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Chapter 7 Activity and the Systematic Treatment of Equilibrium. Water Binding to Ions. Equilibrium Constants with Concentrations and Activities. Fe 3+ + SCN - = Fe(SCN) 2+. Pale yellow. Colorless. Red. 7-1 The Effect of Ionic Strength on Solubility of Salts.

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Equilibrium Constants

with Concentrations and Activities

Fe3+ + SCN- = Fe(SCN)2+

Pale yellow




7-1 The Effect of Ionic Strength on Solubility of Salts

CaSO4(s) = Ca2+ + SO42- Ksp = 2.4 X 10-5 (8-2)

When we add salt to a solution, we say that the ionic strength of the solution increases.

We call this region the ionic atmosphere (Figure 8-2).

The greater the ionic strength of a solution, the higher the charge in the ionic atmosphere. Each ion-plus-atmosphere contains less net charge and there is less attraction between any particular cation and anion.


The effect is to reduce their tendency to come together, thereby increasing the solubility of CaSO4.




Potassium hydrogen tartrate


Box7-1 Salts with Ions of Charge ≥ |2| Do Not Fully Dissociate

Ion pair formation constant: Mn+(aq) + Lm-(aq) = Mn+Lm-(aq)

Ion pair


7-2 Activity Coefficients

To account For the effect of ionic strength, concentrations are replaced by activities:

Activity of C:

The activity of species C is its concentration multiplied by its activity coefficients.

General form of equilibrium constant:


The ionic atmosphere model leads to the extended Debye-Hückel equation, relating activity coefficients to ionic strength:

Extended Debye-Hückel equation:

To find activity coefficients for ionic strengths above 0.1 M(up to molalities of 2-6 mol/kg for many salts), more complicated Pitzer equations are usually used.


Effect of ionic Strength, Ion Charge, and Ion Size of the Activity Coefficient

As ionic strength increases, the activity coefficient decreases (Figure 8-4). The activity coefficient (γ) approaches unity as the ionic strength (µ) approaches 0.

As the magnitude of the charge of the ion increases, the departure of its activity coefficient from unity increases. Activity corrections are more important for ions with a charge of ±3 than for ions with a charge of ±1 (Figure 8-4).

The smaller the ion size (α), the more important activity effects become.


In linear interpolation, we assume that values between two entries of a table lie on a straight line.


That is, the activity of a neutral molecule will be assumed to be equal to its concentration.

AH2 = PH2γH2

The activity of a gas is called its fugacity, and the activity coefficient is called the fugacity coefficient. Therefore, for all gases, we will setA = P(bar).


7-3 pH Revisited

pH = -logAH+ = -log[H+]γH+ (8-8)

When we measure pH with a pH meter, we are measuring the negative logarithm of the hydrogen ion activity, not its concentration.


However, the concentration of H+ in 0.10 M KCl (1.26 X 10-7 M) is 26% greater than the concentration of H+ in pure water (1.00 X 10-7 M).


7-4 Systematic Treatment of Equilibrium

The systematic treatment of equilibrium is a way to deal with all types of chemical equilibria, regardless of their complexity.

The charge balance is an algebraic statement of electroneutrality: The sum of the positive charges in solution equals the sum of the negative charges in solution.

[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-] (8-11)

The coefficient in front of each species always equals the magnitude of the charge on the ion.

[H+] = 5.1 X 10-12 M [H2PO4-] = 1.3 X 10-6 M

[K+] = 0.0550 M [HPO42-] = 0.0220 M

[OH-] = 0.0020 M [PO43-] = 0.0030 M

[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-]

5.1 X 10-12 + 0.0550 = 0.0020 + 1.3 X 10-6 + 2(0.0220) + 3(0.0030)

0.0550 M = 0.0550 M


Charge balance:

Where [C] is the concentration of a cation, n is the charge of the cation, [A] is the concentration of an anion, and m is the magnitude of the charge of the anion.


The mass balance, also called the material balance, is a statement of the conservation of matter. The mass balance states that the quantity of all species in a solution containing a particular atom (or group of atoms) must equal the amount of that atom (or group) delivered to the solution.

CH3CO2H = CH3CO2- + H+

Acetic acid Acetate

Mass balance for 0.050M = [CH3CO2H] + [CH3CO2-]

Acetic acid in water: What we put into Undissociated Dissociated

the solution product product

0.0250 M = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]



La(IO3)3(s) = La3+ + 3IO3-


[IO3-] = 3[La3+]

[Total iodate] = 3[total lanthanum]

[IO3-] + [LaIO32+] = 3{[La3+] + [LaIO32+] + [LaOH2+]}


Systematic Treatment of Equilibrium

Step 1 Write the pertinent reactions.

