Chapter 10Chapter 10

How Equilibrium Calculations Can How Equilibrium Calculations Can Be Applied to Complex SystemsBe Applied to Complex Systems

Jess Sproul

Chapter 10 Notes

10A: Solving Multiple-Equilibrium 10A: Solving Multiple-Equilibrium Problems by a Systematic MethodProblems by a Systematic Method Aqueous solutions encountered in the lab often contain Aqueous solutions encountered in the lab often contain

several species that interact with one another and water several species that interact with one another and water to yield two or more equilibria that function to yield two or more equilibria that function simultaneously.simultaneously.

When water is saturated with barium sulfate, three When water is saturated with barium sulfate, three equilibria evolve:equilibria evolve:

BsSOBsSO4(s)4(s) Ba Ba2+2+ + SO + SO442-2-

SOSO442-2- + H + H33OO++ HSO HSO44

-- + H + H22OO2H2H22O O H H33OO++ + OH + OH--

These equilibria will shift accordingly if ions are added.These equilibria will shift accordingly if ions are added. Solution of a multiple-equilibrium problem requires as Solution of a multiple-equilibrium problem requires as

many independents equations as there are participants in many independents equations as there are participants in the system being studied. In the above system, there are the system being studied. In the above system, there are five species: Bafive species: Ba2+2+, SO, SO44

2-2-, HSO, HSO44--, H, H33OO++, and OH, and OH--. In order . In order

to calculate the solubility of barium sulfate, there needs to calculate the solubility of barium sulfate, there needs to be five independent algebraic equations. to be five independent algebraic equations.

10A-1: Mass-Balance Equation10A-1: Mass-Balance Equation• Mass-balance equations relate the equilibrium Mass-balance equations relate the equilibrium

concentrations of various species in a solution to one concentrations of various species in a solution to one another and to the analytical concentrations of the another and to the analytical concentrations of the various solutes.various solutes.

Example 10-1Write 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

- + H2O2H2O 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.

10A-2: Charge-Balance Equation10A-2: Charge-Balance Equation

Solutions are neutral because the molar concentration of positive Solutions are neutral because the molar concentration of positive charge in an electrolyte solution always equals the molar charge in an electrolyte solution always equals the molar concentration of negative charge.concentration of negative charge.

# mol/L positive charge = # mol/L negative charge# mol/L positive charge = # mol/L negative charge The concentration of charge contributed to a solution by an ion is The concentration of charge contributed to a solution by an ion is

equal to the molar concentration of that ion multiplied by its charge.equal to the molar concentration of that ion multiplied by its charge. For NaFor Na++

mol positive charge/L = (1 mol positive charge/mol Namol positive charge/L = (1 mol positive charge/mol Na++) X (mol Na) X (mol Na++/L)/L)= 1 X [Na+]= 1 X [Na+]

For MgFor Mg2+2+

mol positive charge/L = (2 mol positive charge/mol Mgmol positive charge/L = (2 mol positive charge/mol Mg2+2+) X (mol Mg) X (mol Mg2+2+/L)/L)= 2 X [Mg= 2 X [Mg2+2+]]

For POFor PO443-3-

mol positive charge/L = (3 mol positive charge/mol POmol positive charge/L = (3 mol positive charge/mol PO443-3-) X (mol PO) X (mol PO44

3-3- /L) /L)= 3 X [PO= 3 X [PO44

3-3-]]

10A-2: Charge-Balance Equation 10A-2: Charge-Balance Equation (cont.)(cont.)

• Example 10-3Example 10-3Write a charge-balance equation for the system in Write a charge-balance equation for the system in Example 10-2.Example 10-2.

[Ag[Ag++] + [Ag(NH] + [Ag(NH33))2+2+] + [H] + [H33OO++] + [NH] + [NH44++] = [OH] = [OH--] + [Br ] + [Br --]]

• ExExample 10-4ample 10-4Neglecting the dissociation of water, write a charge-Neglecting the dissociation of water, write a charge-balance equation for a solution that contains NaCl, balance equation for a solution that contains NaCl, Ba(ClOBa(ClO44))22, and Al, and Al22(SO(SO44))33..

[Na[Na++] + 2[Ba] + 2[Ba2+2+] + 3[Al] + 3[Al3+3+] = [Cl] = [Cl--] + [ClO] + [ClO44--] + 2[SO] + 2[SO4422--]]

10A-3: Steps for Solving Problems Involving 10A-3: Steps for Solving Problems Involving Several EquilibriaSeveral Equilibria

Write a set of balanced chemical equations for all pertinent Write a set of balanced chemical equations for all pertinent equilibria.equilibria.

