How to calculate solubility product

The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol Ksp.

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Significance of Solubility Product
Solubility Product Constant
Related Videos
Solubility Product Ksp
Solubility of Ionic Compounds
Practice Questions
Quotient Solubility Product Q
Practice Questions

The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increased solubility.

Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution. The solubility of ionic compounds (which dissociate to form cations and anions) in water varies to a great deal. Some compounds are highly soluble and may even absorb moisture from the atmosphere, whereas others are highly insoluble.

Significance of Solubility Product

Solubility depends on a number of parameters, amongst which the lattice enthalpy of salt and solvation enthalpy of ions in the solution are of most importance.

When a salt is dissolved in a solvent the strong forces of attraction of solute (lattice enthalpy of its ions) must be overcome by the interactions between ions and the solvent.
The solvation enthalpy of ions is always negative which means that energy is released during this process.
The nature of the solvent determines the amount of energy released during solvation that is solvation enthalpy.
Non-polar solvents have a small value of solvation enthalpy, meaning that this energy is not sufficient to overcome the lattice enthalpy.
So the salts are not dissolved in non-polar solvents. Hence, for salt to be dissolved in a solvent, its solvation enthalpy should be greater than its lattice enthalpy.
Solubility depends on temperature and it is different for every salt.

Salts are classified on the basis of their solubility in the following table.

Category I	Soluble	Solubility > 0.1M
Category II	Slightly soluble	0.01M< Solubility<0.1M
Category III	Sparingly soluble	Solubility < 0.1M

Solubility Product Constant

Suppose barium sulphate along with its saturated aqueous solution is taken. The following equation represents the equilibrium set up between the undissolved solids and ions:

The equilibrium constant in the above case is:

In the case of pure solid substances, the concentration remains constant, and so we can say:

= =

Here is known as the solubility product constant. This further tells us that solid barium sulphate when in equilibrium with its saturated solution, the product of concentrations of ions of both barium and sulphate is equal to the solubility product constant.

We now have a brief idea about the solubility of a compound and the factors affecting it. For any further query on this topic, call the mentor support team at BYJU’S.

The solubility of a substance in a solvent is the total amount of the solute that can be dissolved in the solvent at equilibrium. On the other hand, the solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.

When dissolved in polar solvents, one magnesium fluoride molecule dissociates into one magnesium cation and two fluoride anions. This equilibrium reaction can be represented as follows.

MgF2 ⇌ Mg2+ + 2F–

Therefore, the solubility product constant can be expressed as:

Ksp = [Mg2+][F–]2

The chemical formula of common salt is NaCl. When dissolved in polar solvents, one sodium chloride molecule dissociates into one sodium cation and one chloride anion. This equilibrium reaction can be represented as follows.

NaCl ⇌ Na+ + Cl–

Therefore, the solubility product constant can be expressed as:

Ksp = [Na+][Cl–]

The chemical formula of calcium chloride is CaCl2. When dissolved in polar solvents, one calcium chloride molecule dissociates into one calcium cation and two chloride anions. This equilibrium reaction can be represented as follows.

CaCl2 ⇌ Ca2+ + 2Cl–

Therefore, the solubility product constant can be expressed as:

Ksp = [Ca2+][Cl–]2

Some important factors that have an impact on the solubility product constant are:

The common-ion effect (the presence of a common ion lowers the value of Ksp).
The diverse-ion effect (if the ions of the solutes are uncommon, the value of Ksp will be high).
The presence of ion-pairs.

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Solubility Product Ksp

Dissolution of ionic compounds forms an equilibrium between dissolved ions and undissolved solids.
Dissolution of lead (II) fluoride:

$${\rm PbF}_{2(s)}\rightleftharpoons {\rm Pb}_{(aq)}^{2+}+{2F}_{(aq)}^-$$

The expression of equilibrium constant or solubility product for this reaction is:

$$K_{sp}=[{Pb}^{2+}][F^-]^2$$

· The concentration of solids (ionic compounds) is not included in the expression because it stays relatively constant compared to the dissolved ions. The concentrations of dissociated ions are always expressed in mol L-1 (M).

· Solids and liquids are also heterogenous in that they do not disperse throughout the solution.

· At equilibrium, the rates of dissolution and re-formation of ionic compound are equal. This is known as the point of saturation as no more ions can dissolve in the solvent. Adding more solutes to a solution at saturation point (equilibrium) will cause precipitation to happen.

Figure: Dissociation of solid PbI2 in a beaker.

Solubility product constant varies between ionic compounds. It indicates how far the dissolution proceeds at equilibrium.
- A large solubility product value indicates high solubility as a relatively large quantity of ions are dissolved at equilibrium.
- Vice versa, a small solubility product value indicates low solubility as relatively small quantity of ions are dissolved at equilibrium.

Table: Solubility product expressions of selected few ionic compounds.

Ionic compound	Dissolution reaction	Solubility product
Magnesium carbonate	$${\rm MgCO}_{3(s)}\rightleftharpoons{\rm Mg}_{(aq)}^{2+}+{\rm CO}_{3(aq)}^{2-}$$	$$K_{sp}=[{\rm Mg}^{2+}][{\rm CO}_3^{2-}]$$
Iron (II) hydroxide	$${\rm Fe(OH)}_{2(s)}\rightleftharpoons{\rm Fe}_{(aq)}^{2+}+{\rm 2OH}_{(aq)}^-$$	$$K_{sp}=[{\rm Fe}^{2+}][{\rm OH}^-]^2$$
Calcium phosphate	$${{\rm Ca}_3(PO_4)}_{2(s)}\rightleftharpoons{\rm 3Ca}_{(aq)}^{2+}+{\rm 2PO}_{4(aq)}^{3-}$$	$$K_{sp}=[{\rm Ca}^{2+}]^3[{\rm PO}_4^{3-}]^2$$

Example 1

The following table shows the solubility product constants of selected ionic compounds at 25ºC.

