How to Find Which Ion Concentration Reaches Precipitation First
A solution contains calcium ions and magnesium ions. You slowly add sodium fluoride, which supplies fluoride ions.
A solution contains calcium ions and magnesium ions. You slowly add sodium fluoride, which supplies fluoride ions. Both metals can form insoluble fluoride salts.
Which solid appears first?
It is tempting to compare the two \(K_{sp}\) values and choose the smaller one. That can give the wrong answer. The concentration of each metal ion matters, as does the number of ions involved in its dissolution equation.
The reliable method is to calculate the minimum free concentration of the added ion needed to start precipitation for each possible solid. The solid requiring the lower concentration reaches precipitation first.
There is one extra complication: what if some of that ion is already present in the solution? You must account for it before calculating the precipitation thresholds.
01The key idea: calculate the concentration at the precipitation threshold
Imagine slowly adding sodium fluoride, \(\ce{NaF}\), to a solution containing \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\).
The sodium fluoride dissociates:
\[ \ce{NaF(aq) -> Na+(aq) + F-(aq)} \]
The fluoride ions can react with either metal ion to form a precipitate:
\[ \ce{Ca^2+(aq) + 2F-(aq) -> CaF2(s)} \]
\[ \ce{Mg^2+(aq) + 2F-(aq) -> MgF2(s)} \]
Both solids are sparingly soluble.
Before any solid forms, the fluoride concentration increases as more sodium fluoride is added.
Eventually, one of the possible solids reaches its saturation condition.
That is the point at which precipitation can begin.
What does reaching precipitation actually mean?
For a sparingly soluble salt, we compare two quantities:
- \(Q_{sp}\): the ion product calculated using the current ion concentrations.
- \(K_{sp}\): the solubility product at the specified temperature.
The comparison tells us the state of the solution.
| Condition | Meaning |
|---|---|
| \(Q_{sp}<K_{sp}\) | The solution is unsaturated. No precipitation is favoured. |
| \(Q_{sp}=K_{sp}\) | The solution is just saturated. Precipitation can begin. |
| \(Q_{sp}>K_{sp}\) | The solution is supersaturated. Precipitation is thermodynamically favoured. |
For HSC calculations, we generally assume that precipitation begins as soon as saturation is reached. We ignore delays caused by crystal nucleation.
Our job is to find the concentration at which \(Q_{sp}=K_{sp}\) for each possible solid.
The smaller required concentration identifies the first precipitate.
02Start with the balanced dissolution equation
The dissolution equation tells you how many ions are released when one mole of solid dissolves.
This determines the powers in the \(K_{sp}\) expression.
Consider calcium fluoride:
\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]
One mole of calcium fluoride produces:
- One mole of calcium ions.
- Two moles of fluoride ions.
Therefore:
\[ K_{sp}(\ce{CaF2}) = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
Now consider magnesium fluoride:
\[ \ce{MgF2(s) <=> Mg^2+(aq) + 2F-(aq)} \]
Its expression is:
\[ K_{sp}(\ce{MgF2}) = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 \]
Notice that the fluoride concentration is squared in both expressions.
This is not an arbitrary mathematical rule. It comes directly from the balanced dissolution equations.
The general method for a precipitation threshold
For a salt that dissolves according to:
\[ \ce{MX2(s) <=> M^2+(aq) + 2X-(aq)} \]
The solubility product is:
\[ K_{sp}=[\mathrm{M}^{2+}][\mathrm{X}^{-}]^2 \]
At the instant precipitation begins:
\[ Q_{sp}=K_{sp} \]
Therefore:
\[ K_{sp}=[\mathrm{M}^{2+}][\mathrm{X}^{-}]^2 \]
Rearranging:
\[ \boxed{ [\mathrm{X}^{-}]_{\text{threshold}} = \sqrt{\frac{K_{sp}}{[\mathrm{M}^{2+}]}} } \]
This equation gives the minimum free concentration of the added anion needed to start precipitation.
The symbols mean:
- \(K_{sp}\) is the solubility product.
- \([\mathrm{M}^{2+}]\) is the current dissolved metal-ion concentration, in mol L\(^{-1}\).
