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Page ID

105616

1. Solutions of Gases in Liquids

2. Solutions of Liquids in Liquids

3. Solutions of Solids in Liquids

4. Summary

5. Key Equations

6. Glossary

7. Contributors

Skills to Develop

Describe the effects of temperature and pressure on solubility

State Henry's law and use it in calculations involving the solubility of a gas in a liquid

Explain the degrees of solubility possible for liquid-liquid solutions

Imagine adding a small amount of salt to a glass of water, stirring until all the salt has dissolved, and then adding a bit more. You can repeat this process until the salt concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent-solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved salt remains. The concentration of salt in the solution at this point is known as its solubility.

The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium. Referring to the example of salt in water:

\[\ce{NaCl}(s)⇌\ce{Na+}(aq)+\ce{Cl-}(aq) \label{11.4.1}\]

When a solute's concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute's concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.

If we add more salt to a saturated solution of salt, we see it fall to the bottom and no more seems to dissolve. In fact, the added salt does dissolve, as represented by the forward direction of the dissolution equation. Accompanying this process, dissolved salt will precipitate, as depicted by the reverse direction of the equation. The system is said to be at equilibrium when these two reciprocal processes are occurring at equal rates, and so the amount of undissolved and dissolved salt remains constant. Support for the simultaneous occurrence of the dissolution and precipitation processes is provided by noting that the number and sizes of the undissolved salt crystals will change over time, though their combined mass will remain the same.
Video \(\PageIndex{1}\): Watch this impressive video showing the precipitation of sodium acetate from a supersaturated solution.
Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states. For example, the carbonated beverage in an open container that has not yet "gone flat" is supersaturated with carbon dioxide gas; given time, the CO ₂ concentration will decrease until it reaches its equilibrium value.

Solutions of Gases in Liquids

In an earlier module of this chapter, the effect of intermolecular attractive forces on solution formation was discussed. The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl ₃. Considering the role of the solvent's chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C ₆H ₁₄, is approximately 20 times greater than it is in water.

Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure \(\PageIndex{1}\)). This is one of the major impacts resulting from the thermal pollution of natural bodies of water.
Figure \(\PageIndex{1}\): The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa (1 atm) of gas above the solutions.
When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water's ecosystems and, in severe cases, can result in large-scale fish kills (Figure \(\PageIndex{2}\)).
Figure \(\PageIndex{2}\): (a) The small bubbles of air in this glass of chilled water formed when the water warmed to room temperature and the solubility of its dissolved air decreased. (b) The decreased solubility of oxygen in natural waters subjected to thermal pollution can result in large-scale fish kills. (credit a: modification of work by Liz West; credit b: modification of work by U.S. Fish and Wildlife Service)
The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO ₂ at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure \(\PageIndex{3}\)). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become "flat."
Figure \(\PageIndex{3}\): Opening the bottle of carbonated beverage reduces the pressure of the gaseous carbon dioxide above the beverage. The solubility of CO₂ is thus lowered, and some dissolved carbon dioxide may be seen leaving the solution as small gas bubbles. (credit: modification of work by Derrick Coetzee)
For many gaseous solutes, the relation between solubility, C g, and partial pressure, P g, is a proportional one:

where k is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry's law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.

Example \(\PageIndex{1}\): Application of Henry's Law

At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10 −3 mol L −1. Use Henry's law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth's atmosphere.
Solution
According to Henry's law, for an ideal solution the solubility, C g, of a gas (1.38 × 10 −3 mol L −1, in this case) is directly proportional to the pressure, P g, of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both C g and P g, we can rearrange this expression to solve for k.

\[\begin{align*}
C_\ce{g}&=kP_\ce{g}\\[5pt]
k&=\dfrac{C_\ce{g}}{P_\ce{g}}\\[5pt]
&=\mathrm{\dfrac{1.38×10^{−3}\:mol\:L^{−1}}{101.3\:kPa}}\\[5pt]
&=\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}}\\[5pt]
&\hspace{15px}\mathrm{(1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1})}
\end{align*}\]

Now we can use k to find the solubility at the lower pressure.

\[C_\ce{g}=kP_\ce{g}\]

\[\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}×20.7\:kPa\\[5pt]
(or\:1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1}×155\:torr)\\[5pt]
=2.82×10^{−4}\:mol\:L^{−1}}\]

Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.

