Why Undissolved Solid Does Not Change Ksp at Equilibrium

Two beakers contain 250 mL of pure water at 25°C. You add 0.0200 g of silver chloride to the first and 0.200 g to the second. After stirring and allowing both mixtures to reach equilibrium, some white solid remains at the bottom of each beaker.

Would the second beaker contain ten times as much dissolved silver chloride because you added ten times as much solid?

No. If both solutions have reached equilibrium with excess silver chloride present, their dissolved ion concentrations will be the same.

Undissolved solid does not change \(K_{sp}\) because a pure solid has a constant activity, regardless of how much of it is present. At a fixed temperature, \(K_{sp}\) determines the equilibrium product of dissolved ion activities. Adding more of the same solid does not change that equilibrium constant.

The extra solid simply remains undissolved. It can affect how quickly equilibrium is reached, but not the equilibrium ion product.

01How could an experiment establish that extra solid does not change \(K_{sp}\)?

The obvious first experiment is to put different masses of silver chloride into equal volumes of water and compare how much remains.

But there’s a problem. Looking at the amount of solid tells us very little about the concentration of dissolved ions.

A large pile might mean the solution is saturated. It could also mean the solid has not had enough time to dissolve.

We need to measure what is happening in the liquid.

Designing the comparison

Silver chloride is a sparingly soluble ionic compound. When it dissolves, it separates into silver and chloride ions:

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

A useful investigation would compare two mixtures under controlled conditions:

• Beaker A receives 0.0200 g of solid AgCl.
• Beaker B receives 0.200 g of solid AgCl.
• Both contain 250 mL of initially pure water, maintained at 25°C.
• Both are stirred and allowed to reach equilibrium, with undissolved AgCl remaining.

If extra solid really increased the equilibrium ion concentrations, the liquid in Beaker B should contain more dissolved silver ions and chloride ions.

How could we test that prediction?

Following the apparatus

First, place both beakers in a temperature-controlled water bath. This maintains the same temperature throughout the investigation.

Next, add the different masses of AgCl and stir the mixtures. Allow sufficient time for the dissolved ion concentrations to become constant.

Once the solid has settled, withdraw a small sample of the clear liquid from each beaker using a pipette.

Pass each sample through a suitable fine filter to remove suspended AgCl particles. Keep the samples at the same temperature, and avoid introducing any additional chloride.

Finally, measure the dissolved silver ion concentrations using a suitably sensitive, calibrated analytical method.

The original idea of comparing the solid piles could not distinguish saturation from incomplete dissolution. The revised design measures the dissolved ions directly.

It also controls temperature and gives the crystals time to reach equilibrium.

There’s one more detail. We cannot assume that simply seeing undissolved solid proves equilibrium has been reached. The measured ion concentrations must have stopped changing.

02Why the solid is absent from the equilibrium expression

Picture the surface of a silver chloride crystal in water.

Some silver and chloride ions leave the crystal and become surrounded by water molecules. At the same time, dissolved ions can return to the crystal surface.

Initially, when the water contains almost no dissolved ions, dissolution is favoured.

As the concentrations increase, the reverse process becomes more significant.

Eventually, the rates of dissolution and crystallisation become equal. Ions still move between the crystal and solution, but there is no net change in the amount dissolved.

This is dynamic equilibrium.

For the equilibrium

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

the HSC concentration-based solubility product expression is

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

Here:

• \(K_{sp}\) is the solubility product constant.
• \([\mathrm{Ag}^{+}]\) is the equilibrium concentration of dissolved silver ions.
• \([\mathrm{Cl}^{-}]\) is the equilibrium concentration of dissolved chloride ions.
• Concentrations inside square brackets are measured in \(\mathrm{mol\,L^{-1}}\).

Notice that \(\ce{AgCl(s)}\) is missing from the expression.

Why isn’t there an \([\ce{AgCl(s)}]\) term?

Equilibrium constants are based on the activities of the substances involved.

Activity describes how a substance behaves in an equilibrium, accounting for its chemical environment.

For an ordinary pure crystalline solid, its activity is defined as 1 while that solid phase is present.

Doubling the mass of AgCl does not double its activity. The solid is still pure AgCl, and its activity remains 1.

The complete equilibrium expression therefore reduces to a product involving only the dissolved ions.

This is why the mass of undissolved solid does not appear in \(K_{sp}\).

