WEEK 7
EFFECTS OF ELECTROLYTES ON
CHEMICAL EQUILIBRIA, ACTIVITY AND
ACTIVITY COEFFICIENTS,
EQUILIBRIUM IN COMPLEX SYSTEMS:
SOLVING MULTIPLE EQUILIBRIUM
PROBLEMS
Electrolytes
› Most solutes that are discussed in Analytical
Chemistry are electrolytes.
› Electrolytes are formed from solutes when
dissolved in water (or other solvents) and
produce solutions that conduct electricity.
› Strong electrolytes ionize completely in solvent.
› Weak electrolytes ionize only partially.
Classifications of Electrolytes
Strong
1. Inorganic acids such as HNO3, HClO4, H2SO4*,
HCl, HI, HBr, HClO3, HBrO3
2. Alkali and alkaline-earth hydroxides
3. Most salts
H2SO4 is completely dissociated into HSO4- and H3O+ ions
and for this reason is classified as a strong electrolyte.
Note, however, that the HSO4- ion is a weak electrolyte
and is only partially dissociated into SO42- and H3O+
Weak
1. Many inorganic acids, including H2CO3, H3BO3,
H3PO4, H2S, H2SO3
2. Most organic acids
3. Ammonia and most organic bases
4. Halides, cyanides, and thiocyanates of Hg, Zn,
and Cd
Chemical Equilibrium
› Many reactions used in analytical chemistry
never result in the complete conversion of
reactants to products.
› Instead, they proceed to a state of chemical
equilibrium in which the ratio of the
concentration of reactants and products is
constant.
› Equilibrium-constant expressions are algebraic
equations that describe the concentration
relationships among reactants and products at
equilibrium
Equilibria and Equilibrium Constants in Analytical
Chemistry
Type of
Equilibrium
Name and
Symbol
Typical
Example
Dissociation of
water
Ion Product
Constant, Kw
Heterogeneous
equilibrium
between a
slightly soluble
substance and
its ions in a
saturated
solution
Dissociation of
a weak acid or
base
Solubility
Product, Ksp
2H2O =
H3O+ +
OHBaSO4(s) =
Ba2+ +
SO42-
Dissociation
Constant,
Ka or Kb
Formation of a
complex ion
Formation
constant, B4
Oxidation or
reduction
equilibrium
Kredox
Distribution
equilibrium for
a solute
between
immiscible
solvents
Kd
EquilibriumConstant
Expression
Kw = [H3O+][OH-]
Ksp = [Ba2+][SO42-]
CH3COOH
+ H2O =
H3O+ +
CH3COO-
Ka =
[H3O+][CH3COO-]
/[CH3COOH]
CH3COO- +
H2O = OH+
CH3COOH
Ni2+ +
4CN- =
Ni(CN)42MnO4- +
5Fe2+ +
8H+ =
Mn2+ +
5Fe3+ +
4H2O
I2(aq) =
I2(org)
Kb = [OH-]
[CH3COOH]/[CH3
COO-]
B4 = [Ni(CN)42-]
/[Ni2+][4CN-]
Kredox =
[Mn2+][Fe3+]5
/[MnO4-][Fe2+]5
[H+]8
Kd = [I2](org)/[I2](aq)
❒ ION PRODUCT CONSTANT OF WATER
The self-ionization of water (the process in which
water ionizes to hydronium ions and hydroxide ions)
occurs to a very limited extent. When two molecules
of water collide, there can be a transfer of a
hydrogen ion from one molecule to the other. The
products are a positively charged hydronium ion
and a negatively charged hydroxide ion.
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH- (aq)
We often use the simplified form of the reaction:
H2O(l) ⇌ H+(aq) + OH- (aq)
The equilibrium constant for the self-ionization of
water is referred to as the ion-product for water and
is given the symbol Kw
› The general equilibrium constant for such
processes can be written as:
Kc = [My+]x[Ax-]y
The ion-product of water is the mathematical product o
the concentration of hydrogen ions and hydroxide ions.
