Lesson Explainer: Solubility Chemistry

In this explainer, we will learn how to define and measure the solubility of a substance and explain its dependency on solvent, temperature, and pH.

A solution is a homogeneous mixture consisting of one or more solutes, the minor component(s), dissolved in a solvent, the major component. Aqueous solutions are solutions in which the solvent is water. The solute in an aqueous solution may be a solid, liquid, or gas.

Water is known as the universal solvent because it dissolves a great number of substances, but not all substances are able to dissolve in water. The solubility rules summarized in the table below provide general guidelines for determining which ionic substances can and cannot dissolve in water.

Soluble in WaterInsoluble in Water
Alkali metal compounds: Li+, Na+, K+, Rb+, and Cs+Carbonates: CO32
(except for alkali metals, NH4+)
Ammonium: NH4+Phosphates: PO43
(except for alkali metals, NH4+)
Nitrates: NO3Sulfides: S2
(except for alkali metals, NH4+, Ca2+, Ba2+, and Sr2+)
Bicarbonates: HCO3Hydroxides: OH
(except for alkali metals, NH4+, Ca2+, Ba2+, and Sr2+)
Chlorates: ClO3
Halides: Cl, Br, and I
(except for Ag+, Hg22+, and Pb2+)
Sulfates: SO42
(except for Ag+, Ca2+, Sr2+, Ba2+, Hg22+, and Pb2+)
Acetates: CHO232

Substances that can dissolve are called soluble, whereas those that cannot are called insoluble.

Definition: Soluble Substance

A soluble substance is a substance that can be dissolved.

Definition: Insoluble Substance

An insoluble substance is a substance that cannot be dissolved.

Example 1: Using the Water Solubility Rules to Determine Which Chloride Compound Is Insoluble

Which of the following ionic compounds of chloride is insoluble in water?

  1. Potassium chloride
  2. Silver chloride
  3. Aluminum chloride
  4. Magnesium chloride
  5. Sodium chloride

Answer

A water-insoluble compound is a compound that cannot be dissolved in water. The water solubility rules summarized in the table below can be used to determine which ionic substances are soluble and which are insoluble.

Soluble in WaterInsoluble in Water
Alkali metal compounds: Li+, Na+, K+, Rb+, and Cs+Carbonates: CO32
(except for alkali metals, NH4+)
Ammonium: NH4+Phosphates: PO43
(except for alkali metals, NH4+)
Nitrates: NO3Sulfides: S2
(except for alkali metals, NH4+, Ca2+, Ba2+, and Sr2+)
Bicarbonates: HCO3Hydroxides: OH
(except for alkali metals, NH4+, Ca2+, Ba2+, and Sr2+)
Chlorates: ClO3
Halides: Cl, Br, and I
(except for Ag+, Hg22+, and Pb2+)
Sulfates: SO42
(except for Ag+, Ca2+, Sr2+, Ba2+, Hg22+, and Pb2+)
Acetates: CHO232

The easiest way to use this table is to locate the anion. All the answer choices are ionic compounds of chloride (Cl), a halide. We can see in the table that halides are soluble in water unless the cation is Ag+, Hg22+, or Pb2+. Looking at the answer choices, we see that answer choice B is silver chloride, an ionic compound consisting of Ag+ and Cl. Therefore, the ionic compound of chloride that is insoluble in water is answer choice B, silver chloride.

When two aqueous solutions containing dissolved ions are mixed, the resulting solution may produce a precipitate. A precipitate is an insoluble product that separates from a solution. A chemical reaction that produces a precipitate is known as a precipitation reaction.

Definition: Precipitate

A precipitate is an insoluble product that separates from a solution.

Let us consider the following chemical equation: KCl()+AgNO()KNO()+AgCl()aqaqaqs33

We can see from the state symbols that two aqueous solutions, KCl and AgNO3, were combined. This combination resulted in a new solution of KNO3 and a solid, AgCl. This solid silver chloride is a precipitate. The water solubility rules can be used to determine whether a product produced from the combination of two aqueous solutions is an insoluble precipitate.

Example 2: Determining Which Reaction Will Produce a Precipitate

For which of the following reactions is a precipitate likely to form?