A familiar comparison is sugar collecting at the bottom of a cup of saturated tea. At a fixed temperature, adding more sugar does not make the liquid indefinitely sweeter. The excess remains solid.

The comparison has a limit: sugar dissolves as molecules, whereas AgCl separates into ions. For ionic compounds, it is the product of ion activities, rather than a single dissolved concentration, that is fixed at equilibrium.

What the experiment should show

The following table contains calculated, illustrative results. These are predictions from the equilibrium model, not measurements from a reported experiment.

Here, M is shorthand for \(\mathrm{mol\,L^{-1}}\), not a mass unit.

The important result is that the clear liquids have identical predicted equilibrium ion concentrations despite the large difference in undissolved solid.

A real experiment finding this agreement would support the model. Measurement uncertainties, incomplete equilibration, contamination, and temperature variations would still need to be considered.

Does the amount of solid affect anything?

Yes. It can affect how quickly the mixture reaches equilibrium.

Finely divided crystals often provide more exposed surface area than a large crystal of the same mass. This can increase the rate at which ions exchange with the solution.

But increasing the rate of reaching equilibrium is different from changing the equilibrium constant.

If you crush a lump of AgCl into smaller particles without changing its chemical identity or temperature, it may reach equilibrium faster. Once equilibrium is reached, the value of \(K_{sp}\) is unchanged.

This assumes ordinary macroscopic crystals of the same solid phase. Changes in crystal phase or significant nanoscale surface effects require a more advanced treatment.

03The contrasting case: both beakers contain solid, but their dissolved concentrations differ

So far, the two beakers contained pure water.

What happens if we change the liquid instead?

Consider calcium fluoride, \(\ce{CaF2}\), in two different solutions:

• Beaker A contains pure water.
• Beaker B contains sodium fluoride solution, which already supplies dissolved fluoride ions.

Both receive excess solid calcium fluoride, and both remain at 25°C.

Should their dissolved calcium concentrations be identical because both contain undissolved \(\ce{CaF2}\)?

No. Beaker B already contains a common ion, \(\mathrm{F}^{-}\). This changes the equilibrium concentrations and reduces the amount of calcium fluoride that can dissolve.

It does not change \(K_{sp}\) at the same temperature.

Calcium fluoride dissolves according to

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

Unlike AgCl, the ion ratio is \(1:2\).

Every mole of calcium fluoride that dissolves produces one mole of calcium ions and two moles of fluoride ions.

Therefore,

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

The coefficient 2 becomes an exponent in the equilibrium expression.

The fluoride concentration is squared, but the calcium fluoride solid still does not appear.

04Why visible solid does not always prove the solution is saturated

Suppose you drop a large crystal of AgCl into pure water and immediately measure the dissolved silver concentration.

You might see plenty of undissolved solid but find that the ion concentration is still very low.

Has the value of \(K_{sp}\) changed?

No. The mixture simply has not reached equilibrium.

To distinguish the actual ion concentrations from the equilibrium value, chemists use the ion product, \(Q_{sp}\).

For AgCl:

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

The mathematical expression looks identical to the concentration-based \(K_{sp}\) expression.

The difference is when the concentrations are evaluated.

• \(Q_{sp}\) describes the ion product at the time being considered.
• \(K_{sp}\) describes the equilibrium value at the specified temperature.

Comparing them tells us which process is thermodynamically favoured.

These comparisons describe the thermodynamic tendency. Actual precipitation or dissolution can be delayed by kinetic effects, especially when suitable crystal surfaces are absent.

Three cases that are easy to confuse

Case A: Undissolved solid is present, and the system is at equilibrium.

The solution is saturated, and \(Q_{sp}=K_{sp}\).

Adding more of the same pure solid at the same temperature does not change the equilibrium ion product.

Case B: Undissolved solid is present, but equilibrium has not been reached.

For example, a recently added crystal is still dissolving.

The solution may have \(Q_{sp}<K_{sp}\), even though solid is visible.

Case C: No undissolved solid is present.

The solution might be unsaturated, or it might be exactly saturated.

The absence of visible solid is not enough to distinguish the two.

The deciding information is the ion product, together with the relevant equilibrium conditions.

What assumptions must an HSC calculation make?

Before using a concentration-based \(K_{sp}\) calculation, establish that:

1. The temperature is fixed.
2. The correct solid phase and dissolution equation have been identified.
3. Equilibrium has been reached if setting \(Q_{sp}=K_{sp}\).
4. All significant sources of common ions have been included.
5. Any approximation about ion concentrations is justified.
6. Side reactions, complex formation, and activity effects are negligible or appropriately accounted for.

