🎯⭐ INTERACTIVE LESSON

Solubility Equilibria and K_sp

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Solubility Equilibria and K_sp - Complete Interactive Lesson

Part 1: Solubility Product (Ksp)

💎 Dissolution Equilibrium and K_sp

Part 1 of 7 — The Solubility Product Constant

Topics in This Part

Section
⚖️ Dissolution as an Equilibrium
The K_sp Expression
General Form
📌 Common K_sp Expressions
Key Points

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

Understanding the core concepts covered in Part 1
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

⚖️ Dissolution as an Equilibrium

When you add a slightly soluble salt to water:

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

📌 Common K_sp Expressions

Compound	Dissolution	$K_{sp}$ Expression
$\text{AgCl}$	$\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$

K_sp Expressions 🎯

Writing K_sp Expressions 🧮

Write the $K_{sp}$ expression for each compound. Count the total number of ion concentration terms (including exponents) that appear.

1) $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$ . How many ion terms appear in ? (Enter a number)

K_sp Fundamentals 🔍

Exit Quiz — K_sp Basics ✅

Part 2: Molar Solubility from Ksp

💎 Calculating Molar Solubility from K_sp

Part 2 of 7 — From K_sp to Dissolved Concentration

Topics in This Part

Section
📌 The ICE-Table Approach
For a 1:1 salt: $\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)$

Part 3: Common Ion Effect

💎 The Common Ion Effect

Part 3 of 7 — Reduced Solubility in the Presence of a Common Ion

Topics in This Part

Section
🤔 Why Does the Common Ion Reduce Solubility?
Mathematically
🧪 Worked Example
In Pure Water (for comparison)
In 0.10 M NaCl

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

Understanding the core concepts covered in Part 3
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

🤔 Why Does the Common Ion Reduce Solubility?

Consider dissolving $\text{AgCl}$ in a solution that already contains $\text{NaCl}$ (providing ions):

Part 4: Predicting Precipitation (Q vs Ksp)

💎 Predicting Precipitation — Q vs K_sp

Part 4 of 7 — Will a Precipitate Form?

Topics in This Part

Section
⚛️ The Ion Product, Q_sp
Comparing Q_sp to K_sp
Key Steps
📌 Don't Forget Dilution!

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

Understanding the core concepts covered in Part 4
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

⚛️ The Ion Product, Q_sp

$Q_{sp}$ is calculated exactly like , but using the (non-equilibrium) ion concentrations in solution:

Part 5: Selective Precipitation

💎 Selective Precipitation

Part 5 of 7 — Separating Ions by Adding Precipitating Agents

Topics in This Part

Section
📏 The Principle
Which precipitates first?
Strategy
🧪 Worked Example
Step 1: Find $[\text{Cl}^-]$ to begin precipitating each

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

Understanding the core concepts covered in Part 5
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📏 The Principle

Consider a solution containing both and . Adding can precipitate both:

Part 6: Problem-Solving Workshop

🧮 Problem-Solving Workshop: Solubility Equilibria

Part 6 of 7 — Mixed Ksp Problems

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

Working through complete multi-step problems from start to finish
Building problem-solving strategies you can apply on the AP exam
Identifying which concepts to apply and in what order

📂 Problem-Type Identification

Problem Type	Key Clue	Approach
Find $K_{sp}$

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Solubility Equilibria and K_sp

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

Solving AP-style questions that integrate multiple concepts from this unit
Writing clear, concise explanations using proper chemistry terminology
Identifying and avoiding common AP exam traps and mistakes

📋 Complete Solubility Summary

K_sp Expression

For $\text{M}_a\text{X}_b(s) \rightleftharpoons a\,\text{M}^{n+} + b\,\text{X}^{m-}$ :

Forward: Solid dissolves → ions enter solution
Reverse: Ions collide and re-form solid (precipitation)
Equilibrium: Rate of dissolution = Rate of precipitation

Since the solid has a constant concentration (pure solid activity = 1), it is excluded from the equilibrium expression:

$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

Not $\frac{[\text{Ag}^+][\text{Cl}^-]}{[\text{AgCl}]}$ — the solid is omitted!

