Why Diluting a Mixture Can Prevent a Precipitate from Forming

Two clear solutions can turn cloudy as soon as they are mixed. Calcium nitrate supplies calcium ions, \(\mathrm{Ca}^{2+}\), and sodium fluoride supplies fluoride ions, \(\mathrm{F}^{-}\). These ions can combine to form solid calcium fluoride, \(\ce{CaF2}\).

Now imagine adding water to the mixture before any crystals appear. You haven’t removed any calcium or fluoride ions, yet the mixture might remain completely clear.

Why?

Because precipitation depends on the concentrations of the dissolved ions, not simply whether those ions are present. Dilution spreads the same dissolved particles through a larger volume, reducing their concentrations and lowering the ion product, \(Q_{sp}\).

If dilution reduces \(Q_{sp}\) below the solubility product, \(K_{sp}\), the solution becomes unsaturated. Forming a precipitate is no longer thermodynamically favoured.

If a precipitate has already formed, dilution can also cause some or all of it to dissolve again.

01The short answer: dilution lowers the ion product

To understand exactly why dilution works, you need to distinguish two quantities:

• \(Q_{sp}\), the ion product, describes the actual concentrations of the dissolved ions at a particular moment.
• \(K_{sp}\), the solubility product, is the equilibrium constant for a sparingly soluble ionic compound at a particular temperature.

The expressions for \(Q_{sp}\) and \(K_{sp}\) have the same mathematical form. The difference is that \(Q_{sp}\) uses the current ion concentrations, while \(K_{sp}\) describes the equilibrium condition.

Compare them to predict what happens.

The key point is that dilution changes \(Q_{sp}\), but does not change \(K_{sp}\), provided the temperature stays constant.

This gives dilution a useful effect: it can move a supersaturated solution back below its precipitation threshold.

In standard HSC calculations, we assume precipitation occurs when the solution becomes supersaturated. In a real experiment, crystal formation can be delayed by the time needed for nucleation.

02Why the dissolution equation controls how much dilution matters

Consider calcium fluoride dissolving in water:

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

Read the equation carefully.

One formula unit of calcium fluoride produces:

• One calcium ion.
• Two fluoride ions.

This ratio determines the expression for its solubility product:

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

The ion product has exactly the same form:

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

Notice that fluoride concentration is squared.

This happens because two fluoride ions are released for every calcium ion. The exponent comes from the balanced dissolution equation, not from the charge on the ion.

What happens when you double the volume?

Suppose you have a solution containing calcium and fluoride ions. You add enough water to double its original volume.

The number of dissolved moles has not changed. However, those moles now occupy twice as much volume.

Remember:

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

where:

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

If the volume doubles while the number of moles stays constant, both concentrations halve.

Therefore:

\[ [\mathrm{Ca}^{2+}]_{\text{new}} =\frac{[\mathrm{Ca}^{2+}]_{\text{old}}}{2} \]

\[ [\mathrm{F}^{-}]_{\text{new}} =\frac{[\mathrm{F}^{-}]_{\text{old}}}{2} \]

Now substitute these changes into the calcium fluoride ion product:

\[ Q_{sp,\text{new}} = \frac{[\mathrm{Ca}^{2+}]}{2} \left(\frac{[\mathrm{F}^{-}]}{2}\right)^2 \]

Simplifying:

\[ Q_{sp,\text{new}} = \frac{[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2}{8} \]

Therefore:

\[ \boxed{Q_{sp,\text{new}}=\frac{Q_{sp,\text{old}}}{8}} \]

Doubling the volume reduces the calcium fluoride ion product by a factor of eight.

That’s a much bigger change than simply halving the concentration of each ion.

Compare different dissolution ratios

The effect of dilution depends on the stoichiometry of the salt.

For a general dissolution that produces \(m\) ions of one type and \(n\) ions of another, the total power in the ion product is \(m+n\).

If every dissolved ion concentration is reduced by the same dilution factor, \(D\), then:

\[ \boxed{Q_{sp,\text{new}}= \frac{Q_{sp,\text{old}}}{D^{m+n}}} \]

Here, \(D\) is the final volume divided by the initial volume.

