Why Adding a Common Ion Can Reduce Solubility

Adding sodium fluoride to a saturated calcium fluoride solution might seem like it should help more calcium fluoride dissolve. After all, you’ve added another substance containing one of the ions in calcium fluoride.

But the opposite happens. The amount of calcium fluoride that can dissolve decreases.

This is called the common-ion effect. It occurs because adding an ion already involved in a dissolution equilibrium changes the equilibrium position, reducing the solubility of the sparingly soluble compound.

The important detail is that the size of this effect depends on the dissolution equation. A compound that produces two fluoride ions per formula unit responds differently from one that produces only one.

Let’s work through why.

01Why does adding a common ion reduce solubility?

Consider calcium fluoride, \(\ce{CaF2}\), dissolving in water.

Its dissolution equilibrium is:

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

When calcium fluoride dissolves, each formula unit produces:

• One calcium ion, \(\mathrm{Ca}^{2+}\).
• Two fluoride ions, \(\mathrm{F}^{-}\).

Imagine placing excess solid calcium fluoride in water and allowing it to reach equilibrium. Some dissolves, while some remains as a solid.

Now predict what happens when sodium fluoride, \(\ce{NaF}\), is added.

Sodium fluoride is soluble, so it dissociates:

\[ \ce{NaF(aq) -> Na^+(aq) + F^-(aq)} \]

The solution now contains additional fluoride ions.

Will more calcium fluoride dissolve to balance the extra fluoride, or will less dissolve?

It will dissolve less.

The reason is that the dissolution equilibrium shifts to the left, towards solid calcium fluoride. This reduces the amount of calcium fluoride that can remain dissolved at equilibrium.

The role of Le Chatelier’s principle

Le Chatelier’s principle states that when a system at equilibrium is disturbed, it shifts in the direction that partially opposes the disturbance.

For calcium fluoride, adding fluoride ions increases the concentration of a product in the dissolution equilibrium:

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

The system responds by favouring the reverse reaction:

\[ \ce{Ca^2+(aq) + 2F^-(aq) -> CaF2(s)} \]

Some dissolved ions combine to form solid calcium fluoride.

This reduces the dissolved calcium ion concentration and brings the system to a new equilibrium.

There is an important distinction here:

The equilibrium shifts to reduce the effect of adding fluoride, but it does not remove all the fluoride that was added.

The final fluoride concentration is generally still higher than it was before the addition.

Also, adding a common ion does not change the value of the solubility product constant, \(K_{sp}\), provided the temperature remains constant.

It changes the equilibrium concentrations, not the equilibrium constant.

02How does the solubility product explain the effect?

Le Chatelier’s principle tells us which direction the equilibrium shifts. The solubility product expression lets us calculate the new equilibrium concentrations.

For calcium fluoride:

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

The solubility product expression is:

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

Here:

• \(K_{sp}\) is the solubility product constant.
• \([\mathrm{Ca}^{2+}]\) is the equilibrium calcium ion concentration, in mol L\(^{-1}\).
• \([\mathrm{F}^{-}]\) is the equilibrium fluoride ion concentration, in mol L\(^{-1}\).

Notice that fluoride concentration is squared.

This happens because two fluoride ions are produced for every calcium ion in the dissolution reaction.

The exponent comes directly from the balanced dissolution equation. It is not an arbitrary mathematical feature.

Why does the exponent matter?

Suppose the equilibrium fluoride concentration increases.

Because fluoride appears squared in the \(K_{sp}\) expression, even a moderate increase in fluoride concentration has a substantial effect on the calcium ion concentration that can remain in solution.

Rearranging:

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

If the fluoride concentration increases while \(K_{sp}\) stays constant, the equilibrium calcium concentration must decrease.

For example, if the fluoride concentration doubles, the calcium concentration becomes one-quarter of its previous value, provided the assumptions behind the expression remain valid.

This inverse relationship is the mathematical reason that adding fluoride reduces the solubility of calcium fluoride.

03A closer look: what happens when calcium fluoride dissolves?

Before using the equation in calculations, we need to distinguish between two quantities:

• Molar solubility, represented by \(s\), is the number of moles of a solid that dissolve per litre of solution.
• Ion concentration is the concentration of a particular dissolved ion after accounting for the dissolution stoichiometry and any other sources of that ion.

