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# Activity

## Introduction

In all of the equilibria that we've discussed so far, we've assumed that the equilibria occur in ideal solutions. Solutions that have a total ion concentration greater than approximately 0.001 M cannot really be treated as ideal solutions. When ion concentrations become high enough so that ions interact, the interactions affect solution equilibria. Equilibria are affected not only when the concentrations of the ions involved in an equilibrium become high, but also when the concentrations of spectator ions become high. Spectator ions are not involved directly in equilibria, but they affect the environment of the ions that are part of an equilibrium.

To use an analogy, let's say that you and a date plan to meet five other couples at a skating rink. When you and your date skate onto the rink you see the other couples and you could say that the formal concentration was six couples per rink. Everyone skates around in pairs most of the time, but on average one couple is "dissociated" at any given time. (The topic of why couples dissociate and recombine is best left for a physical chemistry class.) You could describe the equilibrium concentrations as five undissociated couples and one dissociated couple per rink and even determine an equilbrium constant for couple dissociation:

```           (1)(1)
Kc = ------ = 0.20
(5)
The c subscript is used to indicate couple dissociation.
```

Now, what happens if one hundred little kids skate onto the rink? The little kids are rather hyperactive and won't skate as couples themselves or with the adult couples, but they do tend to cluster around any individual adult skater. The rate at which couples dissociate is the same as before (for the usual reasons), but as soon as they dissociate the individual skaters now get surrounded by little kids. The little kids shield individual skaters from other individual skaters, making it more difficult for two skaters to recombine to form a couple. Now you might look around and see that there are four undissociated and two dissociated couples per rink, and that the individual skaters tend to be surrounded by little kids.

```            (2)(2)
Kc' = ------ = 1
(4)
We put a ' on the equilibrium constant to specify new conditions.
```

The presence of little kids has drastically affected couple equilibrium! You can imagine that if there were 500 little kids on the skating rink, the shielding between isolated adult skaters would be even greater. Similarly, if there were only ten little kids scattered around the rink, they probably wouldn't affect the equilibrium concentrations of the undissociated and dissociated couples very much. We could call the situation when there were no or very few little kids on the skating rink the dilute or ideal case.

In the above analogy, the undissociated and dissociated couples represented solutes in an equilibrium and the little kids represented spectator ions. Note that for the little kids to affect the equilibrium, there had to be an attraction between the little kids and the individual adult skaters.

Ions in solution have electrostatic interactions with other ions. Neutral solutes do not have such interactions. When the concentrations of ions in a solution is greater than approximately 0.001 M, a shielding effect occurs around ions similar to the little kids around individual adults. Cations tend to be surrounded by nearby anions and anions tend to be surrounded by nearby cations. This shielding effect becomes significant at ion concentrations of 0.01 M and greater. Doubly or triply charged ions "charge up" a solution more than singly charged ions, so we need a standard way to talk about charge concentration. We describe the charge concentration of a solution by the ionic strength, which is calculated from the following expression: where
zi is the charge on an ion.
Ci is the formal concentration of each ion.
Note that the charge is squared so that positive and negative charges do not cancel. The factor of 1/2 is present because for every positive charge there is a negative charge.

Now, we could tabulate equilibrium constants for acid dissociation equilibria at different ionic strengths, but that would be a lot of work and create huge tables. Ka values are usually tabulated for the ideal case of zero ionic strength (and 25 oC). Appendix II of Rubinson & Rubinson specifies if a Ka value was measured at other conditions.

We call the "effective concentration" of ions in solution activity and we relate it to the ideal concentration by the activity coefficient, i: These activity coefficients serve as correction factors so that equilibrium constants measured in an ideal solution can be used to predict equilibria in non-ideal solutions.

Activity coefficients are unitless numbers that are calculated from the Debye-Hückel equation: where
µ is the ionic strength of the solution.
zi is the charge on the ion. i is the effective diameter of the hydrated ion in nanometers (nm), wich can be looked up in tables (see below).
The constants in this equation are for aqueous solution at 25 oC.

Since the Debye-Hückel equation contains the charge and hydrated diameter of a specific ion, each ion in a solution could have a different calculated activity coefficient. (Experimental measurements give only an average activity coefficient for the ions that are present.) The ionic strength is a description of the solution and the same value of µ is used for calculating the activity coefficients of each ion in solution.

Example: What is the pH of 0.0200 M benzoic acid in water and in a solution of 0.10 M CaCl2?

Since benzoic acid is a weak acid, it will exist mostly as C6H5COOH in water, so the concentration of ions will be very small. The ionic strength will be very close to zero and activity coefficients will be one for ionic solutes.  [H+]2 + 6.28x10-5[H+] - 1.26x10-6 = 0

Solve using the quadratic equation:

[H+] = 1.09x10-3

pH = 2.96

Now let's repeat the calculation for benzoic acid in the CaCl2 solution. Since the ionic strength is high we must use activities rather than concentrations.

C6H5COOH (aq) H+(aq) + C6H5COO-(aq)  We'll need the ionic strength, which we calculate using the charge and concentration of Ca2+ and Cl-. As in the case of the water solution, the ion concentrations from the acid dissociation is small and can be neglected. With the ionic strength and the values (see below), we can calculate the activity coefficients for H+ and C6H5COO-: H+ = 0.9 nm C6H5COO- = 0.6 nm   [H+]2 + 1.09x10-4[H+] - 2.19x10-6 = 0

Solve using the quadratic equation:

[H+] = 1.43x10-3

pH = 2.84

We see that the shielding effect of the spectator ions increases the amount of acid dissociation by approximately 30%. the H+ and benzoate ions are stabilized in solution by nearby spectator ions, just like individual skaters were shielded by little kids from recombining into couples.

Effective hydrated diameters of ions in aqueous solution (25 oC).
cationsanions i in nm
Rb+, Cs+, NH4+, Tl+, Ag+   0.25
K+ Cl-, Br-, I-, NO3-, HCOO- 0.3
OH-, F-, HS-, ClO4-, MnO4- 0.35
Na+ HCO3-, H2PO4-, HSO3-, CH3COO- 0.4-0.45
Hg22+ HPO42-, SO42-, CrO42- 0.40
Pb2+ CO32-, SO32- 0.45
Sr2+, Ba2+, Cd2+, Hg2+ S2- 0.5
Li+, Ca2+, Cu2+, Zn2+
Sn2+, Mn2+, Fe2+, Ni2+, Co2+
Phthalate2-, C6H5COO- 0.6
Mg2+, Be2+   0.8
H+, Al3+, Cr3+, Fe3+, La3+   0.9
Th4+, Zr4+, Ce4+, Sn4+   1.1

From J. Kielland, J. Am. Chem. Soc. 1937, 59, 1675.

To summarize: In dilute solutions ions are surrounded only by water molecules and the solution behaves ideally, e.g., equilibrium constants give good predictions of equilibrium concentrations. When ionic strength is larger than approximately 0.01 M, the shielding of ions affects equilibria significantly. To use equilibrium constants that are tabulated for ideal solutions, we multiply concentrations by correction factors called activity coefficients.

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