How to Calculate Qsp After Mixing Solutions with Different Volumes

Two clear solutions can produce a cloudy mixture when you combine them. But how do you predict whether a precipitate will form if the solutions have different volumes?

Suppose you mix \(20.0\ \mathrm{mL}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) silver nitrate with \(80.0\ \mathrm{mL}\) of \(0.00100\ \mathrm{mol\,L^{-1}}\) sodium chloride. Both solutions contain an ion needed to form silver chloride, \(\ce{AgCl}\).

Should you multiply the two concentrations printed on the bottles to calculate the ion product?

No. Those are the concentrations before mixing. Once the solutions are combined, both ions are diluted into the total volume, and the two solutions contribute different numbers of moles.

The reliable method is to calculate the moles of each relevant ion, add contributions from every source, divide by the total volume, and then calculate \(Q_{sp}\).

Let’s work through why this matters, including what changes when a common ion is present.

All numerical values in the worked examples and questions are illustrative teaching data. Unless stated otherwise, assume a temperature of \(25^\circ\mathrm{C}\), complete dissociation of soluble salts, and additive solution volumes.

01Understand what you are calculating

Consider silver chloride dissolving in water:

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

The solubility product expression is:

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

Here, \(K_{sp}\) is the equilibrium solubility product, and the square brackets represent molar concentrations in \(\mathrm{mol\,L^{-1}}\).

But when you first mix two solutions, the ions might not be at equilibrium.

Instead, calculate the ion product, \(Q_{sp}\):

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

The mathematical expression is the same as the \(K_{sp}\) expression. The difference is which concentrations you substitute.

• \(K_{sp}\) uses equilibrium concentrations at a particular temperature.
• \(Q_{sp}\) uses the actual ion concentrations at the moment you are investigating.

Once you calculate \(Q_{sp}\), compare it with \(K_{sp}\).

If \(Q_{sp}>K_{sp}\), the dissolved ions can form a solid until the remaining solution reaches equilibrium.

A visible precipitate might not appear immediately because forming the first stable solid particles can take time. In standard HSC calculations, we generally assume equilibrium is reached.

02Calculate moles before calculating concentrations

This is the most important step when mixing solutions with different volumes.

Remember the relationship:

\[ n=cV \]

where:

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

Always convert millilitres into litres before using this equation.

For example:

\[ 25.0\ \mathrm{mL}=0.0250\ \mathrm{L} \]

Why you need to calculate moles first

Imagine two students pouring cordial into different-sized drink bottles.

One uses a small amount of highly concentrated cordial. The other uses a larger volume of weak cordial.

Which bottle contains more cordial?

You cannot tell from the volume alone. You need to account for both concentration and volume.

The same principle applies to ions.

For example, \(20.0\ \mathrm{mL}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) silver nitrate contains:

\[ n(\mathrm{Ag}^{+})=0.0100\times0.0200 \]

\[ =2.00\times10^{-4}\ \mathrm{mol} \]

Meanwhile, \(80.0\ \mathrm{mL}\) of \(0.00100\ \mathrm{mol\,L^{-1}}\) sodium chloride contains:

\[ n(\mathrm{Cl}^{-})=0.00100\times0.0800 \]

\[ =8.00\times10^{-5}\ \mathrm{mol} \]

Despite having four times the volume, the sodium chloride solution contributes fewer moles of its relevant ion.

The concentration and volume must be considered together.

Finding the final volume

If the question assumes that solution volumes are additive:

\[ V_{\text{total}}=V_1+V_2 \]

For our example:

\[ V_{\text{total}}=20.0+80.0 \]

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

If the question provides the final volume directly, use that value instead.

Finding the concentration after mixing

Once you know the total moles of an ion, divide by the total volume:

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

This gives you the actual concentration to substitute into \(Q_{sp}\).

Notice that you are not simply adding the starting concentrations. You are combining the moles first.

03What happens in an actual precipitation test?

A simple school laboratory investigation can help explain why this calculation is useful.

Imagine preparing two solutions of silver nitrate and sodium chloride.

