Topic 7.11 Notes – Introduction to Solubility Equilibria

1. What Ksp Is and What Solubility Means

Some ionic solids dissolve completely. Others dissolve only a tiny amount and then stop. When that happens, the system reaches a dynamic equilibrium between solid and dissolved ions.

For example:

$$\ce{MX(s) <=> M^{+}(aq) + X^{-}(aq)}$$

At equilibrium:

• Solid is still present.
• Ions are dissolving and re-forming solid at equal rates.
• The solution is saturated.

The Solubility-Product Constant, $K_{sp}$

For a general salt:

$$\ce{A_pB_q(s) <=> pA^{q+}(aq) + qB^{p-}(aq)}$$

The equilibrium expression is:

$$K_{sp} = [\ce{A^{q+}}]^p[\ce{B^{p-}}]^q$$

Key ideas:

• Solids do not appear in the expression (their activity is constant).
• Only aqueous ion concentrations are included.
• Coefficients become exponents.

What Ksp Tells You

• Larger $K_{sp}$ → more product favored → more soluble
• Smaller $K_{sp}$ → less dissolution → less soluble
• If $K_{sp} > 1$, the salt is considered soluble (this connects to solubility rules you learned earlier)

Solubility rules are qualitative shortcuts. $K_{sp}$ gives you the quantitative version.

2. Molar Solubility and How It Relates to Ksp

Molar solubility (s) means how many moles of solid dissolve per liter to make a saturated solution.

The key is always the same:
Write the balanced equation and express ion concentrations in terms of s.

Case 1: 1:1 Salt

Example:

$$\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)}$$

Let molar solubility = $s$

• $[\ce{Ag^{+}}] = s$
• $[\ce{Cl^{-}}] = s$

$$K_{sp} = s^{2}$$

So:

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

If $K_{sp} = 1.8 \times 10^{-10}$:

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

That square root step is common on tests. Don’t rush it.

Case 2: 1:2 or 2:1 Salt

Example:

$$\ce{CaF2(s) <=> Ca^{2+}(aq) + 2F^{-}(aq)}$$

Let molar solubility = $s$

• $[\ce{Ca^{2+}}] = s$
• $[\ce{F^{-}}] = 2s$

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

Notice how the coefficient becomes part of the concentration before you square it.

If $K_{sp} = 4.0 \times 10^{-11}$:

$$4s^{3} = 4.0 \times 10^{-11}$$

$$s^{3} = 1.0 \times 10^{-11}$$

$$s = 2.15 \times 10^{-4} \text{ M}$$

This cube-root algebra shows up often. Practice it cleanly.

The Universal Method

1. Write the dissolution reaction.
2. Let molar solubility = $s$.
3. Express each ion concentration in terms of $s$.
4. Plug into the $K_{sp}$ expression.
5. Solve.

If you skip step 1, mistakes happen.

3. Using Ksp to Compare Solubilities

Students mess this up every year.

You can compare $K_{sp}$ values directly only if the salts have the same ion ratio.

Example:

• Two 1:1 salts → larger $K_{sp}$ = more soluble.
• A 1:1 salt vs. a 1:2 salt → you must calculate molar solubility (s) for each.

Why? Because $K_{sp}$ depends on the exponents. A 1:2 salt has a cubic relationship to $s$, not a square one.

On AP-style questions, they love giving two salts with different formulas and asking which is more soluble. If you just compare $K_{sp}$ values without checking stoichiometry, you’ll get it wrong.

4. Calculating Ksp from Molar Solubility

Sometimes they flip it.

If molar solubility is given, you:

1. Write the dissolution equation.
2. Use the given $s$ to find ion concentrations.
3. Plug directly into the $K_{sp}$ expression.

Example structure:

$$\ce{PbI2(s) <=> Pb^{2+}(aq) + 2I^{-}(aq)}$$

If solubility is $3.0 \times 10^{-3}$ M:

• $[\ce{Pb^{2+}}] = 3.0 \times 10^{-3}$
• $[\ce{I^{-}}] = 6.0 \times 10^{-3}$

$$K_{sp} = (3.0 \times 10^{-3})(6.0 \times 10^{-3})^{2}$$

That’s it. No solving required.

This is algebraically easier than solving for $s$.