In chemistry, solubility is used to describe the properties of solid compounds that are mixed and completely dissolved with a liquid without leaving any insoluble particles. Only ionized (charged) compounds can dissolve. For convenience, you can simply memorize a few rules or refer to a list to see if most solid compounds will remain solid when placed in water or will dissolve in large quantities. In fact, some molecules will dissolve even if you can't see the change. In order for the experiment to take place with precision, you must know how to calculate the amount that is dissolved.
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Method 1 of 2: Using Quick Rules
Step 1. Study ionic compounds
Normally each atom has a certain number of electrons. However, sometimes atoms gain or lose electrons. The result is a ion which is electrically charged. When a negatively charged ion (having one extra electron) encounters a positively charged ion (losing an electron), the two ions bond together like the positive and negative poles of a magnet, producing an ionic compound.
- Negatively charged ions are called anion, while the positively charged ion is called cation.
- Under normal circumstances, the number of electrons is equal to the number of protons in an atom thereby negating its electric charge.
Step 2. Understand the topic of solubility
Water molecules (H2O) has an unusual structure that is similar to a magnet. One end has a positive charge, while the other end is negatively charged. When an ionic compound is placed in water, the water "magnet" will surround it and try to attract and separate the positive and negative ions. The bonds in some ionic compounds are not very strong. Such a compound water soluble because water will separate the ions and dissolve them. Some other compounds have stronger bonds so that not soluble in water despite being surrounded by water molecules.
Many other compounds have internal bonds that are just as strong as the force water pulls on the molecules. Such compounds are called slightly soluble in water because a large part of the compound is attracted by water, but the rest is still fused.
Step 3. Learn the rules about solubility
Interatomic interactions are quite complex. Compounds that are soluble or insoluble in water cannot simply be seen intuitively. Find the first ion in the compound to look for in the list below to determine its behavior. Next, check for any exceptions to make sure that the second ion doesn't have any unusual interactions.
- For example, to check Strontium Chloride (SrCl2), look for Sr or Cl in the steps in bold below. Cl is "usually water soluble," so check the next one for exceptions. Sr is not an exception so SrCl2 definitely soluble in water.
- The most common exceptions to each rule are listed below. There are a few other exceptions, but they probably won't be found in a lab or chemistry class in general.
Step 4. Compounds can be dissolved if they contain alkali metals, including Li+, Na+, K+, Rb+, and Cs+.
These elements are also known as group IA elements: lithium, sodium, potassium, rubidium, and cesium. Almost all compounds containing one of these ions are soluble in water.
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Exception:
Li3PO4 insoluble in water.
Step 5. NO. Compounds3-, C2H3O2-, NO2-, ClO3-, and ClO4- soluble in water.
They are named in order of the nitrate, acetate, nitrite, chlorate, and perchlorate ions. Note that acetate is often shortened to OAC.
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Exception:
Ag(OAc) (Silver acetate) and Hg(OAc)2 (mercury acetate) is insoluble in water.
- AgNO2- and KClO4- only "slightly soluble in water."
Step 6. Cl. compounds-, Br-, and I- usually slightly soluble in water.
Chloride, bromide, and iodide ions always form water-soluble compounds called halide salts.
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Exception:
If one of these ions binds the silver ion Ag+, mercury Hg22+, or lead Pb2+, the resulting compound is insoluble in water. The same is true for the less common compound, namely the Cu. pair+ and thallium Tl+.
Step 7. Compounds containing SO42- generally soluble in water.
The sulfate ion usually forms water-soluble compounds, but there are some exceptions.
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Exception:
Sulfate ion forms insoluble compounds in water with: strontium Sr2+, barium Ba2+, lead Pb2+, silver Ag+, calcium Ca2+, radium Ra2+, and diatomic silver Ag22+. Note that silver sulfate and calcium sulfate will be sufficiently soluble that some call them slightly soluble in water.
Step 8. Compounds containing OH- or S2- insoluble in water.