These quantities are not always equal.

For calcium fluoride, suppose its molar solubility in pure water is \(s\) mol L\(^{-1}\).

From the dissolution equation:

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

The equilibrium concentrations are:

Substituting into the solubility product expression:

\[ K_{sp}=s(2s)^2 \]

Therefore:

\[ \boxed{K_{sp}=4s^3} \]

The factor of four comes from squaring the fluoride concentration, which is twice the molar solubility.

This relationship applies when calcium fluoride dissolves in pure water, under the usual HSC approximation that other reactions and activity effects are negligible.

What changes when a common ion is already present?

Now suppose we add sodium fluoride before allowing the calcium fluoride equilibrium to establish.

Let the initial fluoride concentration supplied by sodium fluoride be \(C\) mol L\(^{-1}\).

If an additional amount \(s\) mol L\(^{-1}\) of calcium fluoride dissolves, the final concentrations become:

The solubility product expression is now:

\[ K_{sp}=s(C+2s)^2 \]

This is different from the pure-water expression.

The crucial point is that the fluoride concentration includes two contributions:

1. Fluoride already supplied by the sodium fluoride.
2. Fluoride produced by the additional dissolution of calcium fluoride.

Ignoring either contribution can lead to an incorrect calculation.

When \(C\) is much larger than \(2s\), we can use the approximation:

\[ C+2s\approx C \]

Therefore:

\[ \boxed{s\approx\frac{K_{sp}}{C^2}} \]

This approximation is useful when the added common-ion concentration is sufficiently large compared with the concentration produced by the dissolving solid.

We will check this assumption in the worked examples.

05Does adding the common ion always have the same effect?

No. The identity of the common ion matters, as does its concentration.

For calcium fluoride, we could add either:

• Sodium fluoride, which supplies \(\mathrm{F}^{-}\).
• Calcium nitrate, which supplies \(\mathrm{Ca}^{2+}\).

Both introduce a common ion, so both reduce the solubility of calcium fluoride.

However, their effects are not identical for the same numerical concentration.

If we add a common ion that appears with a coefficient of two in the dissolution equation, it has a squared concentration term in the \(K_{sp}\) expression.

If we add the ion that appears with a coefficient of one, its concentration has a power of one.

This is why you must start with the balanced dissolution equation rather than memorising a single common-ion formula.

06Testing the common-ion effect experimentally

How could we check that adding a common ion really reduces solubility?

A useful experiment compares the amount of calcium fluoride dissolved under two different conditions.

Consider the following illustrative investigation.

Apparatus and arrangement

Prepare two identical containers, each containing:

• The same volume of water.
• An excess quantity of solid calcium fluoride.
• The same temperature and stirring conditions.

Add sodium fluoride to the second container, but not the first.

Stir both mixtures for the same period, allowing each to approach equilibrium.

Then separate the undissolved solid by filtration. Analyse the clear solutions to determine their dissolved calcium concentrations using an appropriate quantitative method.

What should the original prediction show?

If the common-ion effect is correct, the solution containing sodium fluoride should have a lower equilibrium calcium ion concentration.

Why?

The sodium fluoride supplies additional fluoride ions. These shift the calcium fluoride dissolution equilibrium towards the solid.

The amount of dissolved calcium fluoride should therefore decrease.

Illustrative results

The following results are hypothetical and are included to demonstrate how the experimental comparison would be interpreted. They are not measured experimental data.

The calcium concentration in the sodium fluoride solution is much lower.

This supports the prediction that adding a common ion reduces calcium fluoride solubility.

The water-only result also provides a useful comparison with the theoretical value:

\[ K_{sp}=4s^3 \]

\[ s=\sqrt[3]{\frac{3.9\times10^{-11}}{4}} \]

\[ s\approx2.14\times10^{-4}\text{ mol L}^{-1} \]

For the sodium fluoride solution:

\[ s\approx\frac{3.9\times10^{-11}}{(0.0100)^2} \]

\[ s\approx3.9\times10^{-7}\text{ mol L}^{-1} \]

The theoretical values agree with the illustrative measurements.

What could make the experiment unreliable?

A few things need controlling.