- \([\mathrm{X}^{-}]_{\text{threshold}}\) is the free anion concentration at the onset of precipitation, in mol L\(^{-1}\).
You can use the same approach for other salts, but you must derive the expression from their own dissolution equations.
03Worked example: Which fluoride precipitates first?
A solution contains:
- \(0.00100\ \mathrm{mol\,L^{-1}}\) calcium ions.
- \(0.100\ \mathrm{mol\,L^{-1}}\) magnesium ions.
Sodium fluoride is added slowly.
At \(25^\circ\mathrm{C}\), use the following illustrative solubility products:
| Solid | \(K_{sp}\) |
|---|---|
| \(\ce{CaF2}\) | \(3.9\times10^{-11}\) |
| \(\ce{MgF2}\) | \(6.4\times10^{-9}\) |
Assume the initial solution contains no fluoride ions, the volume change is negligible, and no significant complex ions form.
Determine which solid precipitates first.
Step 1Write the dissolution equations
For calcium fluoride:
\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]
For magnesium fluoride:
\[ \ce{MgF2(s) <=> Mg^2+(aq) + 2F-(aq)} \]
Both salts produce two fluoride ions per formula unit.
Step 2Write the \(K_{sp}\) expressions
\[ K_{sp}(\ce{CaF2}) = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
\[ K_{sp}(\ce{MgF2}) = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 \]
Step 3Calculate the fluoride threshold for calcium fluoride
At the onset of precipitation:
\[ K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
Rearrange:
\[ [\mathrm{F}^{-}] = \sqrt{\frac{K_{sp}}{[\mathrm{Ca}^{2+}]}} \]
Substitute:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{CaF2}} &= \sqrt{\frac{3.9\times10^{-11}}{0.00100}}\\ &=\sqrt{3.9\times10^{-8}}\\ &=1.97\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Calcium fluoride begins precipitating when the free fluoride concentration reaches approximately \(1.97\times10^{-4}\ \mathrm{mol\,L^{-1}}\).
Step 4Calculate the fluoride threshold for magnesium fluoride
Use the same method:
\[ [\mathrm{F}^{-}] = \sqrt{\frac{K_{sp}}{[\mathrm{Mg}^{2+}]}} \]
Substitute:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{MgF2}} &= \sqrt{\frac{6.4\times10^{-9}}{0.100}}\\ &=\sqrt{6.4\times10^{-8}}\\ &=2.53\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Magnesium fluoride begins precipitating at approximately \(2.53\times10^{-4}\ \mathrm{mol\,L^{-1}}\).
Step 5Compare the thresholds
| Solid | Fluoride concentration at onset |
|---|---|
| \(\ce{CaF2}\) | \(1.97\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{MgF2}\) | \(2.53\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
Since calcium fluoride requires the lower fluoride concentration, it precipitates first.
The first precipitate is \(\ce{CaF2}\).
Notice something interesting.
Magnesium ions initially have a much higher concentration than calcium ions. You might expect magnesium fluoride to form first because there are more magnesium ions available.
However, its larger \(K_{sp}\) means it requires a higher fluoride concentration to reach saturation.
The precipitation order depends on both the metal-ion concentration and the solubility product.
04What changes when a common ion is already present?
So far, we assumed the solution initially contained no fluoride ions.
But imagine the original solution already contained sodium fluoride.
You now have two sources of fluoride:
- Fluoride ions already present in the solution.
- Additional fluoride ions supplied by the reagent.
This is called the common-ion situation.
For example:
\[ \ce{NaF(aq) -> Na+(aq) + F-(aq)} \]
If the initial sodium fluoride concentration is \(0.0100\ \mathrm{mol\,L^{-1}}\), it supplies:
\[ [\mathrm{F}^{-}]_{\text{initial}} = 0.0100\ \mathrm{mol\,L^{-1}} \]
The concentration is not zero before you start adding more sodium fluoride.
The important distinction: total source versus free concentration
The \(K_{sp}\) expression uses the concentration of the free dissolved ions.
It does not matter which soluble compound originally supplied those ions.
For example, if a solution contains \(0.0100\ \mathrm{mol\,L^{-1}}\) NaF, the initial fluoride concentration is \(0.0100\ \mathrm{mol\,L^{-1}}\), assuming complete dissociation.
