EQUILIBRIUM & SOLUBILITY: �THE SOLUBILITY PRODUCT CONSTANT, K_SP

JAPANESE FOOT BRIDGE, CLAUDE MONET, 1899

SOLUBILITY

• The maximum amount of substance that will dissolve in a given amount of solvent at a given temperature.
• It may be considered an equilibrium between the solid ionic compound and ions in solution.

Example: Write the equation for the

dissociation of PbI₂.

PbI₂(s) ⇌ Pb²⁺(aq) + 2 I^-(aq)

THE K_SP EXPRESSION

In general, solubility product, K_sp, is the mathematical product of the dissolved ion concentrations of the substance raised to the power of their stoichiometric coefficients (notice this is the same equation as any other equilibrium constant.)

M_yX_z(s) ⇌ yM^z+(aq) + zX^y-(aq)

K_sp = [M^z+]^y[X^y-]^z

solubility product constant

molar solubility of ions

SOLUBILITY PRODUCT

Most salts dissociate into ions (to some extent) when they dissolve. For example: PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl^-(aq)

This equilibrium system may be described by the expression

K_sp = [Pb²⁺][Cl^-]²

We can calculate the molar solubility (the maximum amount of ion that will dissolve in a given amount of solvent) of a compound if we know its K_sp value and using a RICE table.

SOLUBILITY PRODUCT

What is the maximum solubility of PbCl₂?

(Why does it say “maximum” solubility?)

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SOLUBILITY PRODUCT

What is the maximum solubility of PbCl₂?

R PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl^-(aq)

I 0 M 0 M

C +x +2x (x is the solubility!)

E x 2x

K_sp = [Pb²⁺][Cl^-]²

1.7x10^-5 = (x)(2x)²

SOLUBILITY PRODUCT

What is the maximum solubility of PbCl₂?

R PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl^-(aq)

I 0 M 0 M

C +x +2x (x is the solubility!)

E x 2x

K_sp = [Pb²⁺][Cl^-]²

1.7x10^-5 = (x)(2x)² x = 0.016 M

THE COMMON ION EFFECT

When a slightly soluble salt is dissolved in a solution that already contains one of the ions present in the salt (a common ion), the solubility of the salt will be decreased.

THE COMMON ION EFFECT EXAMPLE

What is the solubility of lead(II) chloride in a 0.050 M solution of sodium chloride?

R PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl^-(aq)

I 0 M 0.050 M

C +x +2x

E x 0.050 + 2x

K_sp = [Pb²⁺][Cl^-]²

1.7x10^-5 = (x)(0.050 + 2x)²

x = [Pb²⁺] = 4.79x10^-3 M

THE COMMON ION EFFECT EXAMPLE

What is the solubility of lead(II) chloride in a 0.050 M solution of sodium chloride?

R PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl^-(aq)

I 0 M 0.050 M

C +x +2x

E x 0.050 + 2x

K_sp = [Pb²⁺][Cl^-]²

1.7x10^-5 = (x)(0.050 + 2x)² x = [Pb²⁺] = 4.79x10^-3 M

Notice that this is substantially less than the previous problem without the common ion where [Pb²⁺] = 1.6x10^-2 M.

CALCULATING K_sp

Calculate the K_sp of AgCl if a saturated solution of AgCl is 1.34x10^-5 M in Ag⁺.

AgCl(s) ⇌ Ag⁺(aq) + Cl^-(aq)

K_sp = [Ag⁺][Cl^-]

= (1.34x10^-5)(1.34x10^-5)

= 1.80x10^-10