Partially soluble salts dissociate in equilibrium equations (similar to how weak acids dissociate partially in equilibrium equations). These are salts considered “insoluble” or “slightly soluble” using solubility rules. But even salts considered “insoluble” dissolve a tiny bit — we can determine the degree to which they dissolve using ICE tables. Additionally, since they are equilibrium processes, their behavior can be influences by the presence of a common ion. We will examine the common ion effect with these salts. Finally we will revisit two topics covered in CHEM 151 — precipitation and reaction quotient — and see how they can be used to predict whether precipitation will occur.

Objective 25: Distinguish between solubility and solubility product constant.

Objective 26: Write the dissociation equation and the K_sp expression for a salt given its name or chemical formula.

Solubility

The solubility of a salt is a measure of how much of the salt will dissolve at equilibrium. The units of solubility include g/L and mol/L (molar solubility).

Here is an example of the dissociation reaction for the insoluble salt silver chloride:

$\text{AgCl}{{\text{ }\!\!~\!\!\text{ }}_{(s)}}\text{ }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ A}{{\text{g}}^{+}}_{(aq)}+\text{ }\!\!~\!\!\text{ C}{{\text{l}}^{-}}_{(aq)}$

When enough of the salt has dissolved so the reaction has reached equilibrium, the solution is saturated. Addition of any Ag⁺or Cl^– to a saturated solution will shift the reaction left, and the ions will come out of solution going bask to solid crystal. For that reason, no more salt can dissolve in a saturated solution.

Dissociation Equations and Solubility Product Constant (Ksp) Expressions

As mentioned above, the dissociation equation for silver chloride is:

$\text{AgCl}{{\text{ }\!\!~\!\!\text{ }}_{(s)}}\text{ }\!\!~\!\!\text{ }\rightleftarrows \text{ }\!\!~\!\!\text{ A}{{\text{g}}^{+}}_{(aq)}+\text{ }\!\!~\!\!\text{ C}{{\text{l}}^{-}}_{(aq)}$

The equilibrium constant (K) can be written for the dissociation reaction:

${{\text{K}}_{sp}}=\left[ \text{A}{{\text{g}}^{+}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]$

Notice there is no denominator — the AgCl is not included in K_sp because it is a solid (s). Recall only aqueous (aq) and gaseous (g) species are included in equilibrium constants — liquids and solids are not.

K_sp is called the solubility product constant, and the equation ${{\text{K}}_{sp}}=\left[ \text{A}{{\text{g}}^{+}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]$ is the solubility product constant expression.

K_spvalues of various salts are tabulated in various places, including here. The Ksp value for AgCl is 1.6 X 10^-10. The solubility product constant has one value for a particular salt at a given temperature and does not change as molarities change. Its value will vary with temperature like all equilibrium constants.

K_sp and solubility are often confused with each other — to reiterate, solubility is the amount of salt that will dissolve per liter of solution, while the solubility product constant (K_sp) is the equilibrium constant for the dissociation reaction. Make sure you understand the difference between the two.

Objective 26 example

A video talking through the writing of a dissociation equation and a Ksp expression for the salt barium fluoride (BaF₂) is at this link.

Objective 26 practice

Objective 27: Calculate the K_sp given the solubility of a partially soluble salt or the solubility given K_sp.

Earlier in this unit, you learned that, using an ICE table, you could determine the pH of a weak acid given the K_a of the acid or given the K_a you could calculate the pH. Similarly, using an ICE table, given the solubility of a salt you can calculate the K_sp or vice versa. Lets look at an example:

Objective 27 example

Ag₂S has a solubility of 3.4 X 10^-17 M. Calculate K_sp for Ag₂S. The solution is shown in this video.

Objective 27 practice

CuBr has a solubility of 2.0 X 10^-4 M at 25 degrees C. Calculate K_sp for CuBr at 25 degrees C.

Given the K_sp, you should also be able to determine the molar solubility or the solubility in g/L. In these problems, you will use the same approach initially but will instead be solving for x from the ICE table, as the subtracted x is the solubility in mol/L. Try this one for practice:

The K_sp of PbBr₂ is 4.6 X 10^-6 at 25 °C. Calculate the solubility of PbBr₂ at 25 °C.

Objective 28: Explain, using equations, the effect of a common ion on the solubility of a salt.