  1. Pb(NO)()+MgSO()324aqaq
  2. Ca(OH)()+2HCl()2aqaq
  3. NaCO()+HSO()2324aqaq
  4. HNO()+NH()33aqaq
  5. CuO()+HSO()saq24

Answer

In order to answer this question, we need to identify the products of each reaction. The reactions in answer choices A, B, and E can all be treated as double-replacement reactions. In a double-replacement reaction, the cation and anion in each reactant dissociate in water. If any combination of cation and anion forms an insoluble product, a precipitate forms. An insoluble substance is a substance that does not dissolve. We can determine whether a substance is soluble in water or not using the water solubility rules summarized in the table below.

Soluble in WaterInsoluble in Water
Alkali metal compounds: Li+, Na+, K+, Rb+, and Cs+Carbonates: CO32
(except for alkali metals, NH4+)
Ammonium: NH4+Phosphates: PO43
(except for alkali metals, NH4+)
Nitrates: NO3Sulfides: S2
(except for alkali metals, NH4+, Ca2+, Ba2+, and Sr2+)
Bicarbonates: HCO3Hydroxides: OH
(except for alkali metals, NH4+, Ca2+, Ba2+, and Sr2+)
Chlorates: ClO3
Halides: Cl, Br, and I
(except for Ag+, Hg22+, and Pb2+)
Sulfates: SO42
(except for Ag+, Ca2+, Sr2+, Ba2+, Hg22+, and Pb2+)
Acetates: CHO232

We can begin with answer choice A by identifying the cations and anions in the two reactants. The cations are Pb2+ and Mg2+, and the anions are NO3 and SO42. We need to determine whether any of the combinations of cations and anions will form an insoluble precipitate.

The easiest way to accomplish this is to locate the anions in the table. Nitrates (NO3) are always soluble in water. Sulfates (SO42) are soluble in water unless Ag+, Ca2+, Sr2+, Ba2+, Hg22+, or Pb2+ is present. Pb2+ is present in the solution. It can combine with the sulfate ion to form PbSO4, an insoluble precipitate. Therefore, we identified the reaction that can form an insoluble precipitate.

The complete balanced chemical reactions for answer choices B through E are shown below. We can verify that none of the reactions produce a solid precipitate by comparing the products to the water solubility table. We can also see that all of the products are either aqueous, liquid, or gaseous. Notice that reaction D produces dissolved ions via an acid–base reaction.

B. Ca(OH)()+2HCl()CaCl()+2HO()222aqaqaql

C. NaCO()+HSO(NaSO()+HO()+CO()23242422aqaq)aqlg

D. HNO()+NH()NO()+NH()3334+aqaqaqaq

E. CuO()+HSO()CuSO()+HO()saqaql2442

A precipitate is likely to form via the reaction of lead(II) nitrate with magnesium sulfate, that is, answer choice A.

Even if a solute is soluble, only a certain amount of it can be dissolved in a solvent at a given temperature. Solutions containing the maximum amount of solute that can be dissolved in the solvent at a particular temperature are called saturated solutions.

Definition: Saturated Solution

A saturated solution is a solution that contains the maximum amount of solute dissolved in a solvent at a given temperature.

The composition of a saturated solution expressed in terms of a proportion of solute to solvent is called the solute’s solubility.

Definition: Solubility

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a given temperature.

Solubility is commonly expressed as the amount of solute in grams that can dissolve in 100 grams or 100 mL of water at a given temperature. For example, the solubility of sodium acetate is 119 g per 100 g of water at 0C, whereas 100 g of water at 100C can contain 170.15 g of sodium acetate.

Example 3: Calculating the Solubility of Glucose from Experimental Data

A student prepares a saturated solution of glucose in water at 20C. The student then dries the solution to remove the solvent and obtain the solid. By using the experimental data in the table provided, what is the solubility of glucose at 20C? Give your answer as a whole number.

Mass of evaporating dishMass of evaporating dish + saturated solutionMass of evaporating dish + solid (glucose)
43.56 g56.33 g49.61 g

Answer

The solubility of a substance is the maximum amount of solute that can dissolve in a given amount of solvent at a particular temperature. In this problem, we need to determine both the amount of glucose that was able to dissolve and the amount of water that it was dissolved in at 20C.

By subtracting the mass of the evaporating dish from the mass of the evaporating dish and the mass of glucose, 49.6143.56,gg we can determine the amount of glucose dissolved in the solution to be 6.05 g.