First, measure the required volumes using pipettes or measuring cylinders. Transfer both solutions into a clean receiving beaker and stir with a glass rod.

The question is whether a white silver chloride precipitate will form.

The prediction comes from comparing \(Q_{sp}\) with \(K_{sp}\).

If the calculated ion product exceeds the solubility product, the model predicts that solid silver chloride is thermodynamically favoured.

If the ion product is below the solubility product, precipitation is not favoured.

The visible observation is cloudiness caused by suspended solid particles. However, cloudiness is only a qualitative observation. It does not directly measure \(Q_{sp}\), and a very small amount of precipitate might be difficult to see.

Why the experimental method needs accurate volumes

A first attempt might involve adding an approximate number of drops from each solution and judging whether the mixture becomes cloudy.

There is a problem with this design.

Drop sizes can vary, so the actual volumes and resulting concentrations are uncertain. Comparing the original bottle concentrations also ignores dilution after mixing.

A better design uses measured volumes, maintains a constant temperature, and records the concentration of each starting solution.

You can then calculate the initial ion product and compare it with the appropriate \(K_{sp}\).

The calculation predicts the thermodynamic behaviour. The laboratory observation provides a qualitative check of that prediction.

05What changes when a common ion is present?

So far, each relevant ion has come from one starting solution.

But what if two different dissolved salts supply the same ion?

This is called the common-ion situation.

For example, potassium iodide and sodium iodide both supply iodide ions:

\[ \ce{KI(aq) -> K+(aq) + I-(aq)} \]

\[ \ce{NaI(aq) -> Na+(aq) + I-(aq)} \]

If both salts are present, you must include the iodide ions from both sources.

The total number of moles of iodide is:

\[ n(\mathrm{I}^{-})_{\text{total}} = n(\mathrm{I}^{-})_{\ce{KI}} + n(\mathrm{I}^{-})_{\ce{NaI}} \]

Then divide this total by the final volume.

There is no special extra correction to apply to \(Q_{sp}\). You simply need to count every source of the relevant ion.

This is particularly important when the common ion changes your prediction from no precipitation to precipitation.

06A contrasting case: why the original concentrations can give the wrong answer

Let’s look at a particularly tempting mistake.

Mix \(25.0\ \mathrm{mL}\) of \(2.00\times10^{-5}\ \mathrm{mol\,L^{-1}}\) \(\ce{AgNO3}\) with \(75.0\ \mathrm{mL}\) of \(2.00\times10^{-5}\ \mathrm{mol\,L^{-1}}\) \(\ce{NaCl}\).

Use:

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

A student might multiply the two starting concentrations:

\[ (2.00\times10^{-5})^2 =4.00\times10^{-10} \]

This is greater than \(K_{sp}\), so the student predicts a precipitate.

But that calculation is incorrect.

The two starting concentrations are not the concentrations of the ions in the same solution. They also ignore the dilution caused by mixing.

Let’s calculate the actual ion product.

For silver ions:

\[ n(\mathrm{Ag}^{+}) =2.00\times10^{-5}\times0.0250 \]

\[ =5.00\times10^{-7}\ \mathrm{mol} \]

For chloride ions:

\[ n(\mathrm{Cl}^{-}) =2.00\times10^{-5}\times0.0750 \]

\[ =1.50\times10^{-6}\ \mathrm{mol} \]

The final volume is \(0.1000\ \mathrm{L}\).

Therefore:

\[ [\mathrm{Ag}^{+}]=5.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{Cl}^{-}]=1.50\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

The correct ion product is:

\[ \begin{aligned} Q_{sp} &=(5.00\times10^{-6}) (1.50\times10^{-5})\\ &=7.50\times10^{-11} \end{aligned} \]

Comparing:

\[ 7.50\times10^{-11} < 1.8\times10^{-10} \]

Therefore:

\[ \boxed{Q_{sp}<K_{sp}} \]

No precipitation is thermodynamically favoured.

The important lesson is that even if two solutions contain the ions needed to make an insoluble salt, that does not automatically mean a precipitate will form.

The actual post-mixing concentrations determine the ion product.

07A reliable HSC method

When you see a question asking you to calculate \(Q_{sp}\) after mixing solutions, use this sequence.

