15.1 Solubility & Precipitation

Unit 1: Solubility and Precipitation

Based on solubility rules, we say some ionic compounds are soluble in water, others can be partially soluble, and some compounds are almost insoluble. In a precipitation reaction, we study how some ionic compounds produce a precipitate in the reaction mixture. A precipitate is a solid, insoluble ionic compound that forms from an aqueous solution. These compounds have very low solubility, or you can say they are almost insoluble in water.

To understand this concept, we can look at the example of the dissolution of silver chloride (AgCl). If we take a beaker full of water and add solid AgCl to it, initially, there are no Ag⁺ or Cl^– ions present in water. So, added AgCl will dissociate and produce Ag⁺ and Cl^– ions

AgCl (s)                     Ag⁺ (aq) + Cl^–(aq)

As more and more AgCl will dissolve Ag⁺ and Cl^– ions concentration will increase in solution. An increase in the concentration of Ag⁺ and Cl^– will increase the rate of backward reaction.

Ag⁺ (aq) + Cl^– (aq)                        AgCl (s)

Then a point will come where a dynamic equilibrium will establish, the rate of the forward and the backward reaction will be equal. At this point, we can say the system has reached a state of equilibrium.

AgCl (s)                          Ag⁺ (aq) + Cl^–(aq)

Source: www.openstax.org

Picture here shows a very small solubility of AgCl in solution. Most of it is present as soild on the bottom of container.

Like other equilibrium reactions, we can write an equilibrium constant expression for this reaction.

Ksp = [Ag⁺] [Cl^–]

For precipitation reactions, instead of showing an equilibrium constant with K, we use Ksp. Here sp stands for solubility product. Ksp is the product of the concentration of ions in the saturated solution of the compound. [Ag⁺] and [Cl^–] are the concentration of these ions in mol/L. Since AgCl is solid, it will not appear in this equilibrium constant expression.

This YouTube video illustrates the concept and formula equation for Ksp for different compounds.

Solubility and solubility product constant are two different things. Solubility tells us the maximum amount of a compound that can be dissolved in water at a specific temperature. The solubility of a compound can change if some common ions are already present in the solution. However, solubility product Ksp is the equilibrium constant and has only one value for a compound at a specific temperature. Ksp values can be used to find the solubility, the maximum amount of salt that can be dissolved in water or a solution.

Let’s look at the example of solubility of AgCl.

AgCl will dissociate in solution and produce Ag⁺ and Cl^– ions as per the reaction below.

AgCl (s)                          Ag⁺ (aq) + Cl^–(aq)

If the solubility of AgCl = s mol/L.

Since mole ratio of AgCl to Ag⁺ and Cl^– ions in the reaction above is 1: 1, we can say s mol of AgCl will dissociate in water to produce s mol of Ag⁺ and s mol of Cl^– ion.

Hence, the concentration of [Ag⁺] = s mol/L

 and [Cl^– ] = s mol/L

                        Ksp = [Ag⁺] [Cl^–]

Thus                Ksp = (s) (s)

Or                    Ksp = s²

We can use the similar concept to write the expressions for Ksp for different salts.

Example 1: Write the expression for Ksp for Ca₃(PO₄)₂, if the solubility of Ca₃(PO₄)₂ in water is s mol/L.

First, we will write the equation for the dissociation of Ca₃(PO₄)_2.

Ca₃(PO₄)₂(s)                         3Ca⁺²(aq) + 2PO₄^-3(aq)

The initial concentration of both Ca⁺² and PO₄^-3 ions in the solution will be 0. 

Let’s assume s is the molar solubility of Ca₃(PO₄)₂ in solution. Based on the mole ratio from the dissociation reaction above, we can see that 1 mol of Ca₃(PO₄)₂(s) is producing 3 mols of Calcium ions and 2 mols of phosphate ions,  Hence, the change in concentration of calcium ions will be 3s, and the change in concentration of phosphate ions will be 2s.  Since Ca₃(PO₄)₂ is soild, it will not have any value for concetration in solution.

	Ca3(PO4)2(s)		3Ca+2(aq)	2PO4-3(aq)
I	N/A		0	0
C	N/A		3s	2s
E	N/A		3s	2s

                                    Ksp = [Ca⁺²]³ [PO₄^-3]²

                                     Ksp = (3s)³ (2s)²

                                     Ksp = 108s⁵

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Learning Check:

1.  Find the expression for Ksp for PbCl₂ in the solution.
2. If the molar solubility of PbCl₂ is x mol/L, find the expression that represents the Ksp value for PbCl₂.

Answer:

1. PbCl₂ (s)                     Pb⁺²(aq) + 2Cl^–(aq)

 Ksp =  [Pb⁺²] [Cl^–]²

        2.  

	PbCl2 (s)		Pb+2(aq)	2Cl–(aq)
I	N/A		0	0
C	N/A		x	2x
E	N/A		x	2x

Ksp = [Pb⁺²] [Cl^–]² = 4x³

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Here is a reference table for Ksp values of different salts.

In the last example, we saw how we could relate the solubility of a compound to its solubility product (Ksp). Therefore, we can use Ksp to find the solubility of a compound.

Example 2: Find the solubility of AgCl in a saturated solution at 25 ⁰C.

We know AgCl will dissociate in solution as per the reaction below.

