Introduction
Previously, we learned how to predict solubility - that is, how to predict (qualitatively, at least) whether one substance will dissolve in another. Armed with our knowledge of equilibrium, we'll look at solubility again. This time, we will discuss the solubility of so-called "insoluble" salts as an equilibrium process, and we'll use our understanding of equilibrium to quantify exactly how much of these "insoluble" salts will dissolve.
Terminology review
Let's review some key terms related to solubility. A solution is a homogeneous mixture of substances. Homogeneous mixtures have the same concentration of each component throughout. A solvent is the component of a solution present in the greatest amount, and solutes are components present in lesser amounts that the solute.
A saturated solution contains the maximum amount of solute that it is normally possible to dissolve into the solution. You can tell that a solution is saturated (experimentally) by noting that there will be an amount of solid at the bottom (this solid wasn't able to dissolve in the solution) - even if the solution is stirred or left to stand for a long period.
Saturation and equilibrium
Saturated solutions are, from an equilibrium standpoint, solutions where solid "solute" is at equilibrium with solute in solution. For example, consider a saturated solution of silver chloride (an "insoluble" salt), AgCl, in water:
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Illustration 1 - Cartoon drawing of saturated AgCl solution |
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The equilibrium process pictured above can be represented with a simple equation:
AgCl(s) <--> Ag+(aq) + Cl-(aq)
You would write the equilibrium constant expression like this:
[Note that we leave off AgCl solid because the concentration of this pure solid doesn't change even if the amount of pure solid does.]
Similarly, you could look at barium fluoride, BaF2:
BaF2(s) <--> Ba2+(aq) + 2F-(aq)
The form of these equilibrium constant expressions is interesting. It's merely the product of the concentrations of the ions in solution raised to the power of the number of that ion present in the salt. For this reason, these expressions are called solubility product constant expressions and the constant itself is called the solubility product constant, Ksp.
Let's try another. What is the solubility product constant expression for calcium phosphate?
The chemical formula of calcium phosphate is Ca 3 (PO 4 ) 2 , and when it dissolves in water:
Ca3(PO4)2(s) <--> 3Ca2+(aq) + 2PO43-(aq)
So the solubility product constant is really nothing new. It's just an equilibrium constant that is useful enough to be given a special name. Since it's an equilibrium constant, we can apply all of our knowledge of how equilibrium works to the solubility of salts in water.
A simple calculation: The solubility of a salt in water
Let's say we want to know the solubility of a salt in water. The solubility is, of course, the amount of the salt that dissolves in a given amount of water at equilibrium. In other words, it's the amount of salt that is dissolved in a saturated solution. If we know the solubility product constant (which is tabulated for many salts in various books, and in Appendix G of your textbook), we can easily calculate the solubility of the salt. Let's do an example.
What is the solubility of lead iodide, PbI2, in 25 ° C water? Express your answer in terms of grams per liter. Ksp = 6.5x10-9
To solve this problem, treat it as a standard equilibrium problem. Write out the equilibrium reaction and the expression for Ksp:
PbI 2 (s) <--> Pb 2+ (aq) + 2I-(aq)
Now, we define all concentrations in terms of one variable:
[Pb2+] = x
[I-] = 2x (Note: Every time PbI2 dissociates, two iodide ions are formed!)
Now plug in to the equilibrium expression and solve:
(Note: the 2 in front of the x above is inside the parenthesis. A common mistake in solving one of these problems is to forget to square the 2 when solving.)
6.5x10-9 = 4x3
x = 1.1x10-3 M = [Pb 2+ ]
So how does this number relate to the solubility? For every mole of PbI 2 that dissociates , one mole of Pb2+ ions is produced. So, this number is equivalent to the molar solubility (solubility expressed as a molar concentration). The problem, though, asks us for the solubility expressed in grams per liter. To find the requested answer, simply use the formula weight to convert from grams to moles!
So, the solubility of the PbI2 is 0.51 g/L.
Precipitation and the solubility product constant
An application of the solubility equilibrium is gravimetric analysis .
Gravimetric analysis is a method that lets you determine how much of a given substance is present in a sample by reacting it with a precipitating agent. A precipitating agent is simply a compound that causes another to precipitate.
For example, a common method for determining the amount of chloride ion in a sample involves dissolving the sample in water and adding silver nitrate (AgNO3, a source of the silver ion Ag+) to precipitate the chloride as the silver salt. Silver chloride is an "insoluble' salt:
AgCl(s) <--> Ag+(aq) + Cl-(aq) ; Ksp = 1.8x10-10
To precipitate silver chloride, you must first create a saturated solution. Silver chloride (or any other salt, for that matter), will not precipitate from solution unless that solution is saturated. This is one reason why, in a gravimetric analysis, the procedure will often tell you to add a large amount of the precipitating agent.
How do you know when a precipitate will form, given concentrations of ions in solution? If you realize that precipitation can occur only after the solution has enough dissolved ions to be at equilibrium, then it's easy.
In a previous chapter, you calculated a reaction quotient, Q, to determine whether or not a reaction was at equilibrium. You can use the reaction quotient Q (sometimes called the ion product) to determine whether or not a salt will precipitate from solution. Let's do a quick example.
Will silver chloride precipitate from a solution that is 0.014 M Ag+ and 0.00042 M Cl-?
To answer this question, simply find the reaction quotient. Remember that the reaction quotient has the same form as the equilibrium constant, so:
AgCl(s) <--> Ag+(aq) + Cl-(aq) ; Ksp = 1.8x10-10
What do you do with Q? Compare it with Ksp:
If Q < Ksp , then the reaction will proceed to produce more products - i.e. dissolved ions. Therefore, no precipitation will occur.