• Temperature must remain constant because \(K_{sp}\) depends on temperature.
• Both containers must have excess solid so that the solutions can reach saturation.
• Both mixtures must be allowed enough time to approach equilibrium.
• Filtration and sampling must be consistent.
• The analytical method must be suitable for the very low calcium concentrations involved.

There is also an important limitation.

Adding sodium fluoride changes the ionic strength of the solution. In a real experiment, ion activities may differ from their concentrations, so a calculation using concentrations may not agree perfectly with measured results.

The ideal-solution approximation is useful for HSC calculations, but it is not an exact description of every real solution.

07A common-ion effect is not the same as every solubility change

Here is a misconception worth addressing.

A student might see a substance being added to a saturated solution and assume that any added dissolved material must reduce solubility.

That is not correct.

The effect depends on whether the added substance supplies an ion already involved in the dissolution equilibrium.

For example, consider calcium fluoride again.

Sodium nitrate does not introduce either ion in the calcium fluoride dissolution equation.

Therefore, under the simplified HSC model, it does not produce a common-ion effect.

In real solutions, however, changing ionic strength can affect measured solubility through ion activity coefficients. The table describes the common-ion contribution, not every possible effect of adding an electrolyte.

There is another important exception: acid-base reactions.

Fluoride ions can react with hydrogen ions:

\[ \ce{F^- + H^+ <=> HF} \]

If acid is added, some fluoride ions are converted into hydrogen fluoride.

This reduces the concentration of free fluoride ions and can allow more calcium fluoride to dissolve.

So the simple common-ion prediction must be considered alongside any other reactions involving the dissolved ions.

08A practical method for common-ion calculations

When you see a common-ion question in an HSC Chemistry exam, follow this sequence.

1. Write the balanced dissolution equation.

Do not begin with a memorised formula. The coefficients determine the concentration relationships.

2. Write the \(K_{sp}\) expression.

Use the aqueous ion concentrations and their stoichiometric powers. Do not include the solid.

3. Identify the common ion.

Work out which ion is supplied by the added soluble compound.

4. Write the final equilibrium concentrations.

Include both the initial common-ion concentration and the contribution from further dissolution.

5. Substitute into the \(K_{sp}\) expression.

For example, for calcium fluoride with an initial fluoride concentration \(C\):

\[ K_{sp}=s(C+2s)^2 \]

6. Decide whether an approximation is justified.

If \(2s\) is negligible compared with \(C\), use:

\[ s\approx\frac{K_{sp}}{C^2} \]

If the common ion is calcium instead:

\[ K_{sp}=(C+s)(2s)^2 \]

When \(s\) is negligible compared with \(C\):

\[ s\approx\sqrt{\frac{K_{sp}}{4C}} \]

7. Check the result.

Substitute your calculated value back into the concentration expression. If the contribution from dissolution is not negligible, you may need to solve the full equation.

Remember that all concentrations must refer to the final solution, not the original solutions before mixing.

09Questions and solutions

The following questions move from straightforward common-ion calculations towards situations where more than one equilibrium effect needs to be considered.

Use the following illustrative data unless a question provides different values:

\[ K_{sp}(\ce{CaF2})=3.9\times10^{-11} \]

at 25 °C.

Assume ideal behaviour and complete dissociation of soluble salts unless otherwise stated.

Question 1

A student adds sodium fluoride to a saturated calcium fluoride solution.

(a) Explain why the solubility of calcium fluoride decreases.

(b) State whether the value of \(K_{sp}\) changes, assuming the temperature remains constant.

Solution 1

(a) The solubility of calcium fluoride decreases because adding sodium fluoride increases the concentration of fluoride ions, shifting the dissolution equilibrium to the left.

Calcium fluoride dissolves according to:

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

Sodium fluoride dissociates to produce additional fluoride ions:

\[ \ce{NaF(aq) -> Na^+(aq) + F^-(aq)} \]

The increased fluoride concentration disturbs the equilibrium.

According to Le Chatelier’s principle, the equilibrium shifts towards the reactants, forming more solid calcium fluoride.

Consequently, less calcium fluoride remains dissolved at the new equilibrium.

(b) The value of \(K_{sp}\) remains unchanged.

The solubility product constant depends on temperature, not on the initial concentrations of the ions.

Adding sodium fluoride changes the equilibrium concentrations, but does not change \(K_{sp}\) at constant temperature.

