# Lesson Video: Solubility Product Chemistry

In this video, we will learn how to explain and calculate the solubility product.

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### Video Transcript

In this video, we will learn about the solubility product constant. We will learn how to calculate it from ion concentrations in solution. And we will learn how the value of this constant indicates the degree to which a solute dissolves.

What is the solubility product constant? The solubility product constant, or Ksp, is a special equilibrium constant used when a solid ionic substance dissolves or dissociates in solution. The ionic substance must be a sparingly soluble substance. Sparingly soluble means it dissolves to a very small degree or hardly at all, and the solution must be water based. So for an ionic compound, let’s call it A, in equilibrium in aqueous solution with its ions C+ and D−, the equilibrium equation would be aA solid in water giving cC+ aqueous plus dD− aqueous. Of course, the charges on these ions may vary. They may be one plus or two plus or three minus, depending on the ionic compound. So for simplicity, let’s just remove the plus and minus signs for now.

We can calculate the Ksp in much the same way as a normal equilibrium constant. Ksp is the solubility product constant, and this is equal to the concentration of the products multiplied with each other and each raised to a power equal to its stoichiometric coefficient. We can see how the concentration of product C is raised to a power of c, and the concentration of product D is raised to a power of d because D has a coefficient of value small d. Now, because the starting substance is a solid and because these solids are sparingly soluble, we can eliminate the denominator. Solids do not affect the equation. Their concentrations remain essentially constant, so we can remove them from the Ksp expression.

The concentration of the products are used in molarity units, moles per liter, which is the same as moles per dm cubed. Just as with other equilibrium constants, Ksp is dependent on temperature. This is because temperature affects how much solute dissolves. So we know the formula equation for Ksp and the units for the concentration of each product. But how is Ksp useful to us? Ksp is a measure of how much solute dissolves or the molar solubility of a compound. The more soluble a substance is, the higher its Ksp value. So if there is a relatively high concentration of dissolved particles in solution, Ksp will be higher. But if there’s a relatively lower concentration of these dissolved particles, Ksp will be lower.

Before we do some calculations, I want to point out something very important. Sometimes we talk about calculating the Ksp for an insoluble salt. In reality, all salts will dissolve to some degree. No compound is completely insoluble. The Ksp values of these so-called insoluble salts is very small. Now let’s practice writing some Ksp expressions and calculating some Ksp values.

In this example, the question is, “At 298 kelvin, the solubility of barium sulfate is 1.05 times 10 to the negative five moles per dm cubed. Calculate the solubility product Ksp. The equilibrium equation is BaSO4 solid dissolved in water to give Ba2+ aqueous plus SO42− aqueous.”

The equation is balanced. Ksp is equal to the concentration of the aqueous products multiplied by each other. Notice that the barium ion is raised to a power of one because its coefficient is one, and the sulfate ion is also raised to a power of one. We are not given the concentrations of the ions, but we do know that for every one barium sulfate, there is one barium ion and one sulfate ion. And because they are all together in the same volume of solution, we know that the concentration of the barium and sulfate ions will be the same as each other and will be equal to the solubility of the barium sulfate.

This is a small solubility value because barium sulfate is insoluble in water or, more technically correct, very poorly soluble. So substituting the solubility value into the equation, we’ll get an answer of 1.10 times 10 to the negative 10. Note that the units also multiply with each other, giving us the strange unit in the answer, mole squared per dm to the sixth. The unit for Ksp is different for different substances, and sometimes the unit is left out altogether. In this example, because the ratio of barium to sulfate ions was one as to one, we got this particular unit.

Let’s look at another example.

For the equation given, determine Ksp in terms of 𝑥 where 𝑥 is equal to the concentration of the lead ion or Pb2+ and the equation is PbCl2 solid dissolved in water to give Pb2+ aqueous plus 2Cl− aqueous.

Lead chloride is poorly soluble in water. The Ksp expression for this equilibrium is the concentration of the lead two plus product raised to the power of one multiplied by the concentration of the chloride product raised to the power of two. We are not given the concentrations of these ions, but we are given the concentration of the lead ion in terms of 𝑥. So if lead’s concentration is 𝑥 and for every one lead ion, there are two chloride ions, then the concentration of the chloride ions must be two 𝑥. And we mustn’t forget to square the concentration of the chloride ions. We can solve this different ways, but one way is to multiply everything out and we get an answer of four 𝑥 cubed.

Now we do not know the unit of 𝑥. But imagine the unit was mole per dm cubed. We would have mole per dm cubed multiplied by two mole per dm cubed multiplied by two mole per dm cubed. This would give a final answer with a unit of moles cube per dm to the nine. So remember in Ksp calculations to always write out the units so that you get the correct unit at the end.

