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The solubility product constant, Ksp, is an equilibrium constant that indicates how well an ionic molecule dissolves in water. We can compare the Ksp values of substances that dissolve to produce the same number of ions to assess their respective solubilities. The more soluble a substance is, the higher the solubility product constant. The product of the ion concentrations is the Ksp expression for a specific salt. To obtain the solubility equilibrium, each concentration is raised to a power equal to the coefficient of that ion in a balanced equation. The solubility product constants are used to describe the saturated solutions of low-solubility ionic substances.
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Key Takeaways: Solubility Product, Solubility Product Constant (Ksp), saturated solution, ionic substances, equilibrium
What is a Solubility Product?
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The ability of a substance known as a solute to dissolve in a solvent and create a solution is defined as solubility. Ionic chemicals that dissociate and create cations and anions in water have a wide range of solubility. Some substances are highly soluble, even absorbing moisture from the air, whereas others are highly insoluble.
What is the Solubility Product Constant?
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The simplified equilibrium constant (Ksp) for equilibrium between a solid and its corresponding ions in a solution is the solubility product constant. Its value represents how well a compound dissociates in water. The more soluble a substance is, the higher the solubility product constant.
The Ksp expression for salt is the product of the ion concentrations, each concentration is increased to a power corresponding to the ion's coefficient in the balanced equation for solubility equilibrium.
To characterise saturated solutions of ionic substances with low solubility, solubility product constants are utilised. The dissolved, dissociated ionic component and the undissolved solid are in a condition of dynamic equilibrium in a saturated solution.
Saturated Solution: A saturated solution is one in which the maximal concentration of a solute dissolved in the solvent is present.
We can create a formula for Silver chloride at equilibrium in given conditions, just like we can for any other solution:
Ksp = [Ag+][Cl-]/[AgCl]
generic statement for the solubility equilibrium
Where [Ag+] and [Cl-] are the concentrations of Ag+ and Cl- ions, respectively, and [AgCl] is the number of moles in a litre of solid AgCl. Because [AgCl] is a constant, the following equation can be written:
Kc x [AgCl] = [Ag+][Cl-]
We can see that the product of Ag+ and Cl- equilibrium concentrations equals a constant. The solubility product constant, or Ksp, is the name for this constant.
Ksp = [Ag+][Cl-]
Factors Affecting the Solubility Product Constant Value
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The following are some major elements that influence the solubility product constant:
- The action of the common ion (the presence of a common ion lowers the value of Ksp).
- The ion-diversity effect (if the ions of the solutes are uncommon, the value of Ksp will be high).
- Ion-pairs presence also affect the value of Ksp
Solubility Product Formula
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The saturated solutions of ionic substances with limited solubility are described by the solubility product constant. Between the ionic compound and the undissolved solid, a saturated solution is said to be in a condition of dynamic equilibrium.
The Ksp formula is represented by the following equation:
MxAy(s)→xMy+(aq)+yAx−(aq)
The following is the formula for the global equilibrium constant:
Kc=[My+]x[Ax−]y
Importance of Solubility Product
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The solubility of a substance is determined by a number of factors, including the salt's lattice enthalpy and the solvation enthalpy of the ions in the solution. The most significant factors are these two. Let's take a closer look at the relevance of the solubility product.
- When a salt is dissolved in a solvent, the interactions between the ions and the solvent must overcome the strong forces of attraction of the solute, which are the lattice enthalpy of its ions.
- Ions have a negative solvation enthalpy, which indicates that energy is released throughout the process.
- The quantity of energy released during solvation is solvation enthalpy, which is determined by the composition of the solvent.
- Because the solvation enthalpy of non-polar solvents is low, this energy is insufficient to overcome the lattice enthalpy.
- As a result, the salts do not dissolve in nonpolar liquids. As a result, a salt's solvation enthalpy must be greater than its lattice enthalpy in order for it to dissolve in a solvent.
- The solubility of each salt varies depending on the temperature.
Things to Remember
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- The ability of a substance known as a solute to dissolve in a solvent and create a solution is defined as solubility.
- The simplified equilibrium constant (Ksp) for equilibrium between a solid and its corresponding ions in a solution is the solubility product constant. Its value represents how well a compound dissociates in water.
- The dissolved, dissociated ionic component and the undissolved solid are in a condition of dynamic equilibrium in a saturated solution.
- A saturated solution is one in which the maximal concentration of a solute dissolved in the solvent is present.
- The formula for the global equilibrium constant: Kc=[My+]x[Ax−]y
Also Read:
Sample Questions
Ques. PbCl2(s)\(\rightleftharpoons\)Pb2+(aq) + 2Cl−(aq)
Express the solubility product constant expression for the given reaction. ( 3 marks )
Ans: Ksp=[Pb2+][Cl−]2
Explanation:
The equation:
PbCl2(s)\(\rightleftharpoons\)Pb2+(aq) + 2Cl−(aq)
The solubility product constant (Ksp), on the other hand, informs us how much a solid dissolves in solution. The higher the Ksp, the more water soluble a chemical is. The same rules apply to writing this expression as they do for other equilibrium constant expressions. As a result, solids and water (when it is the solvent) are not included in this equation. The concentration of the compounds involved must be increased to the power of the coefficient.