The common ion effect works in solubility equilibria similarly to the way it works in acid/base equilibria.As an example, consider the effect of the soluble salt NaBr on the solubility of the weak electrolyte PbBr₂.

The sodium bromide dissociates completely. Its equation is $\text{NaBr }\!\!~\!\!\text{ (aq)}\to \text{N}{{\text{a}}^{+}}\text{(aq)+ }\!\!~\!\!\text{ B}{{\text{r}}^{-}}\text{(aq)}$

The dissociation or ionization reaction for the insoluble lead(II) bromide is an equilibrium:

$\text{PbB}{{\text{r}}_{2}}\text{ }\!\!~\!\!\text{ (s)}\rightleftharpoons \text{P}{{\text{b}}^{2+}}\text{(aq)+ }\!\!~\!\!\text{ 2 }\!\!~\!\!\text{ B}{{\text{r}}^{-}}\text{(aq)}$

The common ion is the bromide ion, Br^–. With the common ion effect, the NaBr that is present ADDS Br^–. Considering LeChatlier’s principle, what effect does this addition of Br^– from the sodium salt have on the lead(II) bromide dissociation equilibrium? Answer the two questions.

Shifting the equilibrium left shifts it back toward the solid, reducing solubility. Therefore, just as a common ion makes a weak acid dissociate to a smaller degree, it makes a salt dissolve to a lesser degree.

Objective 29: Calculate the solubility of a salt in the presence of a common ion.

In the Objective 27 practice, you calculated the solubility of PbBr₂ in pure water at 25°C to be 1.0 X 10^-2 M, given the K_sp value of 4.6 X 10^-6. If we instead dissolved it in 0.10 M NaBr (providing a common ion), the common ion effect should reduce the solubility to a lower value. Let’s calculate it:

The dissociation equilibrium is: $\text{PbB}{{\text{r}}_{2}}\text{ }\!\!~\!\!\text{ (s)}\rightleftharpoons \text{P}{{\text{b}}^{2+}}\text{(aq)+ }\!\!~\!\!\text{ 2 }\!\!~\!\!\text{ B}{{\text{r}}^{-}}\text{(aq)}$

The Ksp expression is: ${{K}_{sp}}=[P{{b}^{2+}}]{{[B{{r}^{-}}]}^{^{2}}}$

The ICE table is set up as follows:

	_PbBr2	↔	Pb²⁺	^{2 Br–}
I	pure		0	0.10 (from NaBr)
C	-x		+x	+2x
	—–		x	0.10+2x

Substituting into the K_sp expression

$4.6\times {{10}^{-6}}=[x]{{[0.1+2x]}^{^{2}}}$

Neglecting the +2x since Ksp <<1,

$4.6\times {{10}^{-6}}=[x]{{[0.1]}^{^{2}}}$

solving for x yields $4.6\times {{10}^{-4\text{ }}}M=x=\text{solubility}$

Note the solubility has been decreased significantly from the value in pure water. This is due to the presence of the common ion.

Objective 29 practice

Objective 30: Write the equations for the reaction that occurs to increase the solubility when an acid is added to a salt then predict whether its solubility depends on pH and calculate the molar solubility of a salt at a given pH.

pH effects on solubility of hydroxide salts

Changes in pH can also cause the common ion effect. One example is in insoluble hydroxide salts. For example, consider the solubility of Mg(OH)_2.The dissociation equation is:

$\text{Mg(OH}{{\text{)}}_{2}}\text{ }\!\!~\!\!\text{ (s)}\rightleftharpoons \text{M}{{\text{g}}^{2+}}\text{(aq)+ }\!\!~\!\!\text{ 2 }\!\!~\!\!\text{ O}{{\text{H}}^{-}}\text{(aq)}$

If you consider that increasing pH (making the solution more basic) is, in effect, adding OH^– and decreasing pH (making the solution more acidic) is, in effect, removing OH^–, you can use LeChatlier’s principle to predict the effect on the solubility equilibrium:

pH effects on solubility of non-hydroxide salts

Adding acid can also affect solubility of salts that are not hydroxides. As an example, lowering pH can increase the solubility of calcium carbonate (CaCO₃).