So far, this is what we have learnt. When an ionic solid A dissolves sparingly in aqueous solution to form positive and negative ions C+ and D−, this is the corresponding Ksp expression. And I removed the positive and negative signs because this is just a general equation and a general Ksp expression. We know Ksp is a constant for equilibrium conditions and an equilibrium will only form in a saturated solution. But what if the following occurred? Imagine you multiplied the molar concentrations of ions C and D with each other for a particular solution. And the answer you got was less than the Ksp. This would only happen in the case of an unsaturated solution. This means more of the ionic substance A can dissolve.

Unsaturated solutions are not in equilibrium, and so we cannot use the Ksp expression for an unsaturated solution. In this last scenario, if we multiply the ion concentrations with each other in a particular solution and we get a value greater than the equilibrium solubility constant, this tells us that a precipitate will form. The concentrations of C and D in solution are too high for a stable solution, and solid ionic substance A will begin to form to decrease the concentrations of C and D and bring the system back to equilibrium, which is more stable. Now It’s time for some practice.

Taking the solubility product of aluminum hydroxide to be 1.90 times 10 to the negative 33 at 298 kelvin, what is the concentration of Al3+ ions in a saturated solution? Give your answer in scientific notation to two decimal places. And the answer options are (A) 2.90 times 10 to the negative nine moles per dm cubed. (B) 5.02 times 10 to the negative nine moles per dm cubed. (C) 6.60 times 10 to the negative nine moles per dm cubed. (D) 5.95 times 10 to the negative 12 moles per dm cubed. Or (E) 2.58 times 10 to the negative 11 moles per dm cubed.

We are given a solubility product, or Ksp value, of 1.9 times 10 to the negative 33. This is for aluminum hydroxide or Al(OH)3. And we are asked to find the concentration of aluminum three plus ions in a saturated solution. Now aluminum hydroxide is sparingly soluble. It only partially dissolves in water to give Al3+ ions and OH− or hydroxide ions according to this balanced equation. We are told that the solution is saturated, so we know that it is at equilibrium. We can write the Ksp expression for this sparingly soluble salt. Ksp or the solubility product constant is equal to the molar concentration of the products multiplied with each other. So we can write both ions in square brackets.

Now the stoichiometric coefficient of Al3+ is one. And so we raise the concentration of Al3+ to the power of one. And the stoichiometric coefficient of the hydroxide ion is three. So we raise the concentration of the hydroxide ion to a power of three. Now we are looking for the concentration of the Al3+ ion. So putting in some data that we know, specifically the Ksp value of 1.90 times 10 to the negative 33 — and by the way, in this example, we are not given the units of Ksp though Ksp does have units — and knowing that according to the balanced equation for every one mole of the solid ionic substance, one mole of aluminum ions is formed and three moles of hydroxide ions.

So if we make the concentration of Al3+ ions equal to 𝑥, we could even write let the concentration of Al3+ equal 𝑥, then the concentration of hydroxide ions must be three 𝑥 because of the stoichiometric coefficients. You can write this information down on the side if you wish. We must remember to raise the concentration of hydroxide ions three 𝑥 to a power of three. You don’t have to expand three 𝑥 to the three, but I have here just to make it clear what numbers we are dealing with and to do our calculation correctly. Multiplying everything out, we get 27𝑥 to the power of four. Let’s solve for 𝑥. So simplifying, we get 1.90 times 10 to the negative 33 divided by 27 equals 𝑥 to the four. To get rid of the fourth power, we take the fourth root of both sides, and this gives 𝑥 equals 2.90 times 10 to the negative nine.

Now 𝑥 is equal to the concentration of Al3+, which is what the question was asking for. Since the concentrations of the ions in a Ksp expression are given as molar concentrations, we can write the unit for 𝑥 as moles per dm cubed. The answer is in two decimal places and in scientific notation. So the concentration of Al3+ ions in the saturated solution is 2.9 times 10 to the negative nine moles per dm cubed.

Let’s summarize everything we have learnt. Ksp is the solubility product constant used for sparingly soluble ionic compounds which dissolve in aqueous solution. Ksp is an equilibrium constant. For the general equilibrium equation shown here, we get the following Ksp expression. Ksp is equal to the concentration of ion product C raised to the power c multiplied by the concentration of ion product D raised to the power d.

Ksp concentrations are molar concentrations. We learnt that the more soluble an ionic substance is, the higher its Ksp value and that Ksp can be calculated for an insoluble salt because nothing is truly insoluble in reality. And finally, we learnt that if the concentration of the products multiplied by each other and raised to their respective coefficients is not equal to Ksp, then the system is not in equilibrium.