Common mistakes to avoid

1. Using the original concentrations

The concentrations in the starting beakers are not the concentrations after mixing.

2. Adding concentrations instead of moles

If two solutions contribute the same ion, calculate their moles separately and add them.

3. Forgetting the total volume

Both solutions contribute to the final volume, assuming additive volumes.

4. Ignoring stoichiometric coefficients

For example:

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

The correct expression is:

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

Not simply \([\mathrm{Ca}^{2+}][\mathrm{F}^{-}]\).

5. Confusing \(Q_{sp}\) with \(K_{sp}\)

\(Q_{sp}\) describes the current mixture. \(K_{sp}\) describes the equilibrium condition at a particular temperature.

6. Forgetting that precipitation changes concentrations

The initial \(Q_{sp}\) calculation assumes precipitation has not yet removed any ions. Once solid forms, the dissolved concentrations change.

Important assumptions and limitations

The standard HSC approach generally assumes:

• Soluble salts dissociate completely into their constituent ions.
• Solution volumes are additive unless a final volume is provided.
• No other reactions consume or produce the relevant ions.
• The temperature remains constant, so the same \(K_{sp}\) applies.
• Concentrations can be used in place of activities, as is appropriate for the usual dilute-solution model.

For highly concentrated real solutions, ion activities can differ significantly from concentrations. Complex formation and other chemical reactions can also change the free-ion concentrations. Those effects require a more detailed equilibrium treatment.

If you need to calculate the actual amount of a sparingly soluble salt that dissolves in the presence of a common ion, rather than simply predict whether precipitation occurs, see How to Calculate Solubility with a Common Ion (HSC Chemistry).

If you need to calculate \(K_{sp}\) from a measured solubility, see How to Calculate Ksp from Calcium Fluoride Solubility.

08Questions and solutions

The following questions move from straightforward mixing calculations to more demanding equilibrium reasoning. Use the supplied \(K_{sp}\) values rather than trying to memorise numerical values from a table.

Question 1

At \(25^\circ\mathrm{C}\), \(25.0\ \mathrm{mL}\) of \(0.0200\ \mathrm{mol\,L^{-1}}\) barium chloride, \(\ce{BaCl2}\), is mixed with \(75.0\ \mathrm{mL}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) sodium sulfate, \(\ce{Na2SO4}\).

The solubility product of barium sulfate is:

\[ K_{sp}(\ce{BaSO4})=1.1\times10^{-10} \]

Calculate the initial \(Q_{sp}\) and determine whether barium sulfate precipitation is favoured.

Solution 1

Barium sulfate precipitation is strongly favoured because \(Q_{sp}\) is greater than \(K_{sp}\).

Step 1: Write the dissolution equation and \(Q_{sp}\) expression.

\[ \ce{BaSO4(s) <=> Ba^2+(aq) + SO4^2-(aq)} \]

\[ Q_{sp}=[\mathrm{Ba}^{2+}][\mathrm{SO_{4}}^{2-}] \]

Step 2: Calculate the moles of barium ions.

One mole of \(\ce{BaCl2}\) produces one mole of \(\mathrm{Ba}^{2+}\).

\[ \begin{aligned} n(\mathrm{Ba}^{2+}) &=0.0200\times0.0250\\ &=5.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the moles of sulfate ions.

One mole of \(\ce{Na2SO4}\) produces one mole of sulfate ions.

\[ \begin{aligned} n(\mathrm{SO_{4}}^{2-}) &=0.0100\times0.0750\\ &=7.50\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the total volume and final concentrations.

\[ V_{\text{total}}=0.0250+0.0750 \]

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

Therefore:

\[ [\mathrm{Ba}^{2+}] = \frac{5.00\times10^{-4}}{0.1000} \]

\[ =5.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

And:

\[ [\mathrm{SO_{4}}^{2-}] = \frac{7.50\times10^{-4}}{0.1000} \]

\[ =7.50\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Step 5: Calculate \(Q_{sp}\).

\[ \begin{aligned} Q_{sp} &=(5.00\times10^{-3}) (7.50\times10^{-3})\\ &=3.75\times10^{-5} \end{aligned} \]

Step 6: Compare with \(K_{sp}\).

\[ 3.75\times10^{-5} > 1.1\times10^{-10} \]

Therefore:

\[ \boxed{Q_{sp}>K_{sp}} \]

Barium sulfate precipitation is strongly favoured.

