How to calculate molar solubility from Ksp? – (Ksp to molar solubility), Conversion, Formulas, Equations
Some salts, such as NaCl, are freely soluble in water. They completely dissolve to release their constituent ions in an aqueous solution. However, there are many salts that only partially ionize in water, including AgCl, BaSO4, MgF2 etc.
So we can use Ksp,which stands for the solubility product constant, to find how much of a sparingly soluble salt dissolves in water i.e., its molar solubility.
In this article, you will learn how to find molar solubility from Ksp using a very simple but extremely useful chemical formula.
So without any further delay, let’s start reading!
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 liter of the solution i.e., molarity.
If a chemist says that the molar solubility of a sparingly soluble salt X in water is 1 M, it means 1 mole X gets dissolved in a liter of water.
A saturated solution is a solution in which the maximum amount of solute is dissolved at a particular temperature. Sparingly soluble salts such as silver chloride (AgCl) only partially ionize in water to give a limited number of Ag+ and Cl– ions, as represented by the reversible reaction shown below.
As the solution reaches its saturation point, no more AgCl dissolves. Thus, at the equilibrium stage, some AgCl stays undissociated along with the constituent ions released in the aqueous solution.
As per the balanced chemical equation shown for the dissolution of AgCl, 1 mole AgCl breaks down to release 1 mole of each of the Ag+ and Cl– ions.
If x mol/L of each of Ag+ and Cl– ions are produced at equilibrium, it implies x mol/L AgCl dissolved in water which is referred to as its molar solubility.
What is Ksp?
Ksp stands for the solubility product constant.
It measures the extent of ionization or dissolution of sparingly soluble salt in water.
As per the dissolution of AgCl discussed in the previous section, its solubility product constant (Ksp) can be calculated by applying equation (i).
Ksp = [Ag+] [Cl–]…….Equation (i)
The concentration of AgCl left undissolved at equilibrium is not included in the above equation because it is in solid form. The concentrations ofpuresolids do not affect equilibrium constants.
For a general dissolution reaction, Ksp can be calculated by using the expression shown in equation (ii).
Ksp = [C]x [D]y………. Equation (ii)
In the general formula given above, C and D represent the cation and anion of the salt A respectively. Conversely, the coefficients x and y denote the number of moles of C and D produced from 1 mole of A, according to the balanced dissolution equation.
In equation (i), x = y = 1 as per the balanced chemical equation, so [Ag+] = [Cl–] at the equilibrium point.
As the equilibrium constant (Ksp) is calculated by taking the product of equilibrium molarities thus, the name solubility product constant is given.
Ksp for salt A stays constant at a particular temperature and pressure. Changes in these two variables significantly affect Ksp.
What is the relationship between molar solubility and Ksp?
Molar solubility is directly related to Ksp.
The greater the solubility product constant (Ksp), the given salt ionizes to a large extent in water.
As a result, the equilibrium concentrations of the constituent ions also increase, and so does the amount of salt dissociated at the equilibrium point i.e., molar solubility.
How to find molar solubility from Ksp?
Again taking help from the AgCl example, we said x mol/L of each of Ag+ and Cl– ions are produced at equilibrium. So as per the 1:1 mole ratio, x mol/L AgCl got solubilized in water i.e., molar solubility of AgCl = x.
The value of x can be determined using equation (i) as follows:
Ksp = [Ag+] [Cl–]…….Equation (i)
Ksp = (x) (x)
Ksp = x2
x can be determined by taking the square root on both sides of the above equation.
x = \sqrt{Ksp} ……….Equation (iii)
Once the value of x is determined, we know the molar solubility of AgCl.
Thus, you can find molar solubility from Ksp 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 molar solubility from Ksp
Example # 1: The solubility product constant (Ksp) for AgCl at 25°C is 1.8 x 10-10. What is its molar solubility?
AgCl partially ionizes in water to produce Ag+ and Cl– ions, as shown below.
Ksp expression for the dissolution of AgCl can be written as follows:
Ksp = [Ag+] [Cl–]
Let’s suppose x mol/L of each of Ag+ and Cl– ions are produced at the equilibrium point.
The value of Ksp is already given in the question statement i.e., Ksp = 1.8 x 10-10. Now let’s plug in all the above data into the Ksp expression to find the value of x.
