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Introduction to Solubility Equilibria

Caleb Thomas

Caleb Thomas

7 min read

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Study Guide Overview

This study guide covers solubility equilibria, focusing on the solubility product constant (Ksp). It explains how to calculate Ksp from solubility, and how to determine molar solubility using Ksp and RICE tables. Finally, it shows how to predict precipitation using the ion product (Qsp) compared to Ksp.

Solubility Equilibria: Your Ultimate Guide ๐Ÿš€

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? ๐Ÿค”

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.

Key Concept

Solubility is essentially an equilibrium process, where a solid dissolves into its ions in a solution.

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:

PbI2(s)โ‡ŒPb2+(aq)+2Iโˆ’(aq)PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)

Even though we call PbI2PbI_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.

Quick Fact

Ksp is the product of the ion concentrations raised to their stoichiometric coefficients. Solids are NOT included in the Ksp expression.

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.
Memory Aid

Think of Ksp like a measure of how much a solid 'likes' to dissolve. A big Ksp means it's a social butterfly that loves to dissolve, while a small Ksp means it's a wallflower that prefers to stay solid.

Solubility Equilibrium

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).

Exam Tip

When calculating Ksp, always convert solubility to molarity (mol/L) first!

Example Problem: Lead Chromate

Let's calculate the Ksp of lead chromate (PbCrO4PbCrO_4):

The solubility of PbCrO4PbCrO_4 is 4.5 \times 10^{-5} g/L. What is its solubility product?

  1. Write the dissolution reaction:

    PbCrO4(s)โ‡ŒPb2+(aq)+CrO42โˆ’(aq)PbCrO_4(s) \rightleftharpoons Pb^{2+}(aq) + CrO_4^{2-}(aq)

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

    4.5ร—10โˆ’5gLร—1ย mol323.2ย g=1.4ร—10โˆ’7molL4.5 \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=[Pb2+][CrO42โˆ’]=(1.4ร—10โˆ’7)(1.4ร—10โˆ’7)=1.9ร—10โˆ’14Ksp = [Pb^{2+}][CrO_4^{2-}] = (1.4 \times 10^{-7})(1.4 \times 10^{-7}) = 1.9 \times 10^{-14}

Common Mistake

Don't forget to raise ion concentrations to the power of their stoichiometric coefficients in the Ksp expression!

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 CaCO3CaCO_3 (Ksp=3.36ร—10โˆ’9Ksp = 3.36 \times 10^{-9}).

  1. Write the dissolution reaction:

    CaCO3(s)โ‡ŒCa2+(aq)+CO32โˆ’(aq)CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)

  2. Set up the RICE table:

    ReactionCaCO3CaCO_3Ca2+Ca^{2+}CO32โˆ’CO_3^{2-}
    Initial----0.00 M0.00 M
    Change----+x+x
    Equilibrium----xx
  3. Plug into the Ksp expression and solve:

    Ksp=[Ca2+][CO32โˆ’]=x2=3.36ร—10โˆ’9Ksp = [Ca^{2+}][CO_3^{2-}] = x^2 = 3.36 \times 10^{-9}

    x=3.36ร—10โˆ’9=5.8ร—10โˆ’5Mx = \sqrt{3.36 \times 10^{-9}} = 5.8 \times 10^{-5} M

Key Concept

The molar solubility of CaCO3CaCO_3 is 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).
Memory Aid

Think of Qsp as the current state of the solution, and Ksp as the solution's 'ideal' state. If Qsp is too high, the solution needs to precipitate to get back to Ksp.

Final Exam Focus ๐ŸŽฏ

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 AgClAgCl is 1.3 \times 10^{-5} M. What is the value of the solubility product constant, KspK_{sp}, for AgClAgCl? (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 KspK_{sp} for CaF2CaF_2 is 3.9 \times 10^{-11}. What is the molar solubility of CaF2CaF_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(NO3)2Pb(NO_3)_2 is mixed with 50.0 mL of 0.10 M NaClNaCl. A precipitate of PbCl2PbCl_2 forms. The KspK_{sp} of PbCl2PbCl_2 is 1.6 \times 10^{-5}.

(a) Write the balanced net ionic equation for the precipitation reaction. (b) Calculate the initial concentrations of Pb2+Pb^{2+} and Clโˆ’Cl^- ions in the mixture. (c) Calculate the value of the ion product, QQ, for the initial conditions. (d) Does a precipitate form? Justify your answer. (e) Calculate the equilibrium concentration of Pb2+Pb^{2+} ions in the solution after the precipitate forms.

FRQ Scoring Breakdown:

(a) (1 point)

Pb2+(aq)+2Clโˆ’(aq)โ‡ŒPbCl2(s)Pb^{2+}(aq) + 2Cl^-(aq) \rightleftharpoons PbCl_2(s)

(b) (2 points)

  • [Pb2+]=(0.020M)โˆ—(50.0mL/100.0mL)=0.010M[Pb^{2+}] = (0.020 M) * (50.0 mL/100.0 mL) = 0.010 M (1 point)
  • [Clโˆ’]=(0.10M)โˆ—(50.0mL/100.0mL)=0.050M[Cl^-] = (0.10 M) * (50.0 mL/100.0 mL) = 0.050 M (1 point)

(c) (1 point)

  • Q=[Pb2+][Clโˆ’]2=(0.010)(0.050)2=2.5ร—10โˆ’5Q = [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>KspQ > K_{sp} (1 point)

(e) (3 points)

  • Set up ICE table:

    Pb2+Pb^{2+}2Cl^-
    Initial0.0100.050
    Change-x-2x
    Equil0.010-x0.050-2x
  • Ksp=1.6ร—10โˆ’5=(0.010โˆ’x)(0.050โˆ’2x)2K_{sp} = 1.6 \times 10^{-5} = (0.010-x)(0.050-2x)^2 (1 point)

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

  • Solve for x: x=0.010โˆ’(1.6ร—10โˆ’5)/(0.050)2=0.0036x = 0.010 - (1.6 \times 10^{-5})/(0.050)^2 = 0.0036. [Pb2+]=0.010โˆ’0.0036=6.4ร—10โˆ’3M[Pb^{2+}] = 0.010 - 0.0036 = 6.4 \times 10^{-3} M (1 point)

You've got this! Go ace that exam! ๐Ÿ’ช

Question 1 of 8

Solubility is best described as which type of process? ๐Ÿค”

Irreversible

Equilibrium

Spontaneous

Non-spontaneous