Solubility Equilibria

Chapter 16 focuses on solubility equilibria, quantitating solubility using physical equilibria principles.
In Chapter 4, solubility rules for ionic compounds were introduced qualitatively.
The goal is to provide a quantitative understanding of solubility.

In a supersaturated solution (e.g., silver bromide), solid precipitates at the bottom.
Dynamic Equilibrium: Silver and bromide ions constantly precipitate onto the solid's surface, while the solid continuously dissolves.
This equilibrium is a dynamic process, not static, allowing us to apply Le Chatelier's principle.

The solid is written as the reactant, and the dissolved ions are the products.
Example: AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)
The equilibrium constant, Kc, is expressed as: Kc = [Ag^+][Br^-]
Silver bromide solid is incorporated into the constant due to heterogeneous equilibrium.

Kc is specifically termed K{sp}, the solubility product constant.
K_{sp} represents the concentrations at equilibrium in a saturated solution.
Even in a supersaturated solution, the immediate solution at the solid's surface is saturated.
Analogy to saturated ammonium chloride solution preparation: supersaturate, let solid settle, and pipette the supernatant.

Q is the ion product and it is the initial concentrations of the ions: Q = [Ag^+]0[Br^-]0
Q is analogous to the reaction quotient from chapter 14.
Q and K variations are chemical equilibrium examples, despite different names and symbols.

If Q = K_{sp}, the solution is saturated; no additional solid dissolves.
If Q < K_{sp}, the solution is unsaturated; more solid can dissolve.
- The process shifts forward because there is not enough numerator.
If Q > K_{sp}, the solution is supersaturated; precipitation occurs.
- Excess numerator causes the process to shift in reverse.
- Precipitation continues until equilibrium is reached.

As K_{sp} increases, solubility increases.
Comparison of similar ionic compounds (e.g., 1:1 cation-anion ratio) is valid.
Caution: Comparisons between different ionic compound types (e.g., 1:1 vs. 1:2 ratios) require consideration of stoichiometry.

Dissolution equation: Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)
K_{sp} = [Ca^{2+}][OH^-]^2
Important Chem 1 Reminder: 2OH^- is different from O_2H^-; the latter is non-existent.

Molar Solubility (s): Moles of compound that dissolve in 1 liter of saturated solution.
Solubility (s): Grams of compound that dissolve in 1 liter of saturated solution.
Both are temperature-dependent.
Use the units to identify which one is being referred to!

Problem: Calculate the molar solubility of calcium sulfate (CaSO4), given K{sp} = number.
Write the balanced equation: CaSO4(s) \rightleftharpoons Ca^{2+}(aq) + SO4^{2-}(aq)
Write the K{sp} expression: K{sp} = [Ca^{2+}][SO_4^{2-}]
Let s = molar solubility of CaSO_4.
At equilibrium: [Ca^{2+}] = s and [SO_4^{2-}] = s
Therefore, K_{sp} = s * s = s^2
Solve for s: s = \sqrt{K_{sp}}
To find solubility in grams per liter, convert moles to grams using molecular weight.

K_{sp}'s dependency on s varies with different ionic compounds.
Example: Calculate the solubility product for copper iodate, given its solubility is 1.3 g/L.
Write the balanced equation: Cu(IO3)2(s) \rightleftharpoons Cu^{2+}(aq) + 2IO_3^-(aq)
Write the K{sp} expression: K{sp} = [Cu^{2+}][IO_3^-]^2
Let s = molar solubility of Cu(IO3)2.
Equilibrium concentrations: [Cu^{2+}] = s and [IO_3^-] = 2s
K_{sp} = (s)(2s)^2 = 4s^3
Convert grams per liter to moles per liter before solving for K_{sp}.

Precipitation occurs when Q > K_{sp}.
Example: Silver Bromide - Will a precipitate form if we mix 500 mL of Ag^+ at 10^{-6} M with 500 mL of Br^- at 2 * 10^{-3} M given that the K_{sp} is 1 * 10^{-13}?
Write the Q expression: Q = [Ag^+][Br^-]
Calculations:
- \frac{(10^{-6} M)(0.5 L)}{(0.5 L + 0.5 L)} * \frac{(2 * 10^{-3} M)(0.5 L)}{(0.5 L + 0.5 L)}= 1*10^{-9}
Comparison: Q = 10^{-9} > K_{sp} = 10^{-13}
Conclusion: Yes, a precipitate will form.

Problem: A solution is prepared that is 5.0 * 10^{-4} M in lead (II) ions and 5.0 * 10^{-5} M in chromate ions. Will lead chromate precipitate? K_{sp} = 1.8 * 10^{-14}.
Equation: PbCrO4(s) \rightleftharpoons Pb^{2+}(aq) + CrO4^{2-}(aq)
Q = [Pb^{2+}][CrO_4^{2-}]
Q = (5.0 * 10^{-4})(5.0 * 10^{-5}) = 2.5 * 10^{-8}
Since Q > K_{sp}, a precipitate will form.

Problem: Mixing 1.0 L of 0.0020 M NaOH and 1.0 L of 0.0001 M MgSO_4. Is a precipitate expected? (Ksp magnesium hydroxide is 1.2 * 10^{-11}).
Recognize this is a double displacement reaction; Na2SO4 and Mg(OH)_2 are the products.
Since all sodium salts are soluble, Mg(OH)2 is the potential precipitate (also indicated by the given K{sp}).
Equation: Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)
Q = [Mg^{2+}][OH^-]^2
Initial Concentrations (after mixing):
- [Mg^{2+}] = \frac{(1.0 L)(0.0001 M)}{2.0 L}
- [OH^-] = \frac{(1.0 L)(0.0020 M)}{2.0 L}
Q = (\frac{(1.0 L)(0.0001 M)}{2.0 L})(\frac{(1.0 L)(0.0020 M)}{2.0 L})^2 = 2.5 * 10^{-13}
Q = 2.5 * 10^{-13}, Ksp = 1.2 * 10^{-11}. Since Q < K_{sp}, no precipitate forms. Shift will be forward.