Ksp and Solubility

In this lesson, students, you will learn how chemists describe how much of an ionic solid can dissolve in water and how that connects to equilibrium ⚖️. The big idea is that some solids dissolve only a little, and that “little” can be measured and predicted using the solubility product constant, $K_{sp}$. This topic links directly to Reactivity 2 because it shows how far a reaction goes, how equilibrium is established, and how quantitative chemistry lets us predict the amount of change.

Learning goals

By the end of this lesson, you should be able to:

explain the meaning of solubility, saturation, and $K_{sp}$
write $K_{sp}$ expressions for ionic compounds
use $K_{sp}$ to calculate solubility or ion concentration
predict whether a precipitate will form using the reaction quotient $Q_{sp}$
connect solubility to equilibrium and extent of reaction in real situations 🌊

1. What solubility means

Solubility is the maximum amount of a substance that can dissolve in a given amount of solvent at a specific temperature. For ionic solids, solubility is often discussed in terms of how many moles of the solid dissolve in $1\,\text{dm}^3$ of solution.

When an ionic solid is added to water, two things can happen:

some of the solid dissolves into ions
some solid may remain undissolved if the solution becomes saturated

For example, silver chloride, $\mathrm{AgCl}$, is only slightly soluble in water. Its dissolution can be written as:

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

The double arrow is important. It shows dynamic equilibrium: ions are leaving the solid and entering solution at the same time as ions are coming back together to form the solid. Even though the solution looks unchanged, microscopic change is still happening.

A saturated solution contains the maximum possible amount of dissolved solute at a given temperature. If more solute is added, it does not dissolve and remains as a solid.

2. The solubility product constant, $K_{sp}$

For slightly soluble ionic compounds, equilibrium can be described using an equilibrium constant called the solubility product constant, $K_{sp}$.

For the general salt

$$\mathrm{A_xB_y(s) \rightleftharpoons xA^{m+}(aq) + yB^{n-}(aq)}$$

the expression for $K_{sp}$ is:

$$K_{sp} = [\mathrm{A^{m+}}]^x[\mathrm{B^{n-}}]^y$$

Important points:

the solid is not included in the expression
only aqueous ions appear
exponents come from the balanced equation

For silver chloride:

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

For calcium fluoride:

$$\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)}$$

$$K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2$$

A smaller $K_{sp}$ usually means the solid is less soluble, but students, be careful: the exact solubility also depends on the formula of the compound. You cannot compare $K_{sp}$ values for different compounds without thinking about how many ions each produces.

3. Molar solubility and how to calculate it

Molar solubility is the number of moles of solute that dissolve in $1\,\text{dm}^3$ of solution to make a saturated solution.

Let the molar solubility of $\mathrm{AgCl}$ be $s$. Then at equilibrium:

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

So:

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

If $K_{sp} = 1.8\times 10^{-10}$ for $\mathrm{AgCl}$, then

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

$$s \approx 1.3\times 10^{-5}\,\text{mol dm}^{-3}$$

That means only a tiny amount dissolves.

Now try a more complex case: $\mathrm{CaF_2}$.

$$\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)}$$

If its molar solubility is $s$, then:

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

So:

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

This is a common IB Chemistry HL skill: turning a $K_{sp}$ expression into a solvable algebra problem.

Example

If $K_{sp} = 3.9\times 10^{-11}$ for $\mathrm{CaF_2}$, then:

$$4s^3 = 3.9\times 10^{-11}$$

$$s^3 = 9.75\times 10^{-12}$$

$$s = \sqrt[3]{9.75\times 10^{-12}}$$

$$s \approx 2.1\times 10^{-4}\,\text{mol dm}^{-3}$$

From this, the fluoride ion concentration is:

$$[\mathrm{F^-}] = 2s \approx 4.2\times 10^{-4}\,\text{mol dm}^{-3}$$

4. The common ion effect

The common ion effect happens when a solution already contains one of the ions involved in the dissolving equilibrium. This reduces solubility because the equilibrium shifts to the left, according to Le Châtelier’s principle.

Consider $\mathrm{AgCl}$ in water. Its solubility is limited. If sodium chloride is added, the concentration of $\mathrm{Cl^-}$ increases. Since $\mathrm{Cl^-}$ is a product in the dissolution equilibrium,

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

the equilibrium shifts left, so less $\mathrm{AgCl}$ dissolves.

