15.1 Precipitation and Dissolution

Solubility Equilibria and Precipitation

Solubility equilibria describe the balance between a solid ionic compound dissolving and its ions recombining into a solid. These equilibria let you predict whether a precipitate will form when you mix two solutions, and they let you calculate exactly how much of a compound can dissolve under given conditions.

This section covers how to write solubility equilibrium expressions, calculate solubilities from $K_{sp}$, and use the reaction quotient $Q$ to predict precipitation.

Chemical Equations for Solubility Equilibria

A solubility equilibrium exists when the rate of dissolution of a solid equals the rate of precipitation in a saturated solution. At this point, the solution holds the maximum concentration of dissolved ions it can support, and the system is at dynamic equilibrium.

These equilibria involve slightly soluble (often called "insoluble") ionic compounds dissociating into their ions. For example, silver chloride dissociates as:

$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$

The general form looks like this:

$\ce{A_xB_y(s) <=> xA^{y+}(aq) + yB^{x-}(aq)}$

• $\ce{A_xB_y(s)}$ is the slightly soluble ionic solid
• $\ce{A^{y+}(aq)}$ and $\ce{B^{x-}(aq)}$ are the dissolved ions
• $x$ and $y$ are the stoichiometric coefficients from the balanced equation

The solubility product constant ($K_{sp}$) is the equilibrium constant for this type of reaction. Because the solid doesn't appear in the expression (pure solids have an activity of 1), the $K_{sp}$ expression only includes ion concentrations:

$K_{sp} = [\ce{A^{y+}}]^x[\ce{B^{x-}}]^y$

For the calcium phosphate equilibrium $\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO4^{3-}(aq)}$, the expression would be:

$K_{sp} = [\ce{Ca^{2+}}]^3[\ce{PO4^{3-}}]^2$

A small $K_{sp}$ means very little of the compound dissolves. A larger $K_{sp}$ means more dissolves. You can only compare $K_{sp}$ values directly to rank solubilities when the compounds produce the same number and ratio of ions (e.g., comparing $\ce{AgCl}$ to $\ce{AgBr}$, both 1:1 salts).

Calculations with Solubility Product Constants

Molar solubility is the number of moles of solute that dissolve per liter of saturated solution (units of mol/L or M). You can convert between molar solubility and $K_{sp}$ in either direction.

Calculating molar solubility from $K_{sp}$:

1. Write the balanced dissolution equation.
2. Define the molar solubility as $s$ (the moles of solid that dissolve per liter).
3. Express each ion's equilibrium concentration in terms of $s$, using stoichiometric coefficients.
4. Substitute into the $K_{sp}$ expression and solve for $s$.

Example: Find the molar solubility of $\ce{PbCl2}$ given $K_{sp} = 1.7 \times 10^{-5}$.

The equilibrium is: $\ce{PbCl2(s) <=> Pb^{2+}(aq) + 2Cl-(aq)}$

If $s$ moles of $\ce{PbCl2}$ dissolve per liter, then $[\ce{Pb^{2+}}] = s$ and $[\ce{Cl-}] = 2s$.

$K_{sp} = (s)(2s)^2 = 4s^3$

$1.7 \times 10^{-5} = 4s^3$

$s = \left(\frac{1.7 \times 10^{-5}}{4}\right)^{1/3} = 0.016 \text{ M}$

Calculating ion concentrations from molar solubility is straightforward: multiply the molar solubility by the stoichiometric coefficient of each ion. In the example above, $[\ce{Pb^{2+}}] = 0.016$ M and $[\ce{Cl-}] = 2(0.016) = 0.032$ M.

Prediction of Precipitation Reactions

When you mix two solutions, you need to determine whether the ions present will combine to form an insoluble solid. For example, mixing $\ce{AgNO3(aq)}$ and $\ce{NaCl(aq)}$ introduces $\ce{Ag+}$ and $\ce{Cl-}$ into the same solution, and a white $\ce{AgCl}$ precipitate forms.

To predict whether precipitation occurs, you compare the reaction quotient ($Q$) to $K_{sp}$. The $Q$ expression has the same form as the $K_{sp}$ expression, but you plug in the initial ion concentrations (after mixing, before any reaction):

$Q = [\ce{A^{y+}}]^x[\ce{B^{x-}}]^y$

Be careful here: when you mix two solutions, the total volume increases, so you need to recalculate ion concentrations using dilution before plugging into $Q$.

Comparing $Q$ to $K_{sp}$:

• $Q < K_{sp}$: The solution is unsaturated. No precipitate forms. More solid could still dissolve.
• $Q = K_{sp}$: The solution is exactly saturated. The system is at equilibrium with no net change.
• $Q > K_{sp}$: The solution is supersaturated. A precipitate will form, and ions will leave solution until $Q$ drops to equal $K_{sp}$.

Factors Affecting Solubility

The common ion effect is one of the most tested concepts in this unit. If a solution already contains one of the ions in the $K_{sp}$ expression, the solubility of the compound decreases.

For example, consider dissolving $\ce{AgCl}$ in pure water versus in a 0.10 M $\ce{NaCl}$ solution. In the $\ce{NaCl}$ solution, $[\ce{Cl-}]$ is already 0.10 M before any $\ce{AgCl}$ dissolves. This shifts the equilibrium toward the solid (by Le Chatelier's principle), so far less $\ce{AgCl}$ dissolves than it would in pure water.

Le Chatelier's principle applies to solubility equilibria just like any other equilibrium:

• Adding a common ion shifts equilibrium toward the solid (less dissolves).
• Removing an ion from solution (through a secondary reaction, for instance) shifts equilibrium toward dissolving more solid.
• Temperature changes affect solubility depending on whether dissolution is endothermic or exothermic for that particular compound.

Fractional precipitation takes advantage of different $K_{sp}$ values to selectively remove ions from a mixture. If a solution contains both $\ce{Cl-}$ and $\ce{CrO4^{2-}}$, slowly adding $\ce{Ag+}$ will precipitate $\ce{AgCl}$ first (lower $K_{sp}$) before $\ce{Ag2CrO4}$ begins to precipitate. This technique is useful for separating ions in qualitative analysis.