Exercise \(\PageIndex{1}\)

Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10 −3 g of the solute. Use Henry's law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr).

Answer

7.25 × 10 −3 g in 100.0 mL or 0.0725 g/L

Case Study: Decompression Sickness ("The Bends")

Decompression sickness (DCS), or "the bends," is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver's blood are proportionally higher per Henry's law.

Figure \(\PageIndex{4}\): (a) US Navy divers undergo training in a recompression chamber. (b) Divers receive hyperbaric oxygen therapy. Figure \(\PageIndex{5}\): (a) It is believed that the 1986 disaster that killed more than 1700 people near Lake Nyos in Cameroon resulted when a large volume of carbon dioxide gas was released from the lake. (b) A CO₂ vent has since been installed to help outgas the lake in a slow, controlled fashion and prevent a similar catastrophe from happening in the future. (credit a: modification of work by Jack Lockwood; credit b: modification of work by Bill Evans)
As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver's blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure \(\PageIndex{4}\)).

Deviations from Henry's law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.

Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure \(\PageIndex{5}\)), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO ₂ were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.

Solutions of Liquids in Liquids

We know that some liquids mix with each other in all proportions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in Figure \(\PageIndex{6}\)) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline.

Figure \(\PageIndex{6}\): Water and antifreeze are miscible; mixtures of the two are homogeneous in all proportions. (credit: "dno1967"/Wikimedia commons)
Liquids that mix with water in all proportions are usually polar substances or substances that form hydrogen bonds. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solvent-solvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom "like dissolves like."

Two liquids that do not mix to an appreciable extent are called immiscible. Layers are formed when we pour immiscible liquids into the same container. Gasoline, oil (Figure \(\PageIndex{7}\)), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. The attraction between the molecules of such nonpolar liquids and polar water molecules is ineffectively weak. The only strong attractions in such a mixture are between the water molecules, so they effectively squeeze out the molecules of the nonpolar liquid. The distinction between immiscibility and miscibility is really one of degrees, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility.

Figure \(\PageIndex{7}\): Water and oil are immiscible. Mixtures of these two substances will form two separate layers with the less dense oil floating on top of the water. (credit: "Yortw"/Flickr)
Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure \(\PageIndex{8}\)).
Figure \(\PageIndex{8}\): Bromine (the deep orange liquid on the left) and water (the clear liquid in the middle) are partially miscible. The top layer in the mixture on the right is a saturated solution of bromine in water; the bottom layer is a saturated solution of water in bromine. (credit: Paul Flowers)

Solutions of Solids in Liquids

The dependence of solubility on temperature for a number of inorganic solids in water is shown by the solubility curves in Figure \(\PageIndex{9}\). Reviewing these data indicate a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate.

Figure \(\PageIndex{9}\): This graph shows how the solubility of several solids changes with temperature.
The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure \(\PageIndex{10}\), take advantage of this behavior.
Figure \(\PageIndex{10}\): This hand warmer produces heat when the sodium acetate in a supersaturated solution precipitates. Precipitation of the solute is initiated by a mechanical shockwave generated when the flexible metal disk within the solution is "clicked." (credit: modification of work by "Velela"/Wikimedia Commons) Video \(\PageIndex{2}\): This video shows the crystallization process occurring in a hand warmer.

Summary

The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances' atoms, ions, or molecules. This tendency to dissolve is quantified as substance's solubility, its maximum concentration in a solution at equilibrium under specified conditions. A saturated solution contains solute at a concentration equal to its solubility. A supersaturated solution is one in which a solute's concentration exceeds its solubility-a nonequilibrium (unstable) condition that will result in solute precipitation when the solution is appropriately perturbed. Miscible liquids are soluble in all proportions, and immiscible liquids exhibit very low mutual solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature. The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry's law.

