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Daniel McIntyre Collegiate

Lesson 3 - factors Affecting Solubility

If you ever get a stain in your clothes or on your hands, you want to know how to dissolve that stain. If you want to make you drink sweet, you want to know how to dissolve more sugar. In this lesson we will answer these questions. We will study several factors that affect how much solute can dissolve in a given amount of solvent.

The Nature of Solvent and Solute

From the activity in the last lesson, we learned that the general rule when dissolving substances is “like dissolves like”. This means polar substances dissolve in polar solvents and non-polar substances are soluble in non-polar solvents.

Polar and charged substances dissolve well in polar solvents because of the electrostatic attraction between opposite charges. The attraction between the positive and negative ends of the water molecules and the polar ends of the solute hold the polar solute in solution. Many ionic compounds are very soluble in water because of the electrostatic interaction between water and the positive and negative ions. For example, as we saw earlier, when sodium chloride (NaCl) dissolves in water the positively charged sodium ions (Na+) are attracted to the negative oxygen atom in the water molecule and the negatively charged chloride ions (Cl) are attracted to the positive hydrogens in water.

Non-polar substances, like waxes and oils, are not soluble in water because the forces of attraction between individual water molecules are stronger than the attraction between the water and the oil. Recall that energy is required to overcome the electrostatic forces of attraction between the polar water molecules. That is, the positive hydrogen end of the water molecule is attracted to the negative, or oxygen, end of other water molecules. This is known as cohesion. Since oil is not polar, there is very little attraction between the water and oil molecules. The energy released when the oil mixes with the water is not sufficient to overcome the cohesion between water molecules.

On the other hand, oil does dissolve well in kerosene or turpentine because the forces of attraction between oil molecules are about as weak as the forces of attraction between carbon tetrachloride molecules. It is not difficult to separate the molecules of kerosene to make room for oil molecules.

How to Dissolve Oil in Water

 When cleaning stains in clothing, carpets, etc. we must take into account the nature of the stain. If the stain is of a non-polar nature, water will not remove the stain. We must use a non-polar solvent like carbon tetrachloride or soap. Soap is a compound that helps water dissolve more substances. Soaps can dissolve in both water and oils because they have a polar end and a non-polar tail. The polar head is usually a charged ion, while the non-polar tail is made of fats.

Soap is able to dissolve greasy stains because the non-polar tail digs into oil and grease and pulls it out of the clothing, or off your skin. The oil and soap form a small ball with the polar head on the out side and the oil or grease on the inside. The polar end is attracted to the water and is able to "dissolve" the soap and grease ball. The soap and grease ball is then washed down the drain.

Temperature

In general, a higher temperature increases the amount of solid that can dissolve in a liquid. We noticed this when we made the supersaturated solution. This is due to the energy required during the dissolving process. Adding heat supplies more energy to separate the solute particles and solvent particles. The more heat, the more particles that can enter solution.

Saturated, Unsaturated and Supersaturated Solutions

We will learn several ways to describe a solution. One way is in terms of how much solute is dissolved as compared to how much can dissolve. A solution that contains the maximum amount of solute for that amount of solvent at that particular temperature, is said to be saturated. If you add more solute to the solution, it will not dissolve. For example, if you add more sugar to a saturated sugar solution the added sugar will simply drop to the bottom of the container, or precipitate.

If a solution can hold more solute at that temperature, the solution is unsaturated. If you add more sugar to an unsaturated sugar solution, the sugar will continue to dissolve until the solution is saturated.

A supersaturated solution contains more solute than it normally can at that temperature. A supersaturated solution is very unstable. If you add just one crystal of solid to a supersaturated solution much more solute precipitates out of solution. Prepare a supersaturated solution by first making a saturated solution with excess solute on the bottom of the container, then slowly heating the solution until all solute is dissolved. When all the solute is dissolved, begin slowly cooling the solution. When the solution reaches room temperature, there will be more dissolved than was initially dissolved when the solution is saturated. Add a single crystal of solute or tap the side of the container and watch the excess solute precipitate out of solution. In the photographs below, a crystal is added to a supersaturated solution.

