How to calculate Ksp from molar solubility? – (molar solubility to Ksp), Conversion, Formulas, Equations
Ksp is an equilibrium constant called solubility product constant. It determines the extent to which a sparingly soluble salt ionizes in water i.e., its molar solubility.
If the molar solubility of a partially ionizable salt, such as AgCl, BaSO4, CaCO3, etc., is known, we can easily find its Ksp value at a particular temperature.
Come along and continue reading the article to find out how to calculate Ksp from molar solubility.
It measures the extent of ionization or dissolution of sparingly soluble salt in water in a saturated solution.
A saturated solution is one in which the maximum amount of solute is dissolved at a particular temperature.
1 mole of a salt (A) partially ionizes in water to release x and y moles of its constituent cation (C) and anion (D), respectively. The general dissolution reaction is represented as follows.
The equilibrium constant (Ksp) for the above reaction can be written as:
Ksp = [C]x [D]y………. Equation (i)
Where
Ksp = solubility product constant
[C] = equilibrium concentration of the cation
[D] = equilibrium concentration of the anion
x and y are coefficients obtained from the balanced dissolution equation.
x = number of moles of C produced from 1 mole of A
y = a number of moles of D released from 1 mole of A.
You may note that [A] is not included in the Ksp expression because [A] represents the concentration of salt left undissociated at equilibrium i.e., in solid form, and pure solids do not affect equilibrium constants.
As the equilibrium constant (Ksp) is calculated by taking the product of equilibrium molarities thus, the name solubility product constant is given.
Ksp for any given salt stays constant at a particular temperature, pressure and composition. Changes in any of these variables significantly affect Ksp.
What is molar solubility?
Molar solubility refers to the concentration i.e., the amount of a solid substance dissolved in a saturated aqueous solution.
In this case, the concentration is measured in moles of solute dissolved per litre of the solution i.e., molarity (mol/L or M).
If a chemist says that the molar solubility of a sparingly soluble salt A in water is 1 M, it means 1 mole A gets dissolved in a litre of water.
As per the general dissolution equation shown in the previous section, the concentration of A dissociated till the equilibrium point is reached i.e., [A]dissociated = equilibrium concentrations of C and D.
[A]dissociated = molar solubility. Thus if we know the equilibrium concentrations of salt cation and anion, we can easily find molar solubility and vice versa.
Barium sulfate (BaSO4) is a sparingly soluble salt. Its balanced dissolution equation is shown below.
The Ksp expression for BaSO4 can be derived using equation (i) as follows:
Ksp = [C]x [D]y………. Equation (i)
W.r.t BaSO4, C= Ba2+ and D = SO42- while x = y= 1. So equation (i) is transformed into equation (ii).
Ksp = [Ba2+] [SO42-] …….Equation (ii)
What is the relationship between Ksp and molar solubility?
A higher Ksp value denotes that the particular salt is soluble in water to a greater extent. Therefore, there is a direct relationship between Ksp and molar solubility.
However, you must keep in mind that we cannot compare the solubility of two different salts based on their Ksp values. This is because Ksp is dependent upon mole ratios as per the dissolution equation.
How to find Ksp from molar solubility?
Ksp and molar solubility are interconvertible.
Again taking the example of BaSO4, if we know the molar solubility of BaSO4, we can use its balanced dissolution equation to determine the molarities of Ba2+ and SO42- ions produced at the equilibrium point.
Once we know [Ba2+] and [SO42-], we can conveniently use the Ksp expression given in equation (ii) to find the unknown Ksp value.
Thus, you can find the solubility product constant (Ksp) from molar solubility depending upon the balanced dissolution equation.
You will understand this concept better after you go through the solved examples given below. So let the practice begin.
Solved examples for finding Ksp from molar solubility
Example # 1: The molar solubility of barium sulfate (BaSO4) is 1.05 x 10-5 M at 25°C. Find its Ksp.
1 mole BaSO4 partially ionizes to yield 1 mole Ba2+ and 1 mole SO42- ions as shown below.
The Ksp expression for BaSO4 as per the above-balanced dissolution equation is:
Ksp = [Ba2+] [SO42-]
Molar solubility = [BaSO4] dissociated at equilibrium
As per the balanced dissolution equation; [BaSO4] dissociated = [Ba2+] or [SO42-] at equilibrium
As the molar solubility is given in the question statement, so [BaSO4] dissociated = 1.05 x 10-5 M.