Note that H2O is not included in the ion-product
Kw = [H+] + [OH-] = 1.0 x 10 -14
Example of Solubility Product Expression
1) Generate the solubility product expression for
CuBr.
CuBr ⇌ Cu+ + Br Ksp = [Cu+][Br -]
2) Generate the solubility product expression for
PbCl2.
PbCl2 ⇌ Pb2+ + 2Cl Ksp = [Pb2+][Cl -]2
In pure water, the concentrations of hydrogen and
hydroxide ions are equal to one another. Pure water
on any other aqueous solution in which this ratio
holds is said to be neutral. To find the molarity of
each ion, the square root of Kw is taken.
❒ Calculating for Ksp from solubility data
Example 1: Calculate the solubility product constant
for lead (II) chloride, if 50.0 mL of a saturated
solution of lead (II) chloride was found to contain
0.2207 g of lead (II) chloride dissolved in it.
[H+] = [OH-] = 1.0 x 10 -7
Solution:
First, write the equation for the dissolving of lead (II)
chloride and the equilibrium expression for the
dissolving process.
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
Ksp = [Pb2+] [Cl-]2
expression because it is a pure liquid. The value Kw
is very small, in accordance with a reaction that
favors the reactants. At 25oC, the experimentally
determined value of Kw in pure water is 1.0 x 10 -14.
❒ Calculation using Kw
Example: Calculate the hydronium and hydroxide
ion concentrations of pure water at 25oC (Kw = 1.00
x 10 -14).
Solution:
Because OH- and H3O+ are formed only from the
dissociation of water, their concentration must be
equal.
[H3O+] = [OH-]
Second, convert the amount of dissolved lead (II)
chloride into moles per litre (M).
We substitute this equality into:
Kw = [H3O+]2 = [OH-]2
Or
+
2
[H3O ] = [OH-]2 = √Kw
Third, create an "ICE" table.
Thus at
25oC
:
[H3O+]2 = [OH-]2 = √ 1.00 x 10 -14
= 1.00 x 10 -7 M
❒ SOLUBILITY PRODUCT CONSTANT
› Solubility product constants are used to
describe saturated solutions of ionic compounds
of relatively low solubility. A saturated solution
is in a state of dynamic equilibrium between the
dissolved, dissociated, ionic compound and the
undissolved solid.
MPbCl2 = (0.2207 g PbCl2)(1/50.0 mL solution)(1000
mL/1 L)(1mol PbCl2/278.1 g PbCl2)
= 0.0159 M PbCl2
Initial
Concentration
Change in
Concentration
Equilibrium
Concentration
PbCl2 (s)
All solid
Pb2+ (aq)
0
Cl2- (aq)
0
- 0.0159 M
(dissolves)
Less solid
+ 0.0159 M
+ 0.0318 M
0.0159 M
0.0318M
Fourth, substitute the equilibrium concentrations
into the equilibrium expression and solve for Ksp.
Ksp = [0.0159][0.0318]2
= 1.61 x 10-5 M
❒ Calculating the Solubility of an Ionic Compound
in Pure Water from its Ksp
Example 2: Estimate the solubility of Ag2CrO4 in
pure water if the solubility product constant for
silver chromate is 1.1 x 10-12.
Answers
Solution:
Write the equation and the equilibrium expression.
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-]
❒ EQUILIBRIUM CONSTANTS
› They are not really “constants”.
› They can change with temperature (or other
stress such as pressure, total concentration of
reactant or product).
o Le Chatelier’s principle: states that the
position of chemical equilibrium always
shifts in a direction that tends to relieve
the effect of an applied stress.
› They can change based on the components of
the solution (even the ones that are not involved
in equilibrium).
Make an "ICE" chart.
Let "x" be the number of moles of silver chromate
that dissolve in every liter of solution (its solubility).
