What Does a Saturated Solution Mean at the Particle Level?

You add another spatula of sodium chloride to a beaker of water. Some crystals disappear, but a layer of solid remains at the bottom. Has dissolving stopped?

No. A saturated solution contains the maximum equilibrium concentration of dissolved solute under the specified conditions. At the particle level, ions are still leaving the crystal and returning to it. They simply do so at equal rates, meaning the overall concentration stays constant.

That is the key idea: saturation is a dynamic equilibrium, not a state in which particles stop moving.

Predict what happens if you add another spoonful of salt to the beaker. Will the concentration of dissolved salt increase? If the solution is already saturated, its temperature remains constant, and excess solid is present, the answer is no. More solid will remain undissolved, but particles will continue moving between the crystal and the solution.

01What is happening at the crystal surface?

To understand saturation, picture what happens when an ionic solid enters water.

An ionic crystal, such as sodium chloride, contains positive and negative ions arranged in a repeating lattice. These ions are held together by electrostatic attraction.

Water molecules are polar. Their oxygen ends have a partial negative charge, while their hydrogen ends have partial positive charges.

This allows water molecules to attract ions at the crystal surface.

• The oxygen end of water is attracted towards positive ions such as \(\mathrm{Na^+}\).
• The hydrogen ends are attracted towards negative ions such as \(\mathrm{Cl^-}\).

Water molecules surround the separated ions and help stabilise them in solution. This process is called hydration.

The ions can then move away from the crystal and become distributed throughout the water.

The dissolution of sodium chloride can be represented by:

\[ \ce{NaCl(s) <=> Na+(aq) + Cl-(aq)} \]

There are two opposing processes:

1. Dissolution: ions leave the crystal and enter the solution.
2. Crystallisation: dissolved ions return to the crystal lattice.

Initially, when the solution is unsaturated, dissolution occurs faster than crystallisation. The concentration of dissolved ions increases.

As more salt dissolves, the concentration of ions in the solution increases. This makes it more likely that dissolved ions will collide with the crystal surface and rejoin the lattice.

Eventually, the two processes reach the same rate.

\[ \text{Rate of dissolution} = \text{Rate of crystallisation} \]

This is called dynamic equilibrium.

Think of a revolving door between two rooms. People can continue moving in both directions while the number of people in each room stays constant, provided the same number cross each way.

The analogy captures the equal rates, but actual ions move randomly through the solution and interact through electrostatic forces.

What does dynamic equilibrium actually mean?

At equilibrium:

• Ions are still leaving the crystal.
• Dissolved ions are still returning to the crystal.
• The amount of dissolved solute remains constant.
• The concentration of dissolved ions remains constant.
• There is no overall change in the amount of undissolved solid.

Importantly, equal rates do not mean equal quantities of dissolved and undissolved particles. They describe how quickly particles move between the two states.

A saturated solution can also remain saturated after all visible undissolved solid has been removed by filtration, provided its concentration and temperature remain unchanged. It is the concentration that defines saturation, not the presence of a visible pile of crystals.

02How would you experimentally identify a saturated solution?

You cannot directly watch individual ions moving between a crystal and water using ordinary laboratory equipment.

Instead, you can investigate saturation by measuring how much solute is dissolved in a known volume of solution.

Consider an investigation using potassium chloride, \(\ce{KCl}\).

Setting up the investigation

Place water in a beaker inside a temperature-controlled water bath. Keep the temperature at \(25^\circ\mathrm{C}\), using a thermometer to monitor it.

Add potassium chloride gradually while stirring.

Initially, the crystals dissolve. Continue adding small quantities until some solid remains at the bottom.

Allow the mixture to stand while maintaining the temperature and stirring until the concentration becomes stable.

Now consider what the investigation should show.

If the solution is unsaturated, adding more potassium chloride should increase the amount dissolved.

If the solution is saturated, adding more potassium chloride should not increase the equilibrium concentration. The extra solid remains undissolved.

To test this, collect a sample of the clear liquid above the remaining crystals. Filter it at the same temperature to remove any suspended solid.

Measure a known volume of the filtrate, evaporate the water, and weigh the remaining dry potassium chloride.

Repeat the measurement after allowing more time and adding further solid.

Illustrative experimental data

Suppose each sample has a volume of \(1.00\ \mathrm{mL}\).

These are illustrative measurements, not reported experimental results.

The mass of potassium chloride in each sample has effectively stopped increasing.

This suggests that the solution has reached saturation.