This relationship assumes that all relevant dissolved ions are diluted by the same factor and that no precipitation or other reaction changes their amounts during the dilution.

03Calculate concentrations before comparing Qsp

There is an important calculation step that comes before dilution.

When two solutions are mixed, you cannot simply add their original concentrations. You must account for the number of moles contributed by each solution and the final combined volume.

Use:

\[ n=cV \]

Then calculate:

\[ [\text{ion}]_{\text{mixed}} = \frac{n_{\text{total ion}}}{V_{\text{total}}} \]

If the mixture is subsequently diluted with water:

\[ c_{\text{final}} = c_{\text{initial}} \frac{V_{\text{initial}}}{V_{\text{final}}} \]

Always convert volumes into litres when calculating moles using concentration in \(\mathrm{mol\,L^{-1}}\).

04Why a small amount of dilution may not be enough

There’s a tempting misconception here:

If dilution lowers the ion product, any amount of added water should stop precipitation.

Not quite.

Dilution always reduces the ion product under the assumptions we’ve used, but the reduction might not be large enough to bring it below \(K_{sp}\).

Consider a calcium fluoride mixture with:

\[ [\mathrm{Ca}^{2+}]=1.0\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{F}^{-}]=2.0\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Using the same illustrative \(K_{sp}\):

\[ Q_{sp} =(1.0\times10^{-3})(2.0\times10^{-3})^2 \]

\[ =4.0\times10^{-9} \]

This is 100 times greater than \(K_{sp}\).

Now double the volume.

Because calcium fluoride has a 1:2 dissolution ratio, the ion product decreases by a factor of eight:

\[ Q_{sp,\text{new}} = \frac{4.0\times10^{-9}}{8} \]

\[ =5.0\times10^{-10} \]

But:

\[ 5.0\times10^{-10}>4.0\times10^{-11} \]

The mixture is still supersaturated.

It would need a greater dilution.

Finding the minimum dilution factor

For calcium fluoride:

\[ Q_{sp,\text{new}} = \frac{Q_{sp,\text{initial}}}{D^3} \]

At the exact point of saturation:

\[ K_{sp}=\frac{Q_{sp,\text{initial}}}{D^3} \]

Rearrange:

\[ D^3=\frac{Q_{sp,\text{initial}}}{K_{sp}} \]

Therefore:

\[ D= \sqrt[3]{\frac{Q_{sp,\text{initial}}}{K_{sp}}} \]

Substitute:

\[ D=\sqrt[3]{\frac{4.0\times10^{-9}}{4.0\times10^{-11}}} \]

\[ D=\sqrt[3]{100} \]

\[ D\approx4.64 \]

The final volume must be approximately 4.64 times the initial volume to reach saturation. A slightly greater dilution is needed to make the solution unsaturated.

For example, a fivefold dilution gives:

\[ Q_{sp,\text{new}} = \frac{4.0\times10^{-9}}{5^3} \]

\[ =3.2\times10^{-11} \]

This is now below \(K_{sp}\).

The important distinction is that dilution reduces the ion product, but you must still check whether it reduces it enough.

05What changes when a common ion is present?

A common ion is an ion that is already involved in a dissolution equilibrium and is supplied by another dissolved substance.

For calcium fluoride:

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

Adding sodium fluoride introduces more fluoride ions:

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

The extra fluoride is a common ion.

It increases the fluoride concentration, shifting the calcium fluoride dissolution equilibrium towards the solid. This reduces the amount of calcium fluoride that can remain dissolved at equilibrium.

This is known as the common-ion effect.

However, there is an important distinction when you subsequently dilute the mixture.

A common ion is not automatically held at a constant concentration.

If sodium fluoride is dissolved in the same mixture, adding water dilutes its fluoride ions as well.

For example, suppose the mixture contains fluoride from both sodium fluoride and potassium fluoride. When you double the total volume, both sources are diluted. You must calculate the total fluoride concentration before and after dilution.

There are three useful situations to distinguish:

The third situation is different from simply adding water to a beaker. If a reservoir continually supplies fluoride, you cannot assume its concentration halves.

