Jessica Sproul
Chapter 10 Summary
Dr. Rahni
Analytical Chemistry
How Equilibrium Calculations Can Be Applied to Complex Systems
10A) Solving Multiple-Equilibrium Problems by a Systematic Method
1)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.
2)When water is saturated with barium sulfate, three equilibria evolve:
BsSO4(s) Ba2+ + SO42-
SO42- + H3O+ HSO4- + H2O
2H2O H3O+ + OH-
i)These equilibria will shift accordingly if ions are added.
3)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+, andOH-. In order to calculate the solubility of barium sulfate, there needs to be five independent algebraic equations.
4)There are three types of algebraic equations:
i)Equilibrium-constant expressions
ii)Mass-balance expressions
iii)Charge-balance expressions
10A-1) Mass-Balance Equation
1)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.
Example 10-2
Write mass-balance expressions for the system formed when a 0.010 M NH3 solution is saturated AgBr.
Here, equations for the pertinent equilibria in the solution are
AgBr(s) Ag+ + Br -
Ag+ + 2NH3 Ag(NH3)2+
NH3 + H2O NH4+ + OH-
2H2O H3O+ + OH-
Because the only source of Ag+, Br-, andAg(NH3)2+ is AgBr and because silver and bromide ions are present in a 1:1 ration in that compound, it follows that one mass-balance equation is
[Ag+] + [Ag(NH3)2+] = [Br -]
where the bracketed terms are molar species concentration. Also, we know that the only source of ammonia-containing species is the 0.010 M NH3. Therefore,
c NH3 = [NH3] + [NH4+] + 2[Ag(NH3)2+] = 0.010
From the last two equilibria, we see that the one hydroxide ion is formed for each NH4+ and each hydronium ion. Therefore,
[OH-] = [NH4+] + [H3O+]
10A-2) Charge-Balance Equation
1)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
2)The concentration of charge contributed to a solution by an ion is equal to the molar concentration of that ion multiplied by its charge.
i)For Na+
mol positive charge/L = (1 mol positive charge/mol Na+) X (mol Na+/L)
= 1 X [Na+]
ii)For Mg2+
mol positive charge/L = (2 mol positive charge/mol Mg2+) X (mol Mg2+/L)
= 2 X [Mg2+]
iii)For PO43-
mol positive charge/L = (3 mol positive charge/mol PO43-) X (mol PO43-/L)
= 3 X [PO43-]
3)Now consider how we would write a charge-balance equation for a 0.100 M solution of sodium chloride. Positive charges are supplied by the Na+ and the H3O+ (from the dissociation of water). Negative charges are supplied by the Cl- and the OH-. The molarities of positive and negative charges are
mol positive charge/L = [Na+] + [H3O+] = 0.100 + 1 X 10-7
mol positive charge/L = [Cl-] + [OH-] = 0.100 + 1 X 10-7
We write the charge-balance equation by equating the concentrations of positive and negative charges. That is,
[Na+] + [H3O+] = [Cl-] + [OH-] = 0.100 + 1 X 10-7
4)The charge-balance equations for the aqueous solutions constrain the unknown 's to be such that, when the 's are added to the original data, charge balance is produced in each aqueous solution. The charge-balance equation for an aqueous solution is
, (137)
where is the charge imbalance in aqueous solution q calculated by a speciation calculation and is defined to be the charge on the master species plus the alkalinity assigned to the master species, . For alkalinity, is defined to be -1.0. The summation ranges over all elements or element valence states and includes a term for alkalinity, just as charge balance is commonly calculated by summing over cationic and anionic elements plus a contribution from alkalinity. In the definition of , the alkalinity of the master species is added to the charge for that master species to remove the equivalents for the element or element redox state that are already accounted for in the alkalinity. For example, the contribution of carbonate species in equation 137 is zero with this definition of ( , , ); all of the charge contribution of carbonate species is included in the alkalinity term of the summation.
(
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-]
10A-3) Steps for Solving Problems Involving Several Equilibria
1)Write a set of balanced chemical equations for all pertinent equilibria.
2)Sate in terms of equilibrium concentrations what quantity is being sought.
3)Write equilibrium-constant expressions for all equilibria developed in step 1 and find numerical values for the constants in tables of equilibrium constants.
4)Write mass-balance expressions for the system.
5)If possible, write a charge-balance expression for the system.
6)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)Make suitable approximations to simplify the algebra.
8)Sole the algebraic equations for the equilibrium concentrations needed to give a provisional answer as defined in step 2.
9)Check the validity of the approximations made in step 7 using the provisional concentrations computed in step 8.
10A-4) Making Approximations to Solve Equilibrium Equations
1)When step 6 is completed, there is the mathematical problem of solving several nonlinear simultaneous equations.
2)Without computer help, solving this complex system is formidable, tedious, and time consuming.
3)Approximations can allow a complex system to be solved much more easily.