If it contains \(0.0100\ \mathrm{mol\,L^{-1}}\) \(\ce{MgF2}\) dissolved in solution, the fluoride contribution would be:
\[ [\mathrm{F}^{-}]=2(0.0100) \]
\[ [\mathrm{F}^{-}]=0.0200\ \mathrm{mol\,L^{-1}} \]
The concentration of the compound is not automatically the concentration of its ions.
Always use the balanced dissociation equation.
A common-ion decision rule
Before calculating precipitation thresholds, establish the initial concentration of every relevant ion.
If the solution contains an initial common-ion concentration \(C_0\), the ion product must include that concentration.
For example, if you are considering calcium fluoride:
\[ Q_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
If the initial fluoride concentration is \(C_0\), then before precipitation begins:
\[ Q_{sp}=[\mathrm{Ca}^{2+}](C_0)^2 \]
There are two possibilities:
- If \(Q_{sp}<K_{sp}\), more fluoride is required before precipitation begins.
- If \(Q_{sp}\geq K_{sp}\), the solution has already reached or exceeded saturation. Precipitation may begin immediately.
This second possibility is easy to overlook.
You cannot always assume that adding the first drop of reagent is what causes precipitation. The solution might already be saturated.
05Worked example: Finding the first precipitate after mixing solutions
A student mixes the following solutions:
- \(25.0\ \mathrm{mL}\) of \(0.00400\ \mathrm{mol\,L^{-1}}\) calcium chloride.
- \(75.0\ \mathrm{mL}\) of \(0.0200\ \mathrm{mol\,L^{-1}}\) magnesium nitrate.
- \(10.0\ \mathrm{mL}\) of \(0.00100\ \mathrm{mol\,L^{-1}}\) sodium fluoride.
Sodium fluoride is then added gradually.
At \(25^\circ\mathrm{C}\), use the illustrative values:
\[ K_{sp}(\ce{CaF2})=3.9\times10^{-11} \]
\[ K_{sp}(\ce{MgF2})=6.4\times10^{-9} \]
Determine which fluoride precipitates first.
Assume additive volumes, complete dissociation, negligible complex formation, and no precipitation before the solutions are mixed.
Step 1Calculate the final volume
You must account for dilution before using the initial ion concentrations.
\[ \begin{aligned} V_{\text{total}} &=25.0+75.0+10.0\\ &=110.0\ \mathrm{mL}\\ &=0.1100\ \mathrm{L} \end{aligned} \]
Step 2Calculate the initial calcium concentration after mixing
First calculate the moles of calcium ions.
\[ \begin{aligned} n(\mathrm{Ca}^{2+}) &=CV\\ &=0.00400\times0.0250\\ &=1.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]
Now divide by the total volume:
\[ \begin{aligned} [\mathrm{Ca}^{2+}] &=\frac{1.00\times10^{-4}}{0.1100}\\ &=9.09\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Step 3Calculate the initial magnesium concentration after mixing
\[ \begin{aligned} n(\mathrm{Mg}^{2+}) &=0.0200\times0.0750\\ &=1.50\times10^{-3}\ \mathrm{mol} \end{aligned} \]
Therefore:
\[ \begin{aligned} [\mathrm{Mg}^{2+}] &=\frac{1.50\times10^{-3}}{0.1100}\\ &=1.36\times10^{-2}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Step 4Calculate the initial fluoride concentration
Sodium fluoride dissociates in a \(1:1\) mole ratio:
\[ \ce{NaF(aq) -> Na+(aq) + F-(aq)} \]
Therefore:
\[ \begin{aligned} n(\mathrm{F}^{-}) &=0.00100\times0.0100\\ &=1.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]
The initial fluoride concentration is:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\text{initial}} &=\frac{1.00\times10^{-5}}{0.1100}\\ &=9.09\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
This is our common-ion concentration.
Step 5Check whether either solid has already started precipitating
For calcium fluoride:
\[ \begin{aligned} Q_{sp}(\ce{CaF2}) &=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\\ &=(9.09\times10^{-4}) (9.09\times10^{-5})^2\\ &=7.51\times10^{-12} \end{aligned} \]
Compare:
\[ 7.51\times10^{-12} < 3.9\times10^{-11} \]
Calcium fluoride has not reached saturation.