$\text{CaC}{{\text{O}}_{3}}\text{ }\!\!~\!\!\text{ (s)}\rightleftharpoons \text{C}{{\text{a}}^{2+}}\text{(aq)+ }\!\!~\!\!\text{ C}{{\text{O}}_{3}}^{2-}\text{(aq)}$

The carbonate ion, as the anion of a weak acid, will react with acid to accept H+: $\text{ }\!\!~\!\!\text{ C}{{\text{O}}_{3}}^{2-}\text{(aq) + }{{\text{H}}^{+}}\text{(aq)}\rightleftharpoons \text{HC}{{\text{O}}_{3}}^{-}\text{(aq)}$

Adding acid shifts this second equilibrium to the right, decreasing the CO₃^2- concentration. This decreasing of CO₃^2- shifts the CaCO₃ dissociation equilibrium right, increasing solubility of CaCO₃ . This illustrates that the addition of acid increases solubility of salts when the anion of the salt is a conjugate base of a weak acid.

Objective 31: Predict whether a precipitate will form when two solutions of known concentration of ions are mixed.

In CHEM 151 (or likely in any first semester general chemistry class if you took it elsewhere), you learned about precipitation reactions, in which if an insoluble salt was formed by the reaction of soluble ions, the salt would come out of solution as an insoluble solid precipitate. For example:

$\text{AgN}{{\text{O}}_{\text{3(aq)}}}\text{+NaC}{{\text{l}}_{\text{(aq)}}}\to \text{AgC}{{\text{l}}_{\text{(s)}}}\text{+NaN}{{\text{O}}_{\text{3(aq)}}}$

Since AgCl is an insoluble salt according to solubility rules, it was assumed it would precipitate. You learned to describe this using molecular, ionic, and net ionic equations. But now we realize that AgCl can dissolve slightly (up to its solubility limit). Therefore, AgCl will only precipitate if there are a sufficient number of silver and chloride ions in solution — enough to put it past the saturation point.

Predicting whether a precipitate will form when two solutions of known concentration of ions are mixed requires the concept of comparing the reaction quotient (Q) with the equilibrium constant (K) for an equilibrium reaction (the dissociation reaction). Let’s review that concept.

For any equilibrium reaction such as $a{{A}_{(aq)}}+b{{B}_{(s)}}\rightleftharpoons c{{C}_{(aq)}}+d{{D}_{(aq)}}$ , we can write an expression for Q:

$Q=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}}$

Note the B is omitted as it is a pure solid. The expression for Q looks just like the expression for K, but in the case of Q the concentrations may not be at equilibrium (for K the concentrations are at equilibrium). We can think of Q as a “test” to see whether the reaction is at equilibrium:

If Q>K , the dissociation reaction will shift left to reach equilibrium
If Q<K the dissociation reaction will shift right to reach equilibrium
If Q=K the reaction is at equilibrium

For a slightly soluble salt, we must compare Q with K_sp. Q is the ion product constant – it is the reaction quotient for the dissociation equation

If Q>K_sp , the dissociation reaction will shift left (toward solid) to reach equilibrium and precipitation will occur
If Q<K_sp precipitation will not occur
If Q=K_sp the solution is saturated (contains the maximum amount of dissolved ions) and the introduction of any more of the salt’s ions will result in precipitation.

Let’s look at the example above with AgCl.

Will a precipitate form if 10.0 mL of 0.010M AgNO₃ and 10 mL of 0.00010M NaCl are mixed? Assume that the final volume of the solution is 20.0 mL. Ksp for AgCl is 1.7 x 10^-10.

The solution is shown in this video

Objective 31 Practice

150.0 mL of 0.0100 M Mg(NO₃)₂is added to 250.0 mL of 0.100 M NaF. Will MgF₂precipitate?

(Ksp of MgF₂ = 6.4 X 10^-9)

Objective 32: Predict when ions are added to a solution containing a mixture of several ions, the minimum concentration of the added ion that will cause precipitation and what salt will precipitate.

This objective is best described by an example:

Solid NaI is slowly added to a solution that is 0.0050 M Pb²⁺ and 0.0050 M Ag⁺. Will PbI₂ or AgI precipitate first? Specify the concentration of I^– necessary to begin precipitation.

The K_spof PbI₂ = 1.4 X 10^-8 and the K_spof AgI = 8.3 X 10^-17