PptSep

CHM 1046
General Chemistry II
Dr. Michael Blaber

Additional Aspects of Aqueous Equilibria

Precipitation and Separation of Ions

Solubility, and the establishment of a saturated solution, is an equilibrium process

At equilibrium the rate of solubilization of molecules is equal to the rate of crystallization of the solid

Note: crystallization denotes an orderly lattice arrangement of molecules as they are desolvated. However, desolvation may also result in a disordered arrangement of desolvated molecules - known as a precipitate

The equilibrium condition can be reached from either direction. Consider the establishment of equilibrium for a saturated solution of NaCl:

NaCl(s) ó Na⁺(aq) + Cl^-(aq)

K_sp = [Na⁺(aq)]*[Cl^-(aq)]

We could start with NaCl(s) and add water.

In this case, at the start of the experiment there are no solvated ions of Na⁺ or Cl^-
The reaction proceeds to the right as equilibrium is achieved

In principle, we could also start by combining solutions that contained Na⁺(aq) and Cl^-(aq) ions and no solid NaCl

If the product of the ion concentrations was greater than K_sp then solid NaCl would start to form
The reaction would proceed to the left as equilibrium is achieved

Determination of the reaction quotient, Q, for reactions has previously allowed us to determine the direction that a given reaction will proceed. The same is true for figuring out whether a solution of ions will precipitate

We will compare the value of Q to the value of the solubility product constant, K_sp
For equilibrium expressions involving solubility, Q is calculated omitting the concentrations of solid or liquid components (their concentrations don't change)

In the above case, for example, Q = [Na⁺(aq)]*[Cl^-(aq)]
In other words, the determination of Q (and K_sp) involves only consideration of the concentrations of the soluble ions in the solution. Thus, Q is referred to as the ion product

When comparing the value of the ion product to the solubility product, there are three possible situations and we can draw the following conclusions:

Q > K_sp : in this case we are starting with a situation where the product of the concentrations of the soluble ions is greater than the expected value at equilibrium. Therefore, the concentration of soluble ions must decrease as equilibrium is achieved. In other words, the reaction is driven to the left, and solid compound is produced.
Q = K_sp : in this case we are starting with a situation where the product of the concentrations of the soluble ions is equal to the value expected at equilibrium. In other words, we are already in an equilibrium situation and overall ion concentrations (and mass of solid) will not change.
Q < K_sp : in this case we have a situation where we are starting with insufficient concentration of soluble ions. At equilibrium more soluble ions will be present. Therefore, the reaction must proceed to the right, and some amount of the solid form of the compound must dissolve and release more soluble ions.

How can situation #1 arise? How can we start with Q > K_sp? In short, how can we have essentially a greater concentration of ions than the solubility product constant will allow?

The concentration of ions may arise from different solutions of salts or compounds. For example, when considering the concentration of Na⁺ and Cl^- ions and the possiblity of NaCl precipitating or crystallizing, it may be the case that Na⁺ ions are being provided by another ionic compound (i.e. NaOH) and the Cl^- ions are also being provided by another solution (i.e HCl).
These solutions may be quite soluble as separate solutions, but when mixed, the concentrations of ions may be too great for the solubility of NaCl(s).

Will a precipitate form when 0.15L of 2.5 x 10^-3M of Pb(NO₃)₂ is added to 0.2L of 4.0 x 10^-3M of Na₂SO₄? K_sp for PbSO₄ = 1.6 x 10^-8

The starting solutions are soluble. What are the possible new ion combinations when they are mixed?

Cations are Pb²⁺, Na⁺
Anions are NO₃^-, SO₄^2-
New combinations include: PbSO₄ and NaNO₃
Note that compounds containing NO₃^- ion are freely soluble, as are common compounds of alkali metal ions (pages 122-123 of the text). Thus, NaNO₃ is essentially soluble.
The only question would involve the solubility of PbSO₄. K_sp for PbSO₄ = 1.6 x 10^-8 (appendix D of your text)

The solubility equilibrium we are concerned with is therefore:

PbSO₄(s) ó Pb²⁺(aq) + SO₄^2-(aq)

K_sp = [Pb²⁺] * [SO₄^2-] = 1.6 x 10^-8 at equilibrium

Let's solve for the ion product, Q:

Q = [Pb²⁺] * [SO₄^2-] at the starting concentrations

What are the starting concentrations?