For $\text{M}_a\text{X}_b(s) \rightleftharpoons a\,\text{M}^{b+}(aq) + b\,\text{X}^{a-}(aq)$ :

$\boxed{K_{sp} = [\text{M}^{b+}]^a[\text{X}^{a-}]^b}$

🔑 Key Concept: The $K_{sp}$ expression includes only dissolved ion concentrations, each raised to its stoichiometric coefficient. The pure solid is always excluded (activity = 1).

$K_{sp}$ values are typically very small (e.g., $1.8 \times 10^{-10}$ for AgCl)
Smaller $K_{sp}$ → less soluble (for compounds with the same formula type)

⚠️ Warning: Comparing $K_{sp}$ values directly to rank solubility only works for compounds with the same ion ratio (e.g., both 1:1). For different formula types, calculate and compare molar solubilities.

$K_{sp}$ depends only on temperature (like all equilibrium constants)
The solid must be present for $K_{sp}$ to apply (saturated solution)

2) $\text{Pb(IO}_3)_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{IO}_3^-(aq)$ . The exponent on $[\text{IO}_3^-]$ is? (Enter a number)

3) $\text{Bi}_2\text{S}_3(s) \rightleftharpoons 2\,\text{Bi}^{3+}(aq) + 3\,\text{S}^{2-}(aq)$ . The exponent on $[\text{Bi}^{3+}]$ is? (Enter a number)

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

Understanding the core concepts covered in Part 2
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📌 The ICE-Table Approach

For a 1:1 salt: $\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)$

If molar solubility = $s$ , then:

$[\text{M}^+] = s$ and $[\text{X}^-] = s$

$\boxed{s = \sqrt{K_{sp}}}$

🔑 Key Concept: Set molar solubility = $s$ , express each ion concentration in terms of $s$ using stoichiometry, substitute into $K_{sp}$ , and solve.

For a 1:2 salt: $\text{MX}_2(s) \rightleftharpoons \text{M}^{2+}(aq) + 2\,\text{X}^-(aq)$

If molar solubility = $s$ , then:

$[\text{M}^{2+}] = s$ and $[\text{X}^-] = 2s$

$\boxed{s = \sqrt[3]{\frac{K_{sp}}{4}}}$

For a 2:3 salt: $\text{M}_2\text{X}_3(s) \rightleftharpoons 2\,\text{M}^{3+}(aq) + 3\,\text{X}^{2-}(aq)$

$[\text{M}^{3+}] = 2s$ and $[\text{X}^{2-}] = 3s$

$\boxed{s = \sqrt[5]{\frac{K_{sp}}{108}}}$

🧪 Worked Examples

Problem: Find the molar solubility of AgCl ( $K_{sp} = 1.8 \times 10^{-10}$ ).

Solution:

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

$\boxed{s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} \text{ M}}$

Problem: Find the molar solubility of PbCl₂ ( $K_{sp} = 1.7 \times 10^{-5}$ ).

Solution:

$\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{Cl}^-(aq)$

$K_{sp} = (s)(2s)^2 = 4s^3$

$\boxed{s = \sqrt[3]{\frac{1.7 \times 10^{-5}}{4}} = \sqrt[3]{4.25 \times 10^{-6}} = 1.6 \times 10^{-2} \text{ M}}$

⚠️ Warning: You cannot directly compare $K_{sp}$ values to rank solubility unless the compounds have the same formula type (same ratio of ions). For different types, you must compare molar solubilities.

Molar Solubility 🎯

Practice: Molar Solubility Calculations 🧮

1) $\text{BaSO}_4$ , $K_{sp} = 1.1 \times 10^{-10}$ . (1:1 salt) What is the molar solubility? (Enter in scientific notation, e.g. 1.0e-5)

2) $\text{Ca(OH)}_2$ , $K_{sp} = 5.0 \times 10^{-6}$ . () What is the molar solubility? (Round to 3 significant figures, e.g. 0.01)

3) For the BaSO₄ solution above, what is $[\text{Ba}^{2+}]$ ? (Enter in same format as answer 1)

Solubility Relationships 🔍

Exit Quiz — Molar Solubility ✅

$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

The $\text{Cl}^-$ from NaCl shifts the equilibrium left (Le Chatelier's), reducing the amount of $\text{AgCl}$ that dissolves.