The experiment does not directly prove that individual ions are moving in both directions. Instead, the constant concentration provides macroscopic evidence consistent with dynamic equilibrium.

Why isn’t seeing solid at the bottom enough?

A common experimental mistake is to assume that any solution containing undissolved solid must be saturated.

That conclusion is not always justified.

The solid might simply be dissolving slowly. The solution might not have been stirred properly, or the temperature might not have been controlled.

A better investigation:

• Maintains a constant temperature.
• Allows enough time for equilibrium to establish.
• Stirs the solution to improve mixing.
• Measures the concentration of successive filtrate samples.
• Checks that adding further solute does not increase the measured concentration.

There is another important experimental detail. If the filtrate cools during filtration, some dissolved solute may crystallise before it is measured. This can produce an artificially low concentration.

Using warmed filtration equipment and maintaining the sample temperature helps prevent this problem.

03Concentration units: what are you actually measuring?

Solubility can be expressed using different concentration units. Mixing these up is a common source of errors in HSC calculations.

The important distinction is the denominator.

For molar concentration:

\[ c=\frac{n}{V} \]

where:

• \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\).
• \(n\) is the amount of solute in moles.
• \(V\) is the volume of the solution in litres.

You can calculate the number of moles using:

\[ n=\frac{m}{M} \]

where:

• \(m\) is the mass of solute in grams.
• \(M\) is its molar mass in \(\mathrm{g\,mol^{-1}}\).

Notice that molarity uses the volume of the entire solution, not the volume of the water before dissolving.

Similarly, \(36\ \mathrm{g}\) of solute per \(100\ \mathrm{g}\) of water is not the same as \(36\ \mathrm{g\,L^{-1}}\). The first uses the mass of the solvent, while the second uses the volume of the solution.

04Dissolution stoichiometry: why the ion ratio matters

Here is where many students make an important mistake.

They calculate the molar solubility of an ionic solid and assume that the concentration of every ion is equal to that value.

That only works for certain ionic compounds.

Consider calcium fluoride:

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

One formula unit of calcium fluoride produces:

• One calcium ion.
• Two fluoride ions.

The ratio is therefore \(1:2\).

Suppose the molar solubility of calcium fluoride is \(s\).

This means \(s\) moles of calcium fluoride dissolve per litre of saturated solution.

Consequently:

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

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

The fluoride concentration is twice the calcium concentration.

The same reasoning applies to other ionic compounds.

These relationships assume that the solid dissolves into pure water, there are no other sources of the ions, and the ions remain in solution without significant side reactions.

Connecting solubility to the solubility product

For a sparingly soluble ionic solid, the equilibrium between the solid and its dissolved ions can be represented using a solubility product expression.

For calcium fluoride:

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

The solubility product is:

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

The exponent of each ion concentration comes from its coefficient in the balanced dissolution equation.

Notice that the solid calcium fluoride does not appear in the expression. Its activity is treated as constant while solid is present.

In pure water, we can substitute the molar solubility relationships:

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

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

Therefore:

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

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

This equation allows us to calculate the molar solubility from the solubility product.

05What changes when a common ion is already present?

Now consider a slightly different situation.

You have a saturated solution of calcium fluoride. Instead of adding more calcium fluoride, you add sodium fluoride, \(\ce{NaF}\).

What happens to the solubility of calcium fluoride?

The answer is that it decreases.

To understand why, remember the original equilibrium:

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

Sodium fluoride dissolves readily in water:

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

This introduces additional fluoride ions into the solution.

Fluoride is already a product of the calcium fluoride dissolution equilibrium. It is therefore called a common ion.

The additional fluoride ions increase the concentration of a product.

This temporarily increases the ion product above the value of \(K_{sp}\). Some dissolved calcium and fluoride ions then return to the solid crystal lattice.

The equilibrium shifts towards the undissolved calcium fluoride.

As a result:

• More calcium fluoride remains as solid.
• The equilibrium concentration of calcium ions decreases.
• The molar solubility of calcium fluoride decreases.
• The value of \(K_{sp}\) remains unchanged, provided the temperature is constant.

The important assumption when calculating a common-ion effect

Suppose calcium fluoride dissolves in a solution that already contains \(0.0200\ \mathrm{mol\,L^{-1}}\) of fluoride ions from sodium fluoride.

You cannot simply write:

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

That would ignore the fluoride ions already present.

Instead:

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

The total fluoride concentration contains two contributions:

1. Fluoride supplied by sodium fluoride.
2. Fluoride produced by dissolving calcium fluoride.

The sodium ions do not appear in the \(K_{sp}\) expression because they are not part of the calcium fluoride dissolution equilibrium.