Glossary

Henry's law

law stating the proportional relationship between the concentration of dissolved gas in a solution and the partial pressure of the gas in contact with the solution

immiscible

of negligible mutual solubility; typically refers to liquid substances

miscible

mutually soluble in all proportions; typically refers to liquid substances

partially miscible

of moderate mutual solubility; typically refers to liquid substances

saturated

of concentration equal to solubility; containing the maximum concentration of solute possible for a given temperature and pressure

solubility

extent to which a solute may be dissolved in water, or any solvent

supersaturated

of concentration that exceeds solubility; a nonequilibrium state

unsaturated

of concentration less than solubility

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The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the ...

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Imagine you are drinking a cup of tea. You take a sip, grimace at how bitter it is, then grab some sugar. As you stir in the sugar, you watch it disappear as it dissolves into your now sweeter tea. The ability of sugar to dissolve is based on its solubility .

In this article, we will understand what factors affect solubility and why certain solids are soluble while others aren't.

This article is about solubility.

We will look into how temperature affects solubility based on Le Chatelier's Principle.

Then we will look at how solubility curves graph the change in solubility based on temperature

Then we will review the solubility rules for ionic solids

Lastly, we will calculate the solubility equilibrium constant (Ksp) to understand what we consider "slightly soluble"

Solubility Definition Chemistry

Let's start by looking at the definition of solubility.

Solubility is the maximum concentration of solute (a substance that dissolves in a solvent) that can be dissolved in the solvent (dissolver).

In our tea example, sugar is the solute being dissolved in the solvent (tea). Initially, we have an unsaturated solution, meaning that we haven't met the concentration limit and sugar can still dissolve. Once we add too much sugar, we end up with a saturated solution. This means we have met the limit, so any added sugar won't dissolve, and you'll end up drinking straight sugar granules.

Solubility and Temperature

Solubility is a function of temperature. When a solid is being dissolved, bonds are broken down, which means heat/energy is required. However, heat is also released when new bonds between the solute and solvent are made. Typically, the heat required is greater than the heat released, so it is an endothermic reaction (net gain of heat). However, there are some cases, like in Ca(OH) ₂, where the heat released is greater, so it is an exothermic reaction (net loss of heat).

So, how does this affect solubility? Depending on whether a reaction is endothermic or exothermic, the solubility can change based on Le Chatelier's Principle.

Le Chatelier's Principle states that if a stressor (heat, pressure, concentration of reactant) is applied to a system at equilibrium, the system will shift to try and minimize the effect of the stress.

Back to our tea example for earlier, let's say you really wanted your tea sweet, but aren't a fan of having to drink the solids bits. Would you need to increase or decrease the temperature to increase the solubility of sugar? Let's look at the reaction:

$$C_{12}H_{22}O_{11\,(s)}+\text{solvent}+\text{heat} \rightleftharpoons C_{12}H_{22}O_{aq}$$

The dissolution of sucrose (table sugar), is endothermic, so heat is a reactant. According to Le Chatelier's Principle, the system wants to minimize stress, so if we increase the temperature (i.e. add heat), the system wants to make more product to "use up" the heat added. This means that undissolved sugar will now be able to dissolve. We use solubility curves to graph the change in solubility based on temperature.

The curve above shows how solubility increases with temperature. Curves typically are based on how much solute dissolves in 100g of water, since it is the most common solvent. For solutes that have exothermic dissolving reactions, this curve is flipped.
How many more grams of sucrose can be dissolved if the temperature is increased from 40 to 50 °C? (Assume 100g of water)
Based on our curve, at 40 °C, about 240 g of sucrose can be dissolved. At 50 °C, it is about 260 g. So, we can dissolve ~20 g more sucrose if the temperature is increased by 10°

The fact that more solute can be dissolved at a higher temperature is used to form supersaturated solutions. In a supersaturated solution, a solution has more solute dissolved than its equilibrium solubility. This happens when more solute is dissolved at a higher temperature, then the solution is cooled without precipitating (returning to a solid) the solute.

Solubility Rules

Now that we've covered how solubility changes with temperature, it's now time to look at what makes something soluble in the first place. For ionic solids, there are solubility rules which determine whether they will dissolve or form a precipitate (i.e. stay a solid).

In the next section is a solubility chart with these rules.