Notice that the two chloride ions produced by each formula unit of \(\ce{BaCl2}\) do not enter this particular \(Q_{sp}\) expression. Only the ions in the dissolution equation for barium sulfate are required.

Question 2

A student mixes \(40.0\ \mathrm{mL}\) of \(0.00200\ \mathrm{mol\,L^{-1}}\) calcium nitrate, \(\ce{Ca(NO3)2}\), with \(60.0\ \mathrm{mL}\) of a solution containing:

• \(3.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) \(\ce{NaF}\);
• \(1.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) \(\ce{KF}\).

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

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

Calculate the initial \(Q_{sp}\) and determine whether calcium fluoride precipitation is favoured.

Explain why ignoring the potassium fluoride would change the predicted outcome.

Solution 2

Calcium fluoride precipitation is favoured, and the potassium fluoride is essential to reaching the correct conclusion.

Step 1: Write the dissolution equation.

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

Therefore:

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

Step 2: Calculate the moles of calcium ions.

\[ \begin{aligned} n(\mathrm{Ca}^{2+}) &=0.00200\times0.0400\\ &=8.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the total moles of fluoride ions.

From sodium fluoride:

\[ \begin{aligned} n(\mathrm{F}^{-})_{\ce{NaF}} &=3.00\times10^{-4}\times0.0600\\ &=1.80\times10^{-5}\ \mathrm{mol} \end{aligned} \]

From potassium fluoride:

\[ \begin{aligned} n(\mathrm{F}^{-})_{\ce{KF}} &=1.00\times10^{-4}\times0.0600\\ &=6.00\times10^{-6}\ \mathrm{mol} \end{aligned} \]

Add both contributions:

\[ \begin{aligned} n(\mathrm{F}^{-})_{\text{total}} &=1.80\times10^{-5} +6.00\times10^{-6}\\ &=2.40\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the final concentrations.

The total volume is:

\[ V_{\text{total}}=0.0400+0.0600 \]

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

Therefore:

\[ [\mathrm{Ca}^{2+}] = \frac{8.00\times10^{-5}}{0.1000} \]

\[ =8.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

And:

\[ [\mathrm{F}^{-}] = \frac{2.40\times10^{-5}}{0.1000} \]

\[ =2.40\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Step 5: Calculate \(Q_{sp}\).

\[ \begin{aligned} Q_{sp} &=(8.00\times10^{-4}) (2.40\times10^{-4})^2\\ &=(8.00\times10^{-4}) (5.76\times10^{-8})\\ &=4.608\times10^{-11} \end{aligned} \]

Step 6: Compare with \(K_{sp}\).

\[ 4.608\times10^{-11} > 3.9\times10^{-11} \]

Therefore:

\[ \boxed{Q_{sp}>K_{sp}} \]

Calcium fluoride precipitation is favoured.

Why does ignoring KF change the answer?

If the student only counts fluoride from sodium fluoride:

\[ [\mathrm{F}^{-}] = \frac{1.80\times10^{-5}}{0.1000} \]

\[ =1.80\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

The incorrect ion product would be:

\[ \begin{aligned} Q_{sp} &=(8.00\times10^{-4}) (1.80\times10^{-4})^2\\ &=2.592\times10^{-11} \end{aligned} \]

This is below \(K_{sp}\), so the student would predict no precipitation.

The mistake is failing to count the common fluoride ion supplied by potassium fluoride. Both salts dissociate, and both contribute to the final fluoride concentration.

Question 3

A \(50.0\ \mathrm{mL}\) solution contains:

• \(4.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) \(\ce{Ba(NO3)2}\);
• \(0.0100\ \mathrm{mol\,L^{-1}}\) \(\ce{Sr(NO3)2}\).