Step 2 Write the charge balance equation.

Step 3 Write mass balance equations. There may be more than one.

Step 4 Write the equilibrium constant expression for each chemical reaction. This step is the only one in which activity coefficients appear.

Step 5 Count the equations and unknowns. There should be as many equations as unknowns (chemical species). If not, you must either find more equilibria or fix some concentrations at known values.

Step 6 By hook or by crook, solve for all the unknowns.


Box 7-2 Calcium Carbonate Mass Balance in Rivers

CaCO3(s) + CO2(aq) + H2O = Ca2+ + 2HCO3- (A)

Calcite Bicarbonate


7-5 Applying the Systematic Treatment of Equilibrium

A simple Example: Ionization of Water

Step 1 Pertinent reactions. The only one is Reaction 8-13.

Step 2 Charge balance. The only ions are H+ and OH-, so the charge balance is

[H+] = [OH-] (8-14)

Step 3 Mass balance. Reaction 8-13 creates one H+ for each OH-. The mass balance is simply [H+] = [OH-], which is the same as the charge balance for this system.

Step 4 Equilibrium constant expression.

KW = [H+]γH+[OH-]γOH- = 1.0 X 10-14 (8-15)

This is the only step in which activity coefficients enter the problem.


Step 5 count equations and unknowns. We have two equations, 8-14 and 8-15, and two unknowns, [H+] and [OH-].

Step 6 Solve.

[H+]γH+[OH-]γOH- = 1.0 X 10-14

[H+] · 1 · [H+] · 1 = 1.0 X 10-14

[H+] = 1.0 X 10-7 M

pH = -logAH+ = -log[H+]γH+ = -log(1.0 X 10-7)(1) = 7.00


Solubility of Calcium Sulfate

Step 1 Pertinent reactions. Even in such a simple system, there are quite a few reactions:

There is no way you can be expected to come up with all of these reactions, so you will be given help with this step.

Step 2 Charge balance. Equating positive and negative charges gives

2[Ca2+] + [CaOH+] + [H+] = 2[SO42-] + [HSO4-] + [OH-] (8-21)


Step 3 Mass balance. Reaction 8-16 produces 1mole of sulfate for each mole of calcium. No matter what happens to these ions, the total concentration of all species with sulfate must equal the total concentration of all species with calcium:

[Total calcium] = [total sulfate]

[Ca2+] + [CaSO4(aq)] + [CaOH+] = [SO42-] + [HSO4-] + [CaSO4(aq)] (8-22)

Step 4 Equilibrium constant expressions. There is one for each chemical reaction.

Step 4 is the only one where activity coefficients come in.


Step 5 Count equations and unknowns. There are seven equations (8-21 through 8-27) and seven unknowns: [Ca2+], [SO42-], [CaSO4(aq)], [CaOH+], [HSO4-], [H+], and [OH-]. In principle, we have all the information necessary to solve the problem.

Step 6 Solve. Well, this is not easy! We don’t know the ionic strength, so we cannot evaluate activity coefficients. Also, where do we start when there are seven unknowns?


Solubility of Magnesium Hydroxide

Step 1 Pertinent reactions are listed above.

Step 2 Charge balance: 2[Mg2+] + [MgOH+] + [H+] = [OH-]

Step 3 Mass balance. This is a little tricky. From Reaction 8-30, we could say that the concentrations of all species containing OH- equal two times the concentrations of all magnesium species. However, Reaction 8-32 also creates 1 OH- for each H+. The mass balance accounts for both sources of OH-:

[OH-] + [MgOH+] = 2{[Mg2+] + [MgOH+]} + [H+] (8-34)

Species contatining OH- Species containing Mg2+

After all this work, Equation 8-34 is equivalent to Equation 8-33.


Step 4 Equilibrium constant expressions are in Equations 8-30 through 8-32.

Step 5 Count equations and unknowns. We have four equations (8-30 to 8-33) and four unknowns: [Mg2+], [MgOH+], [H+], and [OH-].

Step 6 Solve.

2[Mg2+]+ [MgOH+] = [OH-] (8-35)

2[Mg2+] + K1[Mg2+][OH-] = [OH-]