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Chapter 18: Solubility Product

August 20, 2021 by Admin

Chapter 18
Solubility Product

This chapter is a Part II continuation of the Ionic Equilibria Series in Chemistry. The first chapter is Acid-Base Equilibria, and Solubility Product is relatively intertwined with the Chemical Equilibria chapter.
There are relatively easy-to-grasp concepts in general for this chapter, however, there are certain things to note, as well as confusion that can easily arise in most students if not carefully revised.

The easiest concept in this chapter would be none other than calculating the solubility and
solubility product of ionic salts. However, there are some important things to note:

1. The higher the Kₛₚ of an ionic salt, the higher the solubility. Is this always true or false?
Answer: We need to take note of the most important criteria in this case - the number of ions present in the solution. With an equal number of ions produced in a solution by the two different ionic salts, we can then compare the solubility using their Kₛₚ values. Otherwise, we cannot do that. Hence, the above statement might not be true all the time.

2. Note that solubility can be affected by:

Common ion effect (explained later on in this article)
Temperature
pH
Complex ion formation

3. Misconception: Solubility is the same as solubility product.
Solubility ≠ Solubility product.
Solubility of a salt is the maximum number of moles of solute that can be dissolved in 1 dm³
of water to form a saturated solution at a fixed temperature.
However, the solubility product of a sparingly soluble salt is the product of the molar concentrations of the constituent ions in a saturated solution, each raised to the power of its stoichiometric coefficient in the balanced equation.

Ionic Product vs Kₛₚ

The following table will summarise the 3 different cases:

I.P. < Kₛₚ	I.P. = Kₛₚ	I.P. > Kₛₚ
Solution is not saturated. No Precipitation.	Solution is saturated. No Precipitation.	Solution is more than saturated. Precipitation occurs, which means that concentration of ions in the solution falls.

I.P. < Kₛₚ

I.P. = Kₛₚ

I.P. > Kₛₚ

Solution is not saturated.

No Precipitation.

Solution is saturated.

No Precipitation.

Solution is more than saturated.

Precipitation occurs, which means that concentration of ions in the solution falls.

Tips to note about Ionic Product and Kₛₚ :

I.P. bears the same expression as the Kₛₚ
Kₛₚ (Solubility product) refers to the concentration of ions (equilibrium system) in a saturated solution but the Ionic Product refers to the concentration of ions in the actual system at the instance of mixing the solution just before the occurrence of precipitation
Kₛₚ value is constant and varies with temperature only, but ionic product is not a constant and varies with a change in concentration

Common Ion Effect

Part I - Understanding the Gist of this Concept
We will look at this crucial concept through an example:

Consider the following:

BaSO₄(s) ⇌ Ba²⁺ (aq) + SO₄²⁻ (aq)

Note: BaSO₄ (s) is a sparingly soluble salt.

Determine the change (Addition of Common Ions).
In this case, if Na₂SO₄ (aq) is added to the solution, more SO₄²⁻ ions are added, and [SO4₄²⁻] in the solution will increase.
Recall the Le Chatelier’s Principle (LCP).
When [SO4₄²⁻] increases, the position of equilibrium shifts leftwards by the Le Chatelier’s Principle in order to remove the extra SO4₄²⁻ ions.
State the resulting effect.
More BaSO₄ (s) is formed, and solubility of BaSO₄ decreases in the solution containing excess SO4₄²⁻ ions, which is the common ion.

Part II - Calculations
We will showcase to you the Common Ion Effect in calculation questions using a sample worked example for a better understanding.

Worked Example
[2017 H2 Chemistry AJC Paper 3 Qn2(iii)]

Silver forms a series of halides of general formula AgX. The chloride, bromide and iodide of silver are sparingly soluble in water at room temperature.

Data about the solubilities in water and the solubility products of the chloride, bromide and iodide of silver at 298 K are given below.

Salt	Solubility / mol dm⁻³	Solubility product / mol² dm⁻⁶
AgCl	1.4 x 10⁻⁵	2.0 x 10⁻¹⁰
AgBr	7.1 x 10⁻⁷	5.0 x 10⁻¹³
AgI	8.9 x 10⁻⁹	7.9 x 10⁻¹⁷

To a 2.0 dm³ of saturated solution of AgI, 0.025 g of AgNO₃ (s) was added. Calculate the mass of precipitate formed.

Solution:

Let the solubility of AgI, after addition of AgNO₃, be 𝑥.
Kₛₚ = [Ag⁺] [I⁻]
7.92 x 10⁻¹⁷ = (7.36 x 10⁻⁵ + 𝑥)(𝑥)
Since solubility of AgI in water is << 10⁻⁵ (due to common ion effect caused by common ion Ag⁺),
(7.36 x 10⁻⁵) + 𝑥 ≈ 7.36 × 10⁻⁵ (Note that this is an assumption that we make theoretically)
𝑥 = 1.08 x 10⁻¹² mol dm⁻³

Mass of AgI precipitated = (8.9 x 10⁻⁹ – 1.08 × 10⁻¹²) x 2 x (107.9 + 126.9)

= 4.2 x 10⁻⁶ g

Here’s an easy tip on how to remember solubility of precipitate < 𝑥:
Use the simple analogy of dissolving sugar (representing precipitate) in sugary water (representing aqueous solution containing common ions) versus pure water. The obvious answer would definitely be sugar being able to dissolve much easier in pure water. Similarly, it would be much more difficult for precipitate to dissolve in aqueous solution containing the common ions.