Sate in terms of equilibrium concentrations what quantity is being Sate in terms of equilibrium concentrations what quantity is being sought.sought.

Write equilibrium-constant expressions for all equilibria developed Write equilibrium-constant expressions for all equilibria developed in step 1 and find numerical values for the constants in tables of in step 1 and find numerical values for the constants in tables of equilibrium constants.equilibrium constants.

Write mass-balance expressions for the system.Write mass-balance expressions for the system. If possible, write a charge-balance expression for the system.If possible, write a charge-balance expression for the system. If the number of unknown concentrations is equal to the number 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 of equations at this point, proceed on to step 7. If the number of unknown concentrations is greater than the number of equations, unknown concentrations is greater than the number of equations, seek additional equations. If there are no possible equations to seek additional equations. If there are no possible equations to be found and suitable assumptions regarding the unknowns be found and suitable assumptions regarding the unknowns cannot be made, the problem cannot be solved.cannot be made, the problem cannot be solved.

10A-3: Steps for Solving Problems Involving 10A-3: Steps for Solving Problems Involving Several Equilibria (cont.)Several Equilibria (cont.)

Make suitable approximations to simplify the algebra.Make suitable approximations to simplify the algebra. Sole the algebraic equations for the equilibrium Sole the algebraic equations for the equilibrium

concentrations needed to give a provisional answer as concentrations needed to give a provisional answer as defined in step 2.defined in step 2.

Check the validity of the approximations made in step 7 Check the validity of the approximations made in step 7 using the provisional concentrations computed in step using the provisional concentrations computed in step 8.Crucibles are used as either containers or filtering 8.Crucibles are used as either containers or filtering devices that allow the supernatant to pass through while devices that allow the supernatant to pass through while retaining the precipitate.retaining the precipitate.

Sometimes, filter paper is used and it must be burned off. Sometimes, filter paper is used and it must be burned off. If this is the case, be sure to handle the heated crucible If this is the case, be sure to handle the heated crucible with care.with care.

10A-4: Making Approximations to 10A-4: Making Approximations to Solve Equilibrium EquationsSolve Equilibrium Equations

When step 6 is completed, there is the mathematical When step 6 is completed, there is the mathematical problem of solving several nonlinear simultaneous problem of solving several nonlinear simultaneous equations.equations.

Without computer help, solving this complex system Without computer help, solving this complex system is formidable, tedious, and time consuming.is formidable, tedious, and time consuming.

Approximations can allow a complex system to be Approximations can allow a complex system to be solved much more easily.solved much more easily.

Only the mass-balance and charge-balance Only the mass-balance and charge-balance equations can be simplified because they involve only equations can be simplified because they involve only sums and differences, not products and quotients.sums and differences, not products and quotients.

With enough knowledge of the chemistry of a system, With enough knowledge of the chemistry of a system, it is possible to assume that a given term in a mass-it is possible to assume that a given term in a mass-balance or charge-balance equation is sufficiently balance or charge-balance equation is sufficiently small that it can be neglected.small that it can be neglected.

10B-2: How pH Influences Stability10B-2: How pH Influences Stability

• The solubility of a The solubility of a precipitate precipitate containing an anion containing an anion with basic with basic properties, a cation properties, a cation with acidic with acidic properties, or both, properties, or both, will depend on pH.will depend on pH.

10B-3: The Solubility of Precipitates in the 10B-3: The Solubility of Precipitates in the Presence of Complexing AgentsPresence of Complexing Agents

Complex Formation with a Common IonComplex Formation with a Common Ion Many precipitates react with the Many precipitates react with the

precipitating agent to form soluble precipitating agent to form soluble complexes.complexes.

Increases in solubility caused by large Increases in solubility caused by large excesses of a common ion are not excesses of a common ion are not unusual.unusual.

Quantitative Treatment of the Effect of Quantitative Treatment of the Effect of Complex Formation on SolubilityComplex Formation on Solubility Solubility calculations for a precipitate in Solubility calculations for a precipitate in

the presence of a complexing reagent are the presence of a complexing reagent are similar in principle to those discussed in similar in principle to those discussed in the previous section.the previous section.

10C: Separating Ions by pH 10C: Separating Ions by pH Control: Sulfide SeparationsControl: Sulfide Separations

Several precipitating agents permit Several precipitating agents permit separation of ions based on solubility separation of ions based on solubility differences. This requires close control differences. This requires close control over the reacting agent at a suitable and over the reacting agent at a suitable and predetermined lever. This is often done predetermined lever. This is often done by controlling the pH with suitable by controlling the pH with suitable buffers.buffers.