In pure water: $K_{sp} = s \cdot s = s^2$ , so $s = \sqrt{K_{sp}}$
In 0.10 M NaCl: $K_{sp} = s \cdot (s + 0.10)$

Since $s \ll 0.10$ :

$\boxed{s \approx \frac{K_{sp}}{[\text{common ion}]}}$

🔑 Key Concept: The common ion effect is Le Chatelier's principle applied to dissolution equilibria — adding an ion that appears in the $K_{sp}$ expression shifts the equilibrium toward the solid, reducing solubility.

💡 Tip: When the common ion concentration is much larger than $s$ , you can ignore $s$ in the sum, greatly simplifying the calculation.

🧪 Worked Example

Problem: Find the molar solubility of $\text{AgCl}$ ( $K_{sp} = 1.8 \times 10^{-10}$ ) in 0.10 M NaCl.

Solution:

In Pure Water (for comparison)

$s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}$ M

In 0.10 M NaCl

The Cl⁻ from NaCl provides an initial $[\text{Cl}^-] = 0.10$ M.

$K_{sp} = [\text{Ag}^+][\text{Cl}^-] = (s)(0.10 + s)$

Since $s \ll 0.10$ : $(s)(0.10) \approx 1.8 \times 10^{-10}$

$\boxed{s = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ M}}$

Comparison

Solution	Molar Solubility
Pure water	$1.3 \times 10^{-5}$ M
0.10 M NaCl	$1.8 \times 10^{-9}$ M

The common ion reduced solubility by a factor of about 7,000!

Common Ion Effect 🎯

Practice: Common Ion Problems 🧮

Find the molar solubility of $\text{BaSO}_4$ ( $K_{sp} = 1.1 \times 10^{-10}$ ) in 0.050 M Na₂SO₄.

$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

$[\text{SO}_4^{2-}]$ from Na₂SO₄ = 0.050 M

$K_{sp} = (s)(0.050 + s) \approx (s)(0.050)$

1) What is the molar solubility $s$ ? (Enter in scientific notation, e.g. 2.2e-9)

2) What is the molar solubility in pure water? (Enter in scientific notation, e.g. 1.0e-5)

3) By what factor did the common ion reduce solubility? (Enter as a whole number, approximately)

Round all answers to 3 significant figures.

Common Ion Concepts 🔍

Exit Quiz — Common Ion Effect ✅

$\boxed{Q_{sp} = [\text{M}^{n+}]^a_{\text{actual}}[\text{X}^{m-}]^b_{\text{actual}}}$

Comparing Q_sp to K_sp

Condition	Meaning	Result
$Q_{sp} < K_{sp}$	Solution is unsaturated	No precipitate; more solid can dissolve
$Q_{sp} = K_{sp}$	Solution is exactly saturated	At equilibrium; no change
$Q_{sp} > K_{sp}$	Solution is supersaturated	Precipitate forms until $Q_{sp} = K_{sp}$

💡 Tip: If $Q_{sp} > K_{sp}$ , a precipitate will form. The system removes ions from solution until $Q_{sp}$ decreases back to $K_{sp}$ .

🔑 Key Concept: Always calculate $Q_{sp}$ using actual ion concentrations (after dilution), then compare to $K_{sp}$ . This three-step method works for any precipitation prediction.

Calculate the ion concentrations after mixing (account for dilution!)
Calculate $Q_{sp}$
Compare $Q_{sp}$ to $K_{sp}$

📌 Don't Forget Dilution!

When mixing two solutions, the total volume increases and concentrations decrease:

$\boxed{[\text{ion}]_{\text{after mixing}} = \frac{[\text{ion}]_{\text{initial}} \times V_{\text{initial}}}{V_{\text{total}}}}$

⚠️ Warning: Always account for dilution when mixing solutions! Forgetting this step overestimates $Q_{sp}$ and can lead to incorrect precipitation predictions.

