Topic 7.12 Notes – Common-Ion Effect

1. The Common-Ion Effect

Consider a slightly soluble salt:

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

Its equilibrium constant is:

$$K_{sp} = [M^{+}][X^{-}]$$

A few reminders you already know but need here:

• Solids are not included in Ksp expressions.
• Ksp depends only on temperature.
• At equilibrium, the ion concentrations must multiply to equal Ksp.

Now imagine dissolving this salt in a solution that already contains $\ce{X^{-}}$.

That $\ce{X^{-}}$ is the common ion. Because it is already a product, adding more of it shifts the equilibrium left, toward solid.

That shift:

• Forms more solid
• Lowers the amount that dissolves
• Decreases the salt’s molar solubility

This is just Le Châtelier’s principle applied to solubility equilibrium.

2. How the Common-Ion Effect Changes Solubility

a. Qualitative Understanding

Take:

$$\ce{PbCl2(s) <=> Pb^{2+} + 2Cl^{-}}$$

If you dissolve it in pure water, both ions start at 0 M.

If you dissolve it in 0.20 M NaCl:

• $[Cl^{-}]$ is already high.
• The system shifts left.
• Less $\ce{PbCl2}$ dissolves.
• Final $[Pb^{2+}]$ is much smaller than in pure water.

Nothing mysterious is happening. You added product. The equilibrium responds.

Students sometimes think the salt becomes “less able” to dissolve. That’s not it. The equilibrium position changes, not the identity of the salt.

b. Quantitative Understanding Using Ksp

This is where most test questions land.

Let’s walk through the structure with a 1:2 salt:

$$\ce{SrF2(s) <=> Sr^{2+} + 2F^{-}}$$

$$K_{sp} = [\ce{Sr^{2+}}][\ce{F^{-}}]^{2}$$

Suppose the solution already contains 0.10 M $\ce{F^{-}}$.

Step-by-step setup

1. Initial concentrations

   • $[\ce{Sr^{2+}}] = 0$
   • $[\ce{F^{-}}] = 0.10$

2. Change

   • $+x$ for $\ce{Sr^{2+}}$
   • $+2x$ for $\ce{F^{-}}$

3. Equilibrium

   • $[\ce{Sr^{2+}}] = x$
   • $[\ce{F^{-}}] = 0.10 + 2x$

Plug into Ksp:

$$K_{sp} = (x)(0.10 + 2x)^{2}$$

Because Ksp values are usually very small, $x$ is tiny compared to 0.10.

So:

$$0.10 + 2x \approx 0.10$$

Then:

$$K_{sp} \approx x(0.10)^{2}$$

Solve for $x$. That $x$ is the molar solubility in the common-ion solution, and it will be much smaller than in pure water.

On FRQs, most lost points here come from:

• Forgetting the coefficient (the 2 in 2x)
• Forgetting to square the fluoride term
• Not including the initial common ion concentration

c. When Coefficients Matter

If the salt produces unequal amounts of ions, coefficients change everything.

Example:

$$\ce{Al(OH)3(s) <=> Al^{3+} + 3OH^{-}}$$

$$K_{sp} = [\ce{Al^{3+}}][\ce{OH^{-}}]^{3}$$

If $\ce{OH^{-}}$ is already present:

• Change in $\ce{OH^{-}}$ is $+3x$
• Expression includes a cube

Small algebra mistakes here completely change the answer. Go slow when writing the equilibrium expression.

3. Solubility With and Without a Common Ion

Here’s the comparison you should have clear in your head:

The crucial point:

• Ksp stays the same
• Equilibrium concentrations change
• Solubility decreases when a common ion is present

If a problem asks which solution dissolves more solid, always look for the one without the common ion.

4. Qsp and Precipitation

The same idea connects to the reaction quotient:

$$Q_{sp} = [M^{+}][X^{-}]$$

• If $Q < K_{sp}$, more dissolves.
• If $Q = K_{sp}$, at equilibrium.
• If $Q > K_{sp}$, precipitation occurs.

Adding a common ion increases Q immediately.

If $Q > K_{sp}$, solid forms until equilibrium is restored.

This shows up in lab-style questions where two solutions are mixed and you must decide whether a precipitate forms.