Solubility Chart

As you can see, there are many solubility rules. When determining whether an ionic solid is soluble, it's important to reference your charts!
Categorize these compounds as either soluble, insoluble, or slightly soluble. a. MgF₂ b. CaSO₄ c. CuS d. MgI₂ e. PbBr₂ f. Ca(CH₃CO₂)2 g. NaOH
a. While fluorides are typically soluble, when it is bonded to Mg, it is insoluble.

b. Sulfates are also typically soluble, but when bonded to Ca, it is slightly soluble.

c. Sulfides are typically insoluble, and Cu is not one of the exceptions, so it is insoluble.

d. Halides are typically soluble, and Mg is not an exception, so it is soluble.

e. Bromine is typically soluble, but with Pb, it is slightly soluble.

f. Acetates are typically soluble, and Ca is not an exception, so it is soluble.

g. Hydroxides are typically insoluble, but when bonded to Na, it is soluble.

Ksp and Temperature

The solubility constant ( is the equilibrium constant for solids dissolving in an aqueous (water solvent) solution. It represents the amount of solute that can dissolve. For a general reaction: $$aA \rightleftharpoons bB + cC$$

The formula for K sp is: $$K_{sp}=[B]^b[C]^c$$

Where [B] and [C] are the concentrations of B and C.

The calculation uses the concentration of the ions, which is called their molar solubility. This is expressed in mol/L (M).

So, when we are referring to something that is "slightly soluble", we mean that it has a very low K sp. Let's look at a problem to further explain.

What is the K sp for PbCl ₂, when the concentration of Pb ²⁺ is 6.7 x 10-5 M?

The first thing we need to do is write out the balanced equation

$$PbCl_2 \rightleftharpoons Pb^{2+} + 2Cl^-$$

Since we know the concentration of Pb ²⁺, we can calculate the concentration of Cl⁻. We do this by multiplying the amount of Pb ²⁺ by the ratio of Pb ²⁺ to Cl⁻.

$$6.7*10^{-5}\,M\,\cancel{Pb^{2+}}*\frac{2\,M\,Cl^-}{1\,M\,\cancel{Pb^{2+}}}=1.34*10^{-4}\'M\,Cl^-$$

Now we can calculate K sp

$$K_{sp}=[Pb^{2+}][Cl^-]^2$$

$$K_{sp}=(6.7*10^{-5})({1.34*10^{-4}})^2$$

$$K_{sp}=1.20*10^{-12}$$

We can also use K sp to see how much how of a solute will dissolve.

The K sp of HgSO ₄ at 25 °C is 7.41 x 10⁻⁷, what is the concentration of SO ₄ ²⁻ that will be dissolved?

We first need to set up the chemical equation, then we can set up the equation for K sp.

$$HgSO_4 \rightleftharpoons 2Hg^+ + SO_4^{2-}$$

$$K_{sp}=[Hg^+]^2[SO_4^{2-}]$$

Now that we have set up our equation, we can solve for the concentration

$$7.41*10^{-7}={[Hg^+]^2}{[SO_4^{2-}]}$$

$$7.41*10^{-7}=[x]^2[x]$$

$$7.41*10^{-7}=x^3$$

$$x=9.05*10^{-3}\,M$$

One thing to note is that even insoluble compounds can have a K sp. The value of K sp is so small, however, that the molar solubility of the ions is negligible in solution. This is why it is considered "insoluble" despite some of it actually dissolving.

Also, K sp, like solubility, is dependent on temperature. It follows the same rules as solubility, so K sp will increase with temperature. It is standard that the K sp is measured at 25 °C (298K).

Solubility - Key takeaways

Solubility is the maximum concentration of solute (dissolvee) that can be dissolved in the solvent (dissolver).

If the dissolution of a compound is exothermic, increasing temperature will decrease solubility. If it is endothermic, an increase in temperature will increase solubility.

Solubility curves graph how solubility changes with temperature.

We can look at the solubility rules to determine if a compound is soluble, slightly soluble, or insoluble.

is the equilibrium constant for solids dissolving in an aqueous (water solvent) solution. It shows how soluble a compound is and can be used to determine molar solubility (concentration of solute dissolved).