For magnesium fluoride:
\[ \begin{aligned} Q_{sp}(\ce{MgF2}) &=[\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2\\ &=(1.36\times10^{-2}) (9.09\times10^{-5})^2\\ &=1.12\times10^{-10} \end{aligned} \]
Compare:
\[ 1.12\times10^{-10} < 6.4\times10^{-9} \]
Magnesium fluoride has not reached saturation either.
Therefore, both solids remain unprecipitated at this stage.
Step 6Calculate the new fluoride threshold for calcium fluoride
Use the diluted calcium concentration:
\[ [\mathrm{F}^{-}]_{\text{threshold}} = \sqrt{\frac{K_{sp}}{[\mathrm{Ca}^{2+}]}} \]
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{CaF2}} &=\sqrt{\frac{3.9\times10^{-11}} {9.09\times10^{-4}}}\\ &=2.07\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Step 7Calculate the new fluoride threshold for magnesium fluoride
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{MgF2}} &=\sqrt{\frac{6.4\times10^{-9}} {1.36\times10^{-2}}}\\ &=6.85\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Step 8Compare the thresholds
| Solid | Free fluoride concentration at onset |
|---|---|
| \(\ce{CaF2}\) | \(2.07\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{MgF2}\) | \(6.85\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
The calcium fluoride threshold is lower.
Calcium fluoride precipitates first.
The initial fluoride concentration was already \(9.09\times10^{-5}\ \mathrm{mol\,L^{-1}}\), so only a relatively small increase was needed to reach its saturation threshold.
The key lesson is that you must use concentrations after mixing, not the original concentrations printed on the reagent bottles.
06The difference between free ion concentration and reagent concentration
This is another common source of incorrect answers.
Suppose a question asks:
What concentration of sodium fluoride must be added to start precipitation?
The threshold calculation gives the required free fluoride concentration.
It does not automatically give the concentration of the sodium fluoride solution you must pour into the beaker.
These are different quantities.
For a solution containing \(0.00100\ \mathrm{mol\,L^{-1}}\) calcium ions, using the illustrative calcium fluoride \(K_{sp}\):
\[ [\mathrm{F}^{-}]_{\text{threshold}} = 1.97\times10^{-4}\ \mathrm{mol\,L^{-1}} \]
This is the required free fluoride concentration in the mixture.
If sodium fluoride is added from a \(0.100\ \mathrm{mol\,L^{-1}}\) stock solution, the amount needed depends on:
- The original solution volume.
- The stock solution concentration.
- The initial fluoride concentration.
- The change in total volume.
- Any fluoride removed through precipitation or other reactions.
You need a separate dilution or mole calculation to determine the volume of stock solution required.
When is the negligible-volume-change assumption acceptable?
In many HSC-style threshold calculations, the question either states that volume changes are negligible or provides concentrations that can be treated as constant until the first precipitate forms.
Under that assumption, you can calculate the required free ion concentration directly.
If the volume change is significant, you must calculate the concentrations after mixing or after each addition.
Do not silently ignore dilution when the question gives enough information to account for it.
07A reliable decision process
When you are given a question asking which ion or compound precipitates first, follow this sequence.
Step 1: Identify the possible precipitates.
Write the chemical formula of each sparingly soluble solid.
Step 2: Write the balanced dissolution equations.
Check the number of each ion produced per mole of solid.
Step 3: Write the correct \(K_{sp}\) expressions.
Use the stoichiometric coefficients as powers.
Step 4: Calculate the initial ion concentrations.
Account for dissociation, dilution, and any common ions already present.
Step 5: Check the initial ion products.
Calculate \(Q_{sp}\) for each possible solid.
If a solution is already saturated or supersaturated, it does not need a higher reagent-ion concentration to reach its threshold.
Step 6: Calculate each precipitation threshold.
Set \(Q_{sp}=K_{sp}\) and rearrange for the free concentration of the ion being added.
Step 7: Compare the thresholds.
The lower required free concentration identifies the first solid to precipitate, provided the reagent can reach that concentration and the equilibrium assumptions remain valid.