For the usual HSC common-ion calculation, assume:

• The added sodium fluoride dissociates completely.
• The temperature remains constant.
• The volume is known and does not change during the calculation.
• No significant complex ions or other side reactions occur.
• Activities can be approximated by concentrations.

These assumptions matter. If the added common ion comes from another solution, you must account for any dilution before calculating the final ion concentrations.

06Saturated, unsaturated, and supersaturated solutions

These three terms describe different relationships between the actual concentration of dissolved solute and its equilibrium solubility.

One particularly important distinction is between saturated and supersaturated solutions.

A supersaturated solution contains more dissolved solute than would normally be stable at that temperature.

It can remain clear for some time because crystallisation does not necessarily begin immediately. The solution is metastable, meaning it can persist temporarily before moving towards equilibrium.

07Why a clear solution can still be supersaturated

Consider potassium nitrate, \(\ce{KNO3}\).

Its solubility increases substantially as temperature increases.

Imagine preparing a saturated potassium nitrate solution at \(60^\circ\mathrm{C}\), then filtering it while hot to remove all undissolved crystals.

Now cool the clear filtrate to \(20^\circ\mathrm{C}\).

What do you expect to happen?

Because potassium nitrate is less soluble at the lower temperature, the original amount of dissolved solute may now exceed the equilibrium solubility at \(20^\circ\mathrm{C}\).

The solution has become supersaturated.

However, it might remain completely clear.

Why?

Crystallisation requires the formation of an initial crystal nucleus. Without a suitable starting point, the dissolved ions or particles can remain in solution temporarily, even though crystallisation is thermodynamically favoured.

Add a small potassium nitrate crystal.

The seed crystal provides a surface onto which dissolved particles can attach. Crystallisation begins, and the concentration decreases towards the equilibrium solubility at \(20^\circ\mathrm{C}\).

This is why visual inspection alone cannot reliably distinguish saturated from supersaturated solutions.

The concentration and temperature must be considered together.

08Questions and solutions

The following questions progress from straightforward dissolution stoichiometry to problems involving common ions, experimental measurements, and competing effects.

Question 1

The molar solubility of barium fluoride, \(\ce{BaF2}\), in pure water at a particular temperature is \(1.5\times10^{-3}\ \mathrm{mol\,L^{-1}}\).

Write the balanced dissolution equation and calculate the equilibrium concentrations of barium ions and fluoride ions.

Solution 1

The equilibrium concentrations are:

\[ \boxed{[\mathrm{Ba^{2+}}]=1.5\times10^{-3}\ \mathrm{M}} \]

\[ \boxed{[\mathrm{F^-}]=3.0\times10^{-3}\ \mathrm{M}} \]

Step 1: Write the balanced dissolution equation.

\[ \ce{BaF2(s) <=> Ba^2+(aq) + 2F^-(aq)} \]

Step 2: Identify the molar solubility.

The molar solubility is:

\[ s=1.5\times10^{-3}\ \mathrm{M} \]

One mole of barium fluoride produces one mole of barium ions and two moles of fluoride ions.

Step 3: Calculate the barium concentration.

\[ [\mathrm{Ba^{2+}}]=s \]

\[ \boxed{[\mathrm{Ba^{2+}}]=1.5\times10^{-3}\ \mathrm{M}} \]

Step 4: Calculate the fluoride concentration.

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

\[ =2(1.5\times10^{-3}) \]

\[ \boxed{[\mathrm{F^-}]=3.0\times10^{-3}\ \mathrm{M}} \]

The fluoride concentration is twice the barium concentration because the formula contains two fluoride ions for every barium ion.

The key is to use the coefficients in the balanced dissolution equation rather than assuming all ion concentrations equal the molar solubility.

Question 2

An unknown ionic solid, \(\ce{M2X3}\), dissolves in pure water according to:

\[ \ce{M2X3(s) <=> 2M^3+(aq) + 3X^2-(aq)} \]

A \(25.0\ \mathrm{mL}\) sample of its saturated solution is filtered and evaporated to dryness.

The remaining solid has a mass of \(0.315\ \mathrm{g}\).

The molar mass of \(\ce{M2X3}\) is \(210.0\ \mathrm{g\,mol^{-1}}\).

Assuming complete dissociation and no side reactions, calculate:

a) The molar solubility of \(\ce{M2X3}\).

b) The equilibrium concentrations of \(\mathrm{M^{3+}}\) and \(\mathrm{X^{2-}}\).

c) The solubility product, \(K_{sp}\).