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Solubility (Chemistry): ✓ Definition ✓ Formula ✓ Rules ✓ Chart ✓ Fiber ✓ Fat ✓ Water ✓ Examples ✓ StudySmarter Original

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Why Do Some Solids Dissolve in Water?

The sugar we use to sweeten coffee or tea is a molecular solid, in which the individual molecules are held together by relatively weak intermolecular forces. When sugar dissolves in water, the weak bonds between the individual sucrose molecules are broken, and these C ₁₂H ₂₂O ₁₁ molecules are released into solution.

It takes energy to break the bonds between the C ₁₂H ₂₂O ₁₁ molecules in sucrose. It also takes energy to break the hydrogen bonds in water that must be disrupted to insert one of these sucrose molecules into solution. Sugar dissolves in water because energy is given off when the slightly polar sucrose molecules form intermolecular bonds with the polar water molecules. The weak bonds that form between the solute and the solvent compensate for the energy needed to disrupt the structure of both the pure solute and the solvent. In the case of sugar and water, this process works so well that up to 1800 grams of sucrose can dissolve in a liter of water.

Ionic solids (or salts) contain positive and negative ions, which are held together by the strong force of attraction between particles with opposite charges. When one of these solids dissolves in water, the ions that form the solid are released into solution, where they become associated with the polar solvent molecules.

We can generally assume that salts dissociate into their ions when they dissolve in water. Ionic compounds dissolve in water if the energy given off when the ions interact with water molecules compensates for the energy needed to break the ionic bonds in the solid and the energy required to separate the water molecules so that the ions can be inserted into solution.

Discussions of solubility equilibria are based on the following assumption: When solids dissolve in water, they dissociate to give the elementary particles from which they are formed. Thus, molecular solids dissociate to give individual molecules

H ₂ O

C ₁₂H ₂₂O ₁₁( s)

C ₁₂H ₂₂O ₁₁( aq)

and ionic solids dissociate to give solutions of the positive and negative ions they contain.

When the salt is first added, it dissolves and dissociates rapidly. The conductivity of the solution therefore increases rapidly at first.

The concentrations of these ions soon become large enough that the reverse reaction starts to compete with the forward reaction, which leads to a decrease in the rate at which Na⁺ and Cl⁻ ions enter the solution.

Eventually, the Na⁺ and Cl⁻ ion concentrations become large enough that the rate at which precipitation occurs exactly balances the rate at which NaCl dissolves. Once that happens, there is no change in the concentration of these ions with time and the reaction is at equilibrium. When this system reaches equilibrium it is called a saturated solution, because it contains the maximum concentration of ions that can exist in equilibrium with the solid salt. The amount of salt that must be added to a given volume of solvent to form a saturated solution is called the solubility of the salt.

There are a number of patterns in the data obtained from measuring the solubility of different salts. These patterns form the basis for the rules outlined in the table below, which can guide predictions of whether a given salt will dissolve in water. These rules are based on the following definitions of the terms soluble, insoluble, and slightly soluble.

A salt is soluble if it dissolves in water to give a solution with a concentration of at least 0.1 moles per liter at room temperature.

A salt is insoluble if the concentration of an aqueous solution is less than 0.001 M at room temperature.

Slightly soluble salts give solutions that fall between these extremes.

Solubility Rules for Ionic Compounds in Water

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Why Do Some Solids Dissolve in Water? The sugar we use to sweeten coffee or tea is a molecular solid, in which the individual molecules are held together by relatively weak intermolecular forces. When sugar dissolves in water, the weak bonds between the individual sucrose molecules are broken, and these C 12H 22O 11 molecules are released into solution.

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Often experiments always involve things that can change, known as variables. Variables need to be identified, so they can then either be changed or controlled.
Plan an experiment to investigate one of the factors that affects solubility.
Make sure that you only change one variable, and keep the others constant for a valid investigation.

What will you measure, known as the ?

Write your method as a numbered list of brief instructions.

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Solutions are mixtures made when a solute dissolves into a solvent. Learn about solutions in this Key Stage 3 chemistry guide aged from BBC Bitesize.