If Q = Ksp , then the reaction is at equilibrium - i.e. you have a saturated solution. Precipitation is just beginning.
If Q > Ksp, then the reaction will proceed to produce more reactants - i.e. precipitate. Therefore, precipitation will occur.
In our example, Q > Ksp, so precipitation of AgCl will occur.
The common-ion effect
We discussed the common-ion effect when we discussed acids and bases. We noted that the presence of an ion that was a product of the dissociation of a weak base or acid suppressed the dissociation of that weak base or acid. Let's use the same logic and apply it to the solution process.
What would be the molar solubility of lead iodide in a solution of 0.14 M sodium iodide?
The solubility equilibrium and Ksp expression are the same as for lead iodide in pure water:
PbI 2 (s) <--> Pb 2+ (aq) + 2I-(aq)
How does the sodium iodide (NaI) figure in? It's a source of the iodide ion , a product in the equilibrium. We'd expect (because of Le Chateleir's principle that if we added sodium iodide to a saturated solution of lead iodide, the equilibrium would shift to the left to produce more PbI 2 . Logically, then, we should expect that less lead iodide would dissolve in a solution of sodium iodide than in pure water . Let's verify this logic by calculation.
Like any equilibrium problem, we express all concentrations in terms of one variable.
[Pb2+] = x
[I - ] = 0.14 + 2 x
Since the amount of lead iodide that dissociates is small (Ksp = 6.5x10-9) - and likely to be even smaller due to the presence of iodide already in solution - we will assume that 2x is small compared to 0.14.
[I-] = 0.14
Now, we plug in and solve just as we did in the case of pure water:
x = 3.3x10-7 M = [Pb 2+ ] = the molar solubility of PbI2 in a 0.14 M NaI solution.
Comparing the molar solubility of PbI2 in NaI (3.3x10 -7 M ) to PbI2 in pure water (1.1x10-3 M ) , we see that the lead iodide is more than 3000 times more soluble in water than it is in 0.14 M NaI. This is, of course, a result of the common-ion effect.
Le Chateleir's principle: How pH and complex formation affect solubility
In addition to the common-ion effect, Le Chaleleir's principle can tell us whether or not other chemicals will change the solubility of a substance. Let's consider pH first.
As we already know, some salts have acidic or basic properties. They have these properties because the ions that make up the salt can be acidic, basic, or neutral. The solubility of acidic or basic salts depends on the pH of the solution you're trying to dissolve them in! As an example, let's look at magnesium hydroxide, a fairly insoluble salt featured in one of the CHM 111 labs.
Mg(OH)2(s) <--> Mg2+(aq) + 2OH-(aq)
The solubility of this salt is very dependent on pH. Why can we say this? Well, the salt forms two ions when it dissolves - magnesium ion (a pH-neutral ion) and hydroxide ion (a basic ion - it can accept a proton to form water). What does Le Chateleir's principle say about the solubility of this salt in solutions of different pH values?
In a basic solution, the concentration of hydroxide is high. Since OH- is a product in the solubility equilibrium, the equilibrium is forced to the left. Less Mg(OH)2 dissolves in a basic solution than in pure water .
In an acidic solution, the concentration of hydroxide is low. In fact, hydroxide ions produced by the dissolution of magnesium hydroxide may be converted to water by the hydronium ions present in the acidic solution. Since OH- is a product in the solubility equilibrium, the equilibrium is forced to the right. More Mg(OH)2 dissolves in an acidic solution than in pure water .
Generalizing to all salts,
Basic salts are more soluble in acidic solutions and less soluble in basic solutions.
Acidic salts are more soluble in basic solutions and less soluble in acidic solutions.
The solubility of a neutral salt is not affected by pH.
The formation of complex ions can also affect the solubility of certain salts. How? Consider the insoluble salt silver bromide, AgBr.
AgBr(s) <--> Ag+(aq) + Br-(aq)
It's known that the silver ion can form a complex with ammonia via a Lewis acid/base reaction:
Ag+(aq) + 2NH3(aq) <--> Ag(NH3)2+(aq)
So what's the effect of the presence of ammonia on the solubility of silver bromide? The solubility goes up, because the ammonia reacts with silver ion and lowers its concentration. Le Chateleir's principle says that the silver bromide solubility equilibrium will shift to the right because the concentration of silver ion has been lowered - so more silver bromide will dissolve.
Silver salts are so much more soluble in the presence of ammonia that a way to remove insoluble silver salts (AgCl, AgBr, etc.) from chemical glassware is to rinse the glassware with concentrated ammonia!
Generalizing, if an ion in a salt forms a complex in solution, the solubility of the salt will be increased.
Summary
In this note pack, we discussed solubility as an equilibrium process. We defined terms related to solubility and learned the names used to describe the equilibrium constant and reaction quotient when dealing with salt solutions. We discussed that saturated solutions were solutions at equilibrium, and we discussed the conditions necessary for the precipitation of salts from a solution. We also discussed Le Chateleir's principle and its applications to solubilty: the common ion effect, pH, and complex formation. You should understand how to do simple solubility calculations (including calculations involving the common-ion effect). You should be able to predict at what pH conditions a given salt is most (or least) soluble. You should also be able to determine whether a precipitate will form given concentration of the ions in a solution.
All original site content ©2007 Charles Taylor. Page updated: December 12, 2007.