Name, formula	Ksp
Aluminium hydroxide, Al(OH)3	`3 xx 10^(-34)`
Lead(II) fluoride, PbF2	`3.6 xx 10^(-8)`
Silver sulfide, Ag2S	`8.0 xx 10^(-48)`
Cobalt(II) carbonate, CoCO3	`1.0 xx 10^(-10)`

Rank the ionic compounds in the table in order of increasing solubility.

Solubility of Ionic Compounds

The solubility of an ionic compound is usually expressed as the amount of solute dissolved per volume of solvent. This can be either expressed as mass per volume e.g. g/100 mL or as molarity, mol L-1 (M) .

When solubility of an ionic compound is known, its Ksp can be determined. For example, at 25ºC, the solubility of PbF2 is found to be 0.64 g/L. Calculate the Ksp of PbF2.

Step 1: Convert solubility into mol L-1 by dividing by the molar mass of the ionic compound. In this case, the molar mass of PbF2 = 245.2 g mol-1

$$n=\frac{0.64\ g}{245.2\ g\ {\rm mol}^{-1}}$$

$$n=0.0026\ mol$$

Therefore, solubility = 0.0026 mol L-1 (M)

Step 2: Write an equation representing the dissociation of the ionic compound.

$${\rm PbF}_{2(s)}\rightleftharpoons{\rm Pb}_{(aq)}^{2+}+{\rm 2F}_{(aq)}^-$$

Step 3: Write an expression for the solubility product of the ionic compound and find the concentration of each ion.

$$K_{sp}=[{\rm Pb}^{2+}][{\rm F}^-]^2$$

$$K_{sp}=[0.0026][2×0.0026]^2$$

$$K_{sp}=7.1\times{10}^{-8}$$

Using Ksp values from the data sheet, solubility of various ionic compounds can be determined.

Step 1: Write an equation representing the dissociation of the ionic compound.	$${\rm PbI}_{2(s)}\rightleftharpoons{\rm Pb}_{(aq)}^{2+}+{\rm 2I}_{(aq)}^-$$
Step 2: Write an expression for the solubility product of the ionic compound.	$$K_{sp}=[{\rm Pb}^{2+}][{\rm I}^-]^2$$
Step 3: Let s be the solubility of the ionic compound. Assign concentration of ions in terms of s.	$${\rm PbI}_{2(s)}\rightleftharpoons{\rm Pb}_{(aq)}^{2+}+{\rm 2I}_{(aq)}^-$$ s s 2s
Step 4: Substitute Ksp (from data sheet) and concentrations of ions in terms of s into Ksp expression.	$$9.8\times{10}^{-9}=[s][2s]^2$$
Step 5: Solve for solubility (s).	$$9.8\times{10}^{-9}=4s^3$$ $$s=1.3\times{10}^{-3} mol L^{-1}$$

Practice Questions

Determining Ksp from solubility

Question 1

(a) Lead(II) sulfate (PbSO4) is a key component in lead acid car batteries.

(b) Its solubility in water at 25°C is 4.25 `xx` 10-3 g/100 mL. What is the Ksp of PbSO4?

(c) In terms of their constituent ions, explain why lead(II) fluoride has a much greater solubility than lead(II) sulfate.

Question 2

The solubility of silver chloride in water at 25ºC is 1.34 `xx` 10-5.What is the Ksp of silver chloride?

Question 3

The solubility of calcium hydroxide in water at 25ºC is 0.074 g/100 mL. What is the Ksp of calcium hydroxide?

Determining solubility from Ksp

Question 1

Ksp of barium sulfate is 11.1 `xx` 10-10. What is the solubility of barium sulfate in moles per litre? What about in grams per mL?

Question 2

Ksp of iron(III) hydroxide is 14.4 `xx` 10-38. What is the solubility of iron(III) hydroxide in moles per litre? What about in grams per mL?

Quotient Solubility Product Q

Similar to equilibrium quotient Q, the solubility quotient is the value assigned to a dissociation reaction when it is not at equilibrium.

When Q < Ksp, the dissociation system has not reached the point of saturation, it is unsaturated. This means more ions will dissolve and become hydrated by solvent molecules e.g. water.

When Q > Ksp, the dissociation system has exceeded the point of saturation, it is saturated. This means precipitate will form as no more ions can be hydrated by solvent molecules.

Important: high Ksp indicates more ions can dissolve before saturation (equilibrium) is reached. This means high solubility. In contrast, low Ksp indicates less ions can dissolve before saturation is reached, low solubility.

Practice Questions

Determining formation of precipitate

Question 1

0.100 g of BaSO4 is added to 500.0 mL of water at 25 ºC. Calculate the Qsp of this barium sulfate solution and thus determine whether a precipitate will form.

Question 2

Will lead(II) chloride precipitate when 50 mL of 0.10 M Pb(NO3)2 solution is mixed with 50 mL of 0.10 M NaCl solution? Support your answer with a balanced chemical equation and calculations.

Question 3

Equal volumes of 0.25 mol L–1 solutions of silver nitrate and calcium chloride are mixed at 25 ºC. Predict whether a precipitate will form. Support your answer with calculations.

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Next section: Common Ion Effect