Solution 2

The molar solubility is \(0.0600\ \mathrm{mol\,L^{-1}}\), the ion concentrations are \(0.120\ \mathrm{M}\) and \(0.180\ \mathrm{M}\), and:

\[ \boxed{K_{sp}=8.40\times10^{-5}} \]

Step 1: Calculate the moles of dissolved ionic solid.

\[ n=\frac{m}{M} \]

\[ n=\frac{0.315}{210.0} \]

\[ n=1.50\times10^{-3}\ \mathrm{mol} \]

Step 2: Convert the sample volume into litres.

\[ V=25.0\times10^{-3} \]

\[ V=0.0250\ \mathrm{L} \]

Step 3: Calculate the molar solubility.

\[ s=\frac{n}{V} \]

\[ s=\frac{1.50\times10^{-3}}{0.0250} \]

\[ \boxed{s=0.0600\ \mathrm{M}} \]

Step 4: Calculate the individual ion concentrations.

From the balanced dissolution equation:

\[ \ce{M2X3(s) <=> 2M^3+ + 3X^2-} \]

Two moles of \(\mathrm{M^{3+}}\) are produced for every mole of solid dissolved.

Therefore:

\[ [\mathrm{M^{3+}}]=2s \]

\[ =2(0.0600) \]

\[ \boxed{[\mathrm{M^{3+}}]=0.120\ \mathrm{M}} \]

Similarly:

\[ [\mathrm{X^{2-}}]=3s \]

\[ =3(0.0600) \]

\[ \boxed{[\mathrm{X^{2-}}]=0.180\ \mathrm{M}} \]

Step 5: Write the solubility product expression.

The coefficients in the dissolution equation become the exponents:

\[ K_{sp}=[\mathrm{M^{3+}}]^2 [\mathrm{X^{2-}}]^3 \]

Substitute the concentrations:

\[ K_{sp}=(0.120)^2(0.180)^3 \]

\[ K_{sp}=0.0144\times0.005832 \]

\[ \boxed{K_{sp}=8.40\times10^{-5}} \]

The important step is distinguishing the molar solubility of the original solid from the concentrations of the ions it produces.

Question 3

At \(25^\circ\mathrm{C}\), excess solid silver chloride, \(\ce{AgCl}\), is added to a solution containing \(0.0500\ \mathrm{mol\,L^{-1}}\) sodium chloride.

The solubility product of silver chloride is:

\[ K_{sp}=1.8\times10^{-10} \]

Calculate the molar solubility of silver chloride in this solution.

Explain why the chloride concentration cannot simply be assumed to equal the molar solubility.

Assume complete dissociation of sodium chloride and ideal dilute-solution behaviour.

Solution 3

The molar solubility of silver chloride is:

\[ \boxed{s=3.6\times10^{-9}\ \mathrm{mol\,L^{-1}}} \]

The important factor is that the solution already contains chloride ions from sodium chloride.

Step 1: Write the dissolution equation.

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

Let the molar solubility of silver chloride be \(s\).

Then:

\[ [\mathrm{Ag^+}]=s \]

However, the chloride concentration has two contributions:

\[ [\mathrm{Cl^-}]=0.0500+s \]

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

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

Substituting:

\[ 1.8\times10^{-10}=s(0.0500+s) \]

Step 3: Apply the common-ion approximation.

Because \(s\) is expected to be very small compared with \(0.0500\), we can approximate:

\[ 0.0500+s\approx0.0500 \]

Therefore:

\[ 1.8\times10^{-10}=s(0.0500) \]

Rearranging:

\[ s=\frac{1.8\times10^{-10}}{0.0500} \]

\[ \boxed{s=3.6\times10^{-9}\ \mathrm{M}} \]

Step 4: Explain the common-ion effect.

The chloride concentration is approximately:

\[ [\mathrm{Cl^-}]=0.0500\ \mathrm{M} \]

It is not equal to \(s\), because most chloride ions came from the dissolved sodium chloride rather than the silver chloride.

The common chloride ions suppress the dissolution of silver chloride, leaving a very small equilibrium concentration of silver ions.

The tempting mistake is to use the 1:1 dissolution ratio and assume both ion concentrations equal \(s\). That relationship only applies when there is no other source of either ion.

Question 4

A saturated potassium nitrate solution is prepared at \(60^\circ\mathrm{C}\).