1.8 x 10-10 = (x) (x)
x2 = 1.8 x 10-10
x = \sqrt{1.8\times 10^{-10}}
= 1.34 x 10-5 mol/L.
As per the balanced chemical equation, [AgCl]dissociated = [Ag+] equilibrium or [Cl–] equilibrium
So [AgCl]dissociated = x = 1.34 x 10-5 mol/L = molar solubility.
Result: The molar solubility of AgCl is 1.34x 10-5 M at 25°C.
Example # 2: The Ksp value for the sparingly soluble magnesium fluoride (MgF2) is 6.6 x 10-9 at normal room temperature. Find its molar solubility using the given Ksp value.
MgF2 partially ionizes in water to produce Mg+ and 2 F– ions, as shown below.
Ksp expression for the dissolution of MgF2 can be written using equation (ii).
Ksp = [C]x [D]y……….Equation (ii)
The above dissolution equation shows x = 1 and y =2, considering C = Mg2+ and D = F–. So w.r.t MgF2, equation (ii) transforms into:
Ksp = [Mg2+] [F–]2
Let’s suppose [Mg2+] = x mol/L at equilibrium. Then [F–] = 2 x.
The value of Ksp is already given in the question statement i.e., Ksp = 6.6 x 10-9. Now let’s plug in all the above data into the Ksp expression to find the value of x.
6.6 x 10-9 = (x) (2x)2
6.6 x 10-9 = (x) (4x2)
6.6 x 10-9 = 4x3
Thus [Mg2+] equilibrium = 1.18 x 10-3 mol/L.
As per the balanced dissolution equation; [MgF2] dissociated = [Mg2+] equilibrium = 1.18 x 10-3 mol/L.
Result: The molar solubility of MgF2 is 1.18 x 10-3 M at room temperature.
Example # 3: Find the molar solubility of barium sulfate at 25°C. Ksp = 1.1 x 10-10.
BaSO4 partially ionizes in water to produce Ba2+ and SO42- ions, as shown below.
Ksp expression for the dissolution of BaSO4 can be derived using the general equation (ii).
Ksp = [C]x [D]y……….Equation (ii)
The balanced dissolution equation shows x = 1 and y = 1, considering C = Ba2+ and D = SO42-. So w.r.t BaSO4, equation (ii) transforms into:
Ksp = [Ba2+] [SO42-]
Let’s suppose [Ba2+] = [SO42-] = x mol/L at equilibrium.
The value of Ksp is already given in the question statement i.e., Ksp = 1.1 x 10-10. Now let’s plug in all the above data into the Ksp expression and find the value of x.
1.1 x 10-10 = (x) (x)
1.1 x 10-10 = x2
x = \sqrt{1.1\times 10^{-10}} = 1.05 x 10-5 mol/L.
As per the balanced chemical equation; [BaSO42-] dissociated = [Ba2+] equilibrium = 1.05 x 10-5 mol/L.
Result: The molar solubility of BaSO4 is 1.05 x 10-5 M at 25°C.
Example # 4: The Ksp value for very weakly soluble calcium phosphate [Ca3(PO4)2] is 2.07 x 10-33. Find the molar solubility of Ca3(PO4)2.
Ca3(PO4)2 partially ionizes in water to produce 3 Ca2+ and 2 PO43- ions, as shown below.
Ksp expression for the dissolution of Ca3(PO4)2 can be derived using the general equation (ii).
Ksp = [C]x [D]y……….Equation (ii)
The balanced dissolution equation shows x = 3 and y = 2, considering C = Ca2+ and D = PO43-. So w.r.t Ca3(PO4)2 equation (ii) transforms into:
Ksp = [Ca2+]3 [PO43-]2
Let’s suppose [Ca2+] = x mol/L at equilibrium.
Then 3 [Ca2+] = 2 [PO43-]
Hence [PO43-] = 2/3 (x).
The value of Ksp is already given in the question statement i.e., Ksp = 2.07 x 10-33. Now let’s plug in all the above data into the Ksp expression and find the value of x.
Result: The molar solubility of Ca3(PO4)2 is 1.14 x 10-7 M at 25°C.
Example # 5: Calculate the water solubility of Ca(IO3 )2 in grams per 100 mL at 25° C. Ksp = 7.1 x 10-7.
To calculate the water solubility of calcium iodate Ca (IO3 )2 in grams per 100 mL, we can first find its molar solubility from the given Ksp value.
