A general chemistry skill set includes writing formula unit, total ionic and net ionic equations starting from a description of the chemical change. One skill that is very important is to be able to predict whether or not a particular salt is considered to be soluble or insoluble. Worth commiting to memory: In general group IA cations and the ammonium ion when paired with any anion make soluble salts. And in general salts of nitrate and acetate are soluble. For all others, you should refer to the solubility table found here to become familar with the solubility rules, which in turn will help you predict when a precipitate is formed.
When dealing with chemistry in aqueous solutions, there are a number of ways in which we can depict the reaction with a chemical equation. For example we might have a reaction in which we mix a solution of sodium chloride and silver nitrate. These two will form a solid precipitate silver chloride. This reaction could be written as
\[ \rm{NaCl(aq) + AgNO_3(aq)} \rightleftharpoons \rm{AgCl(s) + NaNO_3(aq)}\]
Or since we know that the aqueous solutions of the soluble ions will break up into hydrated ions we could write the total ionic equation
\[ \rm{Na^+(aq) + Cl^-(aq) + Ag^+(aq) + NO_3^-(aq)}\] \[\rightleftharpoons \rm{AgCl(s) + Na^+(aq) + NO_3^-(aq)}\]
Better yet, there are ions that are not "doing anything" in the reaction. The aqueous sodium ions appear on both the reactant and product side of the equation. The same is true for the nitrate ions. These ions are called spectator ions since they are just "watching" the action of the other ions. We can remove them from the equation and write the net ionic equation.
\[ \rm{Cl^-(aq) + Ag^+(aq)} \rightleftharpoons \rm{AgCl(s)}\]
The net ionic equation is the one that depicts the change and ignores the spectator ions.
A video on Ionic EquationsA saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. You know you have a saturated solution as it will generally contain some undissolved solid solute. At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.
A saturated solution is at equilibrium. The rate of dissolution and the rate of reforming the solid solute are equal. When the concentration is below the saturation concentration, then any solid will spontaneously dissolve. Once sufficient solute has dissolved and the system is at equilibrium, the net dissolution will stop. Adding further solid solute to the mixture will not change this and the solid will remain undissolved.
We can characterize saturated solutions by their solute concentrations. For ionic solutes that form multiple ions, it turns out that it is the product of the ion concentrations that is fixed (rather than the concentrations themselves). This "ion product" is what we will use to characterize the equilibrium state.
We typically characterize the solubility of a compound in one of two ways. For a solute that is generally soluble, one in which we know it dissolves readily (like sugar in water), the solubility is generally given by the total mass of solute that will dissolve in a given amount of solvent. For example, at 25°C the solubility of sugar (sucrose) in water is 2000g/L. You will also see this as a molar solubility (the number of moles per liter). Alternatively you might find this number given as the number of grams of solute that will dissolve in a certain mass of solvent.
The other way in which we characterize the solubility of a compound is by the ion product. This is the method that is most often used for substances that we describe as "insoluble". Why would anyone talk about the "solubility" of an "insoluble" compound? This is often the case in which we would like to characterize the small concentrations of particular ions in solution.
When a solution is at saturation the concentration of the solute is fixed. For an ionic compound, we need to look at the concentration of all of the ions. In this case, it is the product of the concentration of the ions that is found to be a fixed value. This product is called the ion product and its value at the saturation conditions (equilibrium) is called the solubility product and is given the symbol Ksp.
Ksp is a constant that is specific to the compound (and is temperature dependent). Ksp is also assumed to be for an aqueous solution.
Ksp is the product of all of the ions that are formed by the compound. So for a compound that forms two ions, like AgCl,
\[K_{\rm sp} = \rm{[Ag^+][Cl^-]}\]
where the brackets denote the concentration of the species in units of molarity.
If a compound makes multiple ions (a salt with a ratio other than 1:1), then we need to include them multiple times in the product. This results in some ions being raised to a higher power since each mole of the compound makes multiple moles of some ions. For example, Ksp for lead(II) chloride, PbCl2 is
\[K_{\rm sp} = \rm{[Pb^{2+}][Cl^-][Cl^-]=[Pb^{2+}][Cl^-]^2}\]
It is also important to remember that the concentration of the multiple moles of ions will be higher as well according to the stoichiometry of the salt formula. In the PbCl2 case above, the chloride concentration will always be twice that of the lead(II) concentration when coming from the same source of PbCl2. Algebraically, this often shown as [Pb2+] = x and [Cl-] = 2x.