Question 2

A student compares the solubility of calcium fluoride in two solutions at 25 °C.

• Solution A contains \(0.0200\) mol L\(^{-1}\) sodium fluoride.
• Solution B contains \(0.0200\) mol L\(^{-1}\) calcium nitrate.

Calculate the molar solubility of calcium fluoride in each solution.

Explain why the answers are different.

Solution 2

Solution A has a molar solubility of approximately \(9.75\times10^{-8}\) mol L\(^{-1}\), while Solution B has a molar solubility of approximately \(2.21\times10^{-5}\) mol L\(^{-1}\).

The difference occurs because fluoride and calcium have different stoichiometric coefficients in the dissolution equation.

Step 1: Write the dissolution equation.

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

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

Step 2: Calculate the solubility in Solution A.

Sodium fluoride supplies the common ion \(\mathrm{F}^{-}\).

Let the molar solubility be \(s_A\).

\[ [\mathrm{Ca}^{2+}]=s_A \]

\[ [\mathrm{F}^{-}]=0.0200+2s_A \]

Substitute:

\[ 3.9\times10^{-11} =s_A(0.0200+2s_A)^2 \]

Using the common-ion approximation:

\[ 3.9\times10^{-11} =s_A(0.0200)^2 \]

\[ s_A=\frac{3.9\times10^{-11}}{4.00\times10^{-4}} \]

\[ \boxed{s_A=9.75\times10^{-8}\text{ mol L}^{-1}} \]

Step 3: Calculate the solubility in Solution B.

Calcium nitrate supplies the common ion \(\mathrm{Ca}^{2+}\).

Let the molar solubility be \(s_B\).

\[ [\mathrm{Ca}^{2+}]=0.0200+s_B \]

\[ [\mathrm{F}^{-}]=2s_B \]

Substitute:

\[ 3.9\times10^{-11} =(0.0200+s_B)(2s_B)^2 \]

Using the common-ion approximation:

\[ 3.9\times10^{-11} =0.0200(4s_B^2) \]

\[ s_B^2=\frac{3.9\times10^{-11}}{0.0800} \]

\[ s_B=\sqrt{4.875\times10^{-10}} \]

\[ \boxed{s_B=2.21\times10^{-5}\text{ mol L}^{-1}} \]

Step 4: Compare the results.

\[ \frac{s_B}{s_A} = \frac{2.21\times10^{-5}}{9.75\times10^{-8}} \]

\[ \frac{s_B}{s_A}\approx227 \]

Calcium fluoride is approximately 227 times more soluble in Solution B than in Solution A under these assumptions.

The tempting mistake is to assume that equal initial concentrations of common ions must produce equal reductions in solubility.

They do not.

Fluoride appears squared in the \(K_{sp}\) expression, whereas calcium appears to the first power. This produces different mathematical relationships between the added ion concentration and the molar solubility.

Question 3

A saturated solution of calcium fluoride is prepared by dissolving excess solid in water at 25 °C.

The solution is then mixed with an equal volume of \(0.0200\) mol L\(^{-1}\) sodium fluoride.

Assume that:

• The initial saturated calcium fluoride solution has a molar solubility of \(s_0\).
• The volumes are additive.
• The temperature remains constant.
• The mixture is allowed to reach equilibrium in the presence of excess calcium fluoride solid.
• No other reactions occur.

Calculate the new molar solubility of calcium fluoride.

Explain why using \(0.0200\) mol L\(^{-1}\) as the final fluoride concentration would produce an incorrect answer.

Solution 3

The new molar solubility is approximately \(3.90\times10^{-7}\) mol L\(^{-1}\).

The key is to account for dilution when the two solutions are mixed.

Step 1: Calculate the fluoride concentration after mixing, before further dissolution.

The sodium fluoride solution is diluted to twice its original volume.

Using:

\[ C_1V_1=C_2V_2 \]

where \(C_1\) and \(V_1\) are the initial sodium fluoride concentration and volume, and \(C_2\) and \(V_2\) are the concentration and volume after mixing:

\[ 0.0200V= C_2(2V) \]

Therefore:

\[ C_2=0.0100\text{ mol L}^{-1} \]

Step 2: Establish the final equilibrium concentrations.

Let \(s\) be the new molar solubility.

