How to Calculate Ksp from Calcium Fluoride Solubility

Imagine dissolving calcium fluoride, \(\ce{CaF2}\), in water. You measure its solubility and want to calculate its solubility product, \(K_{sp}\). Should you simply square the solubility to get \(K_{sp}\)? No. Calcium fluoride releases two fluoride ions for every calcium ion, so the stoichiometric ratio matters.

For calcium fluoride dissolving in pure water, the key result is:

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

Here, \(s\) is the molar solubility of \(\ce{CaF2}\), measured in \(\mathrm{mol\,L^{-1}}\).

If the solution already contains calcium ions or fluoride ions from another substance, however, this shortcut no longer works. You must account for those ions separately.

Let’s build the calculation from the dissolution equation, work through both situations, and then tackle some more challenging HSC-style questions.

All numerical values in the worked examples and practice questions are illustrative teaching data.

01Why calcium fluoride gives \(K_{sp}=4s^3\)

Start with the dissolution equation:

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

One formula unit of calcium fluoride produces:

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

This 1:2 ratio is the most important part of the calculation.

Introducing molar solubility

Molar solubility, represented by \(s\), is the number of moles of a substance that dissolve per litre of saturated solution.

Suppose \(s\) mol of calcium fluoride dissolves per litre of pure water.

Using the dissolution equation:

Why is fluoride \(2s\)?

Because every mole of \(\ce{CaF2}\) that dissolves produces two moles of fluoride ions.

This is a stoichiometric relationship, not an assumption about how much of the solid dissolves.

Writing the solubility product expression

The solubility product expression includes the concentrations of the dissolved ions:

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

Substitute the equilibrium concentrations:

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

Simplifying:

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

Notice two things:

• The fluoride concentration is squared because its coefficient in the dissolution equation is 2.
• The solid \(\ce{CaF2}\) does not appear in the expression because the activity of a pure solid is constant.

The cube in \(4s^3\) comes from multiplying \(s\) by \(s^2\). It is not an arbitrary rule to memorise.

02Converting solubility into molar solubility

Before using \(K_{sp}=4s^3\), check the units of the solubility you have been given.

This is where many otherwise correct calculations go wrong.

Solubility might be expressed as:

• \(\mathrm{g\,L^{-1}}\)
• \(\mathrm{mg\,L^{-1}}\)
• \(\mathrm{mol\,L^{-1}}\)
• A mass of solid dissolved in a particular volume of solution.

However, the concentration terms in our calculation must be in \(\mathrm{mol\,L^{-1}}\).

Finding the molar mass of calcium fluoride

Use the periodic table:

\[ \begin{aligned} M(\ce{CaF2})&=40.08+2(19.00)\\ &=78.08\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Therefore, the conversion from mass concentration to molar solubility is:

\[ \boxed{s=\frac{\text{mass concentration}}{78.08}} \]

The mass concentration must be in \(\mathrm{g\,L^{-1}}\).

For other types of data, use these conversions:

Be careful with solubility expressed as grams per 100 g of water. This is not automatically a mass concentration in grams per litre. You need sufficient information about the solution volume or density to convert it accurately.

03Where does the solubility measurement come from?

A solubility value is an experimental measurement, not simply the amount of solid initially placed in water.

To understand why, consider how a saturated calcium fluoride solution could be prepared and analysed.

Setting up the experiment

Place excess solid calcium fluoride into a flask containing deionised water. The flask sits in a constant-temperature water bath, with a stirrer to help the solution reach equilibrium.

Some of the solid dissolves, releasing calcium and fluoride ions.

Eventually, the solution reaches dynamic equilibrium. Dissolution and precipitation continue, but at equal rates, so the concentrations remain constant.

What should happen if you add another spoonful of calcium fluoride?

The dissolved concentration should not increase, provided the solution is already saturated and the temperature remains constant. The additional solid simply remains undissolved.

Separating dissolved material from undissolved solid

The next step is to separate the saturated solution from the remaining solid.

A sample is withdrawn and filtered at the same temperature as the equilibrium vessel. This matters because changing the temperature can change the solubility.