Similarly, if excess solid containing a common ion remains in the mixture, it may dissolve and replenish that ion as water is added. You would need to account for the new equilibrium rather than applying the simple dilution relationship to every ion.

One more point: adding a common ion changes the equilibrium concentrations and solubility, but it does not change \(K_{sp}\) at a constant temperature.

For a fuller explanation of this equilibrium shift, see Why Adding a Common Ion Can Reduce Solubility.

06A reliable method for HSC questions

When a question asks whether dilution prevents a precipitate, work through these steps.

1. Write the balanced dissolution equation. Identify how many ions of each type are produced.

2. Write the \(Q_{sp}\) expression. Use the coefficients in the dissolution equation as the powers.

3. Calculate the initial ion concentrations. If solutions have been mixed, calculate moles from every source before dividing by the total volume.

4. Apply the dilution. Calculate the new concentrations using the initial and final volumes. Include every dissolved source of a common ion.

5. Recalculate \(Q_{sp}\). Do not assume every salt has the same dilution factor for its ion product.

6. Compare with \(K_{sp}\). State whether the final mixture is unsaturated, saturated, or supersaturated, and explain what happens to any existing solid.

If you’re unsure how to calculate the concentrations after combining solutions of different volumes, see How to Calculate Qsp After Mixing Solutions with Different Volumes.

07Questions and solutions

The following questions move from straightforward dilution calculations to situations where common ions, existing solids, and the order of operations affect the final result.

Question 1

An unprecipitated mixture at \(25^\circ\mathrm{C}\) contains:

\[ [\mathrm{Ag}^{+}]=3.0\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{Cl}^{-}]=3.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

The mixture is diluted from \(100.0\ \mathrm{mL}\) to \(800.0\ \mathrm{mL}\) before any crystals appear.

Given:

\[ K_{sp}(\ce{AgCl})=1.8\times10^{-10} \]

Calculate the final ion product and determine whether silver chloride is expected to precipitate.

Solution 1

Silver chloride is not expected to precipitate after dilution. The final ion product is approximately \(1.4\times10^{-10}\), which is below \(K_{sp}\).

Step 1: Calculate the dilution factor.

\[ D=\frac{V_{\text{final}}}{V_{\text{initial}}} \]

\[ D=\frac{800.0}{100.0}=8 \]

Step 2: Calculate the final ion concentrations.

\[ [\mathrm{Ag}^{+}]_{\text{final}} = \frac{3.0\times10^{-5}}{8} \]

\[ =3.75\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Similarly:

\[ [\mathrm{Cl}^{-}]_{\text{final}} = \frac{3.0\times10^{-4}}{8} \]

\[ =3.75\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Step 3: Calculate the final ion product.

For silver chloride:

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

Substitute:

\[ \begin{aligned} Q_{sp} &=(3.75\times10^{-6})(3.75\times10^{-5})\\ &=1.40625\times10^{-10} \end{aligned} \]

Step 4: Compare with \(K_{sp}\).

\[ 1.40625\times10^{-10} < 1.8\times10^{-10} \]

The mixture is unsaturated.

The eightfold increase in volume reduces each concentration by eight, so the 1:1 ion product decreases by \(8^2=64\).

This is why the final solution no longer favours silver chloride precipitation.

Question 2

An unprecipitated mixture contains:

\[ [\mathrm{Ca}^{2+}]=3.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{F}^{-}]=8.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

The fluoride ions are supplied by dissolved sodium fluoride.

At \(25^\circ\mathrm{C}\), use:

\[ K_{sp}(\ce{CaF2})=4.0\times10^{-11} \]

Consider two different situations.

Situation A: The mixture is diluted with water until its volume doubles.

Situation B: The mixture is diluted until its volume doubles, but an external fluoride reservoir maintains the free fluoride concentration at \(8.0\times10^{-4}\ \mathrm{mol\,L^{-1}}\).

For each situation:

a. Calculate the final ion product.

b. Determine whether calcium fluoride precipitation is favoured.

c. Explain why the two situations give different results.