The last point is a limitation of the simple HSC model.

Strictly, thermodynamic \(K_{sp}\) is expressed using dimensionless ion activities. Concentration products approximate those activities in sufficiently dilute solutions.

At higher ionic strengths, the concentration product may differ from the thermodynamic activity product. This does not mean that adding more solid has changed the true equilibrium constant.

05Questions and solutions

The following questions use illustrative teaching data unless otherwise stated. Assume a constant temperature of 25°C and ideal dilute-solution behaviour. Ignore complex formation, other significant ionic equilibria, and volume changes unless specified.

Question 1

A saturated solution of lead(II) iodide contains excess solid \(\ce{PbI2}\).

The equilibrium concentrations are:

\[ [\mathrm{Pb}^{2+}]=1.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

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

Calculate \(K_{sp}\).

An additional 0.500 g of solid lead(II) iodide is added at the same temperature. Explain what happens to \(K_{sp}\) and the equilibrium dissolved ion concentrations.

Solution 1

The solubility product is \(4.00\times10^{-9}\). Adding more pure solid \(\ce{PbI2}\) does not change this value or the equilibrium ion concentrations, provided equilibrium is maintained.

The dissolution equation is

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

The coefficient 2 on iodide becomes an exponent:

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

Substituting the given concentrations in \(\mathrm{mol\,L^{-1}}\):

\[ \begin{aligned} K_{sp}&=(1.00\times10^{-3})\\ &\quad\times(2.00\times10^{-3})^2\\ &=4.00\times10^{-9} \end{aligned} \]

The additional 0.500 g increases the amount of solid present.

It does not increase the activity of the pure solid, so it does not change \(K_{sp}\). At equilibrium, the dissolved ion concentrations remain those given in the question.

A tempting mistake is to include the mass of solid in the equilibrium expression. Only the dissolved ion activities contribute to this solubility product expression.

Question 2

Two flasks each initially contain 10.0 mL of saturated AgCl solution at 25°C.

Flask A also contains 0.200 mg of undissolved AgCl crystals.

Flask B contains only the clear saturated solution, with no solid.

The solubility product is

\[ K_{sp}=1.44\times10^{-10} \]

An additional 90.0 mL of pure water is added to each flask.

Assume volumes are additive, and both are left to reach equilibrium.

Determine:

(a) The ion product immediately after dilution, before any further dissolution.

(b) The final dissolved silver concentration in each flask.

(c) How much solid dissolves in Flask A, and whether any remains.

Use \(M(\ce{AgCl})=143.3\ \mathrm{g\,mol^{-1}}\).

Solution 2

Immediately after dilution, both flasks are unsaturated. Flask A subsequently returns to saturation by dissolving approximately 0.155 mg of AgCl, while Flask B remains unsaturated. The value of \(K_{sp}\) stays \(1.44\times10^{-10}\) in both.

Part (a): Calculate the immediate ion product.

Before dilution, each solution is saturated.

For AgCl in pure water:

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

\[ \begin{aligned} s&=\sqrt{1.44\times10^{-10}}\\ &=1.20\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The volume increases from 10.0 mL to 100.0 mL.

Before any further dissolution, the number of dissolved moles is unchanged. Each concentration therefore falls by a factor of 10:

\[ [\mathrm{Ag}^{+}]=[\mathrm{Cl}^{-}] =1.20\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Consequently,

\[ \begin{aligned} Q_{sp}&=[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\\ &=(1.20\times10^{-6})^2\\ &=1.44\times10^{-12} \end{aligned} \]

Since

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

both solutions are initially unsaturated after dilution.

Part (b): Determine the final concentrations.

Flask A contains a source of additional AgCl.

Its solid can dissolve until the solution returns to saturation, provided enough solid is available.

The final concentration in Flask A is therefore

\[ [\mathrm{Ag}^{+}]_A=1.20\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Flask B has no AgCl crystals left to dissolve.

Assuming no other source of silver or chloride ions, its concentrations remain at their diluted values:

\[ [\mathrm{Ag}^{+}]_B=1.20\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

Part (c): Check the amount of AgCl required in Flask A.