There is also a problem with simply taking a sample from a cloudy mixture: tiny calcium fluoride particles could enter the sample and later dissolve during analysis. The measured amount would then include material that was not dissolved at equilibrium.

Filtering the sample before analysis addresses this problem.

A suitable calibrated analytical method can then determine the dissolved calcium or fluoride concentration.

Connecting the measurement to Ksp

Suppose an illustrative analysis of a saturated solution gives:

\[ [\mathrm{Ca}^{2+}]=2.14\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Assuming the solution contains no other source of calcium ions:

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

The fluoride concentration follows from the dissolution stoichiometry:

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

This gives the same \(K_{sp}\) calculation as before.

We can also estimate the amount of dissolved calcium fluoride in a particular sample.

For a \(25.0\ \mathrm{mL}\) aliquot:

\[ \begin{aligned} n&=cV\\ &=(2.14\times10^{-4})(0.0250)\\ &=5.35\times10^{-6}\ \mathrm{mol} \end{aligned} \]

The corresponding mass of dissolved calcium fluoride is:

\[ \begin{aligned} m&=nM\\ &=(5.35\times10^{-6})(78.08)\\ &=4.18\times10^{-4}\ \mathrm{g} \end{aligned} \]

That is approximately \(0.418\ \mathrm{mg}\) in the \(25.0\ \mathrm{mL}\) sample.

This illustrates why analytical measurements are useful: even a relatively small mass of dissolved calcium fluoride can be converted into a concentration and then into \(K_{sp}\).

04What changes when a common ion is present?

So far, we have assumed that calcium fluoride dissolves in pure water.

Now imagine dissolving it in a solution that already contains sodium fluoride, \(\ce{NaF}\).

Sodium fluoride dissociates:

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

This introduces fluoride ions before any calcium fluoride dissolves.

Does the solubility product change?

No. At a fixed temperature, \(K_{sp}\) remains constant. However, the solubility of calcium fluoride decreases because the solution already contains one of its products.

This is called the common-ion effect.

Case 1: Calcium fluoride in pure water

Let the molar solubility be \(s\).

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

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

Therefore:

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

Case 2: Calcium fluoride in sodium fluoride solution

Suppose the initial sodium fluoride concentration is \(c\).

The sodium fluoride has already supplied \(c\) mol L\(^{-1}\) of fluoride ions.

When calcium fluoride dissolves, it adds another \(2s\) mol L\(^{-1}\) of fluoride ions.

Therefore:

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

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

The new expression becomes:

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

This is different from \(4s^3\) because the initial fluoride concentration cannot be ignored.

What if calcium ions are already present?

The same reasoning applies if the solution contains calcium nitrate, \(\ce{Ca(NO3)2}\).

Calcium nitrate dissociates:

\[ \ce{Ca(NO3)2(aq) -> Ca^2+(aq) + 2NO3^-(aq)} \]

If its initial calcium concentration is \(c_0\), the equilibrium concentrations become:

\[ [\mathrm{Ca}^{2+}]=c_0+s \]

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

Therefore:

\[ \boxed{K_{sp}=(c_0+s)(2s)^2} \]

Notice that the calcium concentration is no longer simply \(s\). It includes the calcium ions already present before dissolution.

General expression for common-ion situations

If a solution initially contains both calcium ions and fluoride ions, define:

• \(c_0\) as the initial calcium ion concentration.
• \(f_0\) as the initial fluoride ion concentration.
• \(s\) as the molar solubility of calcium fluoride.

The equilibrium concentrations are:

\[ [\mathrm{Ca}^{2+}]=c_0+s \]

\[ [\mathrm{F}^{-}]=f_0+2s \]

So:

\[ \boxed{K_{sp}=(c_0+s)(f_0+2s)^2} \]

This general expression is useful because it handles all three situations.

One important distinction: sodium ions and nitrate ions do not appear in the \(K_{sp}\) expression. They are spectator ions. The expression includes only the dissolved ions that come from calcium fluoride.

05A reliable decision process for calculating Ksp

Before starting any calcium fluoride question, work through these steps.

Step 1: Write the balanced dissolution equation.