Problem: 50.0 mL of $0.0020$ M $\text{Pb(NO}_3)_2$ is mixed with 50.0 mL of $0.0040$ M NaCl. Does PbCl₂ precipitate? ()

Solution:

Step 1: Calculate concentrations after mixing

$V_{\text{total}} = 50.0 + 50.0 = 100.0$ mL

$[\text{Pb}^{2+}] = \frac{0.0020 \times 50.0}{100.0} = 0.0010$ M

$[\text{Cl}^-] = \frac{0.0040 \times 50.0}{100.0} = 0.0020$ M

Step 2: Calculate Q_sp

$Q_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (0.0010)(0.0020)^2 = (0.0010)(4.0 \times 10^{-6}) = 4.0 \times 10^{-9}$

Step 3: Compare

$Q_{sp} = 4.0 \times 10^{-9} < K_{sp} = 1.7 \times 10^{-5}$

$Q_{sp} < K_{sp}$ → No precipitate forms.

Precipitation Predictions 🎯

Practice: Will It Precipitate? 🧮

25.0 mL of $0.0010$ M AgNO₃ is mixed with 75.0 mL of $0.0020$ M NaCl.

$K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$

1) What is $[\text{Ag}^+]$ after mixing? (Enter in scientific notation, e.g. 2.5e-4)

2) What is $[\text{Cl}^-]$ after mixing? (Enter in scientific notation, e.g. 1.5e-3)

3) What is $Q_{sp}$ ? (Enter in scientific notation, e.g. 3.8e-7)

Round all answers to 3 significant figures.

Precipitation Concepts 🔍

Exit Quiz — Precipitation ✅

$\text{AgCl}$ : $K_{sp} = 1.8 \times 10^{-10}$
$\text{PbCl}_2$ : $K_{sp} = 1.7 \times 10^{-5}$

Which precipitates first?

The salt requiring the lower $[\text{Cl}^-]$ to reach $Q_{sp} = K_{sp}$ precipitates first.

For $\text{AgCl}$ : $[\text{Cl}^-] = \frac{K_{sp}}{[\text{Ag}^+]}$

For $\text{PbCl}_2$ : $[\text{Cl}^-] = \sqrt{\frac{K_{sp}}{[\text{Pb}^{2+}]}}$

Since $K_{sp}(\text{AgCl})$ is much smaller, AgCl precipitates at a much lower $[\text{Cl}^-]$ — it precipitates first.

Calculate the $[\text{reagent}]$ needed to start precipitating each ion
The ion requiring less reagent precipitates first
Increase reagent until just before the second ion begins to precipitate
Filter to separate the first precipitate

🔑 Key Concept: In selective precipitation, the salt with the smallest $K_{sp}$ generally precipitates first because it reaches saturation ( $Q_{sp} = K_{sp}$ ) at the lowest reagent concentration.

🧪 Worked Example

Problem: A solution contains $[\text{Ag}^+] = 0.010$ M and $[\text{Pb}^{2+}] = 0.010$ M. NaCl is added slowly. Which precipitates first, and can they be separated?

Solution:

Step 1: Find $[\text{Cl}^-]$ to begin precipitating each

AgCl: $[\text{Cl}^-] = \frac{K_{sp}}{[\text{Ag}^+]} = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8}$ M

PbCl₂: $[\text{Cl}^-] = \sqrt{\frac{K_{sp}}{[\text{Pb}^{2+}]}} = \sqrt{\frac{1.7 \times 10^{-5}}{0.010}} = \sqrt{1.7 \times 10^{-3}} = 0.041$ M

Step 2: Order of precipitation

AgCl precipitates first (at $[\text{Cl}^-] = 1.8 \times 10^{-8}$ M). PbCl₂ doesn't start precipitating until $[\text{Cl}^-] = 0.041$ M.

Step 3: Can we separate them completely?

When $[\text{Cl}^-] = 0.041$ M (just before PbCl₂ starts), what is $[\text{Ag}^+]$ ?

$[\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.041} = 4.4 \times 10^{-9}$ M

This is essentially zero compared to the initial 0.010 M — virtually all Ag⁺ has precipitated before any Pb²⁺ does. Excellent separation!