4)Only the mass-balance and charge-balance equations can be simplified because they involve only sums and differences, not products and quotients.
5)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.
i)In a solution containing a reasonable concentration of an acid, the hydroxide concentration will often be negligible with respect to many other species in the solution and the term for the hydroxide concentration can usually be neglected in a mass- or charge-balance expression without introducing a significant error in the calculations.
6)Never be afraid to make an assumption in attempting to solve an equilibrium problem. If the assumption is not valid, you will know it as soon as you have an approximate answer.
10B) Calculating Solubilities by the Systematic Method
10B-1) Solubility Calculations
Example 10-5
Calculate the molar solubility of Mg(OH2) in water.
Step 1.Pertinent Equilibria
Two equilibria that need to be considered are
Mg(OH2)(s) Mg2+ + 2OH-
2H2O H3O+ + OH-
Step 2.Definition of the Unknown
Since 1 mol of Mg2+ is formed for each mole of Mg(OH2) dissolved,
Solubility Mg(OH2) = [Mg2+]
Step 3.Equilibrium-Constant Expression
Ksp = [Mg2+][OH-]2 = 7.1 X 10-12
Ksp = [H3O+][OH-] = 1.00 X 10-14
Step 4.Mass-Balance Expression
As shown by the two equilibrium equations, there are two sources of hydroxide ions: Mg(OH2) and H2O. The hydroxide ion resulting from the dissociation of Mg(OH2) is twice the magnesium ion concentration and that from the dissociation of water is equal to the hydronium ion concentration. Thus,
[OH-] = 2[Mg2+] + [H3O+]
Step 5.Charge-Balance Expression
[OH-] = 2[Mg2+] + [H3O+]
Not that this equation is identical to the mass-balance equation. Often a mass-balance equation and a charge-balance equation are the same.
Step 6.Number of Independent Equations and Unknowns
We have developed three indepencent algebraic equations and have three unknowns([OH-], [Mg2+] and [H3O+]). Therefore, the problem can be solved rigorously.
Step 7.Approximations
We can make approximations only in the mass-balance equation. Since the solubility-product constant for Mg(OH2) is relatively large, the solution will be somewhat basic. Therefore, it is reasonable to assume that [H3O+] is much less than [OH-]. The mass-balance equation then simplifies to
2[Mg2+] = [OH-]
Step 8. Solution to Equations
Substitution of the approximation into the first equilibrium-constant expression gives
[Mg2+](2[Mg2+])2 = 7.1 X 10-12
[Mg2+]3 = (7.1 X 10-12) / 4 = 1.78 X 10-12
[Mg2+] = solubility = 1.21 X 10-4 or 1.2 X 10-4 M
Step 9.Check of Assumptions
Substitution into the approximation equation yields
[OH-] = 2 X 1.21 X 10-4 = 2.42 X 10-4
[H3O+] = (1.00 X 10-14) / (2.42 X 10-4) = 4.1 X 10-11
Thus, our assumption that 4.1 X 10-11 is much less than 2.42 X 10-4is certainly valid.
10B-2) How pH Influences Stability
i)The solubility of a precipitate containing an anion with basic properties, a cation with acidic properties, or both, will depend on pH.
1)Solubility Calculations at Constant pH
Example 10-5
Calculate the molar solubility of calcium oxalate in a solution that has been buffered so that is pH is constant and equal to 4.00.
Step 1.Pertinent Equilibria
CaC2O4(s) Ca2+ + C2O42-
Oxalate ions react with water to form HC2O4-andH2C2O4. Thus, there three other equilibria present in this solution.
H2C2O4+ H2O H3O+ + HC2O4-
HC2O4- + H2O H3O+ + C2O42-
2H2O H3O+ + OH-
Step 2.Definition of the Unknown
Calcium oxalate is a strong electrolyte so that its molar analytical concentration is equal to the equilibrium calcium ion concentration. That is,
solubility = [Ca2+]
Step 3.Equilibrium-Constant Expression
[Ca2+][C2O42-] = Ksp = 1.7 X 10-9
([H3O+][HC2O4-])/[H2C2O4] = K1 = 5.60 X 10-2
([H3O+][C2O42-])/[HC2O4-] = K2 = 5.42 X 10-5
[H3O+][OH-] = Kw = 1.0 X 10-14
Step 4.Mass-Balance Expression
Because CaC2O4is the only source of Ca2+ and the three oxalate species,
[Ca2+] = [H2C2O4]+[HC2O4-] + [C2O42-]
Moreover, the problem states that the pH is 4.00. Thus,
[H3O+] = 1.00 X 10-4 and [OH-] = Ksp/[H3O+] = 1.00 X 10-10
Step 5.Charge-Balance Expression
A buffer is required to maintain th pH at 4.00. The buffer most likely consists of some weak acid HA and its conjugate base, A-. The identities of the three species and their concentrations have not been specified, however, so we do not have enough information to write a charge-balance expression.