Initial
Concentration
Change in
Concentration
Equilibrium
Concentration
Ag2CrO4(s)
Ag+(aq)
CrO42-(aq)
All solid
0
0
-x
(dissolves)
Less solid
+2x
+x
2x
x
1. a) Ksp = 4.4 x 10-11
b) Ksp = 3.5 x 10-10
2. 1.2 x 10-3 M
3. 0.178 g
Equilibria and Equilibrium Constants in Analytical
Chemistry
Substitute the equilibrium amounts and the Ksp into
the equilibrium expression and solve for x.
Ksp = [Ag+]2[CrO42-]
1.1x10-12 = [2x]2[x]
1.1x10-12 = 4x3
1.1x10-12 = 4x3
4
4
-13
2.75 x 10 = x3
√ 2.75 x 10-13
X = 6.50 x 10-5 M
EXERCISES ON SOLUBILITY PRODUCT CONSTANT
1. Calculate the solubility product constant for
each of the following substances, given that the
molar concentrations of their saturated
solutions are as indicated:
o RaSO4 (6.6 x 10-6 M)
o Ce(IO3)3 (1.9 x 10-3 M)
2. Estimate the molar solubility of PbI2 if Ksp of PbI2
is 7.1 x 10-9.
3. What mass in (grams) of Ba(IO3)2 (MW=487
g/mol) can be dissolved in 0.5 L of water at 25oC?
The solubility product constant for Ba(IO3)2 at
25oC is 1.57 x 10 -9?
Ionic Solutions
› Equilibrium expressions for solutions are
calculated assuming the expression applies to an
ideal solution. Ideal solutions assume there is
no interaction between ions once the ions are in
solution.
› But ionic solutions are not ideal solutions
because ions solutions do interact with other
ions and solvents. The ionic strength of the
solution changes the behavior of the ions in the
solution.
Causes of non-ideal behaviour of Ionic Solutions
Ionic Atmosphere
Ions in solution travel near other ions. So, cations
will encounter cations and anions. Cations are more
attracted to anions than cations. Ions of opposite
charges shield the effective charge of an ion as
shown in Figure 1. The effect is to make anions and
cations less attracted to one another.
Hydrated Ions and Activity
› Ions are organized by a hydrated sheath of
water molecules.
› Hydrated radius is greater than the ionic radius.
› The activity of aqueous ions is related to the size
of the hydrated species.
› Small highly charged ions have bigger hydrated
radii.
Aqueous Ions
As the ionic strength of the solution increases,
the ionic atmosphere around an ion becomes
stronger. You can see this visually in Figure 2.
Ionic and Hydrated Radii of Several Ions
Activity coefficients for ions 25oC
5. Less crystallization/precipitation of ions in
solution
6. Therefore, more dissolution
7. Higher solubility with higher ionic strength
The Effect of Electrolytes on Solubility
Why non-ideal behavior occurs…
› The presence of electrolytes alters electrostatic
interactions
(ion-ion
and
ion-solvent
interactions)
› The extent of change in the reaction depends on
the concentration and to a larger extent the
charge of the ions.
› We quantify this value using the concept of ionic
strength.
On Chemical Equilibrium
Where [A] is the molarity of ion A, [B] of ion B and Z is the
charge of the ion.
Ionic Atmosphere and Ionic Strength
1. More ions in the solution (higher ionic
strength)
2. More ions diffuse into the ionic atmosphere
3. Less net charge for each ion in the solution
4. Less attraction between ions in solution
The Salt Effect
› The electrolyte effect, results from the
electrostatic attractive and repulsive forces that
exist between the ions of an electrolyte and the
ions involved in an equilibrium.
› The rate of a chemical reaction can be altered by
the presence of non-reacting or inert ionic
species in the solution. This effect is profound
when the reaction takes place between ions,
even at low concentrations. This influence of
charged species on the rate of the reaction is
referred to as salt effect.
Effect of Electrolytes on Chemical Equilibrium
› The magnitude of the electrolyte effect is highly
dependent on the charges of the participants in
the equilibrium. When only neutral species are
involved, the position of equilibrium is
essentially
independent
of
electrolyte
concentration.