💡 Tip: When $K_{sp}$ values differ by several orders of magnitude, selective precipitation gives nearly complete separation. The greater the difference, the cleaner the separation.

Selective Precipitation 🎯

Practice: Selective Precipitation 🧮

A solution contains $[\text{Ca}^{2+}] = 0.020$ M and $[\text{Ba}^{2+}] = 0.020$ M. Na₂SO₄ is added slowly.

$K_{sp}(\text{CaSO}_4) = 4.9 \times 10^{-5}$

Both salts are 1:1 type: $K_{sp} = [\text{M}^{2+}][\text{SO}_4^{2-}]$

1) $[\text{SO}_4^{2-}]$ needed to start precipitating BaSO₄? (Enter in scientific notation, e.g. 5.5e-9)

2) $[\text{SO}_4^{2-}]$ needed to start precipitating CaSO₄? (Enter in scientific notation, e.g. 2.5e-3)

3) Which precipitates first? (Enter "BaSO4" or "CaSO4")

Round all answers to 3 significant figures.

Separation Concepts 🔍

Exit Quiz — Selective Precipitation ✅

💡 Tip: Read each problem carefully for key clues — "in a solution of..." signals common ion, "mixed with..." signals precipitation, and "separate ions" signals selective precipitation.

Problem 1: Finding K_sp from Solubility 🧮

The molar solubility of $\text{Ag}_2\text{CO}_3$ is $1.3 \times 10^{-4}$ M.

$\text{Ag}_2\text{CO}_3(s) \rightleftharpoons 2\,\text{Ag}^+(aq) + \text{CO}_3^{2-}(aq)$

1) What is $[\text{Ag}^+]$ ? (Enter in scientific notation, e.g. 2.6e-4)

2) What is $[\text{CO}_3^{2-}]$ ? (Enter in scientific notation, e.g. 1.3e-4)

3) What is $K_{sp}$ ? (Enter in scientific notation, e.g. 8.8e-12)

Round all answers to 3 significant figures.

Problem 2: Common Ion Effect 🧮

What is the molar solubility of $\text{SrF}_2$ ( $K_{sp} = 4.3 \times 10^{-9}$ ) in 0.10 M NaF?

$\text{SrF}_2(s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\,\text{F}^-(aq)$

$K_{sp} = (s)(0.10 + 2s)^2 \approx (s)(0.10)^2$

1) What is the molar solubility in 0.10 M NaF? (Enter in scientific notation, e.g. 4.3e-7)

2) What is the molar solubility in pure water? ( $K_{sp} = 4s^3$ ) (Enter in scientific notation, e.g. 1.0e-3)

Round all answers to 3 significant figures.

Problem 3: Precipitation Decision 🎯

Problem 4: Strategy Selection 🔍

Exit Quiz — Workshop ✅

$\boxed{K_{sp} = [\text{M}^{n+}]^a[\text{X}^{m-}]^b}$

(Solid excluded — pure solid activity = 1)

🔑 Key Concept: All solubility equilibria problems start with writing the correct $K_{sp}$ expression. From there, identify the problem type (find $K_{sp}$ , find solubility, common ion, precipitation, or selective precipitation) and apply the appropriate method.

Molar Solubility ( $s$ )

Type	$K_{sp}$ in terms of $s$	Solve for $s$
MX	$s^2$	$s = \sqrt{K_{sp}}$
MX₂ or M₂X	$4s^3$	$s = \sqrt[3]{K_{sp}/4}$
M₂X₃	$108s^5$	$s = \sqrt[5]{K_{sp}/108}$

Dissolving in a solution with a shared ion decreases solubility (Le Chatelier's).

$Q_{sp} > K_{sp}$ → precipitate forms $Q_{sp} < K_{sp}$ → no precipitate (unsaturated) $Q_{sp} = K_{sp}$ → saturated (equilibrium)

Selective Precipitation

Add reagent slowly → ion with smallest $K_{sp}$ precipitates first → filter → continue to next ion.