AgCl (s)                           Ag⁺ (aq) + Cl^–(aq)

If we assume the solubility of AgCl is s mol/L. Since 1 mol of AgCl is producing 1 mol Ag⁺ and 1 mol Cl^– ions in solution, the concentration of Ag⁺ and Cl^– ions will also be s mol/L in solution.

Ksp = [Ag⁺] [Cl^–]

From the table above Ksp for AgCl = 1.7 x 10^-10

We can setup the ice chart.

Ksp = [Ag⁺] [Cl^–]

1.7 x 10^-10 = [Ag⁺] [Cl^–]

1.7 x 10^-10 = (s) (s) = s²

s =

s = 1.30 x 10^-5 M

This value of s shows that the maximum solubility of AgCl in water will be 1.30 x 10^-5 M. If we have more than this amount of AgCl present in a solution, it will precipitate out.

Watch this video to see more problems on how solubility can be calculated based on the Ksp value.

https://www.youtube.com/watch?v=LTYabXISM4M

Learning check:

Find the solubility of CaSO₄ in a saturated solution at 25 ⁰C.

 CaSO₄ (s)                           Ca⁺² (aq) + SO₄^-2(aq)

From the Ksp chart above, Ksp for CaSO₄ = 2.4 x 10^-5

Answer: 0.00490 M

Example 3: Calculating Ksp from Solubility

Find the Ksp value for PbCl₂ if it has a solubility of 0.0159 mol/L at 25 ⁰C.

PbCl₂ will dissociate in solution as per the following equation

 PbCl₂ (s)                           Pb⁺² (aq) + 2Cl^–(aq)

We can set up an ICE chart to find the equilibrium concentration of Pb⁺² and Cl^– ions in the solution. The initial concentration of Pb⁺² and Cl^– ions in the solution will be 0. PbCl₂ is solid, so you do not need to show its concentration in the solution.

Since molar solubility of PbCl₂ is = 0.0159 mol/L,

Based on the dissociation reaction for PbCl₂,mole ratio of PbCl₂: Pb⁺² is 1: 1

Therefore, the change in concentration of Pb⁺²  in solution will be = 0.0159 mol/L

Mole ratio of PbCl₂: Cl^– ions is  = 1: 2,.

Therefore the change in concentration of Cl^– ions will be =  2 x 0.0159 = 0.0318 M

By adding initial concentration and change in concentration, we can get the equilibrium concentration of both Pb⁺² and Cl^– ions in solution.

Ksp = [ Pb⁺² ] [Cl^–]²                                                                  

Putting the concentrations for Pb⁺² and Cl^– ions from the table above;

            Ksp = 0.0159 x (0.0318)²

Ksp = 1.61 x 10^-5

Watch this YouTube video for another example of calculating Ksp from molar solubility.

https://www.youtube.com/watch?v=TJMEmCEWbvQ&t=815s

Learning Check:

Find the Ksp for BaF₂, if it’s molar solubility in a saturated solution at 25 ⁰C is 0.01817 M.

	BaF2 (s)	Ba+2 (aq)	2F–(aq)
Initial		0	0
Change		0.01817 M	2 x 0.01817 = 0.0364 M
Equilibrium		0.0182 M	0.03634 M

Ksp = [ Ba⁺² ] [F^–]² = 0.0182 x 0.03634² = 2.40 x 10^-5

Example 4: Determine Ksp from solubility in g/L

Find Ksp of AgBr at 25 ⁰C, if 0.132 mg of AgBr was dissolved in water to make 1.00 L saturated solution. (Molar mass of AgBr is 187.77 g/mol)

AgBr (s)                           Ag⁺ (aq) + Br^–(aq)

Ksp = [Ag⁺] [Br^–]

To solve for Ksp, we first need to find the compound’s molar solubility in a saturated solution. We are given 0.132 mg of AgBr was dissolved in water to make a saturated solution.

We can follow the following plan to find molar solubility.

mg/L of AgBr             g/L of AgBr                mol/L of AgBr

=  7.03 x 10^-7 mol/L

Mole ratio of Ag⁺: AgBr is = 1:1,

Therefore the concentration of [Ag⁺] = 7.03 x 10^-7 mol/L.

Mole ratio of Br^–: AgBr is = 1:1 :

Therefore the concentration of [Br^–] = 7.03 x 10^-7 mol/L

Putting these values in the formula for Ksp will give us

Ksp = [Ag⁺] [Br^–]

Ksp = 7.03 x 10^-7 mol/L x 7.03 x 10^-7 mol/L

Ksp  = 4.94  x 10^-13

*Remember, just like other equilibrium constant values, Ksp does not have any units.

Learning Check:

• Find the solubility of CaF₂ in a solution at 25 ⁰C, given Ksp for CaF₂ = 3.9 x 10^-9
• If the solubility of SrF₂ in a solution at 25 ⁰C was 5.8 x 10^-4 mol/L, find the Ksp for this compound.
• A saturated solution of MgF₂ was prepared by dissolving 169 mg MgF₂ in 1.00 L solution at 25 ⁰C. Find Ksp for MgF₂ at this temperature.

Answer: 3) 9.9 x 10^-4 mol/L                     4) 7.8 x 10^-10   5) 7.98 x 10^-8