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

Step 2: Convert the given solubility into molar solubility if necessary.

For calcium fluoride:

\[ M=78.08\ \mathrm{g\,mol^{-1}} \]

Step 3: Check the initial solution.

Ask whether it already contains calcium ions or fluoride ions.

Step 4: Calculate the equilibrium concentrations.

Remember the 1:2 ratio between calcium and fluoride.

Step 5: Substitute into the correct expression.

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

Common mistakes to avoid

The calculations above also assume that there are no significant side reactions, such as fluoride reacting with hydrogen ions, and that molar concentrations provide a suitable approximation to ion activities. These are the usual simplifying assumptions for the examples here.

06Questions and solutions

The following questions progress from straightforward calculations to situations where you need to combine solubility, stoichiometry, experimental measurements, and equilibrium reasoning.

Question 1

At \(25.0^\circ\mathrm{C}\), \(0.00400\ \mathrm{g}\) of calcium fluoride dissolves to form \(250.0\ \mathrm{mL}\) of saturated solution in pure water.

Calculate the solubility product, \(K_{sp}\).

Assume there are no common ions present.

Solution 1

The solubility product is \(3.44\times10^{-11}\).

The key is to convert the dissolved mass into molar solubility before using the \(K_{sp}\) expression.

Step 1: Calculate the mass concentration.

Convert the volume into litres:

\[ 250.0\ \mathrm{mL}=0.2500\ \mathrm{L} \]

Therefore:

\[ \begin{aligned} \text{Mass concentration} &=\frac{0.00400}{0.2500}\\ &=0.0160\ \mathrm{g\,L^{-1}} \end{aligned} \]

Step 2: Convert to molar solubility.

Using the molar mass of calcium fluoride:

\[ M=78.08\ \mathrm{g\,mol^{-1}} \]

\[ \begin{aligned} s&=\frac{0.0160}{78.08}\\ &=2.05\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 3: Calculate the equilibrium ion concentrations.

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

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

Step 4: Calculate Ksp.

\[ \begin{aligned} K_{sp} &=s(2s)^2\\ &=4s^3\\ &=4(2.05\times10^{-4})^3\\ &=3.44\times10^{-11} \end{aligned} \]

Therefore:

\[ \boxed{K_{sp}=3.44\times10^{-11}} \]

The fact that the original mass was given in grams does not change the dissolution stoichiometry. It simply adds a conversion step before the equilibrium calculation.

Question 2

A saturated solution of calcium fluoride is prepared in \(0.0100\ \mathrm{mol\,L^{-1}}\) sodium fluoride solution at \(25.0^\circ\mathrm{C}\).

The measured mass solubility of calcium fluoride is:

\[ 3.06\times10^{-5}\ \mathrm{g\,L^{-1}} \]

Calculate \(K_{sp}\).

Explain why using \(K_{sp}=4s^3\) would give an incorrect result.

Solution 2

The solubility product is approximately \(3.92\times10^{-11}\).

The important difference from Question 1 is that the solution already contains fluoride ions from sodium fluoride.

Step 1: Convert the mass solubility into molar solubility.

\[ \begin{aligned} s&=\frac{3.06\times10^{-5}}{78.08}\\ &=3.92\times10^{-7}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 2: Calculate the calcium concentration.

There are no calcium ions initially:

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

\[ [\mathrm{Ca}^{2+}]=3.92\times10^{-7}\ \mathrm{M} \]

Step 3: Calculate the total fluoride concentration.

The initial fluoride concentration is \(0.0100\ \mathrm{M}\).

Dissolving calcium fluoride adds \(2s\):

\[ \begin{aligned} [\mathrm{F}^{-}] &=0.0100+2(3.92\times10^{-7})\\ &=0.010000784\ \mathrm{M} \end{aligned} \]

Step 4: Substitute into the Ksp expression.

\[ \begin{aligned} K_{sp} &=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\\ &=(3.92\times10^{-7})\\ &\quad\times(0.010000784)^2\\ &=3.92\times10^{-11} \end{aligned} \]

Therefore:

\[ \boxed{K_{sp}=3.92\times10^{-11}} \]

Why does \(4s^3\) fail?

