Harris: Quantitative Chemical Analysis, Eight Edition

CHAPTER 07: ACTIVITY AND THE SYSTEMATIC TREATMENT OF EQUILIBRIUM

Ionic and hydrated radii of several ions

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

CaSO 4 (s) = Ca 2+ + SO 4 2 K sp = 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 CaSO 4 .

Thiocyanate Phenol Phenolate Potassium hydrogen tartrate

Ionic strength

, µ, is a measure of the total concentration of ions in solution.

Ionic strength:

Box7-1 Salts with Ions of Charge

≥

|2| Do Not Fully Dissociate

Ion pair formation constant:

M

n

+ (

aq

) + L

m

(

aq

) = M

n

+ L

m

(

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:

K sp = A Ca 2+ A SO 4 2 = [Ca 2+ ] γ Ca 2+ [SO 4 2 ] γ SO 4 2-

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.

In

linear interpolation

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

Interpolation:

7-3 pH Revisited

pH

= -log A H + = -log[H + ] γ H + (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 + greater than the concentration of H in 0.10 M KCl (1.26 X + in pure water (1.00 X 10 -7 10 -7 M) is 26% 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 ] + [H 2 PO 4 ] + 2[HPO 4 2 ] + 3[PO 4 3 ] (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 [H 2 PO 4 ] = 1.3 X 10 -6 [K + ] = 0.0550 M [HPO 4 2 ] = 0.0220 M [OH ] = 0.0020 M [PO 4 3 ] = 0.0030 M M 5.1 X [H + ] + [K + ] = [OH ] + [H 2 PO 4 ] + 2[HPO 4 2 ] + 3[PO 4 3 ] 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.

CH 3 CO 2 H = CH 3 CO 2 + H Acetic acid Acetate +

Mass balance for Acetic acid in water:

0.050M = [CH 3 CO 2 H] + [CH 3 CO 2 ] What we put into Undissociated Dissociated the solution product product 0.0250 M = [H 3 PO 4 ] + [H 2 PO 4 ] + [HPO 4 2 ] + [PO 4 3 ]

La(IO 3 ) 3 (

s K sp

) = La 3+ + 3IO 3 Iodate [IO 3 ] = 3[La 3+ ] [Total iodate] = 3[total lanthanum] [IO 3 ] + [LaIO 3 2+ ] = 3{[La 3+ ] + [LaIO 3 2+ ] + [LaOH 2+ ]}

Box 7-2 Calcium Carbonate Mass Balance in Rivers

CaCO 3 (

s

) + CO 2 (

aq

) + H 2 O = Ca 2+ + 2HCO 3 Calcite Bicarbonate (A)

7-5 Applying the Systematic Treatment of Equilibrium A simple Example: Ionization of Water

Step 1 Step 2

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

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.

K W = [H + ] γ H + [OH ] γ OH = 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 + ] γ H + [OH ] γ OH [H + ] · 1 · [H + ] · = 1.0 X 1 = 1.0 X 10 [H + ] = 1.0 X 10 -14 10 -14 -7 M pH = -log A H + = -log[H + ] γ H + = -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[Ca 2+ ] + [CaOH + ] + [H + ] = 2[SO 4 2 ] + [HSO 4 ] + [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] [Ca 2+ ] + [CaSO 4 (

aq

)] + [CaOH + ] = [SO 4 2 ] + [HSO 4 ] + [CaSO 4 (

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: [Ca 2+ ], [SO 4 2 ], [CaSO 4 (

aq

)], [CaOH + ], [HSO 4 ], [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[Mg 2+ ] + [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{[Mg 2+ ] + [MgOH + ]} + [H + ] (8-34) Species contatining OH Species containing Mg 2+ 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: [Mg 2+ ], [MgOH + ], [H + ], and [OH ].

Step 6

Solve.

2[Mg 2+ ] + [MgOH + ] = [OH ] (8-35) 2[Mg 2+ ] + K 1 [Mg 2+ ][OH ] = [OH ]