Solubility Product Constant (Ksp) - Online Tutor, Practice Problems & Exam Prep

Solubility Product Constant (Ksp) Concept 1

Now recall that solubility is a chemical property that deals with the ability of a solute to become dissolved in a solvent. Now connected to this idea of solubility is a new term, our solubility product constant, which uses the variable \( K_{sp} \). This is the equilibrium constant that deals with the solubility of ionic solids. So if we're looking at the solubility of ionic solids, we're dealing with \( K_{sp} \). We're going to say here it deals with their solubility, and solubility can also be referred to as concentration or molarity. Remember molarity uses capital M. The basic idea is the higher your \( K_{sp} \) value, then the more soluble your ionic solid. And the smaller your \( K_{sp} \) value, then the less soluble your ionic solid; the less of it will dissolve within a solvent. So just keep this in mind when looking at \( K_{sp} \).

Solubility Product Constant (Ksp) Example 1

Here it asks, which substance is the most soluble? So here we have silver chloride, magnesium carbonate, calcium sulfate, and copper(II) sulfide, each of them with their given Ksp value. Remember, we want the most soluble of all of these choices. Remember, we're going to say the greater your Ksp value, then the more soluble you are as an ionic solid. So if we look here, we'd say that the answer has to be option C, since it's 7.1×10-5. It has the smallest negative, so it's the largest value there. It would be the most soluble out of all four choices. So here, option C would be the correct answer.

Solubility Product Constant (Ksp) Concept 2

Now, placing an ionic solid within a solvent involves two competing processes. We have the dissolution reaction, so basically where it's being broken up, and we also have its reverse. Okay? So dissolution is looking at the forward direction of the reaction where ionic solid dissolves into its ions. Here we have a b3a2 solid. So if we were to break this up into its ions, we'd say we have two a's. Remember this 3 didn’t come from b, it came from a. So, this is a 3+ ions in solution are aqueous, plus three b, the 2 that a possesses didn’t come from a, came from b, 2 minus aqueous. And remember, we know that the first ion is positive and the second one is negative because that’s the order in which we write ionic compounds. Now we're going to say that from this equilibrium equation, its equilibrium expression can be determined. The equilibrium expression is just the ratio of concentrations of products over reactants. And we're going to say that recall the equilibrium expression ignores solids and liquids for its ratios. Right? So, it's important that we know how to break up our ionic solids into their respective ions in order to successfully write your equilibrium expression later on. So keep this in mind in terms of the parameters in which it operates. It's products over reactants and it ignores solids and liquids.

Solubility Product Constant (Ksp) Example 2

Here we need to provide the equilibrium expression for calcium nitrate. So step 1 says we need to write the equilibrium equation by breaking up the ionic solid into its aqueous ions. Remember, in solution, our ions are in the aqueous form. Alright. So we have calcium nitrate solid. It breaks up into calcium, which is in group 2A, so it's \( Ca^{2+} \) (that's where this 2 came from). Not only is that where the 2 came from, but that also means that we have 2 nitrate ions, \( 2\,NO_3^{-} \), remember the charge of nitrate ion, aqueous. Now that we've done this, we're going to say step 2, using \( K_{sp} \), write the equilibrium expression based on the equilibrium equation. Since the reactant is a solid, set it equal to 1 within the equilibrium expression. Remember, it's products over reactants. So we'd have calcium in brackets, so \( [Ca^{2+}] \times [NO_3^-]^2 \) (and remember this 2 here, whatever the coefficient is, becomes the power). As the reactant is a solid, we just ignore it and replace it with 1. But realize here that we have this expression over 1, which just translates into those concentrations times each other. And nitrates would be squared. Okay. So that would just be our equilibrium expression for this calcium nitrate solid. So these are the steps you need to take in breaking up your ionic solid into its respective ions, and then correctly giving the equilibrium expression for it.