The ions above are named hydroxide and sulfide.
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Exception:
Remember about the alkali metals (Groups I-A) and how easily ions from elements in those groups form water-soluble compounds? Li+, Na+, K+, Rb+, and Cs+ will form water-soluble compounds with hydroxide or sulfide ions. In addition, hydroxides also form water-soluble salts with alkaline earth ions (Group II-A): calcium Ca2+, strontium Sr2+, and barium Ba2+. Note that compounds produced from hydroxides and alkaline earths still have enough molecules bonded together that they are sometimes called "slightly soluble in water."
Step 9. Compounds containing CO32- or PO43- insoluble in water.
One more check for carbonate and phosphate ions. You should already know what will happen to the compound of the ions.
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Exception:
These ions form water-soluble compounds with alkali metals, namely Li+, Na+, K+, Rb+, and Cs+, as is ammonium NH4+.
Method 2 of 2: Calculating Solubility Through Ksp
Step 1. Find the solubility constant of the product Ksp.
Each compound has a different constant, you'll have to look it up in a table in your textbook or online. Because the values are determined experimentally, different tables can display different constants. It is highly recommended that you use the tables in the textbook if you have them. Unless otherwise specified, most tables assume that the temperature is 25ºC.
For example, if what is dissolved is lead iodide PbI2, write the solubility constant of the product. When referring to the table at bilbo.chm.uri.edu, use the constant 7, 1×10–9.
Step 2. Write down the chemical equation
First, determine the process by which the compound separates into ions when dissolved. Then, write the chemical equation with Ksp on one side and the constituent ions on the other.
- For example, a PbI. molecule2 split into Pb. ions2+, I-, and I. ions-. (You only need to know or look for the charge on one ion because the compound as a whole has a neutral charge.)
- Write the equation 7, 1×10–9 = [Pb2+][I-]2
Step 3. Change the equation to use a variable
Rewrite the equation as a simple algebraic problem using knowledge of the number of molecules and ions. In this equation x is the number of soluble compounds. Rewrite the variables that represent the number of each ion in x form.
- In this example, the equation is rewritten as 7, 1×10–9 = [Pb2+][I-]2
- Because there is one lead ion (Pb2+) in the compound, the number of molecules of the compound dissolved is equal to the number of free lead ions. Now we can write [Pb2+] against x.
- Because there are two iodine ions (I-) for each lead ion, the number of iodine atoms can be written as 2x.
- Now the equation is 7, 1×10–9 = (x)(2x)2
Step 4. Take into account other ions normally present if possible
Skip this step if the compound is dissolved in pure water. If a compound is dissolved in a solution that already contains one or more of the constituent ions ("common ions") its solubility will increase significantly. The general ionic effect is best seen in compounds that are largely insoluble in water. In this case it can be assumed that most of the ions at equilibrium come from ions already present in solution. Rewrite the equation for the reaction to include the known molar concentration (moles per liter or M) of the ion already present in solution, thus replacing the value of x used for the ion.
For example, if the compound lead iodide is dissolved in a solution containing 0.2 M lead chloride (PbCl2) then the equation will be 7, 1×10–9 = (0, 2M+x)(2x)2. Then, since 0.2 M is a more concentrated concentration than x, the equation can be rewritten as 7.1×10–9 = (0, 2M)(2x)2.
Step 5. Solve the equation
Solve x to find out how soluble the compound is in water. Since the solubility constant has already been established, the answer is in terms of the number of moles of the compound dissolved per liter of water. You may need a calculator to calculate the final answer.
- The following answer is for solubility in pure water, without the common ions.
- 7, 1×10–9 = (x)(2x)2
- 7, 1×10–9 = (x)(4x2)
- 7, 1×10–9 = 4x3
- (7, 1×10–9) 4 = x3
- x = ((7, 1×10–9) ÷ 4)
- x = 1, 2 x 10-3 moles per liter will dissolve. This amount is so small that it is essentially insoluble in water.