Common mistakes to avoid
| Mistake | Why it is incorrect |
|---|---|
| Choosing the smallest \(K_{sp}\) automatically | The initial ion concentrations and dissolution stoichiometry also matter. |
| Forgetting to square an ion concentration | The powers come from the balanced dissolution equation. |
| Treating the concentration of NaF as twice its fluoride concentration | NaF produces one fluoride ion per formula unit. |
| Ignoring dilution after mixing | The initial concentrations in the final mixture may be very different from the stock concentrations. |
| Ignoring an existing common ion | It contributes to the initial ion product. |
| Using the total amount of reagent as the threshold | \(K_{sp}\) uses free equilibrium ion concentrations. |
| Assuming the first precipitate tells you how much solid forms | The threshold predicts the onset of precipitation, not the final amount precipitated. |
For a more detailed treatment of the underlying equilibrium calculations, see How to Calculate Solubility with a Common Ion (HSC Chemistry).
For a broader comparison of selective precipitation, see Which Ion Precipitates First? A Worked HSC Chemistry Example.
08Questions and solutions
The following questions build from a straightforward threshold calculation to more challenging situations involving common ions, dilution, and competing precipitation conditions.
All numerical \(K_{sp}\) values are illustrative data supplied for these questions. Assume \(25^\circ\mathrm{C}\), complete dissociation of soluble salts, and concentration-based equilibrium calculations unless otherwise stated.
Question 1
A solution contains \(0.0200\ \mathrm{mol\,L^{-1}}\) of \(\mathrm{Ag}^{+}\) ions.
Silver chloride can precipitate when sodium chloride is added.
Given:
\[ K_{sp}(\ce{AgCl})=1.8\times10^{-10} \]
Calculate the minimum free chloride concentration required to begin precipitation.
Solution 1
The minimum free chloride concentration is \(9.0\times10^{-9}\ \mathrm{mol\,L^{-1}}\).
First, write the dissolution equation:
\[ \ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)} \]
The solubility product is:
\[ K_{sp}=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]
At the onset of precipitation:
\[ K_{sp}=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] \]
Rearrange:
\[ [\mathrm{Cl}^{-}]=\frac{K_{sp}}{[\mathrm{Ag}^{+}]} \]
Substitute:
\[ \begin{aligned} [\mathrm{Cl}^{-}] &=\frac{1.8\times10^{-10}}{0.0200}\\ &=9.0\times10^{-9}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Therefore, silver chloride begins precipitating when the free chloride concentration reaches:
\[ \boxed{9.0\times10^{-9}\ \mathrm{mol\,L^{-1}}} \]
The dissolution equation produces one chloride ion per formula unit, so the chloride concentration appears to the first power.
Do not square it. The exponent must match the balanced dissolution equation.
Question 2
A solution contains:
- \(0.0100\ \mathrm{mol\,L^{-1}}\) of \(\mathrm{Ca}^{2+}\).
- \(0.00500\ \mathrm{mol\,L^{-1}}\) of \(\mathrm{F}^{-}\), supplied by dissolved sodium fluoride.
Additional sodium fluoride is added.
Given:
\[ K_{sp}(\ce{CaF2})=3.9\times10^{-11} \]
Determine whether calcium fluoride begins precipitating immediately.
If it does not, calculate the additional free fluoride concentration required to reach precipitation.
Solution 2
Calcium fluoride does not precipitate immediately. The solution would need an additional \(3.25\times10^{-5}\ \mathrm{mol\,L^{-1}}\) of free fluoride ions to reach saturation.
First, write the dissolution equation:
\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]
The solubility product is:
\[ K_{sp}=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
Before adding more sodium fluoride, calculate the ion product:
\[ \begin{aligned} Q_{sp} &=(0.0100)(0.00500)^2\\ &=2.50\times10^{-7} \end{aligned} \]
Compare:
\[ 2.50\times10^{-7} > 3.9\times10^{-11} \]
The solution is already supersaturated with respect to calcium fluoride.
Therefore, under the equilibrium model, precipitation is already thermodynamically favoured.
There is no additional fluoride concentration required to reach the threshold.
This is an important boundary case. A student who automatically calculates a new threshold and assumes the solution must still be unsaturated has missed the initial \(Q_{sp}\) check.
