From WolfWikis

Solubility

There are materials that can form homogeneous mixtures (solutions) in any ratio. Water and alcohol are a liquid example, but there are also solid examples as e.g. silver and gold that form a complete series of solid solutions

However many solutes like salts in aqueous solutions are only soluble to a certain limit. Their behavior can often be well described by means of a solubility product.

Activities

In thermodynamics a dimensionless quantity is used to describe the behavior of mixtures, it is known as the activity. Removing the dimensions from say a concentration is easy. You just divide two concentrations with the same units! In order to do that a standard concentration is agreed upon to divide everything by. This standard can be anything except zero, because dividing by zero is not allowed. (Notice that this type of standard value is very different from a natural zero point like 0K or 0 mol/l. In a sense it is the opposite!)

Often people take for the standard point o

1. p= 1 Atm,
2. all concentrations at 1 [mol/l]
3. T= whatever the temperature of interest is.

Actually any choice is acceptable -if applied consistently- and sometimes other choices are wise, but the choice must be clearly declared in any published work.

Dilute systems

Adopting the above definition we can say that the activity a of a dilute solution is its molarity divided by a molarity of 1. In other words a is numerically the same as the concentration but you drop the units. So a= [X]¡ (where ¡ is my personal notation for: drop units).

This also holds for dilute vapors (gases) where a= p¡. Partial pressure (drop units) is also called fugacity.

Neat solids and liquid

The above may seem a bit like a silly game, but activities are not always just concentrations without units. For pure solids and pure liquids (in general condensed phases) the activity is not related to any concentration but it is simply equal to one. This also holds for the solvent in the case of a dilute solution.

It is not easy to explain why, but think of two big buckets of soup. To one you add one more drop of soup, to the other you add one drop of poison. Which soup is changed most by the addition? It's not the same is it? Chemically (and thermodynamically) adding another drop of soup does not change anything. This is why we take activity to remain one in that case.

Other systems

How about not-so-dilute solutions? There the activity (of the solute, but even the solvent) is a complicated function of composition and this is way beyond CH 101. Unfortunately, examples of such systems are many, many, many... Shocking examples for anyone who wishes to be involved in life sciences or say, marine geochemistry etc are sea water, blood plasma and many aqueous solutions.... People who work with those need more thermo than this course covers. Students of life sciences do not always realize that.

Equilibrium

Each equilibrium can be described by the activities of the reactants, e.g. for the water equilibrium:

H₂O <~>^[1] H⁺ + OH^-

The equilibrium is described by the constant

K = a(H⁺)a(OH^-) / a(H₂O)

The expression shows that if you change one activity, the other must change too, because K remains constant.

We always write the species on the right as the numerator.

So, if we had written the reaction the other way around:

H⁺ + OH^- <~> H₂O

K would be upside down (the reciprocal). However, for tabulation purposes the first notation is chosen as convention. Thus in a table one would see

K_w = 10^-14 = a(H⁺)a(OH^-)

(The denominator a(H₂O)=1)

In good tables it would also say that the standard state was taken to be

1. p=1 Atm
2. all concentrations 1M^[2]
3. T is e.g. 25°C

but many tabulators are sloppy...

The K_w value corresponds to the product of the concentrations of [H⁺] and [OH^-] (albeit with dropped units: [H⁺]¡ and [OH^-]¡). In pure water ("18 megohm water", after its low conductivity) the value would be 10^-7 for both.

Notice that if one of the concentration is raised, e.g. by adding a bit of acid, the other one must fall. Otherwise K_w, the product of the two would not be constant.

Solubility products

The same arguments hold for the solubility equilibrium. The values are always tabulated for the equilibrium with the solid precipitate is written on the left so that the denominator is one. Of course your actual experiment may go the other way.

Take e.g

AgCl <~> Ag⁺ + Cl^-

K_sp = [Ag⁺]¡[Cl^-]¡=1.8 x 10⁻¹⁰ (taking the above standard state).

Notice that we can rewrite this by dividing by [Ag⁺]¡ as:

[Cl^-]¡=1.8 x 10⁻¹⁰/[Ag⁺]¡

If we take the square root of K_sp we get the concentrations (units dropped) of both species if pure water is equilibrated with an excess of solid AgCl.^[3]

This represents the maximum solubility of the salt.

If we were to add a salt like NaCl this would drive much of the AgCl out of solution.

We can see that graphically. If we plot [Cl^-]¡ against [Ag⁺]¡ the set of points at equilibrium is a hyperbolic curve, because the expression [Cl^-]¡=1.8 x 10⁻¹⁰/[Ag⁺]¡ looks like the function f(x)= 1/x. All combinations of concentrations below the curve (the blue zone) are allowed unsaturated solutions, beyond the curve the product of the concentrations is too large and precipitation must result.

If we start by dissolving AgCl in pure water the concentrations of the two ions are always equal because of our balanced equation. This is represented by the diagonal line through the origin. X marks the highest saturated concentration we can get, the solubility limit.

Adding extra chloride (e.g. by dissolving NaCl) brings us to point X+. Precipitation must take place and both ions leave the solution in equal amounts along the diagonal line leading to Y+. This point represents a new equilibrium with far more chloride than silver in solution. Most of the chloride however is what added extra. Subtracting that bring us to point Y that shows us how little AgCl we still have in solution.

Exercises

Some K_sp values (same standard state)

AgBr 5.2 x 10⁻¹³

AgI 8.3 x 10⁻¹⁷

Exercise 1

1. How much NaCl_(s) (by weight) can be added to a 100 ml solution of 3mM AgNO₃ before AgCl starts precipitating?
2. How much NaBr_(s) (by weight) can be added to a 10 l solution of .2mM AgNO₃ before AgBr starts precipitating?
3. How much NaI_(s) (by weight) can be added to a 300 ml solution of 1μM AgNO₃ before AgI starts precipitating?

Exercise 2

1. What is the mass of silver dissolved in 3000 l of a solution saturated in AgCl in pure water?
2. What is the mass of silver dissolved in 3000 l of a solution saturated in AgI in pure water?
3. Could you advise a mining company what to do with solutions containing low concentrations of silver?

Exercise 3

1. How many moles of nitric acid do we need to add to a 10 liter drum of 18 megohm water to raise the [H⁺] to 10^-5 [mol/l]? What is the resulting [OH^-]? (Hint: First ignore the contribution [H⁺]_native from the water itself. Then do the calculation. Then argue that the assumption was valid because [H⁺]_native is equal to [OH^-]). How many times larger is [H⁺] compared to [H⁺]_native?

References

1. ↑ Unfortunately, this wiki does not allow me to write proper equilibrium arrows, so I will use <~> instead.
2. ↑ Notice how unrealistic this choice is: we can never have 1M for both H⁺ and OH^-, but that does not matter as long as we stick to our convention
3. ↑ In this case the reaction equation tells us that there are equal amounts in solution so that [Ag⁺]=[Cl^-]=x. This means that K_sp =x². Thus x= √K_sp