Solubility Units

 Solubility is the maximum amount of solute that can dissolve in a certain amount of solvent or the amount of solute needed to make a saturated solution, under certain conditions. The units of solubility are usually in terms of mass of solute, in grams, per 100 grams of solvent. Since the density of water is 1.00 g/mL, the solubility in water is sometimes given as grams of solute per 100 mL of water. Another set of units used for solubility is grams of solute per Litre of solvent, for a liquid solvent.

The solubility of a solute must be determined experimentally. One method used to determine the solubility of a substance at a specific temperature is to add a known mass of solute to a known mass, or volume, of solvent. The mixture is heated until all the solute dissolves. When the solute has dissolved, the solution is cooled until the first signs of solid crystals appear. The temperature at this point is recorded and the solubility is converted to the desired solubility units.

Example 1.

If 25.0 g of a solute is the maximum amount of solute that can dissolve in 40.0 g of solvent at a certain temperature, what is the solubility in grams of solute/100 g of solvent?

Solution.

For this type of problem, it is best to use the following:

The first term gives the grams of solute per gram or mL of solvent, then we can multiply by 100, for 100 g or mL of solvent or 1000 mL/L for grams per Litre.

To solve this problem,

You can also solve this problem by using ratios. The equation above is a modified form of a ratio. If you solve this problem using ratios, it would look like this:

You may choose the method that you like best.

Example 2.

If 30.1 g of a solute can dissolve can dissolve in 350.0 mL of water at a certain temperature, what is the solubility of the substance in g/100g water?

Solution.

We are given mL of water, which is volume. We need to convert the volume of water to mass of water. We will always assume the density of water is 1g/mL, unless indicated otherwise. So the mass of water is 350.0 g.

or by using ratios

 

The Solubility Curve

As we saw in the previous pages, solubility is temperature dependent. We can find the solubility of a substance at various temperatures and plot this data on a graph. The result is what is known as a solubility curve. The diagram below shows an example of a solubility curve for potassium nitrate.

 

Each point on the curve represents a saturated solution of potassium nitrate. The area below the curve represents quantities that produce an unsaturated solution at that temperature. The points above the curve indicate either a supersaturated solution or a saturated solution with some remaining undissolved solute.

 

Example 3.

What is the solubility of potassium nitrate at 20°C?

Solution.

Start at 20°C and move up to the curve, then across to the solubility axis.


 


The solubility of potassium nitrate at 20°C is about 33 g/100 g water.

Example 4.

At what temperature is the solubility of potassium nitrate 20 g/25 g water?

Solution.

The units in the graph are g/100 g water. We are given 20 g/25 g water. We need to convert the given amount into g/100g water.

or using ratios

Start at 80 g on the y-axis, move across to the curve, then down to the temperature axis.

The solubility of potassium nitrate is 20 g/25 g water at about 46°C.

Example 5.

Is a solution of 50 g of potassium nitrate in 100 g of water a saturated, unsaturated or supersaturated at 50°C?

Solution.

The y-axis indicates that the units of solubility are g/100g water. Using the graph, is 50g/100g water a saturated solution? Start on the x-axis at 50°C and move up to the 50 g mark.

Do you cross the solubility curve? No

The solution is unsaturated, since 90 g of KNO3 can dissolve in 100 g of water at 50°C.

Example 6.

A 75 mL of a saturated solution of KNO3 at 70°C is cooled to 40°C. How much solid precipitates from the solution?

Solution.

From the graph, a saturated solution of KNO3 at 70°C is 140 g/100 g water.

We only have 75 mL of water. Assuming the density of water is 1 g/mL, the mass of water is 75 g. We must convert the solubility read from the graph into g/75 g water. We do this for both temperatures given in the question: At 70°C:

At 70°C, 105 g of solute will dissolve.

If we cool the solution, the solubility of the solid will decrease. Reading the graph at 40°C, we see the solubility is now about 66 g/100g water. We must convert this value top g/75 g water.

At 40°C, only 49.5 g of solute will dissolve

If only 49.5 g remains in solution and we started with 105 g, then the amount that comes out for solution is

105 g – 49.5 g = 55.5 g

55g will precipitate out of solution upon cooling.

Example 7.

What volume of water is required to dissolve 240 g of of KNO3 at 60°C?

Solution.

From the graph,


at 60°C the solubility of KNO3 is 113 g/100 g water.

Using ratios,

About 212 g of water or 212 mL of water is needed.