Why Molar Solubility Is Not Always an Ion Concentration

A saturated solution has a molar solubility of \(0.600\ \mathrm{mol\,L^{-1}}\). Does that mean every ion in the solution has a concentration of \(0.600\ \mathrm{mol\,L^{-1}}\)?

No. If the dissolved solid is potassium sulfate, \(\ce{K2SO4}\), the solution contains \(1.20\ \mathrm{mol\,L^{-1}}\) of potassium ions but only \(0.600\ \mathrm{mol\,L^{-1}}\) of sulfate ions.

The reason is that one formula unit of potassium sulfate produces two potassium ions and one sulfate ion.

There is a second complication. What if the water already contains some of those ions before the solid dissolves? The final ion concentration must include both the ions already present and those produced by dissolution.

These are the two reasons molar solubility is not always equal to every ion concentration:

• Dissolution stoichiometry: Different ions are produced in different ratios.
• Common ions: Some ions may already be present before the sparingly soluble solid dissolves.

Let’s work through both, starting with how the ions are actually produced.

01Molar solubility counts dissolved formula units, not individual ions

Consider potassium sulfate dissolving in water.

\[ \ce{K2SO4(s) -> 2K+(aq) + SO4^2-(aq)} \]

Before looking at the numbers, predict how many moles of potassium ions are produced when one mole of potassium sulfate dissolves.

The answer is two moles.

The balanced equation tells us that every mole of potassium sulfate produces:

• 2 mol of \(\mathrm{K}^{+}\)
• 1 mol of \(\mathrm{SO_{4}}^{2-}\)

What does molar solubility actually measure?

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

Its unit is:

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

Suppose the molar solubility of potassium sulfate is \(s\).

The dissolution equation gives us:

Therefore:

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

\[ [\mathrm{SO_{4}}^{2-}]=s \]

If \(s=0.600\ \mathrm{mol\,L^{-1}}\):

\[ [\mathrm{K}^{+}]=2(0.600) =1.20\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{SO_{4}}^{2-}]=0.600\ \mathrm{mol\,L^{-1}} \]

The molar solubility is not the potassium concentration. It measures how much potassium sulfate dissolves, while the potassium concentration measures how many potassium ions that dissolution produces.

Why do some salts give equal ion concentrations?

Consider sodium chloride:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

One mole of sodium chloride produces one mole of sodium ions and one mole of chloride ions.

If it dissolves in pure water, with no other sources of either ion:

\[ [\mathrm{Na}^{+}]=s \]

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

Both concentrations happen to equal the molar solubility.

This is where the misconception starts. Students see a 1:1 salt and assume the same relationship applies to every ionic compound.

It doesn’t.

The coefficients in the balanced dissolution equation determine the relationship.

For example:

Notice that the coefficients, rather than the ionic charges, determine the mole ratios.

02How an experiment can reveal the difference

There is a practical measurement problem here.

Suppose you want to determine the molar solubility of potassium sulfate. You cannot simply weigh the solid you initially add to water and assume that all of it dissolves.

If excess solid remains, some of the original mass is still sitting at the bottom of the flask.

You need to measure what actually entered the solution.

Following the experiment

One proposed investigation uses a gravimetric method to measure sulfate ions.

First, add excess solid potassium sulfate to deionised water in a flask. Keep the flask in a constant-temperature water bath at \(25.0^\circ\mathrm{C}\), and stir the mixture.

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

Some solid should remain in the flask. This confirms that the solution has reached saturation.

Next, filter the mixture at the same temperature. Collect the clear filtrate in a clean flask.

Why filter it?

Tiny undissolved crystals could enter the sample and interfere with the measurement. We want to measure dissolved sulfate, not suspended potassium sulfate particles.

What should the experiment show?

Take a measured aliquot of the clear potassium sulfate solution.

Add excess barium chloride solution. The sulfate ions react with barium ions to form insoluble barium sulfate:

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

The barium sulfate precipitate can be filtered, washed, dried, and weighed.

The reaction has a 1:1 mole ratio:

\[ n(\ce{BaSO4})=n(\mathrm{SO_{4}}^{2-}) \]

Once we know the sulfate concentration, the original dissolution equation tells us the potassium concentration.

The prediction is that the potassium concentration should be twice the sulfate concentration, provided potassium sulfate is the only source of these ions.