It is mixed with \(50.0\ \mathrm{mL}\) of \(2.00\times10^{-5}\ \mathrm{mol\,L^{-1}}\) sodium sulfate.

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

\[ K_{sp}(\ce{BaSO4})=1.1\times10^{-10} \]

\[ K_{sp}(\ce{SrSO4})=3.2\times10^{-7} \]

Immediately after mixing, before precipitation occurs:

(a) Calculate the ion product for both possible precipitates.

(b) Determine which precipitate is thermodynamically favoured.

(c) Calculate the sulfate concentration required for \(\ce{SrSO4}\) to just reach saturation if the strontium concentration remains unchanged.

Explain what this tells you about selective precipitation.

Solution 3

Barium sulfate is initially supersaturated, while strontium sulfate is undersaturated. Barium sulfate is therefore the precipitate favoured immediately after mixing.

Step 1: Calculate the post-mixing concentrations of the metal ions.

The total volume is:

\[ V_{\text{total}}=0.0500+0.0500 \]

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

For barium ions:

\[ \begin{aligned} [\mathrm{Ba}^{2+}] &=\frac{4.00\times10^{-4}\times0.0500} {0.1000}\\ &=2.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

For strontium ions:

\[ \begin{aligned} [\mathrm{Sr}^{2+}] &=\frac{0.0100\times0.0500}{0.1000}\\ &=5.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 2: Calculate the sulfate concentration after mixing.

\[ \begin{aligned} [\mathrm{SO_{4}}^{2-}] &=\frac{2.00\times10^{-5}\times0.0500} {0.1000}\\ &=1.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 3: Calculate \(Q_{sp}\) for barium sulfate.

The dissolution equation is:

\[ \ce{BaSO4(s) <=> Ba^2+(aq) + SO4^2-(aq)} \]

Therefore:

\[ \begin{aligned} Q_{sp}(\ce{BaSO4}) &=[\mathrm{Ba}^{2+}][\mathrm{SO_{4}}^{2-}]\\ &=(2.00\times10^{-4})(1.00\times10^{-5})\\ &=2.00\times10^{-9} \end{aligned} \]

Compare:

\[ 2.00\times10^{-9} > 1.1\times10^{-10} \]

Barium sulfate is supersaturated.

Step 4: Calculate \(Q_{sp}\) for strontium sulfate.

\[ \ce{SrSO4(s) <=> Sr^2+(aq) + SO4^2-(aq)} \]

Therefore:

\[ \begin{aligned} Q_{sp}(\ce{SrSO4}) &=[\mathrm{Sr}^{2+}][\mathrm{SO_{4}}^{2-}]\\ &=(5.00\times10^{-3})(1.00\times10^{-5})\\ &=5.00\times10^{-8} \end{aligned} \]

Compare:

\[ 5.00\times10^{-8} < 3.2\times10^{-7} \]

Strontium sulfate is undersaturated.

Therefore, only barium sulfate precipitation is initially favoured.

Step 5: Calculate the sulfate concentration required to saturate strontium sulfate.

At the point of saturation:

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

For strontium sulfate:

\[ K_{sp}=[\mathrm{Sr}^{2+}][\mathrm{SO_{4}}^{2-}] \]

Rearrange:

\[ [\mathrm{SO_{4}}^{2-}] = \frac{K_{sp}}{[\mathrm{Sr}^{2+}]} \]

Substitute:

\[ \begin{aligned} [\mathrm{SO_{4}}^{2-}]_{\text{sat}} &=\frac{3.2\times10^{-7}}{5.00\times10^{-3}}\\ &=6.4\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The initial sulfate concentration is only:

\[ 1.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

It is therefore below the strontium sulfate saturation threshold.

What does this tell us?

Different salts can begin precipitating at different ion concentrations, even when they share a common ion.

Barium sulfate reaches saturation at a much lower sulfate concentration than strontium sulfate under these conditions.

This is the basis of selective precipitation.

The threshold calculated for strontium assumes its concentration remains constant. Once significant precipitation occurs, the concentrations change, so a later prediction must account for the changing amounts of dissolved ions.

