Solubility Product Definition Ksp Expression Calculations and Solved Examples
The solubility product constant is a simplified equilibrium constant denoted as Ksp which is defined for equilibrium between a solid and its respective ions in a given solution. Its value shows the degree to which a compound can dissociate in water. The greater the solubility product constant, the more soluble is the compound. The Ksp expression for a given salt is the product of the concentrations of the ions.
Each concentration is raised to a power that is equal to the coefficient of that ion in a balanced equation to get the solubility equilibrium. The solubility product constants are used for describing the saturated solutions of ionic compounds of relatively low solubility. A saturated solution is in a dynamic equilibrium state between the dissolved and dissociated ionic compound and the undissolved solid. In this article, we will learn about what is the solubility product, what is Ksp, the solubility product definition, and some solubility product numerical.
Solubility Product Definition
The definition of the solubility product is given as follows:
Solubility is defined as the property of a substance known as solute to get dissolved in a solvent for the formation of a solution. The solubility of ionic compounds that dissociate and form cations and anions in water varies to a great extent. Some compounds are highly soluble and can even absorb moisture from the atmosphere whereas, on the other hand, the others are highly insoluble.
Solubility Product Constant
Now, let us understand in detail what is the solubility product constant.
The solubility product is a type of equilibrium constant whose value depends on temperature. Ksp usually increases with the increase in temperature because of the increased solubility.
Most ionic compounds which are insoluble would still dissolve in water to a small extent. These insoluble compounds are considered to be strong electrolytes since whatever portion of the compound is dissolved also dissociates. For example, silver chloride dissociates to a small extent in the silver ions and chloride ions when added to water.
\[AgCl (s) \leftrightarrow Ag^{+}(aq) + Cl^{-}(aq)\]
This process is written as an equilibrium since the dissociation occurs only to a smaller extent. Hence, an equilibrium expression is written for the process. You must keep in mind that the solid silver chloride does not have a variable concentration and is therefore not included in the expression. This is the solubility product principle.
\[K_{sp} =[Ag^{+}][cl^{-}] \]
Solubility Product Formula
The solubility product constant is used to describe the saturated solutions of ionic compounds having relatively low solubility. A saturated solution is said to be in a state of dynamic equilibrium between the ionic compound and the undissolved solid.
The Ksp formula is given in the form of the following equation:
\[M_{x}A_{y} (s) \rightarrow xM^{y+}(aq) + yA^{x-}(aq)\]
The general equilibrium constant is written as follows:
\[Kc = [M^{y+}]^{x} [A^{x-}]^{y} \]
Significance of the Solubility Product
The solubility depends on several parameters which include the lattice enthalpy of salt and solvation enthalpy of ions in the solution. These two factors are the most important. Let us learn about the significance of the solubility product in detail.
Whenever a salt is dissolved in a solvent, the strong forces of attraction of solute that are the lattice enthalpy of its ions need to be overcome by the interactions between the ions and the solvent.
The solvation enthalpy of ions is always negative and this means that energy is released during the process.
The nature of the solvent finds the amount of energy that is released during the solvation which is solvation enthalpy.
Non-polar solvents have a small value of solvation enthalpy which means that this energy is not sufficient for overcoming the lattice enthalpy.
Hence the salts are not dissolved in the non-polar solvents. Therefore, for a salt to be dissolved in a solvent, its solvation enthalpy must be greater than its lattice enthalpy.
The solubility depends on the temperature and it is varied for every salt.
Salts are Classified According to their Solubility which is Given in the Table Below:
Solubility Product Numericals
Let us now take a look at some of the Ksp chemistry examples.
Example 1:
Determine the solubility product constant for \[MgF_{2}\].
Solution:
The relevant equilibrium is given by:
\[MgF_{2} (s) \leftrightarrow Mg^{2+}(aq) + 2F^{-}(aq)\]
Hence, the associated equilibrium constant is given by:
\[K_{sp} =[Mg^{2+}][F^{2-}] \]
Example 2:
Give the solubility product constant for \[Ag_{2}CrO_{4}\].
Solution:
The relevant equilibrium is given by:
\[Ag_{2}CrO_{4} (s) \leftrightarrow 2Ag^{+}(aq) + CrO_{4}^{2-}(aq)\]
The associated equilibrium is therefore given by:
\[K_{sp} =[Ag^{2+}]^{2}[CrO_{4}^{2-}] \]
The Ksp-Solubility Relationship
Because it is basically the product of the solubilities of the ions in moles per liter, Ksp is called the solubility product. A salt's solubility product, or vice versa, can therefore be computed from its solubility.
The sensitivity of AgBr to light is used in photographic films. When light strikes an AgBr crystal, a small percentage of the Ag+ ions are reduced to silver metal. When the film is developed, the remaining Ag+ ions in these crystals are reduced to silver metal. To "fix" the image, AgBr crystals that do not absorb light are removed from the film.
Example: 3
To test if AgBr can be eradicated by merely washing the film, let's calculate the solubility of AgBr in water in grams per liter.
Solution:
We start with the equilibrium equation of this salt in an aqueous solution, which is a balanced equation.