The increase in dissolved silver concentration is

\[ \begin{aligned} \Delta[\mathrm{Ag}^{+}]&=1.20\times10^{-5}\\ &\quad-1.20\times10^{-6}\\ &=1.08\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

With a final volume of 0.100 L, the additional amount of AgCl needed is

\[ \begin{aligned} n&=(1.08\times10^{-5})(0.100)\\ &=1.08\times10^{-6}\ \mathrm{mol} \end{aligned} \]

Its mass is

\[ \begin{aligned} m&=(1.08\times10^{-6})(143.3)\\ &=1.55\times10^{-4}\ \mathrm{g}\\ &=0.155\ \mathrm{mg} \end{aligned} \]

Flask A initially contained 0.200 mg of solid.

Approximately

\[ 0.200-0.155=0.045\ \mathrm{mg} \]

remains, confirming that saturation can be reached.

The tempting prediction is that both flasks must return to the same concentration because they started saturated.

The missing constraint is the supply of solid. Only Flask A has enough AgCl available to replace the dissolved ions lost in concentration through dilution.

Question 3

A beaker contains 50.0 mL of saturated AgCl solution at 25°C, together with excess solid AgCl.

Initially,

\[ [\mathrm{Ag}^{+}]=[\mathrm{Cl}^{-}] =1.20\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Another 50.0 mL of \(0.00100\ \mathrm{mol\,L^{-1}}\) NaCl solution is added.

Assume the final volume is 100.0 mL, NaCl dissociates completely, and no solid is lost.

The solubility product remains

\[ K_{sp}=1.44\times10^{-10} \]

(a) Calculate \(Q_{sp}\) immediately after mixing, before precipitation or dissolution occurs.

(b) Predict whether AgCl precipitates or dissolves.

(c) Estimate the final \([\mathrm{Ag}^{+}]\) and the net mass of AgCl precipitated.

Use \(M(\ce{AgCl})=143.3\ \mathrm{g\,mol^{-1}}\).

Solution 3

AgCl precipitates, even though the original solution is diluted. The initial ion product after mixing is \(3.04\times10^{-9}\), which exceeds \(K_{sp}\). At the new equilibrium, \([\mathrm{Ag}^{+}]\) is approximately \(2.88\times10^{-7}\ \mathrm{mol\,L^{-1}}\), and about \(8.18\times10^{-5}\) g of AgCl has precipitated.

Part (a): Calculate the concentrations immediately after mixing.

Silver ions are supplied only by the original saturated solution.

Their concentration is halved:

\[ \begin{aligned} [\mathrm{Ag}^{+}]&=\frac{(1.20\times10^{-5})(50.0)}{100.0}\\ &=6.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Chloride ions come from both solutions.

\[ \begin{aligned} [\mathrm{Cl}^{-}]&=\frac{(1.20\times10^{-5})(50.0)}{100.0}\\ &\quad+\frac{(0.00100)(50.0)}{100.0}\\ &=5.06\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The volumes in these ratios are both in mL, so the volume units cancel.

Now calculate \(Q_{sp}\):

\[ \begin{aligned} Q_{sp}&=(6.00\times10^{-6})\\ &\quad\times(5.06\times10^{-4})\\ &=3.04\times10^{-9} \end{aligned} \]

This is greater than \(K_{sp}=1.44\times10^{-10}\).

Part (b): Predict the direction of change.

AgCl precipitates.

The solution initially contains too large a dissolved ion product for the new equilibrium.

The tempting prediction is that adding liquid must make more solid dissolve because the solution has been diluted.

But the added liquid contains chloride ions. The common-ion increase is much more important than the dilution of the original ions.

Part (c): Find the final silver concentration and precipitated mass.

At equilibrium,

\[ [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]=1.44\times10^{-10} \]

The added NaCl contributes \(5.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) chloride after dilution.

Because precipitation removes silver and chloride in a \(1:1\) ratio, the difference between their final concentrations remains

\[ [\mathrm{Cl}^{-}]-[\mathrm{Ag}^{+}] =5.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Let final \([\mathrm{Ag}^{+}]=a\), in \(\mathrm{mol\,L^{-1}}\).

Then:

\[ a(a+5.00\times10^{-4}) =1.44\times10^{-10} \]

Since \(a\) is expected to be much smaller than \(5.00\times10^{-4}\), approximate:

\[ \begin{aligned} a(5.00\times10^{-4})&=1.44\times10^{-10}\\ a&=2.88\times10^{-7} \end{aligned} \]

The neglected value is only about 0.058% of the background chloride concentration, so the approximation is appropriate.

