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Solubility Equilibrium

Solubility Equilibrium.

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Solubility Equilibrium

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  1. Solubility Equilibrium In a saturated solution, equilibrium exists between dissolved and undissolved solute. Slightly soluble/insoluble ionic compounds reach equil w/ very little solute dissolved. We can use this equilibrium to determine ion concentrations for these compounds. N.b. for soluble ionic compounds [ion] will be the same as the M. Ex: If [Pb(NO3)2] is 0.010 M then [Pb2+] = 0.010 M; [NO3-] = 0.020 M To determine [ion] for a slightly soluble compound we can use the ion product expression, Qsp, as we would for any equilibrium. PbF2(s)  Pb2+(aq) + 2F-(aq) Qsp = [Pb2+][F-]2

  2. Solubility Equilibrium For a saturated solution, equilibrium exists between dissolved and undissolved solute, so Qsp attains a constant value called the solubility-product constant, Ksp. Ex: Write the solubility-product constant expression for an aq solution of calcium phosphate. The Ksp value indicates how far the dissociation proceeds at equilibrium. That is, the greater the value of Ksp, the more dissociation is likely to occur. Like any Keq it can be calculated from the [equilibrium] of each ion. Ex: The solubilty of lead(II) fluoride in water at 250C is 0.064 g/100 mL of solution. Calculate the Ksp of lead(II) fluoride.

  3. Solubility Equilibrium The Ksp for a substance can also be used to determine its solubility. Ex: calcium hydroxide (lime) is a major component in both plaster and cement. Calculate its molar solubility if the Ksp = 6.5 x 10-6. Ex: The Ksp of copper(II) iodate at 25 0C is 1.40 x 10-7. Calculate the molar solubility. As long as we compare compounds with the same number of ions in the formula, Ksp values indicate the relative solubilties: the higher the Ksp the greater the solubilty (see Table 18.2).

  4. Solubility Equilibrium A particular application of Le Chatelier’s principle is known as the common ion effect. Adding a common ion decreases the solubility of a slighly soluble compound. Imagine a saturated solution of lead(II) chromate with a Ksp = 2.3 x 10-13. If we add Na2CrO4 the [CrO42-] is increased, shifting the equilbrium back to the left decreasing the [Pb2+]. PbF2(s)  Pb2+(aq) + CrO42-(aq) Ex: What is the solubility of calcium hydroxide in 0.10 M calcium nitrate? By comparing the values of Qsp and Ksp it can be determined whether combining solutions of specific concentrations will, or will not for a precipitate.

  5. Solubility Equilibrium If Qsp = Ksp solution is saturated, no change will occur. If Qsp > Ksp a precipitate will form. If Qsp < Ksp solution is unsaturated, no precipitate will form. Ex: Does a precipitate form when 50.0 mL of 0.0010 M BaCl2 is added to 50.0 mL 0.00010 M Na2SO4? Ksp BaSO4 = 1.1 x 10-10 Ex: A solution consists of 0.20 M magnesium chloride and 0.10 M copper(II) chloride. What [OH-] would be necessary to separate the metal ions? Ksp of Mg(OH)2 = 6.3 x 10-10; Ksp Cu(OH)2 = 2.2 x 10-20

  6. Spontaneous vs. Non-Spontaneous • Thermodynamics allows us to predict if a process is spontaneous – now, never use that again; call it Thermodynamically Favored. • TF means that the process will occur under the defined conditions. • Ball rolls downhill, not up • Iron rusts, rust is not spontaneously converted to Fe & O2 • At temperatures above 273K H2O(s) will melt, not freeze • TF (spontaneity) does not refer to how fast a reaction proceeds. • What determines if a process is TF: • Change in enthalpy • Change in entropy • Temperature

  7. Entropy • Entropy (S) is the disorder of a system • dependent upon temperature • +ΔS = more disorder, less order = greater # of microstates • -ΔS = less disorder, more order • Universal tendency greater entropy (probability) • Which has a higher entropy? • Cube of sugar in coffee / cube of sugar • I2(g) / I2(s) • PCl5(g) / PCl3(g) + Cl2(g) • C(s) + CO2(g) / 2 CO(g)

  8. Gibb’s Free Energy • We can relate enthalpy & entropy quantitatively in a relationship known as….. • ΔG = ΔH – T ΔS ΔG = Change in Gibb’s Free Energy [kJ/mol] ΔH = change in Enthalpy (endo is + & exo is -) [kJ/mol] ΔS = change in Entropy [J/mol. K] T = Temperature in Kelvin • -ΔG = Thermodynaically favorable. • +ΔG = not favorable but if reversible the reverse rxn is TF • ΔG = 0 Equilibrium