The mass of the saturated solution is the total mass of the glucose and water. Thus, we can subtract the mass of the evaporating dish and the mass of the glucose from the mass of the evaporating dish and the mass of the saturated solution: 56.3343.566.05.ggg

Therefore, the mass of the water is 6.72 g.

We now know that the solubility of glucose is 6.05 g per 6.72 g of water at 20C. However, solubility is commonly reported per 100 g of water. We can set up a proportion to scale the solubility of glucose in 6.72 g of water to 100 g of water: 6.056.72=𝑥100,gglucosegwatergwater

Where 𝑥 represents the maximum mass of glucose that can dissolve in 100 g of water at 20C. We can rearrange the equation to solve for 𝑥: 0.900297619=𝑥10090.0297619=𝑥.gglucosegwatergwatergglucose

Giving our answer as a whole number, we have determined the solubility of glucose at 20C to be 90 g per 100 g of water.

If less than the maximum amount of solute is dissolved in the solvent at a particular temperature, the solution is considered to be unsaturated. If only 80 g of sodium acetate was dissolved in 100 g of water at 0C, the solution would be unsaturated.

Definition: Unsaturated Solution

An unsaturated solution is a solution that contains less than the maximum amount of solute dissolved in a solvent at a given temperature.

As seen with sodium acetate, the solubility of a solute is affected by temperature. The graph below shows the solubility of several substances at various temperatures. The line representing the solubility of a substance over temperature is called a solubility curve.

We can see from the graph that the solubility of NaNO3 and KNO3 greatly increases with increasing temperature, while the solubility of NaCl is only slightly affected by increasing temperature. In general, the solubility of a solid in water increases with increasing temperature, although we can see from the solubility curve of NaSO24 that this is not always the case.

When looking at the solubility curves of gases, such as NH3 and SO2, we can see that the solubility of gases tends to decrease with increasing temperature.

Example 4: Interpreting a Solubility–Temperature Graph

The graph provided shows the solubility curve for four different substances.

  1. What is the solubility of aluminum nitrate at 12C?
  2. At what temperature is the solubility of ammonium chloride and potassium nitrate the same?
  3. A student wants to dissolve exactly 40 g of potassium chloride in 100 g of water. To what temperature will they need to heat the water?

Answer

Part 1

This question is asking us to determine the solubility of aluminum nitrate at a given temperature. The solubility of a substance is the maximum amount of that substance that can be dissolved in a given amount of water at a specified temperature. The graph provided shows the solubility values of four substances in 100 g of water at various temperatures.

To determine the solubility of aluminum nitrate at 12C, we can draw a line up from 12C on the 𝑥-axis to the aluminum nitrate curve and then a line from this point horizontally across to the 𝑦-axis.

This gives us a value of 68 g per 100 g of water. Therefore, the solubility of aluminum nitrate at 12C is 68 g per 100 g of water.

Part 2

To determine the temperature at which the solubility values of ammonium chloride and potassium nitrate are the same, we need to identify where their solubility curves intersect.

By drawing a line from the point of intersection to the 𝑥-axis, we can see that the curves intersect at 26C. Therefore, the solubility values of ammonium chloride and potassium nitrate are the same at 26C.

Part 3

We need to determine the temperature necessary to dissolve exactly 40 g of potassium chloride in 100 g of water. We can draw a line across from the mark of 40 g per 100 g of water on the 𝑦-axis to the potassium chloride curve and then a line from this point vertically down to the 𝑥-axis.

This gives us a value of 40C. Therefore, the student will need to heat the water to 40C in order to dissolve exactly 40 g of potassium chloride in 100 g of water.

For solutes that exhibit acidic or basic properties, the pH of the solution can affect the solubility of the solute. Magnesium hydroxide (Mg(OH)2) is a base that is insoluble in water, although a very small amount, 1.2 mg/100 g of water, can dissolve to produce Mg2+ and OH ions: Mg(OH)()Mg()+2OH()22+saqaqHO2

The addition of a base to the resulting solution would increase the pH and add additional OH ions to the solution. According to Le Chatelier’s principle, a system at equilibrium responds to a stress by shifting in the direction that minimizes this stress. In response to the additional OH ions, the equilibrium will shift to the left, producing more solid magnesium hydroxide, thus decreasing the solubility of the solute: Mg(OH)()Mg()+2OH()22+saqaqHO2

However, the addition of an acid to this solution would decrease the pH and add H+ ions to the solution. These H+ ions can react with the OH ions to produce water: H()+OH()HO()+2aqaql

The reaction between H+ and OH removes the hydroxide ions from the solution. In response, the magnesium hydroxide equilibrium shifts to the right, producing more OH ions, thus increasing the solubility of the solute: Mg(OH)()Mg()+2OH()22+saqaqHO2

The opposite effect is observed with acids that are insoluble. Therefore, the solubility of insoluble bases tends to increase in acidic solutions (at low pH), whereas the solubility of insoluble acids tends to increase in basic solutions (at high pH).

