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

The solubility product constant ($K_{\mathrm{sp}}$) is a specialized equilibrium constant that measures the solubility of solid ionic compounds in a solvent. $K_{\mathrm{sp}}$ values indicate the degree of solubility: a higher $K_{\mathrm{sp}}$ means greater solubility, while a lower $K_{\mathrm{sp}}$ indicates less. Comparisons of $K_{\mathrm{sp}}$ are only valid between compounds that dissociate into the same number of ions. Solubility is an equilibrium process, often requiring an ICE chart for calculations. The $K_{\mathrm{sp}}$ expression is unique because it only includes the concentrations of the products (ions in solution), as solids and liquids are not included in equilibrium expressions. Thus, $K_{\mathrm{sp}}$ is calculated by taking the product of the ion concentrations, each raised to the power of its coefficient in the balanced chemical equation. This understanding is crucial for predicting solubility and performing related calculations in chemistry.

Solubility Product Constant (Ksp)

Now, the solubility product constant uses the variable KSP and represents yet another type of equilibrium constant. Here it measures solubilities of solid ionic compounds in a solvent at equilibrium. Here we're going to say that solubility is just the maximum amount of solid dissolved in a solvent, and it's usually represented as capital M which stands for molar solubility.

Now the magnitude of KSP value determines the degree of solubility. Basically we're going to say here the greater the KSP value then the more soluble your ionic soluble will be and the lower your KSP then the less soluble your ionic solid will be. Here we're going to say this comparison can only be used between compounds that break up into the same number of ions.

So for example you have any CL which breaks up into two ions, and you have let's say lithium bromide which also breaks up into two ions because they break up into the same number of ions. You can look at their KSPs and directly compare them. So this one has a higher KSP so it's more soluble. If they broke up into different number of ions then more calculations would be needed.

So for example, if you had barium chloride, this breaks up into three ions, so you couldn't just look at its KSP value and compare it to the other two because it breaks up into different numbers of ions. Later on we'll see how to compare them by doing further calculations.

Solubility Product Constant: Ksp Example

Given the following ionic compounds which have the highest will have the highest concentration of hydroxide ions. Concentration. Here it says HINT, which is the most soluble in water. All right, so if we took a look at every single one of these compounds, you will see that they all break up into three ions. They would break up into the respective metal and two hydroxide ions.

Now, because they break up into the same number of ions, we can just look at their KSP values and determine which one would produce the most. OH^-. Remember the grittier KSP value. The more soluble your ionic solid will be, the more ions it will produce. So if we take a look here, 10-17 to the 10-20 to the 10-13 to the 10-27 here, we'd say magnesium hydroxide is our answer.

It has the highest KSP value, therefore it is the most soluble, meaning that it will break up the most and produce the most OH^- ions as products. This in turn will create the highest hydroxide ion concentration.

Ksp Calculations

Now, when it comes to K_sp calculations, realize that solubility is an equilibrium process. Hence calculations will require an ICE chart. So if you see K_sp within a question, that means you're most likely going to have to use an ICE chart. Now here with K_sp we have a formula we need to remember, so let's say that here we have our equation.

Again, K_sp deals with the solubility of ionic solids, which means that you're going to start out with an ionic solid. As you're reacting, we look to see how it breaks up into various ions. So let's say that it breaks up into some unknown number of B ions plus some unknown number of C ions. These are ions that they'd have charges, so I'm just going to write a random charge here for each one. Now, A, B, and C are your coefficients.

K_sp is an equilibrium constant and like all equilibrium constants, it equals products over reactants. However, remember we said K_sp deals with solids. Remember solids and liquids are ignored. So although K_sp equals products over reactants, like other equilibrium constants, the reactant will be a solvent, therefore it can be ignored. So K_sp simplify suggests equal to products.

Now, like other equilibrium constants, we'd have our products within brackets represent concentration, and then remember that their coefficients would become these powers. So that's how we break down the equilibrium expression for a K_sp value. So just remember it's equal to products over reactants. The reactant is a solid, so just drop it. So K_sp equals products.

Solubility Product Constant: Ksp Example

Here we have lead to fluoride is a white solid and has diverse applications in pharmaceuticals, metallurgy and technology. If the concentration of lead to fluoride is 4.2 molar with a K_sp of 3.6×10-8, calculate the molar solvability of the solid at 25°C. All right, So when they say solubility, whether they say calculate solubility or calculate molar solubility, they're just asking you to find X. And when they're asking for the most ability of the ionic solid, that's just equal to X.

Here we're going to follow the steps in order to solve for this particular question. So step one, we set up an ice chart with solid as the only reactant because we were looking at how it breaks up into ions cross out the reactant side because remember in a nice chart we ignore solids and liquids. This is an ice chart, so this is initial change equilibrium using the initial role set products equal to 0. Now remember we lose reactants in order to make products, so using the change row place a + X for the products. We also have to take into account coefficients, so here this would be plus X. There's a 2 here, so this would be plus 2X.

Using the equilibrium row, we're going to set up the equilibrium constant expression with K_sp and solve for X. So now this is plus X and this is +2 X. Now here we're going to say that the variable X in the ice chart represents the molar solubility of my ionic solid. All right, so here we're going to say K_sp equals products over reactants. But again, the reactant is a solid so we ignore it. So it just equals PB²⁺ or plus 2 * F^-. Remember, the coefficient also pops up here too as a power.

