How to Determine Solubility: 14 Steps

2024 Author: Samantha Chapman | [email protected]. Last modified: 2024-01-15 13:47

Solubility is a concept used in chemistry to express the ability of a solid compound to dissolve completely in a liquid without leaving undissolved particles. Only ionic compounds are soluble. To solve practical questions, it is sufficient to memorize some rules or refer to a table of soluble compounds, to know if most of the ionic compound remains solid or if a considerable amount dissolves once immersed in water. In fact, some molecules dissolve even if you can't see any changes, so precise experiments are needed to learn how to calculate these quantities.

Steps

Method 1 of 2: Using Quick Rules

Step 1. Study ionic compounds

Each atom has a certain number of electrons, but sometimes it acquires one more or loses it; the result is one ion which is equipped with an electric charge. When a negative ion (an atom with an extra electron) meets a positive ion (which has lost an electron) a bond is formed, just like the negative and positive poles of magnets; the result is an ionic compound.

• Negatively charged ions are called anions, those with positive charge cations.
• Normally, the number of electrons is equal to that of protons, neutralizing the charge of the atom.

Step 2. Understand the concept of solubility

The water molecules (H.₂O) have an unusual structure that makes them similar to magnets: they have one end with a positive charge and another with a negative charge. When an ionic compound is dropped into water, it is surrounded by these liquid "magnets" which try to separate the cation from the anion.

• Some ionic compounds don't have a very strong bond, so they are soluble, since water can divide and dissolve them; others are more "resistant" e insoluble, because they remain united despite the action of water molecules.
• Some compounds have internal bonds with the same strength as the attractive power of molecules and are said slightly soluble, as a significant part dissolves in water, while the rest remains compact.

Step 3. Study the rules of solubility

Since the interactions between atoms are quite complex, understanding which substances are soluble and which insoluble is not always an intuitive process. Look at the first ion of the compounds described below to find its normal behavior; then, check for exceptions to make sure it doesn't interact in a particular way.

• For example, to find out if strontium chloride (SrCl₂) is soluble, check the behavior of Sr or Cl in the bold steps listed below. Cl is "generally soluble", so you need to check for exceptions; Sr is not on the list of exceptions, so you can say that the compound is soluble.
• The most common exceptions to each rule are written under it; there are others, but they are rarely encountered during a chemistry course or in laboratory experiences.

Step 4. Understand that compounds are soluble if they contain alkali metals

Alkali metals include There⁺, Na⁺, K⁺, Rb⁺ and Cs⁺. These are called Group IA elements: lithium, sodium, potassium, rubidium and cesium; almost all the ionic compounds that contain them are soluble.

Exceptions: There₃BIT₄ it is insoluble.

Step 5. Compounds of NO₃^-, C₂H.₃OR₂^-, NO₂^-, ClO₃^- and ClO₄^- they are soluble.

Respectively, they are the ions: nitrate, acetate, nitrite, chlorate and perchlorate; remember that acetate is often abbreviated to OAc.

• Exceptions: Ag (OAc) (silver acetate) and Hg (OAc)₂ (mercury acetate) are insoluble.
• AgNO₂^- and KClO₄^- they are only "slightly soluble".

Step 6. The compounds of Cl^-, Br^- and I.^- they are normally soluble.

Chloride, bromide and iodide ions almost always form soluble compounds called halides.

Exceptions: if any of these ions bind to the silver ion Ag⁺, mercury Hg₂²⁺ or lead Pb²⁺, the resulting compound is insoluble; the same applies to the less common ones formed by the copper ion Cu⁺ and thallium Tl⁺.

Step 7. Compounds that contain So₄^2- they are generally soluble.

The sulphate ion usually forms soluble compounds, but there are several peculiarities.

Exceptions: the sulphate ion creates insoluble compounds with the ions: strontium Sr²⁺, barium Ba²⁺, lead Pb²⁺, silver Ag⁺, calcium Ca²⁺, radio Ra²⁺ and diatomic silver Hg₂²⁺. Remember that silver and calcium sulfate dissolve just enough for people to find them slightly soluble.

Step 8. Compounds that contain OH^- or S^2- they are insoluble.

These are, respectively, the hydroxide and sulphide ion.