The solution is filtered while hot to remove all undissolved potassium nitrate crystals. It is then cooled to \(20^\circ\mathrm{C}\).

After cooling, the solution remains completely clear for ten minutes.

A student claims that the solution must be unsaturated because there are no visible crystals.

Assess this claim.

Explain:

a) The state of the solution after cooling.

b) Why crystals may not appear immediately.

c) What would happen if a small potassium nitrate seed crystal were introduced.

Solution 4

The solution is likely supersaturated, not unsaturated, provided its dissolved concentration exceeds the equilibrium solubility at \(20^\circ\mathrm{C}\).

Step 1: Consider the original solution.

At \(60^\circ\mathrm{C}\), the potassium nitrate solution was saturated.

It therefore contained the equilibrium concentration of dissolved potassium nitrate at that temperature.

Step 2: Consider what happens during cooling.

Potassium nitrate becomes less soluble as the temperature decreases from \(60^\circ\mathrm{C}\) to \(20^\circ\mathrm{C}\).

The solution now contains more dissolved potassium nitrate than can normally remain in equilibrium at the lower temperature.

It is therefore supersaturated.

Step 3: Explain why no crystals have appeared.

Supersaturated solutions can remain temporarily clear because crystallisation requires an initial nucleation event.

Without a suitable starting surface or nucleus, the dissolved particles can remain in the solution even though the concentration exceeds the equilibrium solubility.

The absence of visible crystals does not prove that the solution is unsaturated.

Step 4: Predict what happens when a seed crystal is added.

The seed crystal provides a surface for dissolved potassium nitrate particles to attach to.

Crystallisation begins, and the concentration of dissolved potassium nitrate decreases.

Eventually, the solution approaches its equilibrium saturation concentration at \(20^\circ\mathrm{C}\).

The important distinction is between the equilibrium state a solution should reach and a metastable state it can temporarily maintain.

Question 5

This final problem combines common-ion effects, dissolution stoichiometry, concentration, and the difference between molar solubility and total dissolved amount.

A \(1.00\ \mathrm{L}\) vessel contains pure water and excess solid calcium fluoride at \(25^\circ\mathrm{C}\).

The system reaches equilibrium.

Use the illustrative solubility product:

\[ K_{sp}=4.0\times10^{-11} \]

Now, \(1.00\ \mathrm{L}\) of \(1.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) sodium fluoride solution is added.

The final solution volume is \(2.00\ \mathrm{L}\), and excess calcium fluoride remains in the vessel throughout.

After equilibrium is re-established, a concentration measurement gives:

\[ [\mathrm{Ca^{2+}}] =1.99\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Two students make different predictions:

• Student A argues that the common-ion effect reduces solubility, so fewer moles of calcium fluoride must now be dissolved.
• Student B argues that the fluoride concentration is higher, so more calcium fluoride must have dissolved.

Determine whether either argument is correct.

Calculate:

a) The initial molar solubility and total moles of dissolved calcium fluoride.

b) The final fluoride concentration, accounting for both sodium fluoride and calcium fluoride.

c) The final total moles of dissolved calcium fluoride.

d) Whether the common-ion effect has actually been contradicted.

Solution 5

Student A is incorrect: the final solution contains more total moles of dissolved calcium fluoride, even though its equilibrium calcium-ion concentration is slightly lower.

Student B reaches the correct direction for the total dissolved amount in this particular scenario, but their reasoning is incomplete. Some of the fluoride ions came from sodium fluoride, not calcium fluoride.

The key is to distinguish concentration from total amount and identify where each ion originated.

Step 1: Calculate the initial molar solubility.

In pure water:

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

Let the initial molar solubility be \(s_0\).

Then:

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

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

Substitute into the \(K_{sp}\) expression:

\[ 4.0\times10^{-11}=s_0(2s_0)^2 \]

\[ 4.0\times10^{-11}=4s_0^3 \]

\[ s_0^3=1.0\times10^{-11} \]

Taking the cube root:

\[ \boxed{s_0=2.15\times10^{-4}\ \mathrm{M}} \]

Step 2: Calculate the initial total amount dissolved.

The initial solution volume is \(1.00\ \mathrm{L}\).

Using:

\[ n=cV \]

\[ n_0=(2.15\times10^{-4})(1.00) \]

\[ \boxed{n_0=2.15\times10^{-4}\ \mathrm{mol}} \]

This is the initial amount of calcium fluoride dissolved.

Step 3: Calculate the fluoride contribution from sodium fluoride.