› With ionic participants, the magnitude of the
electrolyte effect increases with charge
❒ IONIC STRENGTH
The ionic strength of a solution is a measure of the
total electrolyte concentration and is a function of
both the concentration and charge of ions in the
solution.
Ionic strength = µ = ½ ([A]Z2A + [B]Z2B + [C]Z2C + …)
Where [X] is the molarity of ions A, B, C, … and ZA, ZB, ZC
Recognize that the total ionic strength of a solution
is the sum of all ions in the solution.
How to Calculate Ionic Strength
Calculate the ionic strength of (a) 0.1M solution of
KNO3 and (b) 0.1M solution of Na2SO4
(a) For the KNO3 solution [K+] = [NO3-] = 0.1M,
the charge z = +1 and -1. We can substitute:
(b) For the Na2SO4 solution [Na+] = 0.2M and
[SO42-] = 0.1M, the charge z = +1 and -2,
respectively. We can substitute:
Ionic strength = µ = ½ ([A]Z2A + [B]Z2B + [C]Z2C + …)
(a) µ = ½ ([0.1]12 + [0.1](-1)2) = 0.1 M
(b) µ = ½ ([0.2]12 + [0.1](-2)2) = 0.3 M
What is the ionic strength of a solution that is 0.05M
in KNO3 and 0.1M in Na2SO4?
µ = 1/2 (0.05 M x 12 + 0.05 M x -12 + 0.2 M x 12 +
0.1 M x -22 )
= 0.35 M
Ionic Strength = Molar
Monovalent Species
Type
1:1
1:2
1:3
2:2
Concentration
Example
NaCl
Ba(NO3)2 or Na2(SO4)
Al(NO3)3 Na3(PO4)
Mg(SO4)
for
Ionic Strength
c
3c
6c
4c
Cautions in Calculating Ionic Strength
› You can only calculate ionic strength for very
dilute solutions.
› The higher the ionic concentration, the higher
the ionic charge, the more likely it is for ion
pairing (formation of complex ions) to occur
resulting to unreliable calculations
EXERCISES ON IONIC STRENGTH
Calculate the ionic strength of a solution that is:
1) 0.030 M in FeSO4
2) 0.30 M in FeCl3 and 0.20 M in FeCl2
Answers to exercises on ionic strength
1) 0.12 M
2) 2.4 M
❒ ACTIVITY AND ACTIVITY COEFFICIENTS
› Chemists use a term called “activity” to account
for the effects of electrolytes in chemical
equilibrium.
› The activity or effective concentration of
species “X” depends on the ionic strength of the
medium.
› To account for the effects of ionic strength,
concentration are replaced by activities in
equilibrium formulae.
Activity of an Ion is Effective Concentration
In order to take into the effects of electrolytes on
chemical equilibria, we use “activity” instead of
“concentration”.
The activity of an ion is defined as:
Activity of an ion – ai = [Xi] γi – activity coefficient
varies with ionic strength, charge, ion mobility.
Where [Ci] is the molarity of the ith ion and γi is the
ion’s activity coefficient (a dimensionless quantity).
γi depends on the ionic strength of the solution.
dependent interactions between ions in a
solution. Or simply, it is used to show how much
the solution deviates from the ideal.
› In some cases, the activity of a reactant is
essentially equal to its concentration, and we
can write the equilibrium constant in terms of
the concentration of the participating species.
› In the case of ionic equilibrium, activities and
concentrations can be substantially different.
Such equilibria are also affected by the
concentrations of the electrolytes in the
solution that may not participate directly in the
reaction.