Question 3
A solution contains \(0.00400\ \mathrm{mol\,L^{-1}}\) of calcium ions and \(0.0400\ \mathrm{mol\,L^{-1}}\) of magnesium ions.
Sodium fluoride is added slowly.
Use the following illustrative data:
| Solid | \(K_{sp}\) |
|---|---|
| \(\ce{CaF2}\) | \(3.9\times10^{-11}\) |
| \(\ce{MgF2}\) | \(6.4\times10^{-9}\) |
A student argues:
“Magnesium ions are ten times more concentrated than calcium ions, so magnesium fluoride must precipitate first.”
Evaluate this claim using calculations.
Solution 3
The claim is incorrect. Calcium fluoride precipitates first.
The number of fluoride ions released by each solid is two, so both precipitation thresholds can be calculated using:
\[ [\mathrm{F}^{-}]_{\text{threshold}} = \sqrt{\frac{K_{sp}}{[\mathrm{M}^{2+}]}} \]
For calcium fluoride:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{CaF2}} &=\sqrt{\frac{3.9\times10^{-11}}{0.00400}}\\ &=\sqrt{9.75\times10^{-9}}\\ &=9.87\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
For magnesium fluoride:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{MgF2}} &=\sqrt{\frac{6.4\times10^{-9}}{0.0400}}\\ &=\sqrt{1.60\times10^{-7}}\\ &=4.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Compare:
| Solid | Fluoride threshold |
|---|---|
| \(\ce{CaF2}\) | \(9.87\times10^{-5}\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{MgF2}\) | \(4.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
Calcium fluoride needs a much lower free fluoride concentration.
Therefore, it precipitates first.
The student’s reasoning focuses only on the initial metal-ion concentrations. A higher metal-ion concentration does lower the required concentration of the added anion, but it does not determine the outcome by itself.
The \(K_{sp}\) value must also be considered.
Question 4
A student mixes the following solutions:
- \(40.0\ \mathrm{mL}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) calcium nitrate.
- \(60.0\ \mathrm{mL}\) of \(0.0200\ \mathrm{mol\,L^{-1}}\) magnesium nitrate.
- \(50.0\ \mathrm{mL}\) of \(0.00300\ \mathrm{mol\,L^{-1}}\) sodium fluoride.
The student then adds more sodium fluoride.
Use:
\[ K_{sp}(\ce{CaF2})=3.9\times10^{-11} \]
\[ K_{sp}(\ce{MgF2})=6.4\times10^{-9} \]
Assume additive volumes and negligible volume change during the subsequent addition of sodium fluoride.
Determine:
a) The initial concentrations of \(\mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}\), and \(\mathrm{F}^{-}\) after mixing.
b) Whether either fluoride precipitate forms immediately.
c) Which fluoride precipitates first as additional sodium fluoride is added.
Solution 4
Calcium fluoride precipitates first, and neither solid is initially saturated.
The key is to calculate all three concentrations using the total mixed volume.
Part a: Calculate the concentrations after mixing.
First, calculate the total volume:
\[ V=40.0+60.0+50.0 \]
\[ V=150.0\ \mathrm{mL}=0.1500\ \mathrm{L} \]
For calcium ions:
\[ \begin{aligned} [\mathrm{Ca}^{2+}] &=\frac{0.0100\times0.0400}{0.1500}\\ &=2.67\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
For magnesium ions:
\[ \begin{aligned} [\mathrm{Mg}^{2+}] &=\frac{0.0200\times0.0600}{0.1500}\\ &=8.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
For fluoride ions, remember that NaF produces one fluoride ion per formula unit:
\[ \begin{aligned} [\mathrm{F}^{-}] &=\frac{0.00300\times0.0500}{0.1500}\\ &=1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
The initial concentrations are:
| Ion | Concentration after mixing |
|---|---|
| \(\mathrm{Ca}^{2+}\) | \(2.67\times10^{-3}\ \mathrm{mol\,L^{-1}}\) |
| \(\mathrm{Mg}^{2+}\) | \(8.00\times10^{-3}\ \mathrm{mol\,L^{-1}}\) |
| \(\mathrm{F}^{-}\) | \(1.00\times10^{-3}\ \mathrm{mol\,L^{-1}}\) |
Part b: Check whether either solid forms immediately.
