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Common Ion Effect

Emily Wilson

Emily Wilson

7 min read

Next Topic - pH and Solubility
Study Guide Overview

This study guide covers the common ion effect on solubility equilibria. It explains how adding a common ion decreases solubility, using Le Chatelier's Principle. It includes practice problems involving Ksp calculations, ICE tables, and molar solubility with and without common ions. The guide also emphasizes the importance of this topic for the AP exam.

#Solubility Equilibria and the Common Ion Effect

Hey there, future AP Chem master! Let's dive into how solubility changes when we're not just dealing with pure water. It's all about the common ion effect, and it's a game-changer! 💡

This topic is crucial because it connects solubility, equilibrium, and Le Chatelier's principle—all big hitters on the AP exam. Expect to see it in both multiple-choice and free-response questions.

#The Common Ion Effect: What is it?

The common ion effect is a fancy way of saying that if you add a salt that shares an ion with another salt already in solution, the solubility of the second salt goes down. It's like adding more of one ingredient to a recipe – it throws off the balance. Think of it as a crowded dance floor; if there are already a lot of dancers of one type, it's harder for more of that type to join in!

#Key Points

  • A common ion is an ion that is present in two or more different compounds.
  • Adding a common ion to a solution decreases the solubility of a slightly soluble salt.
  • This effect is a direct result of Le Chatelier's Principle.
Key Concept

The common ion effect is a direct application of Le Chatelier's Principle. Remember, systems at equilibrium will shift to relieve stress. Adding a common ion is a stress that the system counteracts by reducing the solubility of the salt.

#How Does it Work? 🤔

Let's say we have a salt, ABABAB, that dissolves like this:

AB(s)⇌A+(aq)+B−(aq)AB(s) \rightleftharpoons A^+(aq) + B^-(aq)AB(s)⇌A+(aq)+B−(aq)

The solubility product, KspK_{sp}Ksp​, is:

Ksp=[A+][B−]K_{sp} = [A^+][B^-]Ksp​=[A+][B−]

Now, if we add a salt like NaBNaBNaB, which also has B−B^-B− ions, we're increasing the concentration of B−B^-B− in the solution. According to Le Chatelier's Principle, the equilibrium will shift to the left, favoring the formation of the solid ABABAB and reducing the solubility of ABABAB.

Memory Aid

Think of it like a seesaw. If you add weight to one side (the common ion), the seesaw tips the other way (less of the salt dissolves).

#Common Ion Effect Practice Problem ✏️

Let's work through an example to make this crystal clear. We'll calculate the molar solubility of AgBrAgBrAgBr (Ksp=7.7×10−13K_{sp} = 7.7 \times 10^{-13}Ksp​=7.7×10−13) in pure water and then in a 0.0010M0.0010 M0.0010M solution of NaBrNaBrNaBr.

#Molar Solubility in Pure Water

First, the equilibrium:

AgBr(s)⇌Ag+(aq)+Br−(aq)AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)AgBr(s)⇌Ag+(aq)+Br−(aq)

Set up an ICE table:

AgBrAgBrAgBrAg+Ag^+Ag+Br−Br^-Br−
Initial-000000
Change-+x+x+x+x+x+x
Equilibrium-xxxxxx

So, Ksp=[Ag+][Br−]=x2=7.7×10−13K_{sp} = [Ag^+][Br^-] = x^2 = 7.7 \times 10^{-13}Ksp​=[Ag+][Br−]=x2=7.7×10−13

Solving for x, we get:

x=7.7×10−13=8.8×10−7Mx = \sqrt{7.7 \times 10^{-13}} = 8.8 \times 10^{-7} Mx=7.7×10−13​=8.8×10−7M

#Molar Solubility in 0.0010M0.0010 M0.0010M NaBrNaBrNaBr

Now, we have an initial concentration of 0.0010M0.0010 M0.0010M of Br−Br^-Br− from NaBrNaBrNaBr. Our ICE table changes:

AgBrAgBrAgBrAg+Ag^+Ag+Br−Br^-Br−
Initial-0000.00100.00100.0010
Change-+x+x+x+x+x+x
Equilibrium-xxx0.0010+x0.0010 + x0.0010+x

Our KspK_{sp}Ksp​ expression becomes:

Ksp=[Ag+][Br−]=x(0.0010+x)≈x(0.0010)=7.7×10−13K_{sp} = [Ag^+][Br^-] = x(0.0010 + x) \approx x(0.0010) = 7.7 \times 10^{-13}Ksp​=[Ag+][Br−]=x(0.0010+x)≈x(0.0010)=7.7×10−13

We can ignore the '+x' because x will be very small compared to 0.0010. Solving for x:

x=7.7×10−130.0010=7.7×10−10Mx = \frac{7.7 \times 10^{-13}}{0.0010} = 7.7 \times 10^{-10} Mx=0.00107.7×10−13​=7.7×10−10M

Notice that the solubility of AgBrAgBrAgBr is much lower in the NaBrNaBrNaBr solution than in pure water! That's the common ion effect in action.

Quick Fact

When solving for solubility with a common ion, you can often ignore the '+x' in the equilibrium expression if the initial concentration of the common ion is significantly larger than the Ksp value.

#Justifying the Common Ion Effect with Le Chatelier's Principle

Let's break down why this happens using Le Chatelier's Principle. Remember, Le Chatelier's Principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

In our example, the stress is the addition of the common ion (Br−Br^-Br−). The system relieves this stress by shifting the equilibrium backwards, towards the solid AgBrAgBrAgBr. This means less AgBrAgBrAgBr dissolves, and the molar solubility decreases.

