A precipitate may form when solutions are mixed which contain the appropriate ions. For example, a solution containing $Ba2+and\; SO$₄^2- ions may form a precipitate of $BaSO$₄.

When will this occur?

In the previous section, the process that occurs when a salt is first put into water to form a solution was described, and the maximum concentration of solute calculated from $K$_sp. But if the the concentration of solute is greater than the maximum, the system will move back to equilibrium by reducing the amount of solute in solution, through precipitation.

Mathematically, this can be described using the Q value:

• Q (ion product quotient): the product of the equilibrium expression when the system is not at equilibrium

If the concentration of solute is greater than the maximum possible concentration, a precipitate will form. Likewise, when substituted into the equilibrium expression, a value larger than $K$_sp results. Thus, when Q is greater than $K$_sp results in precipitation.

For example, for the reaction:

$BaSO$₄ (s) Ba²⁺ (aq) + SO₄^2- (aq)

If Q is larger than K, then the reaction will shift to the left to establish equilibrium. Thus, $BaSO$₄ (s), a precipitate, forms.

Remember:

If Q > $K$_sp a precipitate will form (until Q decreases to $K$_sp.

If Q = $K$_sp, equilibrium is reached - no precipitate will form.

If Q < $K$_sp, any ppt in solution will dissolve until Q increases to $K$_sp.

EXAMPLE

Two possible precipitates could form:

• $NaNO$₃ (highly soluble - thus NO precipitate in solution),
• $PbSO$₄ ( K_sp = 1.6 x $10-8-\; thus\; POSSIBLE\; precipitate\; in\; solution)$

Determine $[Pb2+]$ and $[SO$₄^2-] so that we can determine Q:

The total volume of the solution is 0.500 L. Because $K$_sp is based on molarities, we must determine the new molarity of each substance in solution, based on the new volume. This is a dilution calculation.

Thus:

$[Pb2+]\; =\; \{\; 3.0$ x $10-3M$ x $0.100\; L\; /\; 0.500\; L\; \}\; =\; 6.0$ x $10-4$ M

$[SO$₄^2-] = { 5.0 x $10-3M$ x $0.400\; L\; /\; 0.500\; L\; \}\; =\; 4.0$ x $10-3M$

$Q\; =\; [Pb2+][SO$₄^2-] = (6.0 x $10-4)(4.0$ x $10-3)\; =\; 2.4$ x $10-6$ M

Because $Q\; >\; K$_sp, PbSO₄ will precipitate!

Solubility and pH

The solubility of almost any salt may be affected by the pH of the solution, especially if one or both of the ions are moderately acidic or basic.

If a substance has a basic anion, such as $Mg(OH)$₂ and $CaF$₂, its solubility will be affected by the pH of the solution.

Consider:

A) $Mg(OH)$₂ (s) Mg²⁺ (aq) + 2 OH^- (aq), at a constant pH (a buffered solution) of 9.0

$[OH-]\; =\; 1.0$ x $10-5M$ ( calculated from pH ; therefore it is the EQUILIBRIUM value and can be included in $K$_sp! )

This indicates that a maximum of 0.18 moles of $Mg(OH)$₂ would dissolve in 1.00 L of solution. (All excess would precipitate!)

Note that if the solution were made more acidic, the solubility of $Mg(OH)$₂ would INCREASE because the $OH-$ ions would react with the hydronium ions and be removed, shifting the equilibrium to the right, so more $Mg(OH)$₂ dissolves.

B) $CaF$₂ (s) + 2H⁺ (aq) Ca²⁺ (aq) + 2HF^- (aq)

As in case (A), the presence of the basic anion ($F-$) affects the solubility at different pH's. An increase in acidity will cause more hydronium and $F-$ ions to react, forming HF and removing $F-$'s from solution. This shifts the equilibirum right, increasing the solubility of CaF$$₂ (s).

IN GENERAL, the solubility of slightly soluble salts which contain basic anions, increase as the pH is lowered.