AgBr(s) \[\Longleftrightarrow\] \[Ag^{+}(aq) + Br^{-}(aq) \]
The solubility product equation for this reaction is then written.
\[K_{sp} =[Ag^{+}][Br_{-}] = 5.0 \times 10^{-13}\]
For two unknowns, the Ag+ and Br- ion concentrations, one equation cannot be solved. However, by noting that one Ag+ ion is emitted for every Br- ion, we can derive a second equation. Because there is no alternative source of either ion in this solution, their equilibrium concentrations must be equal.
\[ [Ag^{+}]=[Br^{-}] \]
The following result is obtained by substituting this equation into the Ksp expression.
\[[Ag^{+}]^{2} = 5.0 \times 10^{-13} \]
The equilibrium concentrations of Ag+ and Br- ions can be determined by taking the square root of both sides of this equation.
\[ [Ag^{+}]=[Br^{-}]\] = \[7.1 \times 10^{-7} M \]
We can calculate the solubility in grams per liter after we know how many moles of AgBr dissolve in a liter of water.
\[ \frac{7.1 \times 10^{-7} \text{mol AgBr}}{1 L} \times \frac{187.8 g AgBr}{1 \text{mol AgBr}} = 1.3 \times 10^{-4} g \text{AgBr}/L\]
AgBr dissolves in water at a rate of 0.00013 grams per liter. Attempting to wash the unexposed AgBr off the photographic film with water is therefore impractical.
Calculating the solubility product for 1:1 salts like AgBr is relatively simple. We need to understand the relationship between a salt's solubility and the concentrations of its ions at equilibrium in order to expand such calculations to compounds with more complex formulas. The amount of salt that dissolves in water will be denoted by the symbol Cs.
Get to know more about the concepts of solubility product at Vedantu. The insights of the top subject matter experts will aid you to understand this topic better and use its concepts to solve problems in the exercises and exams.
FAQs on Solubility Product Constant Ksp in Chemical Equilibrium
1. What is the solubility product (Ksp)?
Solubility product (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. It represents the product of the molar concentrations of the ions, each raised to the power of its stoichiometric coefficient, at equilibrium.
For example, for AgCl(s) ⇌ Ag+(aq) + Cl-(aq):
- Ksp = [Ag+][Cl-]
Only dissolved ions appear in the Ksp expression; the solid is not included because its concentration is constant.
2. What is the formula for calculating Ksp?
The formula for Ksp is the product of the equilibrium concentrations of the dissolved ions, each raised to its stoichiometric coefficient.
For a general salt AxBy(s) ⇌ xAy+(aq) + yBx-(aq):
- Ksp = [Ay+]x[Bx-]y
Example: For CaF2(s) ⇌ Ca2+(aq) + 2F-(aq),
- Ksp = [Ca2+][F-]2
3. How do you calculate solubility from Ksp?
To calculate solubility from Ksp, express ion concentrations in terms of molar solubility and substitute into the Ksp expression.
- Step 1: Write the balanced dissolution equation.
- Step 2: Let molar solubility = s mol L-1.
- Step 3: Express ion concentrations in terms of s.
- Step 4: Substitute into the Ksp formula and solve for s.
Example for AgCl: If solubility = s, then [Ag+] = s and [Cl-] = s.
- Ksp = s2
Thus, s = √Ksp.
4. What is the difference between solubility and solubility product?
Solubility is the maximum amount of a substance that dissolves in a solvent, while Ksp is the equilibrium constant for its dissolution.
- Solubility is expressed in mol L-1 or g L-1.
- Ksp is a constant at a given temperature.
- Ksp depends on stoichiometry of ions.
Two salts can have the same solubility but different Ksp values if their ionic ratios differ.
5. What is the common ion effect in relation to Ksp?
The common ion effect is the decrease in solubility of a salt when a common ion is added to the solution.
For example:
- AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Adding NaCl increases [Cl-], shifting equilibrium left according to Le Châtelier’s principle, thus decreasing solubility while Ksp remains constant.
6. How do you know if a precipitate will form using Ksp?
A precipitate forms if the ionic product (Q) is greater than the Ksp value.
- If Q < Ksp: No precipitate forms (unsaturated).
- If Q = Ksp: Solution is saturated.
- If Q > Ksp: Precipitate forms.
Calculate Q using current ion concentrations and compare it to Ksp.
7. Does temperature affect the solubility product (Ksp)?
Yes, Ksp depends on temperature because it is an equilibrium constant.
- If dissolution is endothermic, increasing temperature increases Ksp.
- If dissolution is exothermic, increasing temperature decreases Ksp.
Therefore, Ksp values are always specified at a particular temperature, usually 25°C.
8. What are some examples of Ksp expressions for different salts?
The Ksp expression depends on the dissociation equation of the salt.
- NaCl (highly soluble, no Ksp usually used).
- AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] - CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
Ksp = [Ca2+][F-]2 - Al(OH)3(s) ⇌ Al3+(aq) + 3OH-(aq)
Ksp = [Al3+][OH-]3
9. Why are solids not included in the Ksp expression?
Pure solids are not included in the Ksp expression because their concentration remains constant during the reaction.
In equilibrium expressions:
- Only aqueous (aq) and gaseous (g) species are included.
- Pure solids (s) and pure liquids (l) are omitted.
For example, in AgCl(s) ⇌ Ag+(aq) + Cl-(aq), only the ions appear in Ksp.
10. What is the relationship between Ksp and molar solubility?
The relationship between Ksp and molar solubility depends on the stoichiometry of the salt’s dissociation.
- For MX(s) ⇌ M+ + X-, if solubility = s, then Ksp = s2.
- For MX2(s) ⇌ M2+ + 2X-, if solubility = s, then Ksp = 4s3.
Thus, higher Ksp generally indicates greater solubility, but ionic ratios must always be considered.