AP-Style Multiple Choice — Set 1 🎯

AP Free-Response Style 🧮

$\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\,\text{OH}^-(aq)$ , $K_{sp} = 5.6 \times 10^{-12}$

1) Calculate the molar solubility of $\text{Mg(OH)}_2$ in pure water. ( $K_{sp} = 4s^3$ ) (Enter in scientific notation, e.g. 1.1e-4)

2) Calculate the molar solubility in a solution buffered at pH = 12.0 ( $[\text{OH}^-] = 0.010$ M). (Enter in scientific notation, e.g. 5.6e-8)

3) At pH = 12.0, what fraction of the pure-water solubility remains? (Enter as a percentage, e.g. 0.005)

Final Concept Review 🔍

Final Exit Quiz ✅

K_{s p} = s \cdot s = s^{2}

K_{sp} = (s)(2s)^2 = 4s^3

K_{sp} = (2s)^2(3s)^3 = 4s^2 \cdot 27s^3 = 108s^5

K_{s p} = 1.7 \times 1 0^{- 5}

K_{sp}(\text{BaSO}_4) = 1.1 \times 10^{-10}

Solubility Equilibria and K_sp

Solubility Equilibria and K_sp - Complete Interactive Lesson

Part 1: Solubility Product (Ksp)

💎 Dissolution Equilibrium and K_sp

Topics in This Part

What You'll Master in Part 1

⚖️ Dissolution as an Equilibrium

📌 Common K_sp Expressions

Part 2: Molar Solubility from Ksp

💎 Calculating Molar Solubility from K_sp

Topics in This Part

Part 3: Common Ion Effect

💎 The Common Ion Effect

Topics in This Part

What You'll Master in Part 3

🤔 Why Does the Common Ion Reduce Solubility?

Part 4: Predicting Precipitation (Q vs Ksp)

💎 Predicting Precipitation — Q vs K_sp

Topics in This Part

What You'll Master in Part 4

⚛️ The Ion Product, Q_sp

Part 5: Selective Precipitation

💎 Selective Precipitation

Topics in This Part

What You'll Master in Part 5

📏 The Principle

Part 6: Problem-Solving Workshop

🧮 Problem-Solving Workshop: Solubility Equilibria

Practice Makes Perfect

What You'll Master in Part 6

📂 Problem-Type Identification

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Bringing It All Together

What You'll Master in Part 7

📋 Complete Solubility Summary

K_sp Expression

The K_sp Expression

General Form

Key Points

What You'll Master in Part 2

📌 The ICE-Table Approach

For a 1:1 salt: MX(s)⇌M+(aq)+X−(aq)\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)MX(s)⇌M+(aq)+X−(aq)

For a 1:2 salt: MX2(s)⇌M2+(aq)+2 X−(aq)\text{MX}_2(s) \rightleftharpoons \text{M}^{2+}(aq) + 2\,\text{X}^-(aq)MX2​(s)⇌M2+(aq)+

For a 2:3 salt: M2X3(s)⇌2 M3+(aq)+3 X2−(aq)\text{M}_2\text{X}_3(s) \rightleftharpoons 2\,\text{M}^{3+}(aq) + 3\,\text{X}^{2-}(aq)M2​X3​(s)⇌2

🧪 Worked Examples

Mathematically

🧪 Worked Example

In Pure Water (for comparison)

In 0.10 M NaCl

Comparison

Comparing Q_sp to K_sp

Key Steps

📌 Don't Forget Dilution!

Which precipitates first?

Strategy

🧪 Worked Example

Step 1: Find [Cl−][\text{Cl}^-][Cl−] to begin precipitating each

Step 2: Order of precipitation

Step 3: Can we separate them completely?

Molar Solubility (sss)

Common Ion Effect

Precipitation

Selective Precipitation

For a 1:1 salt: $\text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq)$

For a 1:2 salt: $\text{MX}_2(s) \rightleftharpoons \text{M}^{2+}(aq) + 2\,\text{X}^-(aq)$

For a 2:3 salt: $\text{M}_2\text{X}_3(s) \rightleftharpoons 2\,\text{M}^{3+}(aq) + 3\,\text{X}^{2-}(aq)$

Step 1: Find $[\text{Cl}^-]$ to begin precipitating each

Molar Solubility ( $s$ )