Properties of Activities VS Concentration
For Dilute Solutions (IDEAL SOLUTIONS)
 When µ < 0.01 then γX => then ai => [i]
For NON-IDEAL SOLUTIONS
 When µ < 0.1 then γX < 1 then ai < [i]
 When µ < 0.1 then things get complicated
 In general, when we try to keep ionic
strength < 0.1 so that we can use
concentration instead of activities
Activity Coefficients for Ions 25oC
❒ THERMODYNAMIC SOLUBILITY PRODUCT
› The
thermodynamic
solubility
product
expression is defined by the equation:
𝑲𝒔𝒑
𝒏
𝑲𝒔𝒑 = 𝒂𝒎
𝑿 ∙ 𝒂𝒀
𝒏
𝒎 𝒏
𝒎
𝒏
= 𝜸𝒎
𝑿 𝜸𝒀 ∙ [𝐗] [𝐘] = 𝜸𝑿 𝜸𝒀 ∙ 𝑲′𝒔𝒑
› Here, Ksp is the concentration solubility product
constant and Ksp is the thermo-dynamic
equilibrium constant. The activity coefficients γX
and γY vary with ionic strength in such a way as
to keep Ksp numerically constant and
independent of ionic strength (in contrast to the
concentration constant Ksp).
❒ ACTIVITY AND ACTIVITY COEFFICIENT
› The activity of a chemical species is related to its
concentration by a factor called the activity
coefficient.
› The activity coefficient of an electrolyte
solution is used to factor in the concentration-
Properties of Activity Coefficients
› The activity coefficient of a species is a measure
of the effectiveness with which that species
influences an equilibrium in which it is a
participant. In very dilute solutions in which the
ionic strength is minimal, this effectiveness
becomes constant, and the activity coefficient
is unity. Under these circumstances, the activity
and molar concentration are identical. As the
ionic strength increases, however, an ion loses
some of its effectiveness, and its activity
coefficient decreases.
At moderate ionic strengths, γX < 1; as the
solution approaches infinite dilution, however,
γX → 1 and thus aX → [X] and Ksp → Ksp.
› In solutions that are not too concentrated, the
activity coefficient for a given species is
independent of the nature of the electrolyte and
dependent only on the ionic strength.
› For a given ionic strength, the activity coefficient
of an ion departs farther from unity as the
charge carried by the species increases (as
shown in the figure in the next slide)
Effect of Ionic Strength on Activity Coefficients
Assumptions of Debye-Hückel Theory
› The Debye-Hückel theory is based on three
assumptions of how ions act in solution:
› Electrolytes completely dissociate into ions
in solution.
› Solutions of Electrolytes are very dilute, on
the order of 0.01 M.
› Each ion is surrounded by ions of the
opposite charge, on average.
The Debye and Hückel Equation Calculates Activity
Coefficient
1923-Debye and Hückel derive an expression that
allows calculation of activity coefficients, γX for ions
from knowledge of charge, Z, the ionic strength of
the solution, µ and average diameter of hydrated
ion in nm, ∝ (at 25oC).
𝟎. 𝟓 𝒁𝟐𝒙 √𝝁
−𝒍𝒐𝒈 𝜸𝒙 =
𝟏 + 𝟑. 𝟑 𝜶𝒙 √𝝁
X for ion X = Activity Coefficient
α = diameters of ion in nanometers
2 = charge of the ion, X
õ = ionic strength of the solution
› At any given ionic strength, the activity
coefficients of ions of the same charge are
approximately equal.
› The activity coefficient of a given ion describes
its effective behavior in all equilibria in which it
participates.
For example, at a given ionic strength, a single
activity coefficient for cyanide ion describes the
influence of that species on any of the following
equilibria:
HCN + H2O ⇌ H3O+ + CNAg+ + CN- ⇌ AgCN
Ni2+ + 4CN- ⇌ Ni(CN)4 2-
The Debye and Hückel Equation Calculates Activity
Coefficient
• Over range of ionic strengths from 0 to 0.1M the
effect of each variable on activity coefficient is
as follows:
1. As the ionic strength, μ increases, the
activity coefficient(ϒ) decreases (<1).
2. As the ionic strength approaches 0, the
activity coefficient(ϒ) approaches 1.
3. As the charge of the ion increases, the
departure of away from ideal behavior
(when ϒ = 1). Highly charged ions cause
deviations.