∴ [Ba2+] equilibrium = [SO42-] equilibrium = x= 1.05 x 10-5 M.
Now that we know the required molarities, we can easily substitute them into the Ksp expression as follows:
Ksp = (1.05 x 10-5) (1.05 x 10-5)
Ksp = 1.10 x 10-10.
Result: The Ksp value for BaSO4 is 1.10 x 10-10 at 25°C.
Usually, no units are used for Ksp because the molar concentrations of the reactants and products are different for each dissolution equation.
Example # 2: Determine the Ksp of calcium fluoride (CaF2) given that its molar solubility is 2.14 x 10-4 M at 25°C.
1 mole CaF2 partially ionizes to yield 1 mole Ca2+ and 2 mole F– ions, as shown below.
The Ksp expression for CaF2 can be derived from the general equation (i)
Ksp = [C]x [D]y………. Equation (i)
As per the above-balanced dissolution equation, x = 1 and y =2 considering C= Ca2+ and D= F–, so equation (i) transforms into:
Ksp = [Ca2+] [F–]2
Molar solubility = [CaF2] dissociated at equilibrium
As per the balanced dissolution equation; [CaF2] dissociated = [Ca2+] equilibrium
As the molar solubility is given in the question statement, so [CaF2] dissociated = 2.14 x 10-4 M.
∴ [CaF2] dissociated = [Ca2+] equilibrium = 2.14 x 10-4 M.
∴ [F–]equilibrium = 2 [Ca2+] equilibrium = 2 (2.14 x 10-4) = 4.28 x 10-4 M.
Now that we know the required molarities, we can easily substitute them into the Ksp expression as follows:
Ksp = (2.14 x 10-4) (4.28 x 10-4)2
Ksp = 3.92 x 10-11.
Result: The Ksp value for CaF2 is 3.92 x 10-11 at 25°C.
Example # 3: Calculate the Ksp for magnesium phosphate Mg3(PO4)2 from its molar solubility i.e., 3.57 x 10-6 M at 25°C.
1 mole Mg3(PO4)2 partially ionizes in water to produce 3 moles Mg2+ and 2 moles of PO43- ions, as shown below.
The Ksp expression for Mg3(PO4)2 can be derived from the general equation (i)
Ksp = [C]x [D]y………. Equation (i)
As per the balanced dissolution equation, x = 3 and y = 2 considering C= Mg2+ and D= PO43-, so equation (i) transforms into:
Ksp = [Mg2+]3 [PO43-]2
Molar solubility = [Mg3(PO4)2] dissociated at equilibrium
As per the balanced dissolution equation; [Mg2+] equilibrium = 3 [Mg3(PO4)2] dissociated
As the molar solubility is given in the question statement, so [Mg3(PO4)2] dissociated = 3.57 x 10-6 M.
∴ [Mg2+] equilibrium = 3 [Mg3(PO4)2] dissociated = 3(3.57 x 10-6) = 1.071 x 10-5 M.
∴ [PO43-] equilibrium = 2 [Mg3(PO4)2] dissociated = 2(3.57 x 10-6) = 7.14 x 10-6 M.
Now that we know the required molarities, we can easily substitute them into the Ksp expression as follows:
Ksp = (1.071 x 10-5)3 (7.14 x 10-6)2
Ksp = 6.26 x 10-26.
Result: The Ksp value for Mg3(PO4)2 is 6.26 x 10-26 at 25°C.
Example # 4: Solid Ag2S has a molar solubility of 6.50 x 10-5 M. What is the Ksp of this compound?
1 mole Ag2S partially ionizes to produce 2 moles Ag+ and 1 mole S2- ions, as shown below.
The Ksp expression for Ag2S can be derived from the general equation (i)
Ksp = [C]x [D]y………. Equation (i)
As per the balanced dissolution equation, x = 2 and y = 1 considering C= Ag+ and D= S2-, so equation (i) transforms into:
Ksp = [Ag+]2 [S2-]
Molar solubility = [Ag2S] dissociated at equilibrium
As per the balanced dissolution equation; [Ag+] equilibrium = 2 [Ag2S] dissociated
As the molar solubility is given in the question statement, so [Ag2S] dissociated = 6.50 x 10-5 M.