Question 4

At \(25^\circ\mathrm{C}\), \(40.0\ \mathrm{mL}\) of \(0.00500\ \mathrm{mol\,L^{-1}}\) \(\ce{Pb(NO3)2}\) is mixed with \(60.0\ \mathrm{mL}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) \(\ce{NaI}\).

Immediately after mixing, before any solid has nucleated, pure water is added until the final volume reaches \(300.0\ \mathrm{mL}\).

Use:

\[ K_{sp}(\ce{PbI2})=7.9\times10^{-9} \]

Calculate:

(a) The initial \(Q_{sp}\), immediately after the first mixing.

(b) The new \(Q_{sp}\) after dilution.

(c) Whether lead iodide precipitation is favoured after dilution.

Explain why tripling the volume changes \(Q_{sp}\) by a factor of 27 rather than a factor of 9.

Solution 4

After dilution, lead iodide precipitation is no longer thermodynamically favoured because the new \(Q_{sp}\) is below \(K_{sp}\).

Step 1: Write the dissolution equation and \(Q_{sp}\) expression.

\[ \ce{PbI2(s) <=> Pb^2+(aq) + 2I-(aq)} \]

\[ Q_{sp}=[\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2 \]

Step 2: Calculate the initial moles of each ion.

For lead ions:

\[ \begin{aligned} n(\mathrm{Pb}^{2+}) &=0.00500\times0.0400\\ &=2.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]

For iodide ions:

\[ \begin{aligned} n(\mathrm{I}^{-}) &=0.0100\times0.0600\\ &=6.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the initial concentrations.

The initial total volume is:

\[ V=0.0400+0.0600=0.1000\ \mathrm{L} \]

Therefore:

\[ [\mathrm{Pb}^{2+}] = \frac{2.00\times10^{-4}}{0.1000} \]

\[ =2.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

And:

\[ [\mathrm{I}^{-}] = \frac{6.00\times10^{-4}}{0.1000} \]

\[ =6.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate the initial ion product.

\[ \begin{aligned} Q_{sp} &=(2.00\times10^{-3}) (6.00\times10^{-3})^2\\ &=7.20\times10^{-8} \end{aligned} \]

Since:

\[ 7.20\times10^{-8} > 7.9\times10^{-9} \]

the initial mixture is supersaturated.

Step 5: Calculate the new concentrations after dilution.

The final volume increases from \(100.0\ \mathrm{mL}\) to \(300.0\ \mathrm{mL}\).

This is a threefold increase in volume.

Because no precipitate has formed, the total moles of each dissolved ion remain unchanged.

The new lead concentration is:

\[ [\mathrm{Pb}^{2+}] = \frac{2.00\times10^{-4}}{0.3000} \]

\[ =6.67\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

The new iodide concentration is:

\[ [\mathrm{I}^{-}] = \frac{6.00\times10^{-4}}{0.3000} \]

\[ =2.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Step 6: Calculate the new \(Q_{sp}\).

\[ \begin{aligned} Q_{sp} &=(6.67\times10^{-4}) (2.00\times10^{-3})^2\\ &=2.67\times10^{-9} \end{aligned} \]

Compare:

\[ 2.67\times10^{-9} < 7.9\times10^{-9} \]

Therefore:

\[ \boxed{Q_{sp}<K_{sp}} \]

After dilution, precipitation is no longer favoured.

Why does the volume increase reduce \(Q_{sp}\) by 27?

This is the important conceptual part.

For lead iodide:

\[ Q_{sp}=[\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2 \]

Tripling the volume reduces each concentration to one-third of its original value.

Therefore:

\[ Q_{sp,\text{new}} = \frac{[\mathrm{Pb}^{2+}]}{3} \left(\frac{[\mathrm{I}^{-}]}{3}\right)^2 \]

Simplifying:

\[ Q_{sp,\text{new}} = \frac{Q_{sp,\text{initial}}}{27} \]

The factor of 27 comes from the combined powers in the solubility-product expression:

\[ 3^{1+2}=27 \]

This is why the stoichiometric coefficients in the dissolution equation matter so much.

The assumption that no solid has formed before dilution is also essential. If precipitation had already occurred, you would need to account for the ions removed from solution.