For calcium fluoride:
\[ \begin{aligned} Q_{sp}(\ce{CaF2}) &=(2.67\times10^{-3})(1.00\times10^{-3})^2\\ &=2.67\times10^{-9} \end{aligned} \]
Since:
\[ 2.67\times10^{-9}>3.9\times10^{-11} \]
the solution is supersaturated with respect to calcium fluoride.
For magnesium fluoride:
\[ \begin{aligned} Q_{sp}(\ce{MgF2}) &=(8.00\times10^{-3})(1.00\times10^{-3})^2\\ &=8.00\times10^{-9} \end{aligned} \]
Since:
\[ 8.00\times10^{-9}>6.4\times10^{-9} \]
the solution is also supersaturated with respect to magnesium fluoride.
Therefore, both precipitates are thermodynamically favoured immediately after mixing.
Part c: Identify the first precipitate.
There is an important distinction here.
The solution has already exceeded both saturation thresholds. It is therefore incorrect to describe the subsequent addition of sodium fluoride as the event that first brings either solid to saturation.
To determine which solid is favoured first as the mixture is brought together, compare the fluoride thresholds using the concentrations immediately after mixing, before precipitation alters them.
For calcium fluoride:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{CaF2}} &=\sqrt{\frac{3.9\times10^{-11}}{2.67\times10^{-3}}}\\ &=1.21\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
For magnesium fluoride:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{MgF2}} &=\sqrt{\frac{6.4\times10^{-9}}{8.00\times10^{-3}}}\\ &=8.94\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
The calcium fluoride threshold is lower.
Therefore, calcium fluoride reaches saturation first as the solutions are initially combined.
The final mixture is already supersaturated with respect to both solids, so both precipitations may occur. The precise amounts and subsequent precipitation behaviour require a coupled equilibrium calculation.
This question exposes an important trap: calculating thresholds from the final mixture is useful for comparing the saturation conditions, but it cannot by itself reconstruct the exact chronological sequence once both solids are already supersaturated.
Question 5
A solution initially contains:
\[ [\mathrm{Ca}^{2+}]=0.00200\ \mathrm{mol\,L^{-1}} \]
\[ [\mathrm{Mg}^{2+}]=0.0200\ \mathrm{mol\,L^{-1}} \]
It also contains fluoride ions supplied by sodium fluoride.
The initial free fluoride concentration is:
\[ [\mathrm{F}^{-}]_0=1.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]
Additional sodium fluoride is added slowly.
Use the following illustrative solubility products:
\[ K_{sp}(\ce{CaF2})=3.9\times10^{-11} \]
\[ K_{sp}(\ce{MgF2})=6.4\times10^{-9} \]
Assume:
- The temperature remains constant.
- The solution volume is effectively constant.
- There is no significant complex-ion formation.
- Both metal ions remain dissolved until their respective precipitation thresholds are reached.
- Equilibrium is established without a significant nucleation delay.
A student calculates the two precipitation thresholds and concludes:
“Calcium fluoride precipitates first. Once it starts forming, magnesium fluoride cannot precipitate because the fluoride ions are being removed from the solution.”
Evaluate the student’s conclusion.
Your answer must:
a) Calculate the initial ion products for both possible solids.
b) Determine which solid reaches saturation first.
c) Explain whether magnesium fluoride can subsequently precipitate while calcium fluoride is already present.
d) Explain why the initial threshold calculations alone cannot determine the final amounts of both precipitates.
Solution 5
Calcium fluoride precipitates first, but magnesium fluoride can subsequently precipitate if enough additional fluoride is supplied. The first precipitate does not permanently prevent the second from forming.
The student’s first conclusion is correct, but the final conclusion is not.
The key is that precipitation changes the dissolved ion concentrations. The system must be considered again after the first solid forms.
Part a: Calculate the initial ion products.
