Solubility Equilibria

Solubility Equilibria

Introduction

• 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.

Solubility as an Equilibrium Process

• 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.

Representation of Solubility Equilibria

• 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, $K<em>c$, is expressed as: $K</em>c = [Ag^+][Br^-]$
• Silver bromide solid is incorporated into the constant due to heterogeneous equilibrium.

Solubility Product Constant - Ksp

• $K<em>c$ is specifically termed $K</em>{sp}$, 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.

Ion Product - Q

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

Interpreting Q and K

• 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.

Relationship Between Ksp and Solubility

• 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.

Calcium Hydroxide Example

• 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.

Definitions for Calculations

• 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!

Calculating Molar Solubility Example

• Problem: Calculate the molar solubility of calcium sulfate ($CaSO<em>4$), given $K</em>{sp} = number$.
• Write the balanced equation: $CaSO<em>4(s) \rightleftharpoons Ca^{2+}(aq) + SO</em>4^{2-}(aq)$
• Write the $K<em>{sp}$ expression: $K</em>{sp} = [Ca^{2+}][SO_4^{2-}]$
• 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.

Ksp Dependence on s

• $K_{sp}$'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: $Cu(IO<em>3)</em>2(s) \rightleftharpoons Cu^{2+}(aq) + 2IO_3^-(aq)$
• Write the $K<em>{sp}$ expression: $K</em>{sp} = [Cu^{2+}][IO_3^-]^2$
• Let s = molar solubility of $Cu(IO<em>3)</em>2$.
• 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 $K_{sp}$.

Numerical Example

• $K_{sp} = 4 * (\frac{1.3 \frac{g}{L}}{238.4 \frac{g}{mol}})^3$
• $K_{sp} = 6.5 * 10^{-7}$
  • Remember significant figures!

Aluminum Chloride Dissolution

• Dissolution equation: $AlCl_3(s) \rightleftharpoons Al^{3+}(aq) + 3Cl^-(aq)$
• If s = molar solubility of $AlCl_3$, then $[Al^{3+}] = s$ and $[Cl^-] = 3s$
• $K_{sp} = [Al^{3+}][Cl^-]^3 = s(3s)^3 = 27s^4$
• Observe the varying dependencies of $K_{sp}$ on s, based on salt formula.

Criteria for Precipitate Formation

• 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.

Lead Chromate Example

• 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: $PbCrO<em>4(s) \rightleftharpoons Pb^{2+}(aq) + CrO</em>4^{2-}(aq)$
• $Q = [Pb^{2+}][CrO_4^{2-}]$
• $Q = (5.0 * 10^{-4})(5.0 * 10^{-5}) = 2.5 * 10^{-8}$
• Since Q > K_{sp}, a precipitate will form.

Mixing Solutions - Magnesium Hydroxide Precipitation

• 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; $Na<em>2SO</em>4$ and $Mg(OH)_2$ are the products.
• Since all sodium salts are soluble, $Mg(OH)<em>2$ is the potential precipitate (also indicated by the given $K</em>{sp}$).
• 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.

Skipping Sections

• 16.7 Fractional Precipitation- skip.
• 16.8 and 16.9 Common Ion Effect and pH Effects on Solubility- skip.
• 16.10 and 16.11 Complex Ion Equilibria- skip.
• Qualitative Analysis- skip.