Formation of Complex Ions

Halide ions react with AgNO₃ (aq) to form an AgX precipitate.
The silver halides can be distinguished by observation of colour or the reaction of precipitate with dilute and concentrated NH₃ (aq).

The following table is a quick summary of the reactions of halide ions with aqueous Ag⁺ ions, and then NH₃ (aq).

Halides	Cl⁻(aq)	Br⁻(aq)	I⁻(aq)
Add AgNO₃(aq)	White precipitate (AgCl)	Cream precipitate (AgBr)	Yellow precipitate (AgI)
Effect of Sunlight on AgX	White precipitate turns grey Application: Usage of salts in photographic films (Tested in 2020 H2 Chemistry A Level Application Question Context)	Cream precipitate turns grey	Yellow precipitate turns grey
Solubility of AgX in dilute NH₃ (aq)	Soluble to give colourless solution	Insoluble	Insoluble
Solubility of AgX in conc. NH₃ (aq)	Soluble to give colourless solution	Soluble to give colourless solution	Insoluble

Question: Why do we use solubility in NH₃ to distinguish AgX?
Sometimes, white and cream colour can be very similar, hence solubility in NH₃ helps us to further confirm our answers, especially in practical.

Note this:

Kₛₚ of AgCl > Kₛₚ of AgBr > Kₛₚ of AgI

The explanation for solubility of silver halide is easily tackled using the following simplified checklist:

❏ Firstly, write out the equation for the reaction between the silver halide precipitate and the dilute / concentrated NH₃ (aq) which forms the complex ion, [Ag(NH₃)₂]⁺ (aq):

AgX (s) ⇌ Ag⁺ (aq) + X⁻ (aq)

Ag⁺ (aq) + 2NH₃(aq) → [Ag(NH₃)₂]⁺ (aq)

❏ When dilute / concentrated NH₃ (aq) is added, [Ag⁺] will decrease since it will be used to form the complex ion, [Ag(NH₃)₂]⁺ (aq). (Refer to the above equation to understand better)

❏ Hence, [Ag⁺] decreases to the point where I.P. (AgX) < Kₛₚ (AgX), so it shifts the position of equilibrium of the equation AgX (s) ⇌ Ag⁺ (aq) + X⁻ (aq) rightwards, therefore dissolving AgX precipitate.

In the case of insolubility, the explanation would be as such:

❏ As dilute / concentrated NH₃ (aq) is added to the AgX precipitate, the formation of complex ion [Ag(NH₃)₂]⁺ (aq) would cause the [Ag⁺] to decrease

❏ Lowering the I.P. (AgX)

❏ However, the value of I.P. (AgX) will still be > Kₛₚ (AgX) since the Kₛₚ (AgX) is of a low value

Now, let’s try using a sample question below to test our understanding:

Worked Example

Barium salts are poisonous. If a solution containing barium ions with concentration about 10⁻³ mol dm⁻³ is drunk, it will cause stomach upsets. Yet, patients are given a ‘barium meal’ of barium sulfate, BaSO₄, before doctors take stomach X-rays.

(a) (i) Calculate the concentration of barium ions in a saturated solution of BaSO₄.
[K_sp of BaSO₄ = 1.0 × 10⁻¹⁰ mol² dm⁻⁶ ]

Answer:

Since the K_sp of BaSO₄ is K_sp = [Ba²⁺][SO₄^2-],
And [Ba²⁺] = [SO₄^2-],
[Ba²⁺] = √ ᴷₛₚ

Hence, [Ba²⁺]

= √ 1.0 x 10⁻¹⁰

= 1.0 x 10 ^-5 mol dm^-3

(ii) Would this concentration of barium ions pose a threat to patients taking a barium meal? Explain.

Answer: Since [Ba²⁺] = 1.0 10⁻⁵ mol dm⁻³, which is less than 1.0 10-3 mol dm⁻³ 1.0 × 10 ⁻³, it would not pose a threat to patients taking the barium meal.

(b) A patient has taken a barium meal shortly after taking some ‘Epsom salts’, MgSO₄ and is worried about the effects of mixing the two. A doctor estimates that the sulfate ion concentration in the stomach is about 10⁻² mol dm⁻³.

(i) Calculate the concentration of barium ions in the patient’s stomach.

Answer:

Concentration of Ba²⁺ is no longer equal to concentration of SO₄^2-, since [SO₄^2-]
increased upon addition of MgSO₄.
Since K_sp= [Ba²⁺][SO₄²⁻],
Rearranging the equation,

= 1.0 10⁻⁸ mol dm⁻³

(ii) Is the patient's fear justified?
The patient's fear is not justified since [Ba²⁺] is 1.0 x 10⁻⁸ mol dm⁻³, which is even lower than 1.0 x 10⁻³ mol dm⁻³.

(ii) Would this concentration of barium ions pose a threat to patients taking a barium meal? Explain.

Answer: Since [Ba²⁺] = 1.0 × 10⁻⁵ mol dm⁻³, which is less than 1.0 × 10⁻³ mol dm⁻³, it would not pose a threat to patients taking the barium meal.

Chapter 18: Solubility Product

Chapter 18 Solubility Product

Chapter 18
Solubility Product