Solution 2

Situation A becomes unsaturated, while Situation B remains supersaturated. The difference is whether the fluoride concentration is allowed to decrease during dilution.

Step 1: Calculate the initial ion product.

For calcium fluoride:

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

\[ \begin{aligned} Q_{sp} &=(3.0\times10^{-4})(8.0\times10^{-4})^2\\ &=(3.0\times10^{-4})(6.4\times10^{-7})\\ &=1.92\times10^{-10} \end{aligned} \]

The initial mixture is supersaturated.

Step 2: Calculate Situation A.

Both dissolved ion concentrations halve when the volume doubles.

\[ [\mathrm{Ca}^{2+}]_{\text{final}} = 1.5\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{F}^{-}]_{\text{final}} = 4.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Calculate:

\[ \begin{aligned} Q_{sp} &=(1.5\times10^{-4})(4.0\times10^{-4})^2\\ &=(1.5\times10^{-4})(1.6\times10^{-7})\\ &=2.4\times10^{-11} \end{aligned} \]

Compare:

\[ 2.4\times10^{-11}<4.0\times10^{-11} \]

Situation A is unsaturated. Precipitation is not favoured.

Step 3: Calculate Situation B.

The calcium concentration still halves:

\[ [\mathrm{Ca}^{2+}]_{\text{final}} = 1.5\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

However, the external reservoir maintains:

\[ [\mathrm{F}^{-}]_{\text{final}} = 8.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \begin{aligned} Q_{sp} &=(1.5\times10^{-4})(8.0\times10^{-4})^2\\ &=(1.5\times10^{-4})(6.4\times10^{-7})\\ &=9.6\times10^{-11} \end{aligned} \]

Compare:

\[ 9.6\times10^{-11}>4.0\times10^{-11} \]

Situation B remains supersaturated, so precipitation is thermodynamically favoured.

Step 4: Explain the difference.

In ordinary dilution, both calcium and fluoride concentrations decrease. Since fluoride is squared in the ion product, \(Q_{sp}\) decreases by a factor of eight.

In Situation B, fluoride remains constant. Only the calcium concentration halves, so \(Q_{sp}\) decreases by a factor of two.

The common-ion concentration must be treated according to the actual conditions. Calling an ion a common ion does not automatically mean its concentration stays constant.

Question 3

Two solutions are prepared:

• \(50.0\ \mathrm{mL}\) of \(2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}\) calcium nitrate;
• \(50.0\ \mathrm{mL}\) of \(2.0\times10^{-3}\ \mathrm{mol\,L^{-1}}\) sodium fluoride.

Use:

\[ K_{sp}(\ce{CaF2})=4.0\times10^{-11} \]

Assume complete dissociation and additive volumes.

A student compares two procedures.

Procedure 1: Mix the two original solutions, then immediately add \(200\ \mathrm{mL}\) of water.

Procedure 2: Add \(100\ \mathrm{mL}\) of water to each original solution separately, then combine the diluted solutions.

Determine:

a. Whether the mixture in Procedure 1 is initially supersaturated before water is added.

b. The final calcium and fluoride concentrations for both procedures.

c. Whether either procedure leaves a calcium fluoride precipitate at final equilibrium.

d. Whether the order of dilution and mixing can change what is temporarily observed.

Solution 3

Neither procedure leaves calcium fluoride precipitate at final equilibrium, and both produce the same final dissolved-ion concentrations. However, Procedure 1 initially creates a supersaturated mixture, so temporary precipitation is possible.

Step 1: Calculate the total moles of each ion.

For calcium:

\[ \begin{aligned} n(\mathrm{Ca}^{2+}) &=(2.0\times10^{-4})(0.0500)\\ &=1.0\times10^{-5}\ \mathrm{mol} \end{aligned} \]

For fluoride:

\[ \begin{aligned} n(\mathrm{F}^{-}) &=(2.0\times10^{-3})(0.0500)\\ &=1.0\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Both procedures contain the same total moles of calcium and fluoride.

Step 2: Check the mixture before dilution in Procedure 1.