Exceptions: do you remember alkali metals (of group IA) and how they form soluble compounds? There⁺, Na⁺, K⁺, Rb⁺ and Cs⁺ they are all ions that form soluble compounds with that hydroxide and sulfide. The latter also bind to the alkaline earth ions (group IIA) to obtain soluble salts: calcium Ca²⁺, strontium Sr²⁺ and barium Ba²⁺. The compounds resulting from the bond between the hydroxide ion and the alkaline earth metals have enough molecules to remain compact to the point that they are sometimes considered "slightly soluble".

Step 9. Compounds that contain CO₃^2- or PO₄^3- they are insoluble.

A final check on the carbonate and phosphate ions should allow you to understand what to expect from the compound.

Exceptions: these ions form soluble compounds with alkali metals (Li⁺, Na⁺, K⁺, Rb⁺ and Cs⁺), as well as with the ammonium ion NH₄⁺.

Method 2 of 2: Calculate the Solubility from K._sp

Step 1. Look for the solubility constant K_sp.

This is a different value for each compound, so you must consult a table in the textbook or online. Since these are numbers determined experimentally, they can change a lot according to the table you decide to use; therefore refer to the one you find in the chemistry book, if any. Unless specifically stated, most tables assume you are working at 25 ° C.

For example, if you are dissolving lead iodide PbI₂, note its solubility constant; if this is the reference table, use the value 7, 1 × 10^–9.

Step 2. Write the chemical equation

First, determine how the compound separates into ions when it dissolves and then write the equation with the value of K_sp on one side and the constituent ions on the other.

• For example, the PbI molecules₂ they separate into Pb ions²⁺, I.^- and I.^--. You must know or look for only the charge of an ion, since you know that the overall charge of the compound is always neutral.
• Write the equation 7, 1 × 10^–9 = [Pb²⁺][THE^-]².
• The equation is the solubility constant of the product, which can be found for the 2 ions from a solubity table. There being 2 negative ions I.^-, this value is raised to the second power.

Step 3. Modify it to use variables

Rewrite it as if it were a simple algebra problem, using the values you know of the molecules and ions. Set as unknown (x) the amount of compound that dissolves and rewrite the variables that represent each ion in terms of x.

• In the example considered you have to rewrite: 7, 1 × 10^–9 = [Pb²⁺][THE^-]².
• Since there is a lead atom (Pb) in the compound, the number of dissolved molecules is equal to the number of free lead ions; consequently: [Pb²⁺] = x.
• Since there are two iodine ions (I) for each lead ion, you can establish that the amount of iodine ions is equal to 2x.
• The equation then becomes: 7, 1 × 10^–9 = (x) (2x)².

Step 4. Consider common ions, if any

If you are dissolving the mixture in pure water, you can skip this step; on the other hand, if it has been dissolved in a solution that contains one or more constituent ions ("common ions"), the solubility decreases significantly. The effect of the common ion is most evident in compounds that are mostly insoluble and in this case you can consider that the vast majority of the ions in equilibrium come from the one already present in the solution. Rewrite the equation to include the molar concentration (moles per liter or M) of the ions that are already in the solution and substituting the value of x you used for that specific ion.

For example, if the lead iodide compound was dissolved in a solution with 0.2M, you should rewrite the equation as: 7.1 × 10^–9 = (0, 2M + x) (2x)². Since 0.2M is a far greater concentration than x, you can safely rewrite the equation like this: 7.1 × 10^–9 = (0, 2M) (2x)².

Step 5. Perform the calculations

Solve the equation for x and know how soluble the compound is. Considering the method by which the solubility constant is established, the solution is expressed in moles of dissolved compound per liter of water. You may need to use a calculator for this calculation.

• The calculations described below consider solubility in pure water without any common ion:
• 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 = they will melt 1, 2 x 10^-3 moles per liter. This is a very small amount, so you can say that the compound is essentially insoluble.

Advice

If you have experimental data regarding the quantities of dissolved compound, you can use the same equation to find the solubility constant K_sp.

Warnings

• There is no universally accepted definition for these terms, but chemists agree on most compounds. Some borderline cases in which a considerable amount of dissolved and undissolved molecules remain are described differently by the various solubility tables.
• Some old textbooks list NH₄OH among the soluble compounds. This is a mistake: small amounts of NH can be detected₄⁺ and OH ions^-, but they cannot be isolated to form a compound.