03The second complication: a common ion is already present

So far, we have assumed that the dissolving solid is the only source of its ions.

That assumption is not always valid.

Imagine adding silver chloride to a solution that already contains sodium chloride.

Silver chloride dissolves according to:

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

Sodium chloride dissociates:

\[ \ce{NaCl(aq) -> Na+(aq) + Cl-(aq)} \]

Both compounds supply chloride ions.

The chloride concentration therefore cannot be calculated using the amount of silver chloride dissolved alone.

Separating the two sources of chloride

Suppose the initial sodium chloride solution has a concentration of \(0.0100\ \mathrm{mol\,L^{-1}}\).

Let \(s\) represent the molar solubility of silver chloride in this solution.

For every mole of silver chloride that dissolves:

• One mole of \(\mathrm{Ag}^{+}\) is produced.
• One mole of \(\mathrm{Cl}^{-}\) is produced.

However, the sodium chloride has already supplied \(0.0100\ \mathrm{mol\,L^{-1}}\) of chloride ions.

The equilibrium concentrations are therefore:

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

\[ [\mathrm{Cl}^{-}]=0.0100+s \]

The extra \(0.0100\) is not produced by silver chloride. It comes from the sodium chloride already in the solution.

What exactly is a common ion?

A common ion is an ion that appears in the dissolution equation of a sparingly soluble salt and is already present in the solution from another source.

In this example, chloride is the common ion.

Sodium is not a common ion because it does not appear in the dissolution equation of silver chloride.

This distinction matters when writing a \(K_{sp}\) expression. Only the ions involved in the dissolution equilibrium appear in the expression.

For silver chloride:

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

Substituting the actual equilibrium concentrations:

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

Here, \(c_0\) is the initial chloride concentration supplied by sodium chloride.

It must be the concentration after any mixing or dilution, not necessarily the concentration printed on the original stock solution.

04A reliable method for finding ion concentrations

Before starting a solubility calculation, separate the contributions from the dissolving solid and any ions already present.

A useful general model is:

\[ \mathrm{A}_m\mathrm{B}_n(s) \rightleftharpoons m\mathrm{A}(aq)+n\mathrm{B}(aq) \]

Here, \(s\) is the molar solubility, while \(m\) and \(n\) are the stoichiometric coefficients.

If the initial concentrations of the two ions are \(c_{A,0}\) and \(c_{B,0}\), the equilibrium concentrations are:

\[ \boxed{[\mathrm{A}] = c_{A,0}+ms} \]

\[ \boxed{[\mathrm{B}] = c_{B,0}+ns} \]

These relationships assume that dissolution is the only process changing the ion concentrations and that the solution volume is accounted for.

If an ion is not initially present, its initial concentration is zero.

The decision process

Use this sequence when solving an HSC solubility question.

1. Write the balanced dissolution equation. Identify every ion and its coefficient.

2. Define \(s\). This is the molar solubility, not automatically the concentration of every ion.

3. Multiply by the coefficients. Work out how much of each ion is produced by dissolution.

4. Check for common ions. Look for soluble compounds that have already supplied one or more of the ions.

5. Add the contributions. Calculate the total equilibrium concentration of each relevant ion.

6. Use the correct concentrations in \(K_{sp}\). Do not substitute \(s\) for every ion concentration without checking the dissolution equation and initial conditions.

Do not confuse concentration units

Molar solubility and ion concentration are both expressed in:

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

However, solubility may also be reported as a mass concentration, such as \(\mathrm{g\,L^{-1}}\).

These are not interchangeable.

To convert a mass concentration into a molar concentration:

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

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

If a question gives grams per \(100\ \mathrm{g}\) of water, you cannot directly treat that value as grams per litre. Additional information is needed to convert it accurately.

Common mistakes to avoid

One final limitation: these calculations generally assume complete dissociation of soluble ionic compounds, no significant side reactions, and concentration-based equilibrium expressions. Complex formation, acid-base reactions, and other precipitation reactions can change the actual concentrations. More rigorous treatments use ion activities rather than concentrations.

05Questions and solutions

All numerical values in the following questions are illustrative teaching data unless a value is explicitly supplied.

Question 1

A saturated solution of magnesium fluoride, \(\ce{MgF2}\), in pure water has a measured molar solubility of:

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

Calculate the equilibrium concentrations of:

(a) Magnesium ions.