Ca(IO3)2 partially ionizes in water to produce Ca2+ and IO3– ions, as shown below.
Ksp expression for the dissolution of Ca(IO3)2 can be derived using the general equation (ii).
Ksp = [C]x [D]y……….Equation (ii)
The balanced dissolution equation shows x = 1 and y = 2, considering C = Ca2+ and D = IO3–. So w.r.t Ca(IO3)2 equation (ii) transforms into:
Ksp = [Ca2+] [IO3–]2
Let’s suppose [Ca2+] = x mol/L at equilibrium. As [IO3–] = 2[Ca2+] = 2x
Now we can use the Ksp value given in the question statement i.e., Ksp = 7.1 x 10-7and find the value of x as follows.
Thus [Ca2+] equilibrium = 5.62 x 10-3 mol/L.
As per the balanced chemical equation; molar solubility = [Ca(IO3)2] dissociated = [Ca2+] equilibrium = 5.62 x 10-3 mol/L.
The molar mass of Ca(IO3)2 is 389.88 g/mol. We can find the mass of calcium iodate dissolved per liter of solution as follows:
∴ mass = moles × molar mass
Mass of Ca(IO3)2 dissolved per litre = (molar solubility) x (389.88)
= (5.62 x 10-3) x (389.88) = 2.19 grams
2.19 g Ca(IO3)2 is dissolved per litre of the solution.
Now let’s find out how many grams of calcium iodate are dissolved in 100 mL.
Grams of Ca(IO3)2 dissolved in 100 mL = grams of Ca(IO3)2 dissolved in 0.1 L = 2.19 x 0.1 = 0.219 grams.
Result: The solubility of Ca(IO3)2 in grams/100 mL is 0.219 at 25°C.
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 further at a certain temperature.
What is a supersaturated solution?
A supersaturated solution holds dissolved solute more than its capacity. A super-saturated solution can be prepared from a saturated solution by increasing the temperature and then swiftly cooling it down.
A large amount of undissolved solute precipitates out of the solution and settles at the bottom of the flask containing the supersaturated solution, unlike anything witnessed in a saturated solution.
What are sparingly soluble salts?
Sparingly soluble salts are chemical compounds that only partially ionize in water; therefore, they undergo aqueous dissolution to a limited extent.
What is Ksp?
Ksp stands for the solubility product constant. It is an equilibrium constant that determines the extent of ionization and, thus, the dissolution of 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.
Why the concentration of salt is 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; it thus possesses a higher Ksp value at T2 than T1 where T = temperature and T2 > T1.
What is the relationship between molar solubility and Ksp?
Molar solubility is directly related to Ksp. The higher the solubility product constant of a solute, it undergoes dissolution to a greater extent indicating higher molar solubility.
How to find molar solubility from Ksp?
The following steps can be applied to find the molar solubility of salt from its given Ksp value at a fixed temperature.
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: Suppose [C] = x; determine the relative [D] by using the mole ratio.
Step IV: Substitute the Ksp value into the Ksp expression and find the unknown concentration x.
Step V: Determine molar solubility from x.
Summary
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.
Molar solubility is calculated in moles/liter.
Ksp stands for the solubility product constant. It determines the extent of ionization of sparingly soluble salt in an aqueous solution.
Ksp is temperature dependent.
We can find molar solubility from Ksp using the expression Ksp = [C]x [D]y and the balanced dissolution equation.
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
Sciencing. (n.d.). How to Calculate Molar Solubility From Ksp. https://sciencing.com/how-to-calculate-molar-solubility-from-ksp-13710255.html
CK-12 Foundation. (n.d.). Conversion of Ksp to Solubility. CK-12 Chemistry Flexbook 2.0. https://flexbooks.ck12.org/cbook/ck-12-chemistry-flexbook%202.0/section/19.12/primary/lesson/conversion-of-ksp-to-solubility-chem/
Course Hero. (n.d.). Conversion of Ksp to Solubility. Chemistry Study Guides. https://www.coursehero.com/study-guides/cheminter/conversion-of-ksp-to-solubility/
ChemistrySteps. (n.d.). Ksp and Molar Solubility. https://general.chemistrysteps.com.ksp-and-molar-solubility/
University of Calgary. (n.d.). Ksp and Solubility. Chem 203. https://wpsites.ucalgary.ca/chem-textbook/chapter-15-main/ksp-and-solubility/
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