The added sodium fluoride solution contains:

\[ n(\ce{NaF})=cV \]

\[ n(\ce{NaF})=(1.00\times10^{-4})(1.00) \]

\[ n(\ce{NaF})=1.00\times10^{-4}\ \mathrm{mol} \]

Sodium fluoride dissociates completely, so it supplies the same number of moles of fluoride ions.

The final solution volume is \(2.00\ \mathrm{L}\).

Therefore, the fluoride concentration contributed by sodium fluoride is:

\[ [\mathrm{F^-}]_{\ce{NaF}} =\frac{1.00\times10^{-4}}{2.00} \]

\[ \boxed{[\mathrm{F^-}]_{\ce{NaF}}=5.00\times10^{-5}\ \mathrm{M}} \]

Step 4: Calculate the total final fluoride concentration.

The measured final calcium concentration is:

\[ [\mathrm{Ca^{2+}}]=1.99\times10^{-4}\ \mathrm{M} \]

From the calcium fluoride dissolution equation, every calcium ion produced is accompanied by two fluoride ions.

Therefore, the fluoride contribution from calcium fluoride is:

\[ [\mathrm{F^-}]_{\ce{CaF2}} =2(1.99\times10^{-4}) \]

\[ =3.98\times10^{-4}\ \mathrm{M} \]

The total fluoride concentration is the sum of both contributions:

\[ [\mathrm{F^-}]_{\mathrm{total}} =5.00\times10^{-5} +3.98\times10^{-4} \]

\[ \boxed{[\mathrm{F^-}]_{\mathrm{total}} =4.48\times10^{-4}\ \mathrm{M}} \]

Step 5: Check whether the final concentrations are consistent with \(K_{sp}\).

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

Substitute:

\[ Q_{sp}=(1.99\times10^{-4}) (4.48\times10^{-4})^2 \]

\[ Q_{sp}=3.99\times10^{-11} \]

This agrees with the supplied:

\[ K_{sp}=4.0\times10^{-11} \]

The final concentrations are consistent with saturation, allowing for rounding.

Step 6: Calculate the final total amount of dissolved calcium fluoride.

The final volume is \(2.00\ \mathrm{L}\).

Using the calcium concentration:

\[ n_f=[\mathrm{Ca^{2+}}]V \]

\[ n_f=(1.99\times10^{-4})(2.00) \]

\[ \boxed{n_f=3.98\times10^{-4}\ \mathrm{mol}} \]

Step 7: Compare the initial and final amounts.

Initially:

\[ n_0=2.15\times10^{-4}\ \mathrm{mol} \]

Finally:

\[ n_f=3.98\times10^{-4}\ \mathrm{mol} \]

The ratio is:

\[ \frac{n_f}{n_0} = \frac{3.98\times10^{-4}} {2.15\times10^{-4}} \]

\[ \boxed{\frac{n_f}{n_0}=1.85} \]

The final solution contains approximately \(1.85\) times as many moles of dissolved calcium fluoride as the initial solution.

Why are both students’ arguments incomplete?

Student A has confused molar solubility with total dissolved amount.

The common-ion effect does reduce the equilibrium concentration of calcium ions compared with pure water:

\[ 1.99\times10^{-4} < 2.15\times10^{-4} \]

However, the final volume has doubled.

A lower concentration does not automatically mean fewer total moles.

Student B has overlooked the source of the fluoride ions.

The final fluoride concentration includes fluoride supplied by both sodium fluoride and calcium fluoride. You cannot attribute all the fluoride ions to calcium fluoride dissolving.

The correct approach is to separate the contributions, use the balanced dissolution equation, and account for the final solution volume.

The common-ion effect has not failed. It has lowered the molar solubility of calcium fluoride, while the increased solution volume has allowed a greater total amount of calcium fluoride to dissolve.

This distinction is particularly useful in more demanding equilibrium problems, where a change in concentration does not necessarily indicate a corresponding change in the total amount of substance.

09Where this leads next

Once you understand what saturation means at the particle level, the next step is predicting what happens when two ionic solutions are mixed.

You can calculate the ion product, \(Q_{sp}\), and compare it with \(K_{sp}\):

• \(Q_{sp}<K_{sp}\): the solution is unsaturated with respect to the solid.
• \(Q_{sp}=K_{sp}\): the solution is saturated.
• \(Q_{sp}>K_{sp}\): precipitation is thermodynamically favoured.

That comparison turns the particle-level idea of saturation into a practical method for predicting precipitation reactions, including situations where several different ions are present.