For calcium fluoride:
\[ \ce{CaF2(s) <=> Ca^2+(aq) + 2F-(aq)} \]
Therefore:
\[ Q_{sp}(\ce{CaF2}) = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
Substitute:
\[ \begin{aligned} Q_{sp}(\ce{CaF2}) &=(0.00200)(1.00\times10^{-4})^2\\ &=2.00\times10^{-11} \end{aligned} \]
Compare:
\[ 2.00\times10^{-11} < 3.9\times10^{-11} \]
Calcium fluoride is initially unsaturated.
For magnesium fluoride:
\[ Q_{sp}(\ce{MgF2}) = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 \]
Substitute:
\[ \begin{aligned} Q_{sp}(\ce{MgF2}) &=(0.0200)(1.00\times10^{-4})^2\\ &=2.00\times10^{-10} \end{aligned} \]
Compare:
\[ 2.00\times10^{-10} < 6.4\times10^{-9} \]
Magnesium fluoride is also initially unsaturated.
Neither solid has started precipitating.
Part b: Determine which solid reaches saturation first.
Calculate the fluoride threshold for calcium fluoride:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{CaF2}} &=\sqrt{\frac{3.9\times10^{-11}}{0.00200}}\\ &=\sqrt{1.95\times10^{-8}}\\ &=1.40\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Now calculate the threshold for magnesium fluoride:
\[ \begin{aligned} [\mathrm{F}^{-}]_{\ce{MgF2}} &=\sqrt{\frac{6.4\times10^{-9}}{0.0200}}\\ &=\sqrt{3.2\times10^{-7}}\\ &=5.66\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]
Compare the results:
| Solid | Fluoride threshold |
|---|---|
| \(\ce{CaF2}\) | \(1.40\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{MgF2}\) | \(5.66\times10^{-4}\ \mathrm{mol\,L^{-1}}\) |
Calcium fluoride reaches saturation first.
Part c: Can magnesium fluoride precipitate after calcium fluoride starts forming?
Yes.
Once calcium fluoride begins precipitating, it removes calcium and fluoride ions from the solution.
If solid calcium fluoride remains present at equilibrium, its solubility product must be satisfied:
\[ K_{sp}(\ce{CaF2}) = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
As more sodium fluoride is added, the free fluoride concentration increases.
The calcium-ion concentration must then decrease to maintain the calcium fluoride equilibrium, provided solid \(\ce{CaF2}\) remains present.
Now consider magnesium fluoride.
Its ion product is:
\[ Q_{sp}(\ce{MgF2}) = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 \]
As the free fluoride concentration increases, the magnesium fluoride ion product also increases, unless magnesium is being removed by another process.
Eventually, it can reach:
\[ Q_{sp}(\ce{MgF2})=K_{sp}(\ce{MgF2}) \]
At that point, magnesium fluoride can begin precipitating too.
The presence of one solid does not make the solution immune to forming another solid.
Part d: Why are the original thresholds insufficient to determine the final amounts?
The original threshold calculations use the initial metal-ion concentrations.
They predict when each solid first reaches saturation if no precipitation has yet changed those concentrations.
After calcium fluoride begins forming, however, the dissolved calcium concentration falls.
When magnesium fluoride starts forming, the dissolved magnesium concentration also falls.
The two equilibria are now linked through their shared fluoride concentration.
The final solution must satisfy both solubility-product relationships if both solids remain present:
\[ K_{sp}(\ce{CaF2}) = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]
\[ K_{sp}(\ce{MgF2}) = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 \]
You would also need mass-balance equations to determine how much calcium and magnesium remains dissolved and how much has precipitated.
The amount of sodium fluoride added would be important too.
This is the exact limitation of a precipitation-threshold calculation: it predicts the onset of precipitation, not the final composition of a mixture containing multiple solids.
09Where to go next
Once you can calculate which solid reaches saturation first, the next challenge is working out how much of each ion remains dissolved after precipitation begins.
That requires combining \(K_{sp}\) calculations with mole balances and equilibrium concentrations.
For further practice, see:
- Precipitation Titrations: Calculate an Unknown Concentration for precipitation-based concentration calculations.
- How to Calculate \(K_{sp}\) from Calcium Fluoride Solubility for connecting molar solubility with solubility products.
The distinction to carry into those problems is simple: a precipitation threshold tells you when a solid can start forming; an equilibrium calculation tells you how much remains dissolved once the system adjusts.