Selective Precipitation of Ions

Qualitative Analysis for Metallic Elements

EXAMPLE

If $H$₂S$$ gas is bubbled through a solution of $Cu2+$ and $Zn2+$, it will precipitate ONLY the $Cu2+$ at a certain pH. Find this pH at which the ions will be thus separated ( $[H$₂S] = 0.10 M, [Cu²⁺] = [Zn²⁺] = 0.10 M . ).

STEP 1: Identify the two precipitates:

• ZnS(s) - which we do not want to precipitate, and thus must establish boundaries within which precipitation will not occur.
• CuS(s) - which we do wish to precipitate.

STEP 2: The dissociation of $H$₂S:

$H$₂S (aq) 2H⁺ (aq) + S^2- (aq), K_a = 7.4 x $10-21$

STEP 3: Calculate the MAXIMUM $[S2-]$ that can be present in solution before precipitation of ZnS occurs:

Thus:

STEP 4: Now we can find the MINIMUM $[H+]$ that can be present before precipitation of ZnS occurs ( by step 2! ):

STEP 5: Now the MINIMUM pH for which ZnS will NOT precipitate is:

pH = -log $(2.4$ x $10-1)\; =\; 0.59$!

ZnS will not precipitate if the pH = 0.59 or less.

But will CuS still precipitate at such low pH?

STEP 6: Calculate Q for CuS ( $K$_sp = 6.3 x $10-36$ ):

Thus CuS WILL precipitate at this pH!

Effect of Complex Formation on Solubility

Lewis bases other than water can interact with metal ions, affecting the solubility of a metal salt.

For example, AgCl ( $K$_sp = 1.82 x $10-10$ ) dissolves in the presence of ammonia:

AgCl (s) $Ag+(aq)\; +\; Cl-(aq)$

$Ag+(aq)\; +\; 2\; NH$₃ (aq) Ag(NH₃)₂⁺ (aq)

In the above equilibria, a complex ion is formed:

$Ag(NH$₃)₂⁺(aq) is a metal ion with Lewis bases bonded to it.

The equilibrium constant for the formation of $Ag(NH$₃)₂⁺ is (the second equilibrium):

$K$_f is a formation constant.

As $[NH$₃] increases, the second equilibrium is forced to the right, removing free $Ag+$ ions from solution. This forces the first equilibrium to the right, dissolving more AgCl (s).

EXAMPLE:

We know $K$_f = 1.7 x $107$.

This is so large that we can assume that nearly all the $Ag+$ will be converted to $Ag(NH$₃)₂⁺.

Let $x$ equal $[Ag+]$ at equilibrium (this will be very small).

$Ag+(aq)\; +\; 2NH$₃ (aq) Ag(NH₃)₂⁺ (aq)

Dissolution in Strongly Acidic and Basic Solutions

Metallic elements sometimes dissolve in strongly acidic and strongly basic media.

Some examples are hydroxides and the oxides of $Al3+,\; Cr3+$, and $Zn2+$.

As an example, we will look at $Al3+$.

We have previously discussed how hydroxides such as $Al(OH)$₃ will dissolve in acidic solutions. But this compound will also dissolve in a basic solution, resulting from the formation of a complex anion:

$Al(OH)$₃ (s) + OH^- (aq) Al(OH)₄^- (aq)

One way of interpreting amphoterism is by looking at how water molecules behave when surrounding the metal ion.

For example, $Al3+$ is actually $Al(H$₂O)₆³⁺ (aq) with six water molecules bonded to it.

$Al(H$₂O)₆⁺³ loses protons in a step-wise fashion:

$Al(H$₂O)₆³⁺ (aq) + OH^- (aq) Al(H₂O)₅(OH)⁺² (aq) + H₂O (l)

$Al(H$₂O)₅(OH)⁺² (aq) + OH^- (aq) Al(H₂O)₄(OH)₂⁺ (aq) + H₂O (l)

and so on, causing the hydroxide ion to be removed from solution. Thus the $Al3+$ ion is able to dissolve in strongly basic solutions by the formation of a complex.