Solubility Product Constant (Ksp) Concept 3

Here we can say if the solubility or molar concentrations of ions within an ionic solid are known, the K_sp can be calculated. If we take a look here, we calculate the K_sp value for silver phosphate, which is Ag₃PO₄, which has a solubility of 1.8×10-18 at 25 degrees Celsius. Here we write the equilibrium equation by breaking up the ionic solid into its aqueous ions. Alright. So here we have our silver phosphate solid. It is going to break up into its ions. It is going to break up into 3 silver ions, remember ions are aqueous in solution, plus our phosphate ion aqueous. Next, we write the equilibrium expression based on the equilibrium equation. So remember, the equilibrium expression is K_sp = products over reactants. But your reactant is a solid, so we are going to ignore it. So it is just going to be Ag⁺, remember the coefficient here is going to become the power, so cubed times PO₄^3-. Now some new stuff. Step 3, we are going to make concentrations of the ions equal to their coefficients multiplied by the x variable. Alright. So what do I mean by that? We are going to say this is equal to the coefficient in front of the silver ion in our equation as a 3, so this equals 3x and it is going to be cubed. And there is a coefficient of 1 here that is invisible, so that would be times 1x or just x.

Next, we are going to substitute in the given solubility value for the x variable and solve for K_sp. Alright. So before we do that, we have to work out this algebraic expression here. So 3x cubed means that 3 is going to be cubed and x is going to be cubed. 3 times 3 times 3 equals 27 x cubed times x. So 27 x cubed times x is 27 x to the 4th. So that is what my K_sp is equal to. So now we are going to do what step 4 said. It is that substitute in the given solubility value, we are told the solubility is this 1.8×10-18. So K_sp equals 27 times 1.8×10-18 to the 4th. Order of operations. We do what is in the parentheses first. So 1.8×10-18 to the 4th. When you punch that into your calculator, it is going to give you back 1.04976×10-71. Then that multi gets multiplied by 27. When we do that, we get as our answer 2.8×10-70 as the K_sp for our silver phosphate here. Here our answer has 2 significant figures because the solubility given to us in the beginning also has 2 significant figures.

Solubility Product Constant (Ksp) Concept 4

Conversely, the solubility of an ionic solid can be determined when its Ksp value is already known. So here it says the Ksp value for strontium fluoride, SrF₂, is 7.9 × 10 - 10 at 25 ° C . Calculate its solubility in molarity. Right. So step 1, we're going to write the equilibrium equation by breaking up the ionic solid into its aqueous ions. The part doesn't change. We have strontium fluoride solid which breaks up into strontium ion plus two fluoride ions. Next, we write the equilibrium expression based on the equilibrium equation. So that's Ksp equals our strontium ion times our fluoride ion. And remember the coefficient becomes the power. Next, we solve for the solubility variable x based on the given Ksp value. Alright. So remember, to do that we're going to say, we have our Ksp here, which is 7.9 × 10 - 10 , and that equals x , which stands in for the strontium ion, times, remember the coefficient is going to play a part here, it's going to be 2 x , and it's still squared. So we're going to have 7.9 × 10 - 10 equals x times, 2 squared is 2 times 2, so that's 4, x 2 . We're solving this like a math problem. So it's going to be 7.9 × 10 - 10 equals 4 x 3 . We're going to divide both sides now by 4 in order to isolate x 3 , so when we do that we're going to get here x 3 equals 1.975 × 10 - 10 . Here we need to take the cube root of both sides in order to isolate my x . So we're going to say here, when we take the cube root of both sides, x equals at the end 5.8 × 10 - 4 molar as my final concentration for this strontium fluoride solid.

Determine the equilibrium expression of the barium nitride solid.

• A.
• B.
• C.
• D.

Manganese (V) hydroxide has a measured solubility of 3.4×10^–5 M at 25ºC. Calculate its K_sp value.

• A. 5.7×10^–9
• B. 1.5×10^–27
• C. 7.7×10^–27
• D. 4.8×10^–24

The K_sp value for strontium fluoride, SrF₂, is 7.9×10^–10 at 25ºC. Calculate its solubility in g/L.

• A. 5.8×10^–4 g/L
• B. 7.3x10^-2 g/L
• C. 4.6×10^–6 g/L
• D. 0.092 g/L