4. The smaller the hydrated radius of an
ion, the more it causes non-ideality in
the solution (must use activities).
The Debye–Hückel Limiting Law
› When μ is less than 0.01, 1 + √µ ≈ 1 and then the
equation becomes –logγX = 0.51 Z2X √µ
› This equation is referred to as the Debye–Hückel
limiting law (DHLL). Thus, in solutions of very
low ionic strength, the DHLL can be used to
calculate approximate activity coefficients.
How to calculate for Activity Coefficient using
Debye-Hückel Equation
Calculate the activity coefficients for K+ and SO42- in
a 0.20 M solution of K2SO4. Assume αK+ = 0.3nm and
αSO4 = 4.0nm.
❒ EQUILIBRIUM CALCULATIONS IN COMPLEX
SYSTEMS:
SOLVING
MULTIPLE-EQUILIBRIUM
PROBLEMS BY A SYSTEMATIC METHOD
BaSO4(s) ⇌ Ba2+ + SO42SO42- + H3O+ ⇌ HSO4- + H2O
2H2O ⇌ H3O+ + OH[Ba2+] = [SO42-] + [HSO4-]
Three types of algebraic equations are used in
solving
multiple-equilibrium
problems:
(1)
equilibrium-constant expressions, (2) massbalance equations, and (3) a single charge-balance
equation.
How to Calculate for Activity Coefficient
Example 1: Calculate the activity coefficients for K+
and SO42- in a 0.20 M solution of K2SO4.
1
([𝐴]𝑍𝐴2 + [𝐵]𝑍𝐵2 + [𝐶]𝑍𝐶2 + … )
2
1
𝜇 = ([𝐾]12 + [𝑆𝑂4 ](−2)2 )
2
1
𝜇 = ([2 × 0.020]12 + [0.020](−2)2 ) = 0.060
2
𝜇=
−log γK+ =
0.5 12 √0.060
1+3.3 0.3 √0.060
− log γSO4 =
= 0.1005
0.51 (−2)2 √0.060
1+3.3 0.4 √0.060
= 0.463
γK+ = 10−0.101 = 0.79
γK+ = 10−0.463 = 0.344
Example 2: Use activities to calculate the H3O+
concentration in a 0.120M solution of HNO2 that is
also 0.050M NaCl (The thermodynamic equilibrium
constant Koa = 7.1 x 10-4)
1
([𝐴]𝑍𝐴2 + [𝐵]𝑍𝐵2 + [𝐶]𝑍𝐶2 + … )
2
1
𝜇 = ([0.050](1)2 + [0.050](−1)2 ) = 0.0500
2
We neglect the dissociation of HNO2.
𝜇=
We look up in the table the activity coefficients
based on this ionic strength H3O+ = 0.85 and HNO2 =
1.00 (rule 3)
Solving Multiple-Equilibrium Problems using a
Systematic Method
› To solve a multiple-equilibrium problem, we
must write as many independent equations as
there are chemical species in the system being
studied.
› For example, if our task is to compute the
solubility of barium sulfate in an acidic solution,
we must calculate the concentrations of all the
chemical species in the solution.
› In the given example, there are 5 species: [Ba2+],
[SO4 2-], [HSO4-], [H3O+] and [OH-]. To calculate, it
is necessary to create 5 independent algebraic
equations that can be solved simultaneously to
give the 5 concentrations.
❒ 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.
› Mass balance equations are a direct result of the
conservation
of
mass,
moles
and
concentration.
› To write mass-balance expressions, we must
know the properties and amounts of all solutes
in the solution, how the solution was prepared
and the equilibria in the solution.
Example of Mass Balance Expression
Write mass-balance expressions for a 0.0100 M
solution of HCl that is in equilibrium with an excess
of solid BaSO₄.
Solution:
From our general knowledge of the behavior of
aqueous solutions, we can write equations for three
equilibria that must be present in this solution.
