Introduction to solubility equilibria | Equilibrium | AP Chemistry | Khan Academy

Let's say we have a beaker of distilled water at 25 degrees Celsius, and to the beaker, we add some barium sulfate. Barium sulfate is a white solid. A small amount of the barium sulfate dissolves in the water and forms barium 2 plus ions in solution and sulfate ions in solution. So, let me draw those in on our diagram.

We're going to form some barium 2 plus ions and some sulfate anions, but most of the barium sulfate remains undissolved. With that, we'll draw that here, sitting on the bottom of the beaker. Barium sulfate can dissolve to form barium two plus ions and sulfate anions in solution, and it's possible for the barium two plus ion to combine with the sulfate anion to form a precipitate of barium sulfate.

When the rate of dissolution is equal to the rate of precipitation, the system is at equilibrium. These types of equilibria are referred to as solubility equilibria, and when the system is at equilibrium, the concentrations of barium 2 plus ion and sulfate anion in solution are constant, and the amount of solid is constant too. This forms a saturated solution.

The balanced equation shows the dissolution of a salt, barium sulfate, and from the balanced equation, we can write an equilibrium constant expression. So, we would write the equilibrium constant K is equal to the concentration of barium two plus. Since there's a coefficient of one in the balanced equation, it'd be the concentration raised to the first power times the concentration of sulfate, also raised to the first power. Since pure solids are left out of equilibrium constant expressions, we would not include the solid barium sulfate.

For solubility equilibria, we would write Ksp, where sp stands for solubility product. The solubility product constant Ksp has only one value for a given salt at a specific temperature. That temperature is usually 25 degrees Celsius, and Ksp indicates how much of that salt will dissolve. For example, at 25 degrees Celsius, the Ksp value for barium sulfate is 1.1 times 10 to the negative 10th. When the Ksp value is much less than 1, that indicates the salt is not very soluble.

So, barium sulfate is not a soluble salt. If the Ksp value is greater than one, like it is for something like sodium chloride, that indicates a soluble salt that dissolves easily in water. The solubility of a substance refers to the amount of solid that dissolves to form a saturated solution. Usually, the units for solubility are in grams per liter. Molar solubility refers to the number of moles of the solid that dissolve to form one liter of the saturated solution, and therefore, the units would be moles per one liter, or you could just write molar.

Ksp values can be used to predict the relative solubilities of salts that produce the same number of ions in solution. For example, silver chloride, silver bromide, and silver iodide all produce two ions in solution. Let's look at the dissolution equation for silver chloride to see why this is true. Solid silver chloride turns into Ag plus and Cl minus, so that's one Ag plus ion and one Cl minus ion for a total of two ions in solution.

We could write out similar equations for silver bromide and silver iodide, so they all produce two ions in solution. However, a salt like lead II chloride produces three ions in solution. So, lead II chloride would give one Pb2 plus ion and two chloride anions in solution. One plus two is three ions. Since lead II chloride produces three ions in solution, we can't determine its solubility relative to the other three by comparing Ksp values.

Here are the Ksp values for the three salts at 25 degrees Celsius: for silver chloride, it's 1.8 times 10 to the negative 10th; for silver bromide, it's 5.0 times 10 to the negative 13th; and for silver iodide, it's 8.3 times 10 to the negative 17th. When comparing salts that produce the same number of ions, the higher the value of Ksp, the higher the solubility of the salt. Since silver chloride has the highest Ksp value of these three, silver chloride is the most soluble salt.

For some insight into why this is true, let's look at the Ksp expression for silver chloride, which we can get from the balanced equation. The higher the value for Ksp, the higher the concentration of these ions at equilibrium, which means that more of the solid must have dissolved. Therefore, silver chloride has the highest solubility out of these three salts.

Let's say we have some solid calcium fluoride that we add to pure water at 25 degrees Celsius. Eventually, equilibrium is reached, and the equilibrium concentration of calcium 2 plus ions is measured to be 2.1 times 10 to the negative fourth molar. Our goal is to calculate the Ksp for calcium fluoride at 25 degrees Celsius.

The first step is to write out the dissolution equation for calcium fluoride. So, we would write CaF2 solid, and we know that calcium forms a two plus cation, so we would write Ca2 plus in aqueous solution. To balance everything, we'd need two fluoride anions, so 2 F minus, also in aqueous solution. The next step is to use the balanced equation to write the Ksp expression.

So, Ksp is equal to—there's a one as a coefficient in front of Ca2 plus—it would be the concentration of Ca2 plus raised to the first power times the concentration of fluoride anion, and since there's a 2 as a coefficient, this is the concentration of fluoride anion squared for a Ksp expression. These are equilibrium concentrations, and we already know the concentration of calcium 2 plus at equilibrium is 2.1 times 10 to the negative fourth, so that can be plugged in for the equilibrium concentration of Ca2 plus.

So, here's our expression with 2.1 times 10 to the negative fourth plugged in. Next, we need to plug in the equilibrium concentration of fluoride anion. Well, looking at the dissolution equation, the mole ratio of calcium 2 plus to fluoride anion is one to two. Therefore, at equilibrium, there are twice as many fluoride ions in solution as there are calcium 2 plus ions. Therefore, the equilibrium concentration of the fluoride anion would just be twice this concentration for Ca2 plus.

So the equilibrium concentration of fluoride anion must be 4.2 times 10 to the negative fourth molar. And when you do the math, you get that Ksp for calcium fluoride is equal to 3.7 times 10 to the negative 11th at 25 degrees Celsius. Ksp values can be difficult to measure, and therefore, different sources often give different values for Ksp at the same temperature.

For example, for calcium fluoride at 25 degrees Celsius, one source had Ksp equal to 3.5 times 10 to the negative 11th, another one had 3.9 times 10 to the negative 11th. Since we got 3.7 times 10 to the negative 11th, this sounds like a pretty good calculation for the numbers that we used for our problem.

Finally, let's think about the molar solubility of calcium fluoride. So how many moles of our salt dissolve to form one liter of our saturated solution? Well, the mole ratio of calcium 2 plus ions to calcium fluoride is one to one. Therefore, the concentration of calcium 2 plus ions in solution, 2.1 times 10 to the negative fourth molar, that number must also be the molar solubility of calcium fluoride.

Therefore, for this problem, we could say that we use the molar solubility of calcium fluoride to calculate the Ksp value for calcium fluoride.