  9. Why does the ΔG have to be negative…? 2nd Law of Thermodynamics: In any spontaneous (TF) process, there is always an increase in the entropy of the universe. Significance of ΔH depends on Temp ΔSuniv = ΔSsys + ΔSsurr ΔSsurr depends on direction of heat flow; exoΔSsurr is +; endoΔSsurr is -; magnitude depends on T (effect greater at lower T); ΔSsurr = - ΔH/T ΔG = ΔH – TΔS; -ΔG/T = - ΔH/T + ΔS; -ΔG/T = ΔSsurr + ΔS = ΔSuniv so ΔSuniv = -ΔG/T Therefore if ΔG is negative ΔSuniv is positive and the entropy of the universe increases. Q.E.D.

  10. Possible Combinations of ΔH and ΔS and dependence of TF on temperature Between entropy and enthalpy, there can be four possible outcomes:

  11. Predict the sign for ΔS for each of the following: • The thermal decomposition of calcium carbonate • The oxidation of sulfur dioxide in air to produce sulfur trioxide At what temperature is the following process spontaneous? Br2(l) → Br2(g) ΔH0 = 31.0 kJ/molΔS0 = 93.0 J/mol. K ΔS favors vaporization, ΔH favors the condensation. These opposite tendencies will exactly balance out when they are in dynamic equilibrium. This occurs at the boiling point for bromine. Remember at equilibrium ΔG = 0 so 0 = ΔH – T ΔS ΔH = T ΔS and T = ΔH / ΔS Now solve for T. T = 333 K Sample free energy calculations

  12. Methanol is synthesized by the reaction CO(g) + 2H2(g)  CH3OH(g) a. Calculate the standard free-energy change for this reaction at 25 0C. b. Is the reaction TF at this temp? c. If so, at what temperature does the reverse rxn become TF? n.b. Selected Thermodynamic Values Appendix A in text. CO(g) H2(g)CH3OH(g) ΔH0f (kJ/mol) -110.5 0 -201.2 S0 (J/mol . K) 197.6 130.6 239.0 Sample free energy calculations

  13. STANDARD FREE ENERGIES OF FORMATION ∆Gfө of a substance is the free energy change for the formation of 1 mol of the substance, in its standard state, from the most stable form of the constituent elements in their standard states. Ex: N2(g) + 3H2(g)  2NH3(g) ∆Gө = -33.0 kJ So ∆Gfө = -33.0 kJ/2 mol = -16.5 kJ/mol Note: ∆Gfө for the most stable form of an element = 0. These values allow us to calculate ∆Grxnө the same way we solved for ∆Hrxnө, using ∆Hfө values: ∆Gө = ∆Gfө(products) - ∆Gfө(reactants)

  14. Example: Determine ΔGө for: Fe2O3(s) + 3CO(g)  2 Fe(s) + 3CO2(g) Fe2O3(s) = -742.2 kJ/mol; ΔGөvalues are: CO2 = -394.4 kJ/mol CO(g) = -137.2 kJ/mol; Answer is: -29 kJ Thermodynamically favorable So at 25oC this reaction is

  15. Free energy change and composition Reactions often carried out at non-standard conditions. Problem is overcome using this relationship: ΔG = ΔG0 + RT ln Q (may be Qc, Qp, or Qsp) Example: Calculate the free energy change for ammonia synthesis at 22 0C given the following partial pressures: 0.010 atm N2, 0.030 atm H2, 2.0 atm; ΔG = - 33.0 kJ/mol

  16. Free energy and equilibrium Common use of ΔG is to calculate the value for an equilibrium constant, K. ΔG = ΔG0 + RT ln Q; at equil Q = K and ΔG = 0 so at equilibrium ΔG0= -RT ln K Example: For the reaction 2 CO(g) + O2(g)  2 CO2(g) ΔG0 = -257.2 kJ/mol; What is Kp at 25 0C? Example: Ag+(aq) + NH3(aq)  Ag(NH3)2+(aq) ΔG0 = -41.2 kJ/mol; Calculate the value for Kc. Example: Calculate the equilibrium vapor pressure of H2O(g) at 25 0C. ΔG0f = -237.0 kJ/mol for H2O(l); ΔG0f = -228.4 kJ/mol for H2O(g)

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