Typically we use Ksp for compounds that are not very soluble. Otherwise the solubility is given in terms of the number of moles or gram of a compound that will dissolve. However,it is important to be able to convert back and forth between these two.
To convert from Ksp to solubility, we need to think about what Ksp can tell us about how many moles of a particular compound dissolve. This is simple for a binary compound (with two ions). Taking the AgCl example again, for each mole of the compound that dissolves, we get the same number of moles of silver ions and chloride ions. So if we call the number of moles of AgCl that dissolve in 1 L of water some number x, the concentration of Ag+ and Cl- will both be x. So
\[K_{sp} = {\rm [Ag^+][Cl^-]}=(x)(x)=x^2\]
So in this case, Ksp is the square of the molar solubility and the molar solubility will be the square root of the Ksp.
This is slightly more complicated when there are more than two ions. For example if we dissolve lead(II) chloride, for each mole of the compound we get one mole of lead ions and two moles of chloride ions. So again taking the molar solubility to be some number x, the concentration of Pb2+ will be x, but the concentration of Cl- will be 2x. Thus
\[K_{sp} = {\rm [Pb^{2+}][Cl^-]^2}=(x)(2x)^2=4x^3\]
For example, the solubility of Ag2CO3 is only 32mg/L. For such an insoluble compound, we would typically want Ksp. For this we need to know the molar concentrations at saturation. 32 mg of Ag2CO3 is 1.16 × 10-4moles. If this amount dissolves in a liter of solution we will have a concentration of carbonate ions, [CO32-] = 1.16 × 10-4 M. The concentration of the silver ion will be twice this value since we get two moles of silver ions for each mole of the compound that dissolves. This means that [Ag+] = 2(1.16 × 10-4)= 2.32 × 10-4 M. The ion product is then
\[K_{sp} = {\rm [Ag^+]^2[CO_3^{2-}] = (2.32\times 10^{-4})^2(1.16\times 10^{-4})= 6.2\times 10^{-12}}\]
We can still convert back and forth between molar solubility and Ksp but now the relationship involves a cube-root and a factor of four.
One easy way to estimate solubility is to ignore the small factor of four and simply take the cube root. This will get you close, but it will be only an approximate answer. This will work for compounds with any number of ions. The molar solubility will be the nth-root of Ksp where n is the total number of ions formed (the van't Hoff factor) from the salt.
Here is video about Calculating Molar Solubility when given the value of Ksp.
Here is video about Calculating an Ion Concentration in Solution when both Ksp and the counter ion concentration are given.
Calculating molar solubility using Ksp:
http://laude.cm.utexas.edu/courses/ch302/wsclips/CH302_s11_ws6_q5.swf
Because the solubility of an ionic compound depends on the product of the concentrations of the ions, this solubility can be greatly affected if there are already some of those ions present in the solution. For example, imagine we have a 0.1 molar solution of sodium chloride. This solution has a [Na+] = [Cl-1] = 0.1 M. Now we try to dissolve some AgCl into this same solution. Silver chloride is fairly insoluble. It has a solubility in pure water of about 2 mg/L. That corresponds to a molar solubility of 1.3 x 10-5 moles/L. Given as a Ksp this is 1.8 x 10-10. However, now the amount of silver chloride that will dissolve will be less. This is because there is a common ion in the solution, chloride. The chloride concentration is already 0.1 M. Therefore the silver concentration will be even less than it would be in the solution of pure water. The silver ion concentration will be reflective of the amount of silver chloride that dissolves since this is the only source of silver ions.
We can find the concentration of silver ions at equilibrium from Ksp since we already know the chloride ion concentration. We know that only a tiny amount of silver chloride will dissolve. This means the chloride concentration will stay around 0.1 M. We can then use Ksp to solve for the silver ion concentration.
\[K_{sp} = \rm{[Ag^+][Cl^-] = [Ag^+] = }K_{sp}/\rm{[Cl^-]} = (1.8 x 10^{-10})/(0.1) = 1.8 x 10^{-9}\]
Therefore, in the 0.1 M solution of NaCl only 1.8 x 10-9 moles of AgCl dissolve in one L. This is approximately 10,000 times less than would dissolve in pure water (1.3 x 10-5 M). We would reduce this amount even further by adding more chloride ions to the solution by adding more of the extremely soluble NaCl. In this way, it is possible to manipulate the solubility of different compounds by changing the concentrations of common ions. This can be done to not only change the solubility, but to form precipitates.
