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Chapter 10

Jess Sproul Chapter 10 Notes. Chapter 10. How Equilibrium Calculations Can Be Applied to Complex Systems. 10A: Solving Multiple-Equilibrium Problems by a Systematic Method.

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Chapter 10

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  1. Jess Sproul Chapter 10 Notes Chapter 10 How Equilibrium Calculations Can Be Applied to Complex Systems

  2. 10A: Solving Multiple-Equilibrium Problems by a Systematic Method • Aqueous solutions encountered in the lab often contain several species that interact with one another and water to yield two or more equilibria that function simultaneously. • When water is saturated with barium sulfate, three equilibria evolve: BsSO4(s) Ba2+ + SO42- SO42- + H3O+ HSO4- + H2O 2H2O  H3O+ + OH- These equilibria will shift accordingly if ions are added. • Solution of a multiple-equilibrium problem requires as many independents equations as there are participants in the system being studied. In the above system, there are five species: Ba2+, SO42-, HSO4-, H3O+, and OH-. In order to calculate the solubility of barium sulfate, there needs to be five independent algebraic equations.

  3. 10A-1: Mass-Balance Equation • Mass-balance equations relate the equilibrium concentrations of various species in a solution to one another and to the analytical concentrations of the various solutes. Example 10-1 Write mass-balance expressions for a 0.0100 M solution of HCl that is in equilibrium with an excess of solid BaSO4. From our general knowledge of the behavior of aqueous solutions, we can write equations for three equilibria that must be present in this solution. BsSO4(s) Ba2+ + SO42- SO42- + H3O+ HSO4- + H2O 2H2O  H3O+ + OH- Because the only source for the two sulfate species is the dissolved BsSO4, the barium ion concentration must equal the total concentration of sulfate-containing species, and a mass-balance equation can be written that expresses this equality. Thus [Ba2+] = [SO42-] + [HSO4-] The hydronium ion concentration in this solution has two sources: one from the HCl and the other from the dissociation of the solvent. A second mass-balance equation expression is this [H3O+] + [HSO4-] = cHCl + [OH-] = 0.0100 + [OH-] Since the only source of hydroxide is the dissociation of water, [OH-] is equal to the hydronium ion concentration from the dissociation of water.

  4. 10A-2: Charge-Balance Equation • Solutions are neutral because the molar concentration of positive charge in an electrolyte solution always equals the molar concentration of negative charge. # mol/L positive charge = # mol/L negative charge • The concentration of charge contributed to a solution by an ion is equal to the molar concentration of that ion multiplied by its charge. • For Na+ mol positive charge/L = (1 mol positive charge/mol Na+) X (mol Na+/L) = 1 X [Na+] • For Mg2+ mol positive charge/L = (2 mol positive charge/mol Mg2+) X (mol Mg2+/L) = 2 X [Mg2+] • For PO43- mol positive charge/L = (3 mol positive charge/mol PO43-) X (mol PO43- /L) = 3 X [PO43-]

  5. 10A-2: Charge-Balance Equation (cont.) • Example 10-3 Write a charge-balance equation for the system in Example 10-2. [Ag+] + [Ag(NH3)2+] + [H3O+] + [NH4+] = [OH-] + [Br -] • Example 10-4 Neglecting the dissociation of water, write a charge-balance equation for a solution that contains NaCl, Ba(ClO4)2, and Al2(SO4)3. [Na+] + 2[Ba2+] + 3[Al3+] = [Cl-] + [ClO4-] + 2[SO42-]

  6. 10A-3: Steps for Solving Problems Involving Several Equilibria • Write a set of balanced chemical equations for all pertinent equilibria. • Sate in terms of equilibrium concentrations what quantity is being sought. • Write equilibrium-constant expressions for all equilibria developed in step 1 and find numerical values for the constants in tables of equilibrium constants. • Write mass-balance expressions for the system. • If possible, write a charge-balance expression for the system. • If the number of unknown concentrations is equal to the number of equations at this point, proceed on to step 7. If the number of unknown concentrations is greater than the number of equations, seek additional equations. If there are no possible equations to be found and suitable assumptions regarding the unknowns cannot be made, the problem cannot be solved.

  7. 10A-3: Steps for Solving Problems Involving Several Equilibria (cont.) • Make suitable approximations to simplify the algebra. • Sole the algebraic equations for the equilibrium concentrations needed to give a provisional answer as defined in step 2. • Check the validity of the approximations made in step 7 using the provisional concentrations computed in step 8.Crucibles are used as either containers or filtering devices that allow the supernatant to pass through while retaining the precipitate. • Sometimes, filter paper is used and it must be burned off. If this is the case, be sure to handle the heated crucible with care.

  8. 10A-4: Making Approximations to Solve Equilibrium Equations • When step 6 is completed, there is the mathematical problem of solving several nonlinear simultaneous equations. • Without computer help, solving this complex system is formidable, tedious, and time consuming. • Approximations can allow a complex system to be solved much more easily. • Only the mass-balance and charge-balance equations can be simplified because they involve only sums and differences, not products and quotients. • With enough knowledge of the chemistry of a system, it is possible to assume that a given term in a mass-balance or charge-balance equation is sufficiently small that it can be neglected.

  9. 10B-2: How pH Influences Stability • The solubility of a precipitate containing an anion with basic properties, a cation with acidic properties, or both, will depend on pH.

  10. 10B-3: The Solubility of Precipitates in the Presence of Complexing Agents • Complex Formation with a Common Ion • Many precipitates react with the precipitating agent to form soluble complexes. • Increases in solubility caused by large excesses of a common ion are not unusual. • Quantitative Treatment of the Effect of Complex Formation on Solubility • Solubility calculations for a precipitate in the presence of a complexing reagent are similar in principle to those discussed in the previous section.

  11. 10C: Separating Ions by pH Control: Sulfide Separations • Several precipitating agents permit separation of ions based on solubility differences. This requires close control over the reacting agent at a suitable and predetermined lever. This is often done by controlling the pH with suitable buffers.

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