BaSO4(s) ⇌ Ba2+ + SO42SO42- + H3O+ ⇌ HSO4- + H2O
2H2O ⇌ H3O+ + OHBecause the only source for the two sulfate species
is the dissolved BaSO₄, the barium ion concentration
must equal the total concentration of sulfatecontaining species, and a mass-balance equation
can be written that expresses this equality.
Thus, [Ba2+] = [SO42-] + [HSO4-]
According to the second reaction above, hydronium
ions in the solution are either free H3O+ or they react
with SO42- to form HSO4-. This can be expressed as:
[H3O+]total = [H3O+] + [HSO4-]
O+]
Where [H3 total is the hydronium concentration
from all sources and [H3O+] the free equilibrium
concentration of hydronium.
The protons that contribute to [H3O+]total have 2
sources: aqueous HCl and the dissociation of water.
Thus, [H3O+]total = [H3O+]HCl + [H3O+]H2O
And from the previous equation, we get:
[H3O+]total = [H3O+] + [HSO4-] = [H3O+]HCl + [H3O+]H2O
But [H3O+]HCl = HCl concentration and since the only
source of hydroxide is the dissociation of water,
[OH¯] is equal to the hydronium ion concentration
from the dissociation of water, we may write this as
[H3O+ ] H2O = OH- . By substituting these 2 quantities
into the above equation, we get:
[H3O+]total = [H3O+] + [HSO4-] = HClconc + [OH-]
Therefore, the mass-balance equation is:
[H3O+] + [HSO4- ] = 0.0100 + [OH-]
❒ CHARGE-BALANCE EQUATION
We know that electrolyte solutions are electrically
neutral even though they may contain many
millions of charged ions. Solutions are neutral
because the molar concentration of positive charge
in an electrolyte solution always equals the molar
concentration of negative charge. Thus,
no. mol/L positive charge = no. mol/L negative charge
This equation represents the charge-balance
condition and is called the charge-balance
equation.
n1[C1] + n2[C2] + ….. = m1[A1] + m2[A2] + …..
Examples on Charge Balance Equation
1) Write a charge-balance equation for sodium
chloride.
NaCl ⇌ Na+ + Cl2H2O ⇌ H3O+ + OH –
mol/L positive charge = [Na+] + [H3O+]
mol/L negative charge = [Cl-] + [OH-]
Therefore:
[Na+] + [H3O+ ] = [Cl-] + [OH-]
2) Write a charge-balance equation for BaSO4.
2[Ba2+] + [H3O] = 2[SO4 2- ] + [HSO4-] + [OH-]
BaSo4(s) ⇌ Ba2+ + SO42SO42- + H3O+ ⇌ HSO4- + H2O
2H2O ⇌ H3O+ + OHEXERCISES ON MASS BALANCE AND CHARGE
BALANCE EQUATION
1) Write the mass balance equation for a solution
that is:
a) 0.2 M in HF
b) 0.10 M in H3PO4
2) Write the charge-balance equation for above.
ANSWERS
Chemical reaction (a):
HF + H2O ⇌ H3O+ + FH2O ⇌ H3O+ + OH –
Answers:
1a) 0.2 M = [HF] + [F-]
2a) [H3O+] = [F-] + [OH-]
Chemical reaction (b):
H3PO4 + H2O ⇌ H2PO4 - + H3O+
H2PO4- + H2O ⇌ HPO42- + H3O+
HPO42- + H2O ⇌ PO43- + H3O+
H2O ⇌ H3O+ + OH–
Answers:
1b) 0.10 M = [H3PO4] + [H2PO4 -] + [HPO42-] +
[PO43-]
2b) [H3O+] = [H2PO4 -] + 2[HPO42-] + 3[PO43-] +
[OH-]
STEPS FOR SOLVING PROBLEMS INVOLVING
SEVERAL EQUILIBRIA
Step 1: Write a set of balanced chemical equations
for all pertinent equilibria.
Step 2: State in terms of equilibrium concentrations
what quantity is being sought.