The total volume will be 0.15L + 0.2L = 0.35L
The total amount of Pb²⁺ ion = (0.15L)*2.5 x 10^-3moles/L = 3.75 x 10^-4 moles

Thus the initial molar concentration of Pb²⁺ = 3.75 x 10^-4 moles/0.35L = 1.07 x 10^-3M

The total amount of SO₄^2- ion = (0.20L)*4.0 x 10^-3moles/L = 8.0 x 10^-4 moles

Thus the initial molar concentration of SO₄^2- ion = 8.0 x 10^-4 moles/0.35L = 2.29 x 10^-3M

Therefore:

Q = 1.07 x 10^-3M * 2.29 x 10^-3M = 2.45 x 10^-6

Q > K_sp

Therefore, precipitation of PbSO₄ will occur when the solutions are mixed

And as a result, Pb²⁺ and SO4^2- ions are "selectively removed" from solution due to this precipitation

Ions can be separated from each other based on the solubilities of their salts

AgCl is not particularly soluble, whereas, CuCl₂ is highly soluble
If Cl^- ions are added to a solution of Ag⁺(aq) and Cu²⁺(aq) ions, the Ag⁺(aq) ions will selectively precipitate as AgCl(s)

Separation of ions in an aqueous solution by using a reagent that forms a precipitate with one of the ions is called selective precipitiation

A solution contains both Ag⁺(aq) and Pb²⁺(aq) ions. The Ag⁺(aq) ion concentration = 1.5 x 10^-2M and the Pb²⁺(aq) ion concentration = 0.5 x 10^-2M. If Cl^-(aq) ion is added to the solution we can selectively precipitate the Ag⁺ and Pb²⁺ ions. The K_sp for AgCl = 1.8 x 10^-10, and the K_sp for PbCl₂ = 1.6 x 10^-5. What concentration of Cl^-(aq) is required to precipitate each ion? Which ion (i.e. which salt) will precipitate first?

From the values of the solubility product constants we can set up the conditions at which precipitation of the two salts will occur:

K_sp for AgCl = 1.8 x 10^-10 = [Ag⁺(aq)] * [Cl^-(aq)]

K_sp for PbCl₂ = 1.6 x 10^-5 = [Pb²⁺(aq)] * [Cl^-(aq)]²

We want to know the conditions for precipitation in terms of added Cl^-(aq) ions, therefore, we will rearrange the above equations and solve for the [Cl^-(aq)] concentration for the known concentrations of both Ag⁺(aq) ion and Pb²⁺(aq) ion:

[Cl^-(aq)] = 1.8 x 10^-10 / 1.5 x 10^-2 = 1.2 x 10^-8 M for precipitation of AgCl

[Cl^-(aq)] = (1.6 x 10^-5 / 0.5 x 10^-2)^0.5 = 5.66 x 10^-2 M for precipitation of PbCl₂

Thus, upon addition of 1.2 x 10^-8M Cl^-(aq) ion the Ag⁺(aq) ion will precipitate as the AgCl salt. Upon further addition of Cl^-(aq) ion to a concentration of 5.66 x 10^-2 M, the Pb²⁺(aq) will precipitate as the PbCl₂ salt. Thus, the Ag⁺(aq) will precipitate first.

Note that the results indicate that if Cl^-(aq) ion is added to a concentration greater than 1.2 x 10^-8M, but less than 5.66 x 10^-2 M, we can selectively precipitate the Ag⁺(aq) ion from this solution

Selective precipitation can also be used to find out whether an ion is present at some minimum level. For example, let's say the maximum acceptable concentration of silver ion in a water sample is 1.0 x 10^-10M.

K_sp for AgCl = 1.8 x 10^-10 = [Ag⁺(aq)] * [Cl^-(aq)]

1.8 x 10^-10 = 1.0 x 10^-10 * [Cl^-(aq)]

[Cl^-(aq)] = 1.8 x 10^-10/1.0 x 10^-10

[Cl^-(aq)] = 1.8M

Therefore, if we add Cl^-(aq) to a concentration of 1.8M and see no precipitation form, then we can conclude that the Ag⁺(aq) concentration must be less than 1.0 x 10^-10M