💎 Dissolution Equilibrium and K_sp

Part 1 of 7 — The Solubility Product Constant

When an ionic compound dissolves in water, it establishes an equilibrium between the solid and its dissolved ions. The equilibrium constant for this process is called the solubility product constant, $K_{sp}$.

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

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

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
• $K_{sp}$ 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

Molar solubility ($s$) is the number of moles of solute that dissolve per liter to form a saturated solution. We can calculate it directly from $K_{sp}$.

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

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

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

Worked Examples

Example 1: AgCl ($K_{sp} = 1.8 \times 10^{-10}$)

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

$K_{sp} = s^2 \implies s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}$ M

Example 2: PbCl₂ ($K_{sp} = 1.7 \times 10^{-5}$)

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

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

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

Important Note

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 BaSO₄ 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

When a slightly soluble salt dissolves in a solution that already contains one of its ions, its solubility decreases. This is the common ion effect — a direct application of Le Chatelier's principle.

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 $s \ll 0.10$: $K_{sp} \approx s \cdot 0.10$, so $s \approx \frac{K_{sp}}{0.10}$

This gives a much smaller $s$ than in pure water!

Worked Example

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

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

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

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 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 ✅

💎 Predicting Precipitation — Q vs K_sp

Part 4 of 7 — Will a Precipitate Form?

When two solutions containing ions are mixed, will a precipitate form? The answer depends on whether the ion product ($Q_{sp}$) exceeds the solubility product ($K_{sp}$).

The Ion Product, Q_sp

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

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

Comparing Q_sp to K_sp

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:

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

Worked Example

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? ($K_{sp} = 1.7 \times 10^{-5}$)

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 ✅

💎 Selective Precipitation

Part 5 of 7 — Separating Ions by Adding Precipitating Agents

When a solution contains multiple ions that can form insoluble salts with the same reagent, you can selectively precipitate them by controlling the reagent concentration. The ion with the smallest $K_{sp}$ precipitates first.

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

Worked Example

A solution contains $[\text{Ag}^+] = 0.010$ M and $[\text{Pb}^{2+}] = 0.010$ M. NaCl is added slowly.

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!

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.

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

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

1. $[\text{SO}_4^{2-}]$ needed to start precipitating 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 ✅

🧮 Problem-Solving Workshop: Solubility Equilibria

Part 6 of 7 — Mixed Ksp Problems

This workshop combines all solubility skills: writing $K_{sp}$ expressions, calculating molar solubility, applying the common ion effect, predicting precipitation, and selective precipitation.

Problem-Type Identification

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

This final part reviews all solubility concepts: $K_{sp}$ expressions, molar solubility, common ion effect, precipitation predictions ($Q_{sp}$ vs $K_{sp}$), and selective precipitation. Questions mirror AP Chemistry exam formats.

Complete Solubility Summary

K_sp Expression

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

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

(Solid excluded — pure solid activity = 1)

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 ✅