity coefficients are smaller, and thus activity effects are more important, for ions with higher charges and smaller effective diameters. Finally, the extended Debye–Hückel equation provides reasonable activity coefficients for ionic strengths of less than 0.1. Modifications to the extended Debye–Hückel equation, which extend the calculation of activity coefficients to higher ionic strength, have been proposed.

  25. The pH of the Solution An ionic compound that contains a basic anion becomes more soluble as the acidity of the solution increases. The solubility of CaCO3, for example, increases with decreasing pH because the CO32- ions combine with protons to give HCO3- ions. As CO32- ions are removed from the solution, the solubility equilibrium shifts to the right, as predicted by Le Châtelier’s principle. The net reaction is dissolution of CaCO3, in acidic solution to give Ca2+ ions and HCO3- ions:

  26. Formation of Complex Ions The solubility of an ionic compound increases dramatically if the solution contains a Lewis base that can form a coordinate covalent bond to the metal cation. Silver chloride, for example, is insoluble in water and in acid, but it dissolves in an excess of aqueous ammonia, forming the complex ion [Ag(NH3)2]+. A complex ion is an ion that contains a metal cation bonded to one or more small molecules or ions, such as NH3, CN- or OH-. In accord with Le Châtelier’s principle, ammonia shifts the solubility equilibrium to the right by tying up the Ag+ ion in the form of the complex ion:

  27. Formation of Complex Ions Silver chloride is insoluble in water (left) but dissolves on addition of an excess of aqueous ammonia (right).

  28. The solubility of AgCl in aqueous ammonia at 25°C increases with increasing ammonia concentration owing to formation of the complex ion [Ag(NH3)2]+. Note that the solubility is plotted on a logarithmic scale.

  29. Separation of Ions by Selective Precipitation • A convenient method for separating a mixture of ions is to add a solution that will precipitate some of the ions but not others. The anions SO42- and Cl- for example, can be separated by addition of a solution of Ba(NO3)2. Insoluble BaSO4 precipitates, but Cl- remains in solution because BaCl2 is soluble. • Similarly, the cations Ag+ and Zn2+ can be separated by addition of dilute HCl. Silver chloride, AgCl, precipitates, but Zn2+ stays in solution because ZnCl2 is soluble.

  30. Separation of Ions by Fractional Precipitation • Ions Ba2+ and Ca2+ can be separates if concentration of SO42- ions is controlled. • BaSO4has Ksp = 110–10andCaSO4 hasKsp = 2,310–5

  31. That precipitate of BaSO4 have been removed out, his ionic product must be greater Ksp, but, that precipitate of CaSO4 did not removed out it is simultaneously necessary, that ionic product [Ca2+][SO42-] KspCaSO4, but [Ba2+][SO42-] KspBaSO4 • Therefore, if concentrations both ions are 10-2 mol/L, concentration SO42- ions must be between and

  32. Conditions of dissolution of precipitation It is necessary for dissolution of sediment, that his ionic product became more small constants of solubility product: [Kat+][An–] Ksp_KatAn

  33. Decrease of ions concentration it can be carried out the followings methods: 1. strong dilution of solution

  34. The term 'partly soluble' is used to describe a mixture of which only some of the components dissolve.

  35. Decrease of ions concentration it can be carried out the followings methods: 2. The ions of precipitate are connected in compounds which well water-soluble Co(OH)Cl + HCl = CoCl2 + H2O 3. The ions of precipitate are connected in compounds which give gas ZnS + 2HCl = ZnCl2 + H2S

  36. Formation and dissolution of Cr(OH)3

  37. Decrease of ions concentration it can be carried out the followings methods: 3. The ions of precipitate are connected in compounds which are complex AgCl + 2NH3 = [Ag(NH3)2]Cl 4. Oxidation and reduction of ions of precipitatein others compounds MnO(OH)2 + H2C2O4 + H2SO4 = MnSO4 + 2CO2 + 3H2O

  38. Dissolution of sulfatic-precipitate sodium carbonate extraction is a translation of sulfates of second analytical group in carbonates • BaSO4Ksp=1,1  10-10 • BaCO3 Ksp=5,1  10-9

  39. Aplication of Precipitation equilibrium • Gravimertic analysis: - Particulate gravimetry - Precipitation gravimetry

  40. Thanks for your attention!

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