The initial calcium fluoride solution contributes a small amount of calcium and fluoride ions.

However, the calcium fluoride solid remains present, and the system is allowed to reach a new equilibrium.

The initial calcium concentration after mixing is:

\[ \frac{s_0}{2} \]

The initial fluoride concentration after mixing is:

\[ 0.0100+s_0 \]

As additional calcium fluoride dissolves or precipitates to establish the final equilibrium, the final concentrations are:

\[ [\mathrm{Ca}^{2+}]=\frac{s_0}{2}+x \]

\[ [\mathrm{F}^{-}]=0.0100+s_0+2x \]

Here \(x\) is the net change in dissolved calcium fluoride concentration relative to the diluted starting solution.

The final total molar solubility is:

\[ s=\frac{s_0}{2}+x \]

For the final equilibrium, the \(K_{sp}\) expression becomes:

\[ 3.9\times10^{-11} = s(0.0100+2s)^2 \]

The initial fluoride contribution from the diluted saturated solution is negligible at the precision required here.

Step 3: Apply the common-ion approximation.

Since the common-ion concentration is much larger than the additional fluoride produced by dissolution:

\[ 3.9\times10^{-11}=s(0.0100)^2 \]

\[ s=\frac{3.9\times10^{-11}}{1.00\times10^{-4}} \]

\[ \boxed{s=3.90\times10^{-7}\text{ mol L}^{-1}} \]

Step 4: Explain the dilution error.

Using \(0.0200\) mol L\(^{-1}\) as the final sodium fluoride concentration ignores the fact that the sodium fluoride solution has been mixed with an equal volume of another solution.

Its concentration is halved.

Because fluoride concentration is squared in the \(K_{sp}\) expression, this error would produce a fourfold difference in the calculated molar solubility.

The important lesson is that common-ion calculations after mixing must use final concentrations.

Question 4

A student claims:

“If adding sodium fluoride reduces the solubility of calcium fluoride, adding hydrochloric acid should reduce its solubility even further because hydrochloric acid also adds ions to the solution.”

Evaluate this claim using relevant chemical equations.

Assume that the solution is sufficiently acidic for hydrogen ions to react with fluoride ions, and that hydrogen fluoride formation is significant.

Solution 4

The claim is incorrect. Adding hydrochloric acid can increase the solubility of calcium fluoride because hydrogen ions react with fluoride ions, reducing the concentration of free fluoride ions.

The student is assuming that adding any ions to a solution produces the common-ion effect.

However, hydrochloric acid supplies hydrogen ions, not calcium or fluoride ions.

Step 1: Identify the calcium fluoride equilibrium.

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

The solubility product expression is:

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

Step 2: Identify the reaction involving the added hydrogen ions.

Hydrochloric acid dissociates in water:

\[ \ce{HCl(aq) -> H+(aq) + Cl^-(aq)} \]

Hydrogen ions react with fluoride ions:

\[ \ce{H+(aq) + F^-(aq) <=> HF(aq)} \]

This removes some free fluoride ions from the dissolution equilibrium.

Step 3: Predict the effect on calcium fluoride dissolution.

The decrease in free fluoride concentration disturbs the calcium fluoride equilibrium.

The equilibrium shifts to the right, allowing more calcium fluoride to dissolve and produce additional fluoride ions.

Therefore, the solubility can increase.

Step 4: Explain the limitation.

The overall result depends on the extent of fluoride protonation and the solution conditions.

It is not enough to say that acid always increases the solubility of every ionic compound. The relevant reactions must be considered.

This question demonstrates why the common-ion effect must be considered alongside acid-base equilibria.

Question 5

A student prepares two separate saturated calcium fluoride solutions at 25 °C.

Both contain excess solid calcium fluoride.

The student then makes the following changes:

• In Beaker A, sodium fluoride is added until the free fluoride concentration is \(0.0100\) mol L\(^{-1}\).
• In Beaker B, calcium nitrate is added until the calcium ion concentration is \(0.0100\) mol L\(^{-1}\).

Assume ideal behaviour, complete dissociation, no other reactions, and that excess calcium fluoride solid remains present.

The student confidently predicts that Beaker A must contain less dissolved calcium fluoride because fluoride appears twice in the dissolution equation.