The initial combined volume is:

\[ V=50.0+50.0=100.0\ \mathrm{mL} \]

Therefore:

\[ [\mathrm{Ca}^{2+}] = \frac{1.0\times10^{-5}}{0.1000} = 1.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{F}^{-}] = \frac{1.0\times10^{-4}}{0.1000} = 1.0\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Calculate the ion product:

\[ \begin{aligned} Q_{sp} &=(1.0\times10^{-4})(1.0\times10^{-3})^2\\ &=1.0\times10^{-10} \end{aligned} \]

Since:

\[ 1.0\times10^{-10}>4.0\times10^{-11} \]

the initial mixture is supersaturated.

Precipitation is thermodynamically favoured before the additional water is added.

Step 3: Calculate the final volume for both procedures.

For Procedure 1:

\[ V_{\text{final}}=100.0+200.0 \]

\[ =300.0\ \mathrm{mL}=0.3000\ \mathrm{L} \]

For Procedure 2, each solution has a final volume of:

\[ 50.0+100.0=150.0\ \mathrm{mL} \]

Combining them gives:

\[ V_{\text{final}}=300.0\ \mathrm{mL} \]

Both procedures therefore have the same final volume.

Step 4: Calculate the final concentrations.

For calcium:

\[ [\mathrm{Ca}^{2+}] = \frac{1.0\times10^{-5}}{0.3000} \]

\[ =3.33\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

For fluoride:

\[ [\mathrm{F}^{-}] = \frac{1.0\times10^{-4}}{0.3000} \]

\[ =3.33\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

These concentrations are identical for both procedures.

Step 5: Calculate the final ion product.

\[ \begin{aligned} Q_{sp} &=(3.33\times10^{-5})(3.33\times10^{-4})^2\\ &\approx3.70\times10^{-12} \end{aligned} \]

Compare:

\[ 3.70\times10^{-12}<4.0\times10^{-11} \]

Both final mixtures are unsaturated.

If calcium fluoride formed during Procedure 1 and remained in the vessel, it would dissolve because the final solution can accommodate all the original ions without exceeding \(K_{sp}\).

Step 6: Explain the effect of the order of operations.

The final equilibrium depends on the total moles of dissolved ions and the final volume, not simply the order in which water was added.

However, the temporary observations can differ.

In Procedure 1, the concentrated solutions are initially mixed at a supersaturated composition. Crystals may begin forming before dilution.

In Procedure 2, the solutions are diluted separately, so their concentrations are lower before they meet. The combined mixture is unsaturated.

Under the standard HSC equilibrium approximation, both procedures end with the same solution if no ions are physically removed. In an actual experiment, temporary cloudiness and the time needed for crystals to dissolve can affect what you observe.

Question 4

A \(100.0\ \mathrm{mL}\) saturated silver chloride solution at \(25^\circ\mathrm{C}\) contains:

\[ [\mathrm{Ag}^{+}] = 1.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{Cl}^{-}] = 1.8\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

The solution is in contact with \(1.0\times10^{-6}\ \mathrm{mol}\) of solid silver chloride.

Given:

\[ K_{sp}(\ce{AgCl})=1.8\times10^{-10} \]

Water is added until the total solution volume reaches \(1.00\ \mathrm{L}\).

Determine whether all the silver chloride solid dissolves, some remains, or none dissolves.

Assume no ions are lost from the vessel.

Solution 4

All the silver chloride solid dissolves. If all the solid dissolves, the resulting ion product is still below \(K_{sp}\), so the solution does not become saturated again.

Step 1: Confirm the initial solution is saturated.

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

\[ \begin{aligned} Q_{sp} &=(1.0\times10^{-4})(1.8\times10^{-6})\\ &=1.8\times10^{-10} \end{aligned} \]

This equals the supplied \(K_{sp}\), so the solution is initially saturated.

Step 2: Calculate the total moles of silver if all solid dissolves.

