Solubility Equilibria: Your Ultimate Guide 🚀

Hey there, future AP Chem master! Let's dive into solubility equilibria, a topic that might seem tricky but is totally conquerable with the right approach. Think of this as your pre-exam cheat sheet, designed to make everything click.

Introduction to Solubility Equilibria

What is Solubility Anyway? 🤔

We often talk about substances being "soluble" or "insoluble," but the truth is, everything dissolves to some extent. It's all about how much dissolves.

Instead of thinking of reactions as 'happening' or 'not happening', we should think of them in terms of how far they go. Even so-called "insoluble" compounds like PbI2 do dissolve, just not very much.

The Dissolution Reaction

For example, let's look at lead iodide:

$$PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$$

Even though we call $PbI_2$ "insoluble," this reaction does occur, just with a tiny equilibrium constant (K). This special K for dissolution is called the solubility product constant, or Ksp.

Ksp: The Magic Number

• High Ksp: Indicates a highly soluble compound (like NaCl or KOH). The reaction goes far to the right.
• Low Ksp: Indicates a poorly soluble compound (like PbI2). The reaction barely proceeds to the right.

Image from Purdue University: Visualizing the dynamic equilibrium of dissolution.

Calculating Ksp: Step-by-Step

Grams per Liter vs. Molarity

Solubility is often given in grams per liter (g/L) or moles per liter (mol/L, or molarity, M).

Example Problem: Lead Chromate

Let's calculate the Ksp of lead chromate ($PbCrO_4$):

The solubility of $PbCrO_4$ is 4.5 \times 10^{-5} g/L. What is its solubility product?

1. Write the dissolution reaction:

   $$PbCrO_4(s) \rightleftharpoons Pb^{2+}(aq) + CrO_4^{2-}(aq)$$

2. Convert g/L to mol/L (Molarity):

   $$4.5 \times 10^{-5} \frac{g}{L} \times \frac{1 \ mol}{323.2 \ g} = 1.4 \times 10^{-7} \frac{mol}{L}$$

3. Plug into the Ksp expression:

   $$Ksp = [Pb^{2+}][CrO_4^{2-}] = (1.4 \times 10^{-7})(1.4 \times 10^{-7}) = 1.9 \times 10^{-14}$$

Calculating Molar Solubility

Molar solubility is the concentration of the metal cation in a saturated solution. It is essentially solving for equilibrium concentrations, but we don't include the solid reactant in our calculations. Time to bring back our trusty RICE table!

Example Problem: Calcium Carbonate

Calculate the molar solubility of $CaCO_3$ ($Ksp = 3.36 \times 10^{-9}$).

1. Write the dissolution reaction:

   $$CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$$

2. Set up the RICE table:

   Reaction	$CaCO_3$	$Ca^{2+}$	$CO_3^{2-}$
   Initial	----	0.00 M	0.00 M
   Change	----	+x	+x
   Equilibrium	----	x	x

3. Plug into the Ksp expression and solve:

   $$Ksp = [Ca^{2+}][CO_3^{2-}] = x^2 = 3.36 \times 10^{-9}$$

   $$x = \sqrt{3.36 \times 10^{-9}} = 5.8 \times 10^{-5} M$$

Relating Qsp and Ksp

Just like with regular equilibrium, we can compare the ion product (Qsp) to Ksp to predict if a precipitate will form:

• Qsp > Ksp: Precipitation occurs; the solution is supersaturated.
• Qsp < Ksp: No precipitation; the solution is unsaturated.
• Qsp = Ksp: The solution is at equilibrium (saturated).

Final Exam Focus 🎯

High-Priority Topics

• Calculating Ksp from solubility data.
• Calculating molar solubility using Ksp and RICE tables.
• Predicting precipitation using Qsp and Ksp.
• Understanding the relationship between solubility, Ksp, and equilibrium.

Common Question Types

• Multiple-choice questions involving Ksp calculations and comparisons.
• Free-response questions requiring you to set up RICE tables and calculate molar solubility.
• Questions that combine solubility with other equilibrium concepts.

Last-Minute Tips

• Time Management: Start with the questions you know best. Don't get bogged down on one problem.
• Common Pitfalls: Pay close attention to stoichiometric coefficients and units. Double-check your calculations.
• Challenging Questions: Break down complex problems into smaller steps. Use RICE tables to organize your thoughts.

Practice Question

Practice Questions

Multiple Choice Questions

1. The solubility of $AgCl$ is 1.3 \times 10^{-5} M. What is the value of the solubility product constant, $K_{sp}$, for $AgCl$? (A) 1.3 \times 10^{-5} (B) 2.6 \times 10^{-5} (C) 1.7 \times 10^{-10} (D) 1.3 \times 10^{-10}

2. The $K_{sp}$ for $CaF_2$ is 3.9 \times 10^{-11}. What is the molar solubility of $CaF_2$? (A) 1.95 \times 10^{-11} (B) 3.39 \times 10^{-6} (C) 2.14 \times 10^{-4} (D) 4.26 \times 10^{-4}

Free Response Question

A 50.0 mL sample of 0.020 M $Pb(NO_3)_2$ is mixed with 50.0 mL of 0.10 M $NaCl$. A precipitate of $PbCl_2$ forms. The $K_{sp}$ of $PbCl_2$ is 1.6 \times 10^{-5}.

(a) Write the balanced net ionic equation for the precipitation reaction. (b) Calculate the initial concentrations of $Pb^{2+}$ and $Cl^-$ ions in the mixture. (c) Calculate the value of the ion product, $Q$, for the initial conditions. (d) Does a precipitate form? Justify your answer. (e) Calculate the equilibrium concentration of $Pb^{2+}$ ions in the solution after the precipitate forms.

FRQ Scoring Breakdown:

(a) (1 point)

$$Pb^{2+}(aq) + 2Cl^-(aq) \rightleftharpoons PbCl_2(s)$$

(b) (2 points)

• $[Pb^{2+}] = (0.020 M) * (50.0 mL/100.0 mL) = 0.010 M$ (1 point)
• $[Cl^-] = (0.10 M) * (50.0 mL/100.0 mL) = 0.050 M$ (1 point)

(c) (1 point)

• $Q = [Pb^{2+}][Cl^-]^2 = (0.010)(0.050)^2 = 2.5 \times 10^{-5}$ (1 point)

(d) (1 point)

• Yes, a precipitate forms because $Q > K_{sp}$ (1 point)

(e) (3 points)

• Set up ICE table:

  	$Pb^{2+}$	2Cl^-
  Initial	0.010	0.050
  Change	-x	-2x
  Equil	0.010-x	0.050-2x

• $K_{sp} = 1.6 \times 10^{-5} = (0.010-x)(0.050-2x)^2$ (1 point)

• Since $K_{sp}$ is small, assume x is small: 1.6 \times 10^{-5} = (0.010)(0.050)^2 (1 point)

• Solve for x: $x = 0.010 - (1.6 \times 10^{-5})/(0.050)^2 = 0.0036$. $[Pb^{2+}] = 0.010 - 0.0036 = 6.4 \times 10^{-3} M$ (1 point)

You've got this! Go ace that exam! 💪