Solubility Equilibria and K_sp

💎 Dissolution Equilibrium and K_sp

Part 1 of 7 — The Solubility Product Constant

Topics in This Part

🔑 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:

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

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

The K_sp Expression

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!

General Form

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).

📌 Common K_sp Expressions

Key Points

• $K_{sp}$ 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)

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 $K_{sp}$? (Enter a number)

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)

K_sp Fundamentals 🔍

Exit Quiz — K_sp Basics ✅

💎 Calculating Molar Solubility from K_sp

Part 2 of 7 — From K_sp to Dissolved Concentration

Topics in This Part

🔑 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 $[\text{X}^-] = s$
• $K_{sp} = s \cdot s = s^2$

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

🔑 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 $[\text{X}^-] = 2s$
• $K_{sp} = (s)(2s)^2 = 4s^3$

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

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 $[\text{X}^{2-}] = 3s$
• $K_{sp} = (2s)^2(3s)^3 = 4s^2 \cdot 27s^3 = 108s^5$

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

🧪 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 $PbCl_{2}$ ($K_{sp} = 1.7 \times 10^{-5}$).

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) $\text{Ca(OH)}_2$, $K_{sp} = 5.0 \times 10^{-6}$. ($K_{sp} = 4s^3$) What is the molar solubility? (Round to 3 significant figures, e.g. 0.01)

3) For the $BaSO_{4}$ solution above, what is $[\text{Ba}^{2+}]$? (Enter in same format as answer 1)

Solubility Relationships 🔍

Exit Quiz — Molar Solubility ✅

💎 The Common Ion Effect

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

Topics in This Part

🔑 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 $\text{Cl}^-$ ions):

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

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

Mathematically

• 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

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 $Na_{2}SO_{4}$.

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

$[\text{SO}_4^{2-}]$ from $Na_{2}SO_{4}$ = 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 ✅

💎 Predicting Precipitation — Q vs K_sp

Part 4 of 7 — Will a Precipitate Form?

Topics in This Part

🔑 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 $K_{sp}$, but using the actual (non-equilibrium) ion concentrations in solution:

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

Comparing Q_sp to 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.

Key Steps

1. Calculate the ion concentrations after mixing (account for dilution!)
2. Calculate $Q_{sp}$
3. 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 $PbCl_{2}$ precipitate? ($K_{sp} = 1.7 \times 10^{-5}$)

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 $AgNO_{3}$ 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 ✅

💎 Selective Precipitation

Part 5 of 7 — Separating Ions by Adding Precipitating Agents

Topics in This Part

🔑 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 $\text{Ag}^+$ and $\text{Pb}^{2+}$. Adding $\text{Cl}^-$ can precipitate both:

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

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.

Strategy

1. Calculate the $[\text{reagent}]$ needed to start precipitating each ion
2. The ion requiring less reagent precipitates first
3. Increase reagent until just before the second ion begins to precipitate
4. 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

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

Step 2: Order of precipitation

AgCl precipitates first (at $[\text{Cl}^-] = 1.8 \times 10^{-8}$ M). $PbCl_{2}$ 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 $PbCl_{2}$ 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 $Pb^{2+}$ 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. $Na_{2}SO_{4}$ is added slowly.

$K_{sp}(\text{CaSO}_4) = 4.9 \times 10^{-5}$ $K_{sp}(\text{BaSO}_4) = 1.1 \times 10^{-10}$

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

1) $[\text{SO}_4^{2-}]$ needed to start precipitating $BaSO_{4}$? (Enter in scientific notation, e.g. 5.5e-9)

2) $[\text{SO}_4^{2-}]$ needed to start precipitating $CaSO_{4}$? (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 ✅

🧮 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

💡 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 ✅

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

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

(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 ($s$)

Common Ion Effect

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

Precipitation

$Q_{sp} > K_{sp}$ → 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 ✅