It assumes the fluoride concentration is \(2s\). That would only be correct if no fluoride ions were present initially.

Here, most of the fluoride ions come from sodium fluoride, so ignoring them dramatically underestimates the solubility product.

Question 3

A saturated solution of calcium fluoride is prepared in \(0.00500\ \mathrm{mol\,L^{-1}}\) calcium nitrate solution at \(25.0^\circ\mathrm{C}\).

An analysis of the solution gives an equilibrium fluoride concentration of:

\[ [\mathrm{F}^{-}]=8.90\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Calculate:

(a) The molar solubility of calcium fluoride.

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

Assume there are initially no fluoride ions in the solution.

Solution 3

The molar solubility is \(4.45\times10^{-5}\ \mathrm{mol\,L^{-1}}\), and \(K_{sp}=4.00\times10^{-11}\).

The interesting part is that we are given the fluoride concentration directly, rather than the solubility.

We need to work backwards from the stoichiometric relationship.

Step 1: Calculate the molar solubility.

Each mole of calcium fluoride produces two moles of fluoride ions.

Therefore:

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

Rearranging:

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

Substitute:

\[ \begin{aligned} s&=\frac{8.90\times10^{-5}}{2}\\ &=4.45\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 2: Calculate the equilibrium calcium concentration.

The initial calcium concentration comes from calcium nitrate.

\[ [\mathrm{Ca}^{2+}]_{\text{initial}}=0.00500\ \mathrm{M} \]

Calcium fluoride contributes another \(s\):

\[ \begin{aligned} [\mathrm{Ca}^{2+}] &=0.00500+4.45\times10^{-5}\\ &=0.0050445\ \mathrm{M} \end{aligned} \]

Step 3: Calculate Ksp.

\[ \begin{aligned} K_{sp} &=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\\ &=(0.0050445)(8.90\times10^{-5})^2\\ &=4.00\times10^{-11} \end{aligned} \]

Therefore:

\[ \boxed{K_{sp}=4.00\times10^{-11}} \]

A tempting mistake would be to use \(s\) as the calcium concentration.

That would give:

\[ K_{sp}=(4.45\times10^{-5})(8.90\times10^{-5})^2 \]

This is incorrect because the calcium concentration includes the \(0.00500\ \mathrm{M}\) already supplied by calcium nitrate.

The original concentration of a common ion must never disappear from your equilibrium calculation.

Question 4

An experiment is conducted to determine the solubility of calcium fluoride in pure water.

An experimenter:

• Adds \(0.5000\ \mathrm{g}\) of calcium fluoride to \(250.0\ \mathrm{mL}\) of water at \(25.0^\circ\mathrm{C}\).
• Allows the mixture to reach equilibrium.
• Filters the mixture to collect the undissolved calcium fluoride.
• Dries the recovered solid to constant mass.

The recovered solid has a mass of \(0.4960\ \mathrm{g}\).

Assume no solid is lost during transfer, the final solution volume remains \(250.0\ \mathrm{mL}\), and there are no common ions.

(a) Calculate the mass of calcium fluoride that dissolved.

(b) Calculate the molar solubility.

(c) Calculate \(K_{sp}\).

(d) Explain why using the original \(0.5000\ \mathrm{g}\) as the dissolved mass would produce an incorrect result.

Solution 4

The mass dissolved is \(0.0040\ \mathrm{g}\), the molar solubility is \(2.05\times10^{-4}\ \mathrm{mol\,L^{-1}}\), and \(K_{sp}=3.44\times10^{-11}\).

The key is to distinguish between the amount of solid added and the amount that actually dissolves.

Step 1: Calculate the mass dissolved.

\[ \begin{aligned} m_{\text{dissolved}} &=m_{\text{initial}}-m_{\text{recovered}}\\ &=0.5000-0.4960\\ &=0.0040\ \mathrm{g} \end{aligned} \]

Only \(0.0040\ \mathrm{g}\) entered the solution.

Most of the calcium fluoride remained as undissolved solid.

Step 2: Calculate the mass concentration.