Step 3: Write equilibrium-constant expressions for
all equilibria developed in step 1 and find numerical
values for the constants in tables of equilibrium
constants.
Step 4: Write mass-balance expressions for the
system
Step 5: If possible, write a charge-balance
expression for the system.
Step 6: Count the number of unknown
concentrations in the equations developed in steps
3, 4, and 5, and compare this number with the
number of independent equations. If the number of
equations is equal to the number of unknowns,
proceed to step 7. If the number is not, seek
additional equations. If enough equations cannot be
developed, try to eliminate unknowns by suitable
approximations regarding the concentration of one
or more of the unknowns. If such approximations
cannot be found, the problem cannot be solved.
Step 7: Make suitable approximations to simplify
the algebra.
Step 8: Solve the algebraic equations for the
equilibrium concentrations needed to give a
provisional answer as defined in step 2.
Step 9: Check the validity of the approximations
made in step 7 using the provisional concentrations
computed in step 8.
Systematic Method for Solving Multiple Equation
Problems
1. Write balanced chemical equations
2. Set up equation for unknown quantity
3. Write equilibrium constant expressions
4. Write mass balance expressions
5. Write charge balance equation
6. Count no. of equations and no. of unknowns.
Is no. equations ≥ no. unknowns? If no, stop.
The problem is unsolvable. If yes,
7. Make suitable approximations
8. Solve equations for unknown
9. Were approximations valid? No? Repeat
step 7. Yes? Problem solved!
Example for Calculating Solubilities by the
Systematic Method
Calculate the molar solubility of Mg(OH)₂ in water.
Step 1. Pertinent Equilibria
Two equilibria that need to be considered are:
Mg(OH)2 (s) ⇌ Mg2+ + 2OH2H2O ⇌ H3O+ + OHStep 2. Definition of the Unknown
Since 1 mol of Mg²⁺ is formed for each mole of
Mg(OH)₂ dissolved,
Solubility Mg(OH)2 = [Mg2+]
Step 3. Equilibrium-Constant Expressions
Ksp = [Mg2+][OH-]2 = 7.1 x 10-12
Kw = [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(OH)₂ and
H₂O. The hydroxide ion resulting from dissociation
of Mg(OH)₂ 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+]
Note that this equation is identical to Equation (C).
Often a mass-balance and a charge-balance
equation are the same.
Step 6. Number of Independent Equations and
Unknowns
We have developed three independent algebraic
equations (A),(B), and (C) and have three unknowns
([Mg²⁺], [OH¯], and [H₃O⁺]).Therefore, the problem
can be solved rigorously.
Step 7. Approximations
We can make approximations only in Equation (C).
Since the solubility-product constant for Mg(OH)₂ is
relatively large, the solution will be somewhat basic.
Therefore, it is reasonable to assume that
[H₃O⁺]<<[OH¯]. Equation (C) then simplifies to.
2[Mg2+] ≈ [OH-]
Step 8. Solution to Equations
Substitution of Equation (D) into Equation (A) gives
[Mg 2+ ](2[Mg 2+ ])2 = 7.1 × 10−12
7.1×10−12
[Mg 2+ ]3 =
= 1.78 × 10−12
4
[Mg 2+ ]3 = solubility = 1.21 × 10−4 or 1.2 × 10−4 M
Step 9. Check of Assumptions
Substitution into Equation (D) yields
[OH-] = 2 x 1.21 x 10-4 = 2.42 x 10-4
And from Equation (B)
H3 O+ =
1.00 × 10−14
= 4.1 × 10−4
2.42 × 10−4
Thus, our assumption that
certainly valid.
[H3O+] << [OH-] is
EXERCISE ON SOLVING PROBLEMS INVOLVING
SEVERAL EQUILIBRIA
Calculate the solubility of Fe(OH)3 in water.
Ksp = [Fe 3+] + [OH-] = 2 x 10 -39
Kw = [H3O+] + [OH-] = 1.00 x 10 -14
ANSWER
Solubility = 2 x 10 -18 M
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