Question 5

At \(25^\circ\mathrm{C}\), \(40.0\ \mathrm{mL}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) \(\ce{AgNO3}\) is mixed with \(60.0\ \mathrm{mL}\) of a solution containing:

• \(0.00150\ \mathrm{mol\,L^{-1}}\) \(\ce{NaCl}\);
• \(0.000500\ \mathrm{mol\,L^{-1}}\) \(\ce{KCl}\).

Assume that all soluble salts dissociate completely, volumes are additive, and the resulting silver chloride precipitate reaches equilibrium.

Use:

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

A student argues:

“Once precipitation stops, the silver and chloride concentrations must both equal \(\sqrt{K_{sp}}\).”

Determine whether this statement is correct.

Calculate:

(a) The initial \(Q_{sp}\), immediately after mixing.

(b) The equilibrium concentrations of silver and chloride ions after precipitation.

(c) The amount, in moles, of silver chloride precipitated.

Explain why simply taking the square root of \(K_{sp}\) gives the wrong equilibrium concentrations in this situation.

Solution 5

The student’s statement is incorrect. At equilibrium, silver ions remain in substantial excess, while the chloride concentration becomes extremely small.

This is a more demanding problem because we need to consider two different stages:

• The initial ion product, before precipitation.
• The final equilibrium concentrations, after precipitation has removed ions.

Step 1: Calculate the initial moles of silver ions.

Silver nitrate supplies one silver ion per formula unit.

\[ \begin{aligned} n(\mathrm{Ag}^{+}) &=0.0100\times0.0400\\ &=4.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 2: Calculate the total initial moles of chloride ions.

From sodium chloride:

\[ \begin{aligned} n(\mathrm{Cl}^{-})_{\ce{NaCl}} &=0.00150\times0.0600\\ &=9.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]

From potassium chloride:

\[ \begin{aligned} n(\mathrm{Cl}^{-})_{\ce{KCl}} &=0.000500\times0.0600\\ &=3.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Add both contributions:

\[ \begin{aligned} n(\mathrm{Cl}^{-})_{\text{total}} &=9.00\times10^{-5} +3.00\times10^{-5}\\ &=1.20\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the initial concentrations.

The total volume is:

\[ V_{\text{total}}=0.0400+0.0600 \]

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

Therefore:

\[ [\mathrm{Ag}^{+}]_{\text{initial}} = \frac{4.00\times10^{-4}}{0.1000} \]

\[ =4.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

And:

\[ [\mathrm{Cl}^{-}]_{\text{initial}} = \frac{1.20\times10^{-4}}{0.1000} \]

\[ =1.20\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate the initial \(Q_{sp}\).

For silver chloride:

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

Substituting:

\[ \begin{aligned} Q_{sp} &=(4.00\times10^{-3}) (1.20\times10^{-3})\\ &=4.80\times10^{-6} \end{aligned} \]

Since:

\[ 4.80\times10^{-6} > 1.8\times10^{-10} \]

precipitation is strongly favoured.

We now need to determine what remains dissolved after equilibrium is reached.

Step 5: Identify the limiting ion and the amount of excess silver.

Silver chloride forms according to:

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

The mole ratio is 1:1.

Initially, there are:

\[ 4.00\times10^{-4}\ \mathrm{mol\ Ag+} \]

and:

\[ 1.20\times10^{-4}\ \mathrm{mol\ Cl-} \]

Chloride is the limiting ion.

If precipitation removes one mole of silver for every mole of chloride, the difference between the dissolved silver and chloride moles remains constant.

The excess silver moles are:

\[ \begin{aligned} n_{\text{excess Ag+}} &=4.00\times10^{-4} -1.20\times10^{-4}\\ &=2.80\times10^{-4}\ \mathrm{mol} \end{aligned} \]

This difference cannot disappear through the precipitation reaction because silver and chloride are removed in equal mole amounts.

Divide by the fixed total volume to express this difference as a concentration:

\[ \begin{aligned} [\mathrm{Ag}^{+}]_{\text{eq}} - [\mathrm{Cl}^{-}]_{\text{eq}} &=\frac{2.80\times10^{-4}}{0.1000}\\ &=2.80\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This is our mass-balance constraint.