The solution volume is:

\[ V=0.2500\ \mathrm{L} \]

Therefore:

\[ \begin{aligned} \text{Mass concentration} &=\frac{0.0040}{0.2500}\\ &=0.0160\ \mathrm{g\,L^{-1}} \end{aligned} \]

Step 3: Calculate molar solubility.

\[ \begin{aligned} s&=\frac{0.0160}{78.08}\\ &=2.05\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 4: Calculate Ksp.

Because the solution contains no common ions:

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

\[ \begin{aligned} K_{sp} &=4(2.05\times10^{-4})^3\\ &=3.44\times10^{-11} \end{aligned} \]

Therefore:

\[ \boxed{K_{sp}=3.44\times10^{-11}} \]

Why is using the initial mass wrong?

The initial mass tells us how much calcium fluoride was placed in the flask, not how much dissolved.

At saturation, additional solid remains undissolved. Treating all \(0.5000\ \mathrm{g}\) as dissolved would produce a much higher calculated concentration and a completely incorrect \(K_{sp}\).

This is why experimental measurements must distinguish the dissolved substance from the excess solid.

Question 5

Extension challenge

Two clear, saturated calcium fluoride solutions are prepared at \(25.0^\circ\mathrm{C}\).

Both solutions are filtered to remove excess undissolved calcium fluoride.

• Solution A contains \(0.0100\ \mathrm{mol\,L^{-1}}\) calcium nitrate.
• Solution B contains \(0.0100\ \mathrm{mol\,L^{-1}}\) sodium fluoride.

The two solutions are then mixed in equal volumes.

At this temperature, assume:

\[ K_{sp}=4.00\times10^{-11} \]

Determine:

(a) Whether a precipitate forms when the solutions are mixed.

(b) The approximate mass of calcium fluoride precipitated per litre of final mixture after the system reaches equilibrium.

Assume:

• Both solutions are at the same temperature.
• The salts dissociate completely.
• The final volume is the sum of the original volumes.
• Concentrations can be used in place of activities.
• No other reactions occur.

This question contains a trap: both original solutions are saturated with respect to calcium fluoride. Does that guarantee that mixing them cannot produce a precipitate?

Solution 5

Yes, calcium fluoride precipitates. Approximately \(0.192\ \mathrm{g}\) of \(\ce{CaF2}\) precipitates per litre of final mixture.

The important idea is that two solutions can each be saturated separately but become supersaturated when mixed.

The reason is that one solution supplies calcium ions while the other supplies fluoride ions.

When they are combined, the ion product increases dramatically.

Step 1: Calculate the dissolved calcium fluoride in Solution A.

Solution A initially contains:

\[ [\mathrm{Ca}^{2+}]=0.0100+s_A \]

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

Therefore:

\[ K_{sp}=(0.0100+s_A)(2s_A)^2 \]

Since \(s_A\) is much smaller than \(0.0100\), we can approximate:

\[ 4.00\times10^{-11}=0.0100(4s_A^2) \]

Rearranging:

\[ \begin{aligned} s_A &\approx\sqrt{\frac{4.00\times10^{-11}}{0.0400}}\\ &=3.16\times10^{-5}\ \mathrm{M} \end{aligned} \]

Step 2: Calculate the dissolved calcium fluoride in Solution B.

Solution B contains an initial fluoride concentration of \(0.0100\ \mathrm{M}\).

Therefore:

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

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

The solubility expression is:

\[ K_{sp}=s_B(0.0100+2s_B)^2 \]

Since \(2s_B\) is very small compared with \(0.0100\):

\[ 4.00\times10^{-11}\approx s_B(0.0100)^2 \]

Therefore:

\[ \begin{aligned} s_B &\approx\frac{4.00\times10^{-11}}{1.00\times10^{-4}}\\ &=4.00\times10^{-7}\ \mathrm{M} \end{aligned} \]

Step 3: Calculate the concentrations immediately after mixing.

This is the critical step.

Equal volumes are mixed, so each original concentration is halved.

Let \(C_0\) represent the calcium concentration immediately after mixing, before any precipitation occurs.

Solution A contributes \(0.0100+s_A\), while Solution B contributes \(s_B\).