(b) Fluoride ions.

Solution 1

The concentrations are:

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

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

The important step is recognising that magnesium fluoride produces two fluoride ions for every magnesium ion.

The dissolution equation is:

\[ \ce{MgF2(s) <=> Mg^2+(aq) + 2F-(aq)} \]

Therefore:

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

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

Substitute the given molar solubility:

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

\[ \begin{aligned} [\mathrm{F}^{-}] &=2(1.00\times10^{-3})\\ &=2.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The fluoride concentration is twice the magnesium concentration because of the 1:2 stoichiometric ratio.

Question 2

A solution initially contains \(0.00500\ \mathrm{mol\,L^{-1}}\) sodium chloride.

Magnesium chloride is dissolved into this solution. After mixing, the measured equilibrium chloride concentration is:

\[ [\mathrm{Cl}^{-}]=0.0200\ \mathrm{mol\,L^{-1}} \]

Assume the sodium chloride concentration remains \(0.00500\ \mathrm{mol\,L^{-1}}\) in the final solution and there are no other sources of chloride ions.

Calculate:

(a) The concentration of dissolved magnesium chloride, expressed in \(\mathrm{mol\,L^{-1}}\).

(b) The magnesium ion concentration.

Solution 2

The magnesium chloride concentration is:

\[ \boxed{0.00750\ \mathrm{mol\,L^{-1}}} \]

The magnesium ion concentration is also:

\[ \boxed{[\mathrm{Mg}^{2+}]=0.00750\ \mathrm{mol\,L^{-1}}} \]

The key is to subtract the chloride ions already supplied by sodium chloride before using the 1:2 dissolution ratio.

The dissolution equation is:

\[ \ce{MgCl2(s) -> Mg^2+(aq) + 2Cl-(aq)} \]

Let \(c\) represent the concentration of dissolved magnesium chloride.

Each mole produces two moles of chloride ions, so:

\[ [\mathrm{Cl}^{-}]=0.00500+2c \]

Substitute the measured chloride concentration:

\[ 0.0200=0.00500+2c \]

Rearrange:

\[ \begin{aligned} 2c&=0.0200-0.00500\\ &=0.0150 \end{aligned} \]

\[ c=\frac{0.0150}{2} \]

\[ \boxed{c=0.00750\ \mathrm{mol\,L^{-1}}} \]

Each mole of magnesium chloride produces one mole of magnesium ions.

Therefore:

\[ [\mathrm{Mg}^{2+}]=c \]

\[ \boxed{[\mathrm{Mg}^{2+}]=0.00750\ \mathrm{mol\,L^{-1}}} \]

A tempting mistake would be to divide the total chloride concentration by two:

\[ \frac{0.0200}{2}=0.0100 \]

That incorrectly attributes all the chloride ions to magnesium chloride.

The initial sodium chloride concentration must be removed first.

Note that \(c\) here is the actual dissolved formula concentration. It would only be called the molar solubility if the solution were saturated.

Question 3

A \(20.0\ \mathrm{mL}\) aliquot of a filtered, saturated silver sulfate solution is diluted to a final volume of \(100.0\ \mathrm{mL}\).

Analysis of the diluted sample gives:

\[ [\mathrm{Ag}^{+}]=0.0100\ \mathrm{mol\,L^{-1}} \]

Assume silver sulfate is the only source of silver and sulfate ions.

Calculate:

(a) The original molar solubility of silver sulfate.

(b) The original sulfate ion concentration.

(c) Explain why additional silver sulfate does not dissolve during the dilution.

Solution 3

The original molar solubility is:

\[ \boxed{s=0.0250\ \mathrm{mol\,L^{-1}}} \]

The original sulfate concentration is:

\[ \boxed{[\mathrm{SO_{4}}^{2-}]=0.0250\ \mathrm{mol\,L^{-1}}} \]

The crucial point is that the aliquot was filtered before dilution. No solid remains to dissolve further.

Step 1: Calculate the original silver ion concentration.

The dilution relationship is:

\[ C_1V_1=C_2V_2 \]

Rearrange:

\[ C_1=\frac{C_2V_2}{V_1} \]

Substitute:

\[ \begin{aligned} [\mathrm{Ag}^{+}]_{\text{original}} &=\