Step 6: Apply the equilibrium condition.

At equilibrium, solid silver chloride is present, so:

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

We already know that:

\[ [\mathrm{Ag}^{+}]_{\text{eq}} = [\mathrm{Cl}^{-}]_{\text{eq}}+2.80\times10^{-3} \]

Let:

\[ x=[\mathrm{Cl}^{-}]_{\text{eq}} \]

Therefore:

\[ [\mathrm{Ag}^{+}]_{\text{eq}} =x+2.80\times10^{-3} \]

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

\[ x(x+2.80\times10^{-3}) =1.8\times10^{-10} \]

Because \(K_{sp}\) is very small compared with the excess silver concentration, we expect \(x\) to be much smaller than \(2.80\times10^{-3}\).

We can therefore use the approximation:

\[ x+2.80\times10^{-3} \approx2.80\times10^{-3} \]

So:

\[ x(2.80\times10^{-3}) =1.8\times10^{-10} \]

Rearranging:

\[ \begin{aligned} x&=\frac{1.8\times10^{-10}}{2.80\times10^{-3}}\\ &=6.43\times10^{-8}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{[\mathrm{Cl}^{-}]_{\text{eq}} =6.43\times10^{-8}\ \mathrm{mol\,L^{-1}}} \]

The equilibrium silver concentration is:

\[ [\mathrm{Ag}^{+}]_{\text{eq}} = 6.43\times10^{-8}+2.80\times10^{-3} \]

Therefore:

\[ \boxed{[\mathrm{Ag}^{+}]_{\text{eq}} \approx2.80\times10^{-3}\ \mathrm{mol\,L^{-1}}} \]

Our approximation is valid because the chloride concentration is tiny compared with the excess silver concentration.

Step 7: Calculate the amount of silver chloride precipitated.

Almost all the initial chloride ions have been removed from solution.

The remaining chloride moles are:

\[ \begin{aligned} n(\mathrm{Cl}^{-})_{\text{eq}} &=[\mathrm{Cl}^{-}]_{\text{eq}}V\\ &=(6.43\times10^{-8})(0.1000)\\ &=6.43\times10^{-9}\ \mathrm{mol} \end{aligned} \]

Since each mole of chloride removed produces one mole of silver chloride:

\[ \begin{aligned} n(\ce{AgCl}) &=n(\mathrm{Cl}^{-})_{\text{initial}} -n(\mathrm{Cl}^{-})_{\text{eq}}\\ &=1.20\times10^{-4} -6.43\times10^{-9}\\ &=1.19994\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \boxed{n(\ce{AgCl})\approx1.20\times10^{-4}\ \mathrm{mol}} \]

Why was the square-root approach wrong?

For silver chloride:

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

Taking the square root gives:

\[ \sqrt{K_{sp}} \]

But this only gives the concentration of each ion if the two equilibrium concentrations are equal.

That would require the system to have equal amounts of silver and chloride available, as well as the appropriate equilibrium conditions.

In this question, silver was initially present in substantial excess.

The precipitation reaction removes silver and chloride in a 1:1 ratio, so that excess remains even after equilibrium is established.

The final concentrations must satisfy two conditions:

1. The product of the concentrations equals \(K_{sp}\).
2. The difference between the concentrations equals the amount of excess silver divided by the solution volume.

Ignoring the second condition produces an incorrect answer, even if you use the \(K_{sp}\) equation correctly.

This is the key distinction between calculating an initial ion product and calculating the final equilibrium composition.

09Where to go next

Calculating \(Q_{sp}\) tells you whether precipitation is thermodynamically favoured immediately after mixing.

If precipitation occurs, the next question is how much solid forms and what concentrations remain once equilibrium is reached. That requires combining the solubility-product expression with mole ratios and mass balance.

The common-ion effect becomes especially important at that stage because an ion already present in solution can significantly change the amount of a sparingly soluble salt that dissolves.

Those are the next steps beyond the initial \(Q_{sp}\) calculation.