This matters in real life and in labs. For example, adding a chloride solution to a mixture containing silver ions can cause silver chloride to precipitate. This is useful in qualitative analysis and in controlling ion concentrations.

Why this matters quantitatively

If there is already $[\mathrm{Cl^-}] = 0.10\,\text{mol dm}^{-3}$ in solution, then for $\mathrm{AgCl}$:

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

So the concentration of $\mathrm{Ag^+}$ at equilibrium becomes much smaller than it would be in pure water.

5. Predicting precipitation with $Q_{sp}$

To decide whether a precipitate will form, chemists compare the reaction quotient $Q_{sp}$ with $K_{sp}$.

For the salt $\mathrm{A_xB_y}$:

$$Q_{sp} = [\mathrm{A^{m+}}]^x[\mathrm{B^{n-}}]^y$$

Use the same form as $K_{sp}$, but with the current ion concentrations.

Compare:

if $Q_{sp} < K_{sp}$, the solution is unsaturated and more solid can dissolve
if $Q_{sp} = K_{sp}$, the solution is saturated and at equilibrium
if $Q_{sp} > K_{sp}$, a precipitate forms until equilibrium is reached

Example

Suppose you mix solutions so that:

$[\mathrm{Ag^+}] = 1.0\times 10^{-4}\,\text{mol dm}^{-3}$
$[\mathrm{Cl^-}] = 1.0\times 10^{-4}\,\text{mol dm}^{-3}$

Then:

$$Q_{sp} = [\mathrm{Ag^+}][\mathrm{Cl^-}] = 1.0\times 10^{-8}$$

For $\mathrm{AgCl}$, $K_{sp} = 1.8\times 10^{-10}$.

Since

$$Q_{sp} > K_{sp}$$

a precipitate of $\mathrm{AgCl(s)}$ will form 🧪

This is a clear example of how equilibrium predicts the extent of reaction. The system changes until the ion concentrations reach the point where $Q_{sp} = K_{sp}$.

6. Connecting solubility to Reactivity 2

This topic fits neatly into Reactivity 2 because it combines three major ideas:

How much change occurs

only a limited amount of an ionic solid dissolves
$K_{sp}$ lets us calculate the maximum dissolved concentration

How far a reaction goes

dissolution of a sparingly soluble salt reaches equilibrium before all solid disappears
the value of $K_{sp}$ tells us the position of equilibrium

Dynamic equilibrium

dissolution and precipitation happen at the same rate at saturation
the system appears stable even though particles are moving constantly

This is the same style of reasoning used in other equilibrium topics in IB Chemistry HL. Whether studying acids, gases, or solubility, the core idea is that chemical systems settle into equilibrium rather than going to completion.

Real-world applications include:

water treatment, where ion concentrations must be controlled
geology, where mineral deposits form from saturated solutions
medicine, where some salts must stay dissolved and others may form deposits
analysis of blood and body fluids, where precipitation can be harmful or useful

Conclusion

students, $K_{sp}$ is a powerful tool for understanding the solubility of ionic compounds. It gives a quantitative way to describe a saturated solution, predict whether a precipitate will form, and calculate ion concentrations at equilibrium. The key idea is that many ionic solids do not dissolve completely; instead, they establish a dynamic equilibrium between solid and dissolved ions. This makes solubility an important example of how chemistry answers the question of how far a reaction proceeds. 🌟

Study Notes

Solubility is the maximum amount of a substance that can dissolve in a given amount of solvent at a specific temperature.
A saturated solution is at equilibrium with undissolved solute.
$K_{sp}$ is the equilibrium constant for the dissolution of a sparingly soluble ionic solid.
In a $K_{sp}$ expression, do not include the solid; include only aqueous ions.
Write $K_{sp}$ from the balanced dissociation equation.
Use $s$ to represent molar solubility when solving equilibrium problems.
For salts with more than one ion from each formula unit, relate ion concentrations to $s$ using stoichiometric coefficients.
The common ion effect decreases solubility.
Compare $Q_{sp}$ with $K_{sp}$ to predict precipitation.
If $Q_{sp} < K_{sp}$, more solid can dissolve.
If $Q_{sp} = K_{sp}$, the solution is saturated.
If $Q_{sp} > K_{sp}$, a precipitate forms.
Solubility and $K_{sp}$ are examples of dynamic equilibrium and are part of the broader IB Chemistry HL theme of reactivity, extent, and equilibrium.