The dissolved silver ions initially contribute:

\[ n(\mathrm{Ag}^{+}) =(1.0\times10^{-4})(0.1000) \]

\[ =1.0\times10^{-5}\ \mathrm{mol} \]

The solid contributes another:

\[ 1.0\times10^{-6}\ \mathrm{mol} \]

Therefore:

\[ n(\mathrm{Ag}^{+})_{\text{total}} = 1.1\times10^{-5}\ \mathrm{mol} \]

Step 3: Calculate the total moles of chloride if all solid dissolves.

Initially dissolved chloride contributes:

\[ n(\mathrm{Cl}^{-}) =(1.8\times10^{-6})(0.1000) \]

\[ =1.8\times10^{-7}\ \mathrm{mol} \]

The solid contributes another:

\[ 1.0\times10^{-6}\ \mathrm{mol} \]

Therefore:

\[ n(\mathrm{Cl}^{-})_{\text{total}} = 1.18\times10^{-6}\ \mathrm{mol} \]

Step 4: Calculate the concentrations if all solid dissolves.

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

Therefore:

\[ [\mathrm{Ag}^{+}]_{\text{final}} = \frac{1.1\times10^{-5}}{1.00} \]

\[ =1.1\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Similarly:

\[ [\mathrm{Cl}^{-}]_{\text{final}} = \frac{1.18\times10^{-6}}{1.00} \]

\[ =1.18\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Step 5: Test the all-dissolved condition.

\[ Q_{sp} = (1.1\times10^{-5})(1.18\times10^{-6}) \]

\[ =1.298\times10^{-11} \]

Compare:

\[ 1.298\times10^{-11} < 1.8\times10^{-10} \]

The hypothetical all-dissolved solution is still unsaturated.

Therefore, all the solid can dissolve without reaching the saturation threshold.

The important lesson is that you cannot simply halve the original ion concentrations when a solid is already present. You must account for the extra ions released as the solid dissolves.

Question 5

At \(25^\circ\mathrm{C}\), two identical batches are prepared using the following quantities:

• \(50.0\ \mathrm{mL}\) of \(2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}\) calcium nitrate;
• \(25.0\ \mathrm{mL}\) of \(2.0\times10^{-3}\ \mathrm{mol\,L^{-1}}\) sodium fluoride;
• \(25.0\ \mathrm{mL}\) of \(2.0\times10^{-3}\ \mathrm{mol\,L^{-1}}\) potassium fluoride.

Use:

\[ K_{sp}(\ce{CaF2})=4.0\times10^{-11} \]

Assume complete dissociation, additive volumes, no complex formation, and that both batches are allowed to reach equilibrium before the following procedures.

Batch A: Add \(100.0\ \mathrm{mL}\) of water without removing any solid.

Batch B: Filter off all calcium fluoride precipitate, then add \(100.0\ \mathrm{mL}\) of water to the filtrate.

Answer the following:

a. Calculate the initial ion product and establish whether calcium fluoride precipitation occurs.

b. Determine whether all the precipitate dissolves in Batch A after dilution.

c. Determine whether calcium fluoride precipitates again in Batch B.

d. Are the final dissolved calcium and fluoride concentrations identical in the two batches? Explain your reasoning.

Assume filtration removes all the solid without removing dissolved ions.

Solution 5

Neither batch contains calcium fluoride solid at the end, but their final dissolved-ion concentrations are different. Batch A dissolves all its precipitate, whereas Batch B loses some calcium and fluoride ions during filtration.

This distinction is easy to miss because both final mixtures can appear clear.

Step 1: Calculate the initial moles of calcium ions.

\[ \begin{aligned} n(\mathrm{Ca}^{2+}) &=(2.0\times10^{-4})(0.0500)\\ &=1.0\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Step 2: Calculate the total moles of fluoride ions.

From sodium fluoride:

\[ \begin{aligned} n(\mathrm{F}^{-})_{\ce{NaF}} &=(2.0\times10^{-3})(0.0250)\\ &=5.0\times10^{-5}\ \mathrm{mol} \end{aligned} \]

From potassium fluoride:

\[ \begin{aligned} n(\mathrm{F}^{-})_{\ce{KF}} &=(2.0\times10^{-3})(0.0250)\\ &=5.0\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Total fluoride:

\[ n(\mathrm{F}^{-})_{\text{total}} = 1.0\times10^{-4}\ \mathrm{mol} \]

Both fluoride sources must be included.