Therefore:

\[ \begin{aligned} C_0 &=\frac{(0.0100+s_A)+s_B}{2}\\ &=0.0050158\ \mathrm{M} \end{aligned} \]

Now calculate the initial fluoride concentration, \(F_0\).

Solution A contributes \(2s_A\), while Solution B contributes \(0.0100+2s_B\).

\[ \begin{aligned} F_0 &=\frac{2s_A+(0.0100+2s_B)}{2}\\ &=0.0050316\ \mathrm{M} \end{aligned} \]

Step 4: Determine whether precipitation occurs.

Calculate the initial ion product:

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

\[ \begin{aligned} Q&=(0.0050158)(0.0050316)^2\\ &=1.27\times10^{-7} \end{aligned} \]

Compare this with the solubility product:

\[ Q=1.27\times10^{-7} \]

\[ K_{sp}=4.00\times10^{-11} \]

Therefore:

\[ Q>K_{sp} \]

The mixture is supersaturated, so calcium fluoride precipitates.

Notice the important difference: each original solution had an ion product equal to \(K_{sp}\), but the concentrations changed when the solutions were mixed.

The fact that each solution was individually saturated does not mean the mixture must also be saturated.

Step 5: Calculate the amount of calcium fluoride precipitated.

Let \(x\) be the concentration of calcium fluoride precipitated, in \(\mathrm{mol\,L^{-1}}\).

From the dissolution stoichiometry, each mole of calcium fluoride that precipitates removes:

• One mole of calcium ions.
• Two moles of fluoride ions.

Therefore, at the new equilibrium:

\[ [\mathrm{Ca}^{2+}]=C_0-x \]

\[ [\mathrm{F}^{-}]=F_0-2x \]

The new equilibrium must satisfy:

\[ K_{sp}=(C_0-x)(F_0-2x)^2 \]

Solving this directly is awkward.

Instead, let the final equilibrium fluoride concentration be \(y\).

Then:

\[ y=F_0-2x \]

Rearranging:

\[ x=\frac{F_0-y}{2} \]

The final calcium concentration is:

\[ [\mathrm{Ca}^{2+}]=C_0-\frac{F_0-y}{2} \]

Substituting the initial concentrations:

\[ [\mathrm{Ca}^{2+}]=0.00250+\frac{y}{2} \]

The equilibrium expression becomes:

\[ 4.00\times10^{-11} = \left(0.00250+\frac{y}{2}\right)y^2 \]

Solving gives approximately:

\[ y=1.25\times10^{-4}\ \mathrm{M} \]

Therefore, the amount precipitated is:

\[ \begin{aligned} x&=\frac{F_0-y}{2}\\ &=\frac{0.0050316-0.000125}{2}\\ &=2.45\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 6: Convert the precipitated amount into mass.

Using the molar mass of calcium fluoride:

\[ M=78.08\ \mathrm{g\,mol^{-1}} \]

\[ \begin{aligned} m&=nM\\ &=(2.45\times10^{-3})(78.08)\\ &=0.192\ \mathrm{g\,L^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{m=0.192\ \mathrm{g\,L^{-1}}} \]

The final equilibrium still contains dissolved calcium and fluoride ions. The precipitation does not remove every fluoride ion; it removes just enough to bring the ion product back to \(K_{sp}\).

This is the deeper point of the question.

A tempting approach is to see that both original solutions were saturated and conclude that nothing should happen. But saturation is determined by the product of the ion concentrations at equilibrium, not by the history of the individual solutions.

Mixing changes those concentrations, and the equilibrium responds.

07Where to go next

You now have the method for calculating \(K_{sp}\) from the measured solubility of calcium fluoride.

The reverse calculation is equally useful: given \(K_{sp}\), determine the molar solubility of a sparingly soluble salt.

For calcium fluoride in pure water, that means rearranging:

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

For a solution containing a common ion, you must instead solve the appropriate common-ion expression.

Work through Calculate Solubility from Ksp for HSC Chemistry to practise that reverse process.

The decision to make before either calculation is always the same: identify which ions are already present, then use the actual equilibrium concentrations rather than assuming every ion comes from the dissolving solid.