Calculation of Kps. How to perform the Kps calculation

O Kps calculation (solubility product) is related to two chemical equilibria that occur when a electrolyte (a salt, a base or an acid) sparingly soluble forms a saturated solution with water background. The two balances are:

• dissolution equilibrium

X_TheY_b(here) → X_TheY_{b (ppt)}

In this balance, the speed at which the electrolyte dissolves in water is equal to the speed at which it precipitates. The equilibrium constant (Kc) is:

Kc = 1
X_TheY_b(here)

• balance of dissociation

XaY_b(here) → aX⁺_(here) + bY^-_(here)

as the electrolyte dissolves in the water, automatically he if dissociates, releasing cation and anion. In this case, the equilibrium constant (Kc) is:

Kc = [X⁺]^The. [Y^-]^B
[X_TheY_b(here)]

The molarity of the electrolyte in the solution is always constant, so we can include it in Kc:

Kc. Shah_{Yb (aq)} = X^+a. Y^-B

By including the molarity of the electrolyte in Kc, it is called Kps, and the molarities (solubility or solubility coefficient) of the ions are raised to their respective exponents:

Kps = [X^+a]. [Y^-B]

So, as Kps is related to the ions released by the electrolyte, to develop the calculation of this constant, it is important to know that the molarity of the cation and the anion always obeys a relation in moles with the molarity of the electrolyte of origin, that is:

CaCl₂ → Ca⁺² + 2 Cl^-1

Observing the electrolyte dissociation equation, we have 1 mol of CaCl₂ is for 1 mol of Ca⁺² and 2 moles of Cl^-1. Thus, if the concentration of CaCl₂ for x, that of Ca⁺² will be x and that of Cl^-1 will be 2x.

♦ Kps Calculation Examples

1) (UFRJ) What will be the expression of Kps of CaF₂, using x as the salt molarity?

Resolution:

Initially, it is necessary to set up the salt dissociation equation:

CaCl₂ → Ca⁺² + 2 Cl^-1

In the equation, we have 1 mole of CaF₂ releases 1 mole of CaF₂ and 2 moles of F^-1. Therefore, if the molarity of salt is x, the molarity of Ca⁺² will be x and the molarity of F^-1 will be 2x.

With these data, we can assemble the expression of the Kps of salt:

Kps = [Ca⁺²]. [F^-1]

Kps = x. (2x)²

Kps = x. 4x²

Kps = 4x³

2) (Mackenzie-SP) Determine the solubility product (Kps) of calcium carbonate (CaCO₃) which has a solubility of 0.013g/L, at 20^OÇ. Data: Ca=40; C=12; O = 16.

Resolution:

We have to transform the concentration provided by the exercise from g/L to mol/L, as this is the concentration unit used in the Kps calculations. To do this, calculate the molar mass of the salt and then divide the concentration given by the molar mass:

- Calculation of molar mass:

MCACO₃ = 40 + 12 + 3.(16)

MCACO₃ = 40 + 12 + 48

MCACO₃ = 100g/mol

Conversion of concentration (C) from g/L to mol/L (M):

M =  Ç
M_CaCO3

M = 0,013
100

M = 1.3.10^-4 mol/L

Having in hand the molarity of the salt, it is necessary to know the concentration of each of its ions based on their dissociation:

CaCO₃ → Ca⁺² + CO₃^-2

As a mole of CaCO₃ releases 1 mole of Ca⁺² and 1 mole of CO₃^-2, the concentration of each ion will be equal to that of the salt, that is, 1.3.10^-4. Finally, just calculate the Kps from the expression assembled by the salt dissociation equation:

Kps = [Ca⁺²]. [CO₃^-2]

Kps = 1.3.10^-4. 1,3.10^-4.

Kps = 1.69.10^-8 (mol/L)²

3) (F.C. Chagas-BA) The solubility of a certain MCl chloride₂ in water is 1.0. 10^-3 mol/L. What will the value of your solubility product be:

Resolution:

The exercise has already provided us with the molarity of the electrolyte, so it is enough to carry out its dissociation to determine the molar concentration of each ion and the Kps.

MCI₂ → M⁺² + 2 Cl^-1

As 1 mol of MCl₂ gives 1 mol of M⁺² and 2 moles of Cl^-1, the molarity of M⁺² will be equal to 1.0.10^-3, and the one from Cl^-1 will be double, that is, 2.0.10^-3. Finally, just calculate the Kps from the expression assembled by the electrolyte dissociation equation:

Kps = [M⁺²]. [Cl^-1]²

Kps = 1.0.10^-3. (2,0.10^-3)².

Kps = 1.0.10^-3. 4,0.10^-6

Kps = 4.10^-9 (mol/L)²

4) (OSEC-SP) The solubility product of silver bromide is 5.2×10^-13. If the solution contains 2.0×10^-2 mol of Br^-, what will be the maximum concentration of Ag ions⁺_(here) needed to not precipitate silver bromide (AgBr)?

Resolution:

The data provided by the exercise are:

Kps: 5.2.10^-13

[Br^-1] = 2.10^-2

[Ag⁺¹] = ?

Let's analyze the dissociation of the supplied salt:

AgBr → Ag⁺¹ + Br^-1

We have that 1mol of salt gives rise to 1mol of Ag⁺¹ and 1 mole of Br^-1. Thus, by assembling the Kps expression from these data, we can find the maximum concentration of Ag ions⁺¹:

Kps = [Ag⁺¹].[Br^-1]

5,2.10^-13 = [Ag⁺¹].2,0.10^-2

[Ag⁺¹] = 5,2.10^-13
2,0.10^-2

[Ag⁺¹] = 2,6.10^-11 mol/L

By Me. Diogo Lopes Dias