Step 3: Calculate the initial concentrations and ion product.

The initial volume is:

\[ V=50.0+25.0+25.0 \]

\[ =100.0\ \mathrm{mL}=0.1000\ \mathrm{L} \]

Calcium concentration:

\[ [\mathrm{Ca}^{2+}] = \frac{1.0\times10^{-5}}{0.1000} \]

\[ =1.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Fluoride concentration:

\[ [\mathrm{F}^{-}] = \frac{1.0\times10^{-4}}{0.1000} \]

\[ =1.0\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Calculate:

\[ \begin{aligned} Q_{sp} &=(1.0\times10^{-4})(1.0\times10^{-3})^2\\ &=1.0\times10^{-10} \end{aligned} \]

Since:

\[ 1.0\times10^{-10}>4.0\times10^{-11} \]

the initial mixture is supersaturated.

Calcium fluoride precipitation is thermodynamically favoured, and solid forms as the mixture reaches equilibrium.

Step 4: Consider Batch A.

In Batch A, the solid is retained while \(100.0\ \mathrm{mL}\) of water is added.

The final volume is:

\[ V_{\text{final}}=200.0\ \mathrm{mL} \]

Test what happens if all the original ions are dissolved in this final volume.

The calcium concentration would be:

\[ [\mathrm{Ca}^{2+}]_{\text{final}} = \frac{1.0\times10^{-5}}{0.2000} \]

\[ =5.0\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

The fluoride concentration would be:

\[ [\mathrm{F}^{-}]_{\text{final}} = \frac{1.0\times10^{-4}}{0.2000} \]

\[ =5.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Calculate the ion product:

\[ \begin{aligned} Q_{sp,\text{final}} &=(5.0\times10^{-5})(5.0\times10^{-4})^2\\ &=1.25\times10^{-11} \end{aligned} \]

Compare:

\[ 1.25\times10^{-11}<4.0\times10^{-11} \]

Even with every original calcium and fluoride ion dissolved, the mixture remains unsaturated.

Therefore, all the calcium fluoride precipitate dissolves in Batch A.

Step 5: Consider Batch B.

In Batch B, the precipitate is allowed to form and the mixture reaches equilibrium before filtration.

Because solid calcium fluoride is present at equilibrium:

\[ Q_{sp}=K_{sp} \]

The filter then physically removes the calcium fluoride.

This is a crucial change. The calcium and fluoride ions contained in the filtered solid are no longer available to dissolve back into the solution.

The filtrate initially remains saturated, but its volume is now doubled by adding \(100.0\ \mathrm{mL}\) of water.

Both dissolved ion concentrations halve.

For calcium fluoride, the ion product therefore decreases by a factor of:

\[ 2^3=8 \]

Thus:

\[ Q_{sp,\text{final}} = \frac{K_{sp}}{8} \]

\[ =\frac{4.0\times10^{-11}}{8} \]

\[ =5.0\times10^{-12} \]

Since:

\[ 5.0\times10^{-12}<4.0\times10^{-11} \]

the final filtrate is unsaturated.

No calcium fluoride precipitates again.

Step 6: Compare the final compositions.

Both batches finish without visible calcium fluoride solid.

However, their dissolved-ion concentrations are not equal.

In Batch A, the original precipitate remained in the vessel and dissolved again. All the original calcium and fluoride ions are therefore present in the final solution.

In Batch B, filtration physically removed some calcium and fluoride ions as solid calcium fluoride. Adding water cannot replace those removed ions.

Consequently, Batch B contains fewer dissolved calcium and fluoride ions than Batch A.

This is the important distinction: dilution can reverse precipitation, but it cannot recover ions that have already been physically removed from the mixture.

The same final appearance does not necessarily mean the same chemical composition.

The next useful step is to consider mixtures containing more than one possible precipitate. Once you can calculate how dilution changes each ion product, you can determine which solid remains favoured and which one dissolves first. That is the foundation of selective precipitation.