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+ + SO42SO42- + H3O+  HSO4- + H2O
2H2O  H3O+ + OHi) 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+, and OH-.
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+ + SO42SO42- + H3O+  HSO4- + H2O
2H2O  H3O+ + OHBecause 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 massbalance 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+ + OH2H2O  H3O+ + OHBecause the only source of Ag+, Br -, and Ag(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 PO43mol 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 chargebalance 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.
(http://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/html/final-26.html)
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+ + 2OH2H2O  H3O+ + OHStep 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 massbalance 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 10-4 is 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+ + C2O42Oxalate ions react with water to form HC2O4- and H2C2O4. Thus, there three
other equilibria present in this solution.
H2C2O4 + H2O  H3O+ + HC2O4HC2O4- + H2O  H3O+ + C2O422H2O  H3O+ + OHStep 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 CaC2O4 is 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 equilibriumconstant 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 Cd 2+ 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.