All right. So now we're going to plug in values that we have. So we're going to say here K_sp is 3.6×10-8 at equilibrium led to his X fluoride is 2X, and that's still squared. So we're going to say this is X times 2² is 4, X² is X², so four X² * X comes out to 4X³, and that's still equal to my K_sp. So all we have to do now is solve for X. So divide both sides by 4 and we'll get X³ = 9.0×10-9. Take the cube root of both sides. When we do that, we get X = 2.08×10-3 and that'll be molarity. So this represents the molar solubility of my ionic solid.

So again, if any time they're asking for the molar solubility or solubility of your ionic compound is just equal to X. Now if they were asking for one of the ions though, it wouldn't just simply be X. What you would need to do is you need to look at the equilibrium row and then look to see what is that ion represent at equilibrium. Here led to his ex so it would still be this number but fluoride ion is 2X which would mean that you take this X answer and plug it into this 2X so it would be 2 * X and that would be the true molar solubility of this fluoride ion. So again be careful with when it comes to the ion. For the ions you have to check the equilibrium row and see what their equation is for the solid when you solve for X. That is your final answer.

Solubility of Sn(OH)₂ was found to be 1.11 x 10^-9 M; calculate K_sp of this compound.

• A. 5.47 × 10⁻²⁷
• B. 4.44 × 10⁻⁹
• C. 1.23 × 10⁻¹⁸
• D. 1.37 × 10⁻²⁷

If a saturated solution of Ag₂CO₃ contains 2.56 × 10⁻⁴ M of Ag⁺ ions, determine its solubility product constant.

• A. 4.30 × 10⁻¹⁵
• B. 8.39 × 10⁻¹²
• C. 1.68 × 10⁻¹¹
• D. 6.55 × 10⁻⁸

What is the solubility of CN⁻ ions in a solution of 5.5 M Hg₂(CN)₂, with a K_sp of 5.0 × 10⁻⁴⁰?

• A. 4.5× 10⁻²⁰ M
• B. 5.0 × 10⁻¹⁴ M
• C. 1.1 × 10⁻²⁰ M
• D. 1.0 × 10⁻¹³ M

Here’s what students ask on this topic:

What is the solubility product constant?

The solubility product constant, often denoted as K_sp, is a numerical value that represents the equilibrium between a solid and its respective ions in a saturated solution. When a sparingly soluble ionic compound dissolves in water, it dissociates into its constituent ions up to a certain extent. The equilibrium established between the undissolved solid and the dissolved ions can be represented by a chemical equation.

For example, if we have a generic salt AB that dissociates into A⁺ and B^- ions, the equilibrium equation would be:

AB(s) → A⁺(aq) + B^-(aq)

The solubility product constant for this reaction would be expressed as:

K_sp = [A⁺][B^-]

Here, [A⁺] and [B^-] are the molar concentrations of the ions at equilibrium. The K_sp is unique for each compound at a given temperature and is used to predict whether a precipitate will form under certain conditions. A higher K_sp value indicates a more soluble compound. It's important to note that the solubility product does not give the actual solubility in moles per liter, but rather the extent to which a compound can produce ions in solution.

How can the solubility product constant be found?

The solubility product constant, or K_sp, is a measure of the solubility of a compound under specific conditions in a saturated solution. To find the K_sp of a sparingly soluble ionic compound, follow these steps:

1. Prepare a saturated solution of the compound by adding it to a solvent, usually water, until no more dissolves and some solid remains.
2. Allow the solution to reach equilibrium, where the rate of dissolving equals the rate of precipitation.
3. Filter or decant the solution to remove any undissolved solids.
4. Measure the concentration of the ions in the saturated solution using analytical techniques such as titration, spectrophotometry, or conductivity measurements.
5. Use the balanced chemical equation for the dissolution of the compound to determine the molar ratio of the ions.
6. Calculate the K_sp by multiplying the molar concentrations of the ions, each raised to the power of their stoichiometric coefficients in the balanced equation.

For example, if the compound AB dissolves to form A⁺ and B^- ions, and the concentrations are [A⁺] = a and [B^-] = b, then

K_sp = [A⁺]^m * [B^-]ⁿ

where m and n are the stoichiometric coefficients of A⁺ and B^- respectively in the dissolution equation.

Which of the following best defines the solubility product constant Ksp of a salt?

The solubility product constant, denoted as K_sp, is a numerical value that represents the equilibrium between a solid salt and its ions in a saturated solution. It specifically applies to sparingly soluble salts. The K_sp is the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced equation for the dissolution of the salt. For example, if we have a salt AB that dissociates into A⁺ and B^- ions, the K_sp expression would be:

K_sp = [A⁺][B^-]

Here, [A⁺] and [B^-] are the molar concentrations of the ions at equilibrium in the solution. A higher K_sp value indicates a more soluble salt, meaning it dissociates more in solution to produce a greater concentration of ions. Conversely, a lower K_sp value suggests the salt is less soluble. The K_sp is a constant at a given temperature and is unique for each salt. It's important to note that the K_sp does not give the solubility directly but rather the extent to which a salt can dissolve to reach equilibrium.