Step 6.Number of Independent Equations and Unknowns
We have four unknowns ([Ca2+], [H2C2O4], [HC2O4-], and [C2O42-]) as well as four independent algebraic relationships. Therefore, an exact solution can be obtained, and the problem becomes one of algebra.
Step 7.Approximations
An exact solution is so readily obtained in this case that we will not bother with approximations.
Step 8. Solution to Equations
A convenient was to solve the problem is to substitute two of the equilibrium-constant expressions into the mass-balance expression in such a way as to develop a relationship between. Thus, we rearrange an equilibrium-constant expression to give
[HC2O4-] = ([H3O+][C2O42-]) / K2
Substituting numerical values for [H3O+] and K2 gives
[HC2O4-] = (1.00 X 10-4[C2O42-]) / 5.42 X 10-5 = 1.85[C2O42-]
Substituting this relationship into another mass-balance expression and rearranging gives
[H2C2O4] = ([H3O+][C2O42-]•1.85) / K1
Substituting numerical values for [H3O+] and K1 gives
[H2C2O4] = (1.85 X 10-4[C2O42-]) / 5.60 X 10-2 = 3.30 X 10-3 [C2O42-]
Substituting these expressions for [HC2O4-] and [H2C2O4] into the mass-balance equation gives
[Ca2+] = 1.85[C2O42-]+3.30 X 10-3 [C2O42-] + [C2O42-] = 2.85 [C2O42-]
or
[C2O42-] = [Ca2+] / 2.85
Substituting into the first equilibrium-constant expression gives
([Ca2+][Ca2+]) / 2.85 = 1.7 X 10-9
[Ca2+] = solubility = sqrt (2.85 X 1.7 X 10-9) = 7.0 X 10-5 M
2)Solubility Calculations When the pH is Variable
i)Solving the system when the pH is unknown is considerably more difficult. Since the concentrations of [H3O+] and [OH-] are unknown, there would be two more unknowns. To solve this. Two additional expressions would be needed. To solve this system would be tedious and time-consuming.
10B-3) The Solubility of Precipitates in the Presence of Complexing Agents
i)The solubility of a precipitate may increase dramatically in the presence of reagents that form complexes with the anion or the cation of the precipitate.
1)Complex Formation with a Common Ion
i)Many precipitates react with the precipitating agent to form soluble complexes.
ii)Increases in solubility caused by large excesses of a common ion are not unusual.
2)Quantitative Treatment of the Effect of Complex Formation on Solubility
i)Solubility calculations for a precipitate in the presence of a complexing reagent are similar in principle to those discussed in the previous section.
10C) Separating Ions by pH Control: Sulfide Separations
1)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.
Example 10-7
Calcium sulfide is less soluble than thallium sulfide (I). Find the condition under which Cd2+ and Tl+ can, in theory, be separated quantitatively with H2S from a solution that is 0.1 M in each cation.
The constants for the two solubility equilibria are
CdS(s) Cd2+ + S2- [Cd2+][S2-] = 1 X 10-27
Tl2S(s) 2Tl+ + S2- [2Tl+][S2-] = 6 X 10-22
Since CdS precipitates at a lower [S2-] than does Tl2S, we first compute the sulfide ion concentration necessary for quantitative removal of Cd2+ from solution. To make such a calculation, we must first speciful what constitutes a quantitative removal. The decision is arbitrary and depends on the purpose of the separation. In this example, we shall consider a separation to be quantitative when all but 1 part in 1000 of the Cd2+ has been removed; that is, the concentration of the cation has been lowered to 1.00 X 10-4 M. Substituting this into the solubility-product expression gives
10-4 [S2-] = 1 X 10-27
[S2-] = 2 X 10-23
Thus, if we maintain the sulfide concentration at this level or greater, we may assume that quantitative removal of the cadmium will take place. Next, we compute the [S2-] needed to initiate precipitation of Tl2S from a 0.1 M solution. Precipitation will begin when the solubility product is just exceeded. Since the solution is 0.1 M in Tl+,
(0.1)2[S2-] = 6 X 10-22
[S2-] = 6 X 10-20
These two calculations show that quantitative precipitation of Cd2+ takes place if [S2-] is made greater than 1 X 10-23. No precipitation of Tl+ occurs, however, until [S2-] becomes greater than 6 X 10-20 M.
Substituting these two values for [S2-] into a previous equation permits the calculation of the [H3O+] range required for separation.
[H3O+]2 = (1.2 X 10-22) / (1 X 10-23) = 12
[H3O+] = 3.5
and
[H3O+]2 = (1.2 X 10-22) / (6 X 10-20) = 2.0 X 10-3
[H3O+] = 0.045
By maintaining [H3O+] between approximately 0.045 and 3.5 M, we can in theory separate CdS quantitatively from Tl2S.