For the chemical reaction given, the Ksp is:
Ksp=[Pb2+][Cl−]2
Ques. How many grams of BaF2 are dissolved in 75.0 mL of a saturated solution of BaF2? ( 4 marks)
BaF2(s)\(\rightleftharpoons\)Ba2+(aq) + 2 F -(aq)
Given: Ksp=1.00 x 10−6
Ans: 0.0829 g BaF2
Explanation:
First, calculate the molarity of the solution
Using the dissociation equation
BaF2(s)\(\rightleftharpoons\)Ba2+(aq) + 2 F -(aq)
The equation for the solubility product constant is
Ksp=[Ba2+][F -]2
1.00 x 10−6=[Ba2+][F -]2
Since, there are 2 fluoride ions for every barium ion, we can rewrite the equation as
[Ba2+]=x
1.00 x 10−6=(x)(2x)2
On Solving for x
1.00 x 10−6=(x)(4x2)
1.00 x 10−6=4x3
x3=1.00 x 10−64=2.50 x 10−7
x=2.50 x 10−7−−−−−−−−−√3=6.30 x 10−3
[Ba2+]=6.30 x 10−3M
On Solving for concentration of dissolved BaF2
[BaF2]=[Ba2+]=6.30 x 10−3M
Now grams dissolved,
(6.30 x 10−3M)(0.0750 L)=4.73 x 10−4 mol
(4.73 x 10−4 mol)(175.34gmol)=0.0829 g BaF2
Ques. CaF2(s)\(\rightleftharpoons\) Ca2+(aq) + 2F−(aq)
Express the solubility product constant expression for the given reaction. ( 3 marks)
Ans: Ksp=[Ca2+][F−]2
The equilibrium given tells us how the solid dissolves in solution:
CaF2(s)\(\rightleftharpoons\)Ca2+(aq) + 2F−(aq)
The solubility product constant (Ksp), on the other hand, informs us how much a solid dissolves in solution. The higher the Ksp, the more water soluble a chemical is. The same rules apply to writing this expression as they do for other equilibrium constant expressions. As a result, solids and water (when it is the solvent) are not included in this equation. The concentration of the compounds involved must be increased to the power of the coefficient.
For the chemical reaction given, the Ksp is:
So, Ksp=[Ca2+][F−]2
Ques. Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it. ( 4 marks)
Ans: The equation for the dissolving of lead(II) chloride
PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)
Ksp = [Pb2+][Cl-]2
Then, convert the amount of dissolved lead(II) chloride into moles per liter.
(0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2
| PbCl2 (s) | Pb2+(aq) | Cl-(aq) | |
|---|---|---|---|
| Initial Concentration | All solid | 0 | 0 |
| Change in Concentration | - 0.0159 M (dissolves) | + 0.0159 M | + 0.0318 M |
| Equilibrium Concentration | Less solid | 0.0159 M | 0.0318 M |
On Solving, Ksp = [0.0159][0.0318]2
= 1.61 x 10-5
Ques. Estimate the solubility of Ag2CrO4 in pure water if the solubility product constant for silver chromate is 1.1 x 10-12. ( 3 marks)
Ans: Equation: Ag2CrO4(s) --> 2 Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-]
On solving,
Let "x" be the number of moles of silver chromate that dis soluble in 1 litre of solution
| Ag2CrO4(s) | Ag+(aq) | CrO42-(aq) | |
|---|---|---|---|
| Initial Concentration | All solid | 0 | 0 |
| Change in Concentration | - x dissolves | + 2 x | + x |
| Equilibrium Concentraion | Less solid | 2 x | x |
On Solving for x.
1.1 x 10-12 = [2x]2[x]
x = 6.50 x 10-5 M
Ques. Estimate the solubility of barium sulfate in a 0.020 M sodium sulfate solution. The solubility product constant for barium sulfate is 1.1 x 10-10. (4 marks)
Ans: Equation: BaSO4(s) --> Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][SO42-]
Let "x" be the amount of barium sulfate that dissolves in the sodium sulfate solution expressed in moles 1 litre of solution.
| BaSO4(s) | Ba2+(aq) | SO42-(aq) | |
|---|---|---|---|
| Initial Concentration | All solid | 0 | 0.020 M (from Na2SO4) |
| Change in Concentration | - x dissolves | + x | + x |
| Equilibrium Concentration | Less solid | x | 0.020 M + x |
On solving for x,
1.1 x 10-10 = [x][0.020 + x] = [x][0.020]
x = 5.5 x 10-9 M
Ques. 25.0 mL of 0.0020 M potassium chromate are mixed with 75.0 mL of 0.000125 M lead(II) nitrate. Will a precipitate of lead(II) chromate form. Ksp of lead(II) chromate is 1.8 x 10-14. ( 4 marks)
Ans: Net Ionic Equation -
K2CrO4(aq) + Pb(NO3)2(aq) --> 2 KNO3(aq) + PbCrO4(s)
Pb2+(aq) + CrO42-(aq) --> PbCrO4(s)
The latter reaction can be written in terms of Ksp as:
PbCrO4(s) --> Pb2+(aq) + CrO42-(aq)
Ksp = [Pb2+][CrO42-]
Using the dilution equation, C1V1 = C2V2, determine the initial concentration of individual species,
(0.0020 M K2CrO4)(25.0 mL) = (C2)(100.0 mL)
C2 for K2CrO4 = 0.00050 M
Similar calculation for the lead(II) nitrate yields:
C2 for Pb(NO3)2 = 0.0000938 M
Now, Calculate for Q,
Q = (0.0000938 M Pb2+)(0.00050 M CrO42-) = 4.69 x 10-8
Q is greater than Ksp so a precipitate of lead(II) chromate will form.
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