Exam Tip

When explaining the common ion effect on the AP exam, always refer to Le Chatelier's Principle. It's your go-to explanation for equilibrium shifts.

#Another Example

Consider the dissolution of CaSO4CaSO_4CaSO4​ (Ksp=2.4×10−5K_{sp} = 2.4 \times 10^{-5}Ksp​=2.4×10−5):

CaSO4(s)⇌Ca2+(aq)+SO42−(aq)CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)CaSO4​(s)⇌Ca2+(aq)+SO42−​(aq)

If we dissolve CaSO4CaSO_4CaSO4​ in a solution containing CuSO4CuSO_4CuSO4​, the presence of the common ion, SO42−SO_4^{2-}SO42−​, will shift the equilibrium to the left, reducing the solubility of CaSO4CaSO_4CaSO4​.

Common Mistake

Don't forget to consider the stoichiometry when dealing with polyatomic ions. For example, if you have Ca3(PO4)2Ca_3(PO_4)_2Ca3​(PO4​)2​, the concentration of Ca2+Ca^{2+}Ca2+ will be 3x, and the concentration of PO43−PO_4^{3-}PO43−​ will be 2x.

#Final Exam Focus

  • Master the ICE table: It's your best friend for solving solubility problems.
  • Le Chatelier's Principle is key: Always use it to explain equilibrium shifts.
  • Common ion effect: Understand how it impacts solubility.
  • Don't forget KspK_{sp}Ksp​: It's the foundation for all solubility calculations.
Exam Tip

When tackling free-response questions, show all your work. Even if you make a calculation error, you can still earn points for setting up the problem correctly.

# Practice Questions

Practice Question

#Multiple Choice Questions

  1. The molar solubility of AgClAgClAgCl in pure water is 1.3×10−51.3 \times 10^{-5}1.3×10−5 M. What is the molar solubility of AgClAgClAgCl in a 0.100.100.10 M NaClNaClNaCl solution? (A) 1.3×10−51.3 \times 10^{-5}1.3×10−5 M (B) 1.3×10−61.3 \times 10^{-6}1.3×10−6 M (C) 1.7×10−91.7 \times 10^{-9}1.7×10−9 M (D) 1.7×10−101.7 \times 10^{-10}1.7×10−10 M

  2. Which of the following will decrease the solubility of PbI2PbI_2PbI2​ in a saturated solution? (A) Adding NaNO3NaNO_3NaNO3​ (B) Adding Pb(NO3)2Pb(NO_3)_2Pb(NO3​)2​ (C) Adding water (D) Increasing the temperature

#Free Response Question

Consider the dissolution of CaF2CaF_2CaF2​ in water. The KspK_{sp}Ksp​ for CaF2CaF_2CaF2​ is 3.9×10−113.9 \times 10^{-11}3.9×10−11.

(a) Write the balanced chemical equation for the dissolution of CaF2CaF_2CaF2​ in water.

(b) Calculate the molar solubility of CaF2CaF_2CaF2​ in pure water.

(c) Calculate the molar solubility of CaF2CaF_2CaF2​ in a 0.0100.0100.010 M solution of NaFNaFNaF.

(d) Explain why the solubility of CaF2CaF_2CaF2​ is different in pure water versus in the NaFNaFNaF solution, using principles of equilibrium.

#Scoring Breakdown

(a) (1 point)

CaF2(s)⇌Ca2+(aq)+2F−(aq)CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)CaF2​(s)⇌Ca2+(aq)+2F−(aq)

(b) (2 points)

  • Setting up the ICE table correctly (1 point)
  • Correctly solving for x (1 point)

Ksp=[Ca2+][F−]2=(x)(2x)2=4x3=3.9×10−11K_{sp} = [Ca^{2+}][F^-]^2 = (x)(2x)^2 = 4x^3 = 3.9 \times 10^{-11}Ksp​=[Ca2+][F−]2=(x)(2x)2=4x3=3.9×10−11

x=3.9×10−1143=2.1×10−4Mx = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = 2.1 \times 10^{-4} Mx=343.9×10−11​​=2.1×10−4M

(c) (2 points)

  • Setting up the ICE table correctly with the common ion (1 point)
  • Correctly solving for x (1 point)

Ksp=[Ca2+][F−]2=(x)(0.010+2x)2≈x(0.010)2=3.9×10−11K_{sp} = [Ca^{2+}][F^-]^2 = (x)(0.010 + 2x)^2 \approx x(0.010)^2 = 3.9 \times 10^{-11}Ksp​=[Ca2+][F−]2=(x)(0.010+2x)2≈x(0.010)2=3.9×10−11

x=3.9×10−11(0.010)2=3.9×10−7Mx = \frac{3.9 \times 10^{-11}}{(0.010)^2} = 3.9 \times 10^{-7} Mx=(0.010)23.9×10−11​=3.9×10−7M

(d) (2 points)

  • Mentioning the common ion effect (1 point)
  • Explaining that the presence of F−F^-F− shifts the equilibrium to the left, decreasing solubility (1 point)

#Answers

  1. (D)
  2. (B)

You've got this! Remember to breathe, stay focused, and apply these concepts. You're going to do great! 💪

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Question 1 of 7

What is a common ion? 🤔

An ion that is present in only one compound

An ion that is present in two or more different compounds

An ion that is insoluble in water

An ion that is only found in acidic solutions