SolEquil

CHM 1046
General Chemistry II
Dr. Michael Blaber

Additional Aspects of Aqueous Equilibria

Solubility Equilibria

It is time to consider another type of equilibrium: solubility equilibrium

An ionic solid may dissolve in water, but, how much will dissolve?
If enough ionic solid is added to a solvent, some will dissolve but some will be left as undissolved solid
Is there a way to predict how much of a particular ionic solid will dissolve in a solution?

The Solubility-Product Constant, K_sp

Barium sulfate (BaSO₄) is an ionic solid that has a certain solubility in H₂O(l). If enough BaSO₄ is added to a solution of H₂O(l) the solution will become saturated, and some solid BaSO₄ will remain.

In the case of the saturated solution, an equilibrium is established between the solid and dissolved ions of BaSO₄

BaSO₄(s) ó Ba²⁺(aq) + SO₄^2-(aq)

This is an example of a heterogeneous equilibrium (the reactants and products are not all in the same phase)

In writing the equilibrium expression we omit the "concentration" of the solid component(s)

K_sp is the solubility product constant (or just the solubility product)

Note that although the concentration for the solid component(s) is omitted, for the equilibrium to exist some undissolved BaSO₄ must be present

In other words, K_sp = [Ba²⁺(aq)][SO₄^2-(aq)] is true only when some solid BaSO₄ is present in the solution

K_sp describes the equilibrium concentrations of dissolved ions that exists under conditions of saturation with the solid form

The smaller the value of K_sp, the lower the solubility of the ions of an ionic solid
Writing the expression of K_sp is similar for other equilibrium constants; it is equal to the product of the concentrations of the dissolved ions raised to the power of the coefficients from the balanced equation

The ionic solid Ca₃(PO₄)₂ will dissolve to yield calcium ions and phosphate ions:

Ca₃(PO₄)₂ ó 3Ca²⁺(aq) + 2PO₄^3-(aq)

K_sp = [Ca²⁺(aq)]³ * [PO₄^3-(aq)]²

Solubility and K_sp
Solubility

The amount of a substance that dissolves when producing a saturated solution

Solubility can be expressed in g/L or as molar solubility (i.e. mol/L)

Solubility product constant (K_sp)

Describes the concentration(s) of dissolved ions, or substance(s), at saturation equilibrium

The solubility of a substance may change as the concentrations of various ions change (including H⁺), however, the value of K_sp is unique for a given solute at a specified temperature.

A solution of Copper(I)chloride (CuCl) is made such that a solid amount remains after equilibrium (i.e. after a couple of days some solid remains undissolved). The concentration of Cu⁺(aq) ion is determined to be 1.10 x 10^-3M. What is the value of the solubility product constant, K_sp?

CuCl ó Cu⁺(aq) + Cl^-(aq)

K_sp = [Cu⁺][Cl^-]

From the stoichiometry of the balanced equation, at equilibrium [Cu⁺] = [Cl^-]. Therefore, [Cu⁺] = [Cl^-] = 1.10 x 10^-3M. Thus:

K_sp = (1.10 x 10^-3) * (1.10 x 10^-3)

K_sp = 1.21 x 10^-6

What is the solubility of CuCl in g/L?

At equilbrium the molar concentration of Cu+(aq) will be 1.10 x 10^-3M, and the molar concentration of Cl^-(aq) ion will be 1.10 x 10^-3M

For Cu⁺(aq): 1.10 x 10^-3mol/L * (63.5g/mol) = 0.0699g/L

For Cl^-(aq): 1.10 x 10^-3mol/L * (35.4g/mol) = 0.0389g/L

Thus, a total of (0.0699 + 0.0389) = 0.109g/L of CuCl solid will dissolve (i.e. the solubility of CuCl is 0.109g/L)

Here is an example of how the solubility can change, but the solubility product constant is the same:

What is the solubility of CuCl in an aqueous solution of 0.01M NaCl?

CuCl ó Cu⁺(aq) + Cl^-(aq)

K_sp = [Cu⁺][Cl^-] = 1.21 x 10^-6

XM = amount of CuCl that dissolves in the NaCl solution

Therefore at equilibrium we have XM of Cu⁺(aq) and (0.01 + XM) of Cl^-(aq)

K_sp = [Cu⁺][Cl^-] = 1.21 x 10^-6 = (X)(0.01 + X)

X² + 0.01X - 1.21 x 10^-6 = 0

This is a quadratic with a solution X = 1.20 x 10^-4M

This is the solubility of CuCl in the NaCl solution. At equilibrium we would therefore have 1.20 x 10^-4M of Cu⁺(aq) and (0.01 + 1.20 x 10^-4) = 1.01 x 10^-2M of Cl^-(aq). Checking the value for K_sp with these concentrations:

K_sp = [Cu⁺][Cl^-] = (1.20 x 10^-4)( 1.01 x 10^-2) = 1.21 x 10^-6

Thus, the presence of the NaCl has changed the solubility of the CuCl, but the K_sp is an intrinsic constant.