Precipitation is when a solid solute spontaneously forms from a solution. This might occur if we mixed two solutions together that contained a mixture of ions that form an insoluble compound. For example, imagine we mixed a solution of BaCl2 and solution of NaNO3. This would result in a solution with Na+ ions, Ba2+, Cl-1, and NO3- ions. As a result of mixing these solutions we could form Ba(NO3)2 a compound with a low solubility. We could write this reaction as
\[ \rm{BaCl_2(aq) + 2NaNO_3(aq)} \rightleftharpoons \rm{Ba(NO_3)_2(s) + 2NaCl(aq)}\]
We would also write this as a total ionic equation as
\[ \rm{Ba^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2NO_3^-(aq)}\] \[\rightleftharpoons \rm{Ba(NO_3)_2(s) + 2Na^+(aq) + 2Cl^-(aq)}\]
Or ignoring the spectator ions (sodium and chloride) we would simply write the net ionic equation
\[ \rm{Ba^{2+}(aq) + 2NO_3^-(aq)} \rightleftharpoons \rm{Ba(NO_3)_2(s)}\]
You can see the net ionic reaction in this case is essentially simply the dissolution of barium nitrate but written in reverse. Therefore this "reaction" will only happen at certain concentrations. There are concentrations of Ba2+ and NO3- at which we would expect to form some solid precipitate, while at other concentrations we would not (since the solid would be completely dissolved under these conditions). We can predict these concentration from the solubility product.
The ion product is not only useful for predicting equilibrium conditions; it is also useful to determine if a system is in or out of equilibrium. We therefore need to distinguish between the ion product itself and its value at equilibrium. At equilibrium the ion product is the solubility product Ksp. However, we can calculate this product for any solution (or any imagined solution). For any given set of conditions we can calculate the ion product and denote it with the letter Q. To note we are looking at solubility it would be given the symbol Qsp. For example above imagine we used a 0.1M BaCl2 solution and a 1M Na(NO3) solution. This would yield a mixture with a concentration of barium ions, [Ba2+] = 0.1 M, and the concentration of nitrate ions, [NO3-] = 1 M. At these concentrations Ba(NO3)2 would we form a precipitate? To answer this question we need to compare the ion product for the given conditions, Qsp, to that at the equilibrium conditions, Ksp.
If we find
\[ Q_{sp} > K_{sp} \hskip24pt {\rm then\;precipitate\;will\;form}\]
This is because the concentrations in the given solution are higher than we would find at equilibrium. When the concentrations are higher than the equilibrium concentrations, then solid will precipitate until the concentrations are reduced to the equilibrium concentrations. However, if we find
\[ Q_{sp} < K_{sp} \hskip24pt {\rm then\;no\;precipitate\;will\;form}\]
This is because the concentrations are below the equilibrium concentrations. Any solid would dissolve spontaneously at these concentrations. Therefore no solid will form. If we find that
\[ Q_{sp} = K_{sp} \hskip24pt {\rm then\;the\;system\;is\;at\;equilibrium}\]
No solid is precipitate. No more solid can dissolve into this solution. We have formed a saturated solution.
Returning to the Ba(NO3)2 solution, does a precipitate form when [Ba2+] = 0.1 M and [NO3-] = 1 M? We need to compare Qsp and Ksp.
\[ Q_{sp} =\rm{[Ba^{2+}][NO_3^-]^2} = (0.1)(1)^2 = 0.1.\]
A table of Ksp values shows that Ksp for Ba(NO3)2 is 4.6 x 10-3. Since Qsp > Ksp a precipitate will form. Note: nitrate salts are general soluble. However, even with soluble salts it is possible to get the concentrations so high that they will precipitate. It all depends on what you have now (Q) vs what you would have at equilibrium (K).
It is possible to increase the concentration beyond this equilibrium point. A solution with a concentration higher than the saturation concentration is called a super-saturated solution. In such a condition the system is no longer at equilibrium and will spontaneously form a solid precipitate since Qsp > Ksp . However, as the formation of the solid crystals takes time the solution can stay trapped in this super-saturated condition for a long time. Typically it is possible to make a super-saturated solution by first making a saturated solution at higher temperature where the solubility is higher. The solution can then be slowly cooled to a lower temperature where the solubility is lower.
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