Unit 7: Topic 11 - Introduction to Solubility Equilibria

Question 1

What is the Ksp equation?

Answer 1

Let’s say that one day, I’m making pasta. I would need to dissolve some salt (NaCI) in my water.

The dissolution equation would be : NaCI (s) <-> Na+(aq) + CI-(aq)

(Notice that this reaction is reversible)

The Ksp equation is equal to the concentration of products over reactants at equilibrium. However, NaCI is a solid so it isn’t included in the Ksp equation

Ksp = [Na+]*[CI-]

A Ksp&raquo_space; 1 means that the compound is very soluble while a Ksp &laquo_space;1 means that the compound is very insoluble.
A Ksp close to 1 means that the compound is relatively soluble.

The larger the Ksp, the more soluble the compound (note that an exact Ksp value is only really given to insoluble/slightly soluble compounds).

Question 2

What is the solubility of AB if AB dissolves into A- and B+ in solution and the reaction has a Ksp of 1.5*10^-4.

Answer 2

We can write out the reaction equation : AB(s) <-> A-(aq) + B+(aq)

Ksp = [A-][B+], for every molecule of AB(s) that dissolves, we get one molecule of A- and B+, therefore [A-] = [B+], if we let [A-] = x, then Ksp = xx or x^2.

We can now write that 1.5*10^-4 = x^2, and solve for the positive root of x (since negative concentration doesn’t make sense).

We get that x = 0.01224744871 or roughly 1.22*10^-2.

Therefore, the solubility of AB is equal to 1.22*10^-2 moles/liter.

Question 3

Which compound is more soluble : AB₃ with Ksp = 6.210⁻⁷ or C₂D with Ksp = 8.410⁻⁴

Answer 3

We have to do some calculations :

Firstly for AB₃, the dissolution eq. is AB₃(s) <-> A³⁺(aq) + 3B⁻(aq).
Therefore, Ksp for AB₃ = [A³⁺][B⁻]³.
Let [A³⁺] = x, where x is also the solubility of AB₃
So Ksp = x(3x)³ = 27x⁴ = 6.210⁻⁷.
Solving gives that x = 1.2310⁻² so the solubility of AB₃ = 1.23*10⁻² moles/liter

Now, let’s do the same for C₂D :
C₂D(s) <-> 2C⁻(aq) + D²⁺(aq) and so Ksp = [D²⁺][C⁻]². Let [D²⁺] = y so Ksp = y(2y)² = 4y³ = 8.4710⁻⁵
Solving for y gives that y = 2.7710⁻² so the solubility of C₂D = 2.77*10⁻² moles/liter

y > x so C₂D is more soluble

Question 4

Given the following 3 compounds, will Ksp be greater than or less than 1 :

1) (NH₄)CI
2) AgCI
3) PbSO₄

Answer 4

1) (NH₄)⁺ is soluble with every anion so (NH₄)CI will be very soluble in water and Ksp&raquo_space; 1

2) With the exception of fluorine, halides (group 17), such as CI-, are insoluble with Ag⁺ so AgCI will be insoluble in water. Therefore Ksp &laquo_space;1

3) Sulfate is insoluble with lead so PbSO₄ will be insoluble in water. Once again, Ksp &laquo_space;1