Soluble solutes dissolve in the solvent on their own over time, given that the maximum amount of solute that can dissolve is not exceeded. However, the rate of dissolution, the rate at which the solute dissolves, can be affected by the solute’s particle size, stirring, or temperature.

Let us consider sugar, which has a solubility of 200 g per 100 g of water at 25C. If we were to place a 50 g sugar cube, 50 g of granulated sugar, and 50 g of powdered sugar into separate containers each containing 100 g of water at 25C, we would expect that the sugar would be able to dissolve in each case.

Sugar cubes in white bowl
Sugar in bowl
Bowl of powder sugar

However, we will find that although each type of sugar dissolved in each container, the powdered sugar dissolved the fastest and the sugar cube dissolved the slowest. This is because the finer particles of the powdered sugar have a higher total surface area in contact with the solvent than that of the sugar cube. Hence, decreasing the particle size of the solute or increasing the surface area in contact with the solvent increases the rate of dissolution.

Imagine that we placed 50 g of granulated sugar into two containers each containing 100 g of water at 25C and then we stirred one of the containers.

We will find that the sugar that has been stirred dissolved faster than the sugar that has not been stirred. This is because stirring increases the number of interactions between the solute and the solvent. Hence, stirring a solute in a solvent increases the rate of dissolution.

Imagine that we placed 50 g of granulated sugar into two containers each containing 100 g of water at 25C and then we heated one of the containers.

We will find that the sugar in the container that has been heated dissolved faster than the sugar that remained at 25C. This is because heating the solute and solvent increases the kinetic energy of the particles, which in turn increases the number of solute–solvent interactions. Therefore, in general, increasing the temperature increases the rate of dissolution.

Example 5: Ranking the Rates of Dissolution of Sugar in Tea

The diagram below shows five beakers containing tea at various temperatures. An equal mass of sugar is placed into each beaker. Some of the beakers are stirred. What is the most likely order, from quickest to slowest, for the sugar in each beaker to completely dissolve?

Answer

The rate of dissolution, or the rate at which a solute dissolves, is affected by the solute’s particle size, stirring, and temperature. In general, the rate of dissolution increases with decreasing particle size, increased stirring, and increasing temperature.

The temperature difference from 20C to 80C will have a significant effect on the rate of dissolution. Increasing the temperature increases the kinetic energy of the particles, which increases the number of solute–solvent interactions. Thus, the sugar is solutions C, D, and E will dissolve faster than the sugar in solutions A and B.

Comparing solutions A and B, we can see that solution B contains powdered sugar, whereas solution A contains a sugar cube. Powdered sugar has a smaller particle size. This smaller particle size increases the total surface area in contact with the solvent. Thus, powdered sugar dissolves faster than a sugar cube. This tells us that the sugar in solution A will dissolve the slowest, followed by that in solution B.

Solutions C, D, and E are all at 80C. Solution E contains powdered sugar and is stirred. The increased temperature, stirring, and small particle size of the solute mean that the sugar in solution E should dissolve the fastest.

Solutions D and E are both at the same temperature, and both contain a cube of sugar. However, solution C is stirred. Stirring increases the number of solute–solvent interactions, which increases the rate of dissolution. Therefore, the sugar in solution C will dissolve faster than that in solution D.

The most likely order, from quickest to slowest, for the sugar in each beaker to completely dissolve is E C D B A.

Key Points

  • The rules of solubility can be used to determine whether a solute is soluble in water.
  • A precipitate is an insoluble product that separates from a solution.
  • Increasing the temperature tends to increase the solubility of solid solutes and decrease the solubility of gaseous solutes.
  • The solubility of insoluble bases tends to increase with decreasing pH, whereas the solubility of insoluble acids tends to increase with increasing pH.
  • The rate of dissolution tends to increase with decreasing particle size, increased stirring, and increasing temperature.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.