∴ [Ag+] equilibrium = 2 [Ag2S] dissociated = 2(6.50 x 10-5) = 1.3 x 10-4 M.
∴ [S2-] equilibrium = [Ag2S] dissociated = 6.50 x 10-5 M
Now that we know the required molarities, we can easily substitute them into the Ksp expression as follows:
Ksp = (1.3 x 10-4)2 (6.50 x 10-5)
Ksp = 1.10 x 10-12.
Result: The Ksp value for Ag2S is 1.10 x 10-12 at 25°C.
Ksp stands for the solubility product constant. It is an equilibrium constant that determines the extent of ionization and, thus, the dissolution of a sparingly soluble salt in water.
For a sparingly soluble salt A that disintegrates into C and D where C = salt cation, D= salt anion,
Ksp can be calculated by the formula: Ksp = [C]x [D]y.
x and y are coefficients that represent the moles of C and D produced per mole of A in the dissolution equation.
What is molar solubility?
Molar solubility is the maximum amount of solute dissolved in a saturated solution.
What is a saturated solution?
A saturated solution is one in which the given solute is dissolved to a maximum extent. Therefore, no more solute can get dissolved in it at a certain temperature.
Why is the concentration of salt not included in the Ksp expression?
Pure solids, such as undissolved salt, do not affect equilibrium constants, so the concentration of salt is not included in the Ksp expression.
Is Ksp temperature dependent?
Yes. Variables such as temperature and pressure significantly affect the Ksp value for a salt.
The higher the temperature, the salt undergoes ionization to a greater extent. The given salt thus possesses a higher Ksp value at T2 than T1 where T = temperature and T2 > T1.
What is the relationship between Ksp and molar solubility?
A higher Ksp value denotes that the particular salt is more soluble in water. Therefore, there is a direct relationship between Ksp and molar solubility.
How to find Ksp from molar solubility?
The following steps can be applied to find the solubility product constant (Ksp) for salt at a given temperature if we know its molar solubility.
Step I: Write the balanced chemical equation for salt dissolution.
Step II: Write the Ksp expression using the general formula Ksp = [C]x [D]yand the dissolution equation.
Step III: Find [C] and [D] using mole ratios from the balanced dissolution equation and the given molar solubility.
Step IV: Substitute [C] and [D] into the Ksp expression and find Ksp.
Summary
Ksp stands for the solubility product constant. It determines the extent of ionization of sparingly soluble salt in an aqueous solution.
Ksp is a temperature, pressure, and composition-dependent chemical entity.
A sparingly soluble salt only partially ionizes in water to liberate its constituent cation and anion.
The molar solubility of a sparingly soluble salt refers to the amount of salt dissolved in an aqueous solution at a fixed temperature.
Molar solubility is calculated in moles/liter.
If the molar solubility of a salt is given, we can find its Ksp using the balanced dissolution equation and the formula Ksp = [C]x [D]y.
In the Ksp expression, C and D represent the constituent cation and anion produced by salt dissolution, while x and y are coefficients determined from the balanced dissolution equation. x = the number of moles of C produced from 1 mole of salt. D = the number of moles of D produced from 1 mole of salt.
References
Topper. (n.d.). How do you calculate Ksp from molar solubility? Topper. https://www.topper.com/ask/question/how-do-you-calculate-ksp-from-molar-solubility
Khan Academy. (n.d.). Solubility product constant. AP Chemistry Beta. https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:equilibrium/x2eef969c74e0d802:solubility-equilibria/v/solubility-product-constant
Study.com. (n.d.). Using the Solubility of a Compound to Calculate Ksp: Explanation. Study.com. https://www.study.com/skill/learn/using-the-solubility-of-a-compound-to-calculate-ksp-explanation.html
Socratic. (n.d.). How do you calculate Ksp from molar solubility? Socratic. https://socratic.org/questions/how-do-you-calculate-ksp-from-molar-solubility
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
Topblogtenz is a website dedicated to providing informative and engaging content related to the field of chemistry and science. We aim to make complex subjects, like chemistry, approachable and enjoyable for everyone.
∴ [Ba2+] equilibrium = [SO42-] equilibrium = x= 1.05 x 10-5 M.
∴ [F–]equilibrium = 2 [Ca2+] equilibrium = 2 (2.14 x 10-4) = 4.28 x 10-4 M.
