#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.

#How Does it Work? 🤔

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

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

The solubility product, $K_{sp}$, is:

$$K_{sp} = [A^+][B^-]$$

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

#Common Ion Effect Practice Problem ✏️

Let's work through an example to make this crystal clear. We'll calculate the molar solubility of $AgBr$ ($K_{sp} = 7.7 \times 10^{-13}$) in pure water and then in a $0.0010 M$ solution of $NaBr$.

#Molar Solubility in Pure Water

First, the equilibrium:

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

Set up an ICE table:

So, $K_{sp} = [Ag^+][Br^-] = x^2 = 7.7 \times 10^{-13}$

Solving for x, we get:

$$x = \sqrt{7.7 \times 10^{-13}} = 8.8 \times 10^{-7} M$$

#Molar Solubility in $0.0010 M$ $NaBr$

Now, we have an initial concentration of $0.0010 M$ of $Br^-$ from $NaBr$. Our ICE table changes:

Our $K_{sp}$ expression becomes:

$$K_{sp} = [Ag^+][Br^-] = x(0.0010 + x) \approx x(0.0010) = 7.7 \times 10^{-13}$$

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

$$x = \frac{7.7 \times 10^{-13}}{0.0010} = 7.7 \times 10^{-10} M$$

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

#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 ($Br^-$). The system relieves this stress by shifting the equilibrium backwards, towards the solid $AgBr$. This means less $AgBr$ dissolves, and the molar solubility decreases.

#Another Example

Consider the dissolution of $CaSO_4$ ($K_{sp} = 2.4 \times 10^{-5}$):

$$CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)$$

If we dissolve $CaSO_4$ in a solution containing $CuSO_4$, the presence of the common ion, $SO_4^{2-}$, will shift the equilibrium to the left, reducing the solubility of $CaSO_4$.

#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 $K_{sp}$: It's the foundation for all solubility calculations.

# Practice Questions

Practice Question

#Multiple Choice Questions

1. The molar solubility of $AgCl$ in pure water is $1.3 \times 10^{-5}$ M. What is the molar solubility of $AgCl$ in a $0.10$ M $NaCl$ solution? (A) $1.3 \times 10^{-5}$ M (B) $1.3 \times 10^{-6}$ M (C) $1.7 \times 10^{-9}$ M (D) $1.7 \times 10^{-10}$ M

2. Which of the following will decrease the solubility of $PbI_2$ in a saturated solution? (A) Adding $NaNO_3$ (B) Adding $Pb(NO_3)_2$ (C) Adding water (D) Increasing the temperature

#Free Response Question

Consider the dissolution of $CaF_2$ in water. The $K_{sp}$ for $CaF_2$ is $3.9 \times 10^{-11}$.

(a) Write the balanced chemical equation for the dissolution of $CaF_2$ in water.

(b) Calculate the molar solubility of $CaF_2$ in pure water.

(c) Calculate the molar solubility of $CaF_2$ in a $0.010$ M solution of $NaF$.

(d) Explain why the solubility of $CaF_2$ is different in pure water versus in the $NaF$ solution, using principles of equilibrium.

#Scoring Breakdown

(a) (1 point)

$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$

(b) (2 points)

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

$K_{sp} = [Ca^{2+}][F^-]^2 = (x)(2x)^2 = 4x^3 = 3.9 \times 10^{-11}$

$x = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = 2.1 \times 10^{-4} M$

(c) (2 points)

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

$K_{sp} = [Ca^{2+}][F^-]^2 = (x)(0.010 + 2x)^2 \approx x(0.010)^2 = 3.9 \times 10^{-11}$

$x = \frac{3.9 \times 10^{-11}}{(0.010)^2} = 3.9 \times 10^{-7} M$

(d) (2 points)

• Mentioning the common ion effect (1 point)
• Explaining that the presence of $F^-$ 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! 💪