Determine the molar solubility in each beaker and evaluate the student’s prediction.

Solution 5

Beaker A has a molar solubility of approximately \(3.90\times10^{-7}\) mol L\(^{-1}\), while Beaker B has a molar solubility of approximately \(3.12\times10^{-5}\) mol L\(^{-1}\). The student’s prediction is correct, but the reasoning is incomplete: the two solubilities must be calculated using the full stoichiometric relationships, rather than simply counting the number of ions in the dissolution equation.

This is a useful example of why an apparently obvious prediction should be checked mathematically.

Step 1: Write the equilibrium expression.

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

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

Step 2: Calculate the solubility in Beaker A.

The free fluoride concentration is maintained at:

\[ [\mathrm{F}^{-}]=0.0100\text{ mol L}^{-1} \]

Let the molar solubility be \(s_A\).

Then:

\[ [\mathrm{Ca}^{2+}]=s_A \]

Substitute:

\[ 3.9\times10^{-11} =s_A(0.0100)^2 \]

\[ s_A=\frac{3.9\times10^{-11}}{1.00\times10^{-4}} \]

\[ \boxed{s_A=3.90\times10^{-7}\text{ mol L}^{-1}} \]

Step 3: Calculate the solubility in Beaker B.

The calcium ion concentration is maintained at:

\[ [\mathrm{Ca}^{2+}]=0.0100\text{ mol L}^{-1} \]

Let the molar solubility be \(s_B\).

From the dissolution equation:

\[ [\mathrm{F}^{-}]=2s_B \]

Substitute:

\[ 3.9\times10^{-11} =0.0100(2s_B)^2 \]

\[ 3.9\times10^{-11}=0.0400s_B^2 \]

\[ s_B^2=\frac{3.9\times10^{-11}}{0.0400} \]

\[ s_B=\sqrt{9.75\times10^{-10}} \]

\[ \boxed{s_B=3.12\times10^{-5}\text{ mol L}^{-1}} \]

Step 4: Compare the results.

\[ \frac{s_B}{s_A} = \frac{3.12\times10^{-5}}{3.90\times10^{-7}} \]

\[ \frac{s_B}{s_A}=80 \]

Beaker B allows approximately 80 times as much calcium fluoride to dissolve as Beaker A.

The student’s prediction is correct.

However, simply counting the two fluoride ions and one calcium ion is not a complete explanation.

The actual difference comes from how each concentration enters the \(K_{sp}\) expression.

In Beaker A, the fixed fluoride concentration is squared:

\[ s_A=\frac{K_{sp}}{[\mathrm{F}^{-}]^2} \]

In Beaker B, the fixed calcium concentration appears to the first power, while the fluoride concentration produced by dissolution is \(2s_B\):

\[ s_B=\sqrt{\frac{K_{sp}}{4[\mathrm{Ca}^{2+}]}} \]

The different relationships produce different numerical solubilities.

There is a further subtlety: the stated concentrations refer to the final equilibrium conditions. If the problem instead described adding fixed quantities of salts to a closed mixture, the common-ion concentrations would not necessarily remain fixed as dissolution or precipitation occurred. You would need to account for the changing concentrations using mass balance as well as \(K_{sp}\).

This is why the strongest approach is to identify exactly what the question gives you: an initial concentration, a final concentration, or a fixed amount added.

10Where to go next

The common-ion effect explains why adding an ion can reduce the solubility of an ionic compound. The next step is learning how to use the same equilibrium expression to predict when a precipitate will actually form.

That requires comparing the ion product, \(Q_{sp}\), with the solubility product, \(K_{sp}\).

• If \(Q_{sp}<K_{sp}\), the solution is unsaturated.
• If \(Q_{sp}=K_{sp}\), the solution is at equilibrium with the solid.
• If \(Q_{sp}>K_{sp}\), precipitation is thermodynamically favoured.

For a worked explanation of how added chloride changes silver chloride solubility, see Common-Ion Effect: Why Added Chloride Reduces AgCl Solubility.

To apply these ideas to selective precipitation, continue with Which Ion Precipitates First? A Worked HSC Chemistry Example.

The key decision is now clear: use the balanced dissolution equation to establish the ion concentrations, then use \(K_{sp}\) to determine what can remain dissolved at equilibrium.