The decrease in dissolved silver concentration is

\[ \begin{aligned} \Delta[\mathrm{Ag}^{+}]&=6.00\times10^{-6}\\ &\quad-2.88\times10^{-7}\\ &=5.71\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

In 0.100 L:

\[ \begin{aligned} n&=(5.71\times10^{-6})(0.100)\\ &=5.71\times10^{-7}\ \mathrm{mol} \end{aligned} \]

This represents the additional AgCl formed.

\[ \begin{aligned} m&=(5.71\times10^{-7})(143.3)\\ &=8.18\times10^{-5}\ \mathrm{g} \end{aligned} \]

Approximately 0.0818 mg of AgCl precipitates.

The amount of solid increases, while \(K_{sp}\) stays constant. The dissolved concentrations adjust to restore the equilibrium ion product.

Question 4

A flask initially contains 250 mL of pure water and 1.00 mg of calcium fluoride, \(\ce{CaF2}\), at 25°C.

Use

\[ K_{sp}(\ce{CaF2})=4.00\times10^{-11} \]

and

\[ M(\ce{CaF2})=78.08\ \mathrm{g\,mol^{-1}} \]

The flask undergoes the following changes:

1. The contents are allowed to reach equilibrium.
2. Water is evaporated until the total liquid volume becomes 40.0 mL. No dissolved ions or solid are lost, and the contents are again allowed to equilibrate at 25°C.
3. Exactly 20.0 mL of clear liquid is carefully withdrawn without removing any solid.
4. Another 20.0 mL of pure water is added, returning the volume to 40.0 mL. The mixture is allowed to reach equilibrium again.

Determine whether solid CaF₂ is present after stages 1, 2, and 4.

Calculate the mass of solid after stage 2 and the final ion product after stage 4.

Solution 4

All the calcium fluoride dissolves at stage 1. About 0.327 mg is solid after stage 2. Surprisingly, no solid remains after stage 4: the final solution is slightly unsaturated.

The important constraint is that removing saturated liquid also removes dissolved calcium fluoride. The amount left after the replacement is no longer sufficient to saturate 40.0 mL of solution.

Stage 1: Check the initial amount against saturation.

For calcium fluoride dissolving in pure water:

\[ K_{sp}=s(2s)^2=4s^3 \]

Thus the saturated molar solubility is

\[ \begin{aligned} s&=\sqrt[3]{\frac{4.00\times10^{-11}}{4}}\\ &=2.15\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

If all 1.00 mg dissolves in 0.250 L, its concentration is

\[ \begin{aligned} s_1&=\frac{0.00100}{(78.08)(0.250)}\\ &=5.12\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This is less than the saturated molar solubility.

Equivalently,

\[ \begin{aligned} Q_{sp}&=4s_1^3\\ &=5.38\times10^{-13} \end{aligned} \]

Since \(Q_{sp}<K_{sp}\), the entire 1.00 mg dissolves and the liquid remains unsaturated.

Stage 2: Determine what happens during evaporation.

At 40.0 mL, a saturated solution can contain

\[ \begin{aligned} n&=(2.15\times10^{-4})(0.0400)\\ &\approx8.62\times10^{-6}\ \mathrm{mol} \end{aligned} \]

Using the unrounded solubility, the corresponding mass is approximately

\[ m_{\text{dissolved}}=0.673\ \mathrm{mg} \]

The original 1.00 mg of calcium fluoride is still in the flask, but only 0.673 mg can remain dissolved at equilibrium.

Therefore,

\[ \begin{aligned} m_{\text{solid}}&=1.00-0.673\\ &=0.327\ \mathrm{mg} \end{aligned} \]

Solid forms as the liquid becomes concentrated.

Stages 3 and 4: Track the amount removed with the clear liquid.

The 40.0 mL solution is saturated when 20.0 mL is withdrawn.

Half the saturated liquid is removed, so half its dissolved calcium fluoride is also removed:

\[ \begin{aligned} m_{\text{removed}}&=\frac{0.673}{2}\\ &\approx0.336\ \mathrm{mg} \end{aligned} \]

The total calcium fluoride left in the flask, counting solid and dissolved material, is approximately

\[ \begin{aligned} m_{\text{total}}&=1.00-0.336\\ &=0.664\ \mathrm{mg} \end{aligned} \]

But 40.0 mL of saturated solution requires approximately 0.673 mg of dissolved calcium fluoride.

There is now less material in the flask than the amount required to reach saturation.

Consequently, all remaining crystals dissolve.

Calculate the final ion product.

Using the unrounded values from the previous calculations, the remaining mass is approximately \(6.64\times10^{-4}\) g.

The final calcium concentration is

\[ \begin{aligned} [\mathrm{Ca}^{2+}]&=\frac{6.64\times10^{-4}}{(78.08)(0.0400)}\\ &\approx2.12\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Because calcium and fluoride remain in a \(1:2\) ratio:

\[ [\mathrm{F}^{-}]\approx4.25\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Therefore,

\[ \begin{aligned} Q_{sp}&=[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\\ &\approx3.84\times10^{-11} \end{aligned} \]

This is slightly below \(K_{sp}=4.00\times10^{-11}\).

The final solution is unsaturated.

The tempting mistake is to see crystals after evaporation and assume they must remain after the water replacement.

That ignores the calcium fluoride removed in the 20.0 mL of clear liquid.

The equilibrium constant never changed. What changed was the total amount of material available to produce a saturated solution.

Question 5

Two hypothetical sparingly soluble salts, MX and MY, share the same monovalent cation \(\mathrm{M}^{+}\).

Their dissolution equilibria are

\[ \mathrm{MX(s)}\rightleftharpoons \mathrm{M}^{+}(aq)+\mathrm{X}^{-}(aq) \]

\[ \mathrm{MY(s)}\rightleftharpoons \mathrm{M}^{+}(aq)+\mathrm{Y}^{-}(aq) \]

At 25°C, their illustrative solubility products are

\[ K_{sp}(\mathrm{MX})=1.00\times10^{-8} \]

\[ K_{sp}(\mathrm{MY})=1.00\times10^{-10} \]

A 1.00 L vessel initially contains an equilibrium solution in contact with both solids.

Its dissolved concentrations are

\[ [\mathrm{M}^{+}]=1.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{X}^{-}]=1.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{Y}^{-}]=1.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

There is plenty of solid MX, but only \(2.00\times10^{-5}\) mol of solid MY.

A source of \(\mathrm{X}^{-}\), accompanied by an inert counter-ion, is introduced so that the final equilibrium concentration of \(\mathrm{X}^{-}\) is maintained at exactly \(0.0100\ \mathrm{mol\,L^{-1}}\).

Assume the volume remains 1.00 L, temperature stays constant, no \(\mathrm{M}^{+}\) or \(\mathrm{Y}^{-}\) enters or leaves from the external source, and all reactions reach equilibrium. Spectator ions maintain charge balance without participating in these equilibria. Both solids retain their original crystal structures.

Determine:

(a) Whether both solids can remain at the final equilibrium.

(b) The final concentrations of \(\mathrm{M}^{+}\) and \(\mathrm{Y}^{-}\).

(c) The net amount of MX solid formed.

(d) Whether either \(K_{sp}\) has changed.

Solution 5

Only MX remains as a solid at the final equilibrium. All the MY dissolves, while \(1.19\times10^{-4}\) mol of additional MX precipitates. Both solubility product constants remain unchanged.

The surprising part is that increasing the concentration of one anion causes one solid to precipitate while the other solid dissolves. The reason is the shared \(\mathrm{M}^{+}\) ion, together with the limited amount of MY available.

Part (a): Test the assumption that both solids remain.

Suppose both solids were still present at equilibrium.

For MX:

\[ K_{sp}(\mathrm{MX}) =[\mathrm{M}^{+}][\mathrm{X}^{-}] \]

Therefore,

\[ \begin{aligned} [\mathrm{M}^{+}]&=\frac{1.00\times10^{-8}}{0.0100}\\ &=1.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

If MY also remained solid, its equilibrium would require

\[ K_{sp}(\mathrm{MY}) =[\mathrm{M}^{+}][\mathrm{Y}^{-}] \]

Therefore,

\[ \begin{aligned} [\mathrm{Y}^{-}]&=\frac{1.00\times10^{-10}}{1.00\times10^{-6}}\\ &=1.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This would require the dissolved \(\mathrm{Y}^{-}\) concentration to increase from \(1.00\times10^{-6}\) to \(1.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\).

In the 1.00 L vessel, the extra amount of \(\mathrm{Y}^{-}\) required would be

\[ \begin{aligned} \Delta n_Y&=(1.00\times10^{-4})\\ &\quad-(1.00\times10^{-6})\\ &=9.90\times10^{-5}\ \mathrm{mol} \end{aligned} \]

But there are only \(2.00\times10^{-5}\) mol of solid MY available.

That is not enough to supply the required \(\mathrm{Y}^{-}\).

Our assumption that both solids remain is impossible.

All the MY must dissolve.

Part (b): Calculate the actual final concentrations.

Since excess MX remains in equilibrium with the solution, its solubility product still applies:

\[ [\mathrm{M}^{+}] =1.00\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

All \(2.00\times10^{-5}\) mol of MY dissolves into 1.00 L, supplying an additional \(2.00\times10^{-5}\ \mathrm{mol\,L^{-1}}\) of \(\mathrm{Y}^{-}\).

Thus,

\[ \begin{aligned} [\mathrm{Y}^{-}]&=1.00\times10^{-6}\\ &\quad+2.00\times10^{-5}\\ &=2.10\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

We must now check whether the absence of solid MY is consistent with equilibrium.

Its final ion product is

\[ \begin{aligned} Q_{sp}(\mathrm{MY})&=[\mathrm{M}^{+}][\mathrm{Y}^{-}]\\ &=(1.00\times10^{-6})\\ &\quad\times(2.10\times10^{-5})\\ &=2.10\times10^{-11} \end{aligned} \]

Since

\[ Q_{sp}(\mathrm{MY})<K_{sp}(\mathrm{MY}) \]

the solution is unsaturated with respect to MY.

It is therefore consistent that no MY solid remains.

Part (c): Calculate the net amount of MX precipitated.

The initial amount of dissolved \(\mathrm{M}^{+}\) in 1.00 L is

\[ 1.00\times10^{-4}\ \mathrm{mol} \]

Dissolving all the MY releases another

\[ 2.00\times10^{-5}\ \mathrm{mol} \]

The final solution contains only

\[ 1.00\times10^{-6}\ \mathrm{mol} \]

of dissolved \(\mathrm{M}^{+}\).

Since the external source supplies no \(\mathrm{M}^{+}\), and MX is the only remaining solid that can take up these ions, the amount incorporated into additional MX is

\[ \begin{aligned} n_{\text{MX formed}} &=1.00\times10^{-4}\\ &\quad+2.00\times10^{-5}\\ &\quad-1.00\times10^{-6}\\ &=1.19\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Thus \(1.19\times10^{-4}\) mol of MX precipitates.

Part (d): Explain what happened to the equilibrium constants.

Neither solubility product changed.

The final solution remains saturated with respect to MX:

\[ \begin{aligned} Q_{sp}(\mathrm{MX})&=(1.00\times10^{-6})(0.0100)\\ &=1.00\times10^{-8}\\ &=K_{sp}(\mathrm{MX}) \end{aligned} \]

But it is unsaturated with respect to MY:

\[ Q_{sp}(\mathrm{MY}) =2.10\times10^{-11}<K_{sp}(\mathrm{MY}) \]

The extra \(\mathrm{X}^{-}\) favours formation of MX and lowers the dissolved \(\mathrm{M}^{+}\) concentration.

That lower \(\mathrm{M}^{+}\) concentration favours dissolution of MY.

If there were unlimited MY solid, enough could dissolve to maintain both saturation conditions. But the actual supply of MY is finite, so it runs out first.

The tempting shortcut was to impose both \(K_{sp}\) equations at the end simply because both solids were present at the beginning.

That ignores conservation of matter.

An equilibrium expression can tell us the concentrations required while a particular solid phase remains present. It cannot manufacture extra solid when the available amount is insufficient.

06The next decision: is the solution actually at equilibrium?

When you see undissolved solid in an HSC Chemistry question, don’t immediately calculate \(K_{sp}\) from whatever concentrations are given.

First decide whether the system has reached equilibrium, whether the correct solid is still present, and whether other substances have supplied common ions.

At fixed temperature, adding more of the same pure solid will not change \(K_{sp}\). But dilution, mixing, precipitation, or removing part of the solution can change \(Q_{sp}\), sometimes enough to make the last crystals disappear or new solid form.

That leads directly to the next useful skill: calculating \(Q_{sp}\) after mixing solutions with different volumes. The key is to calculate the actual dissolved ion concentrations first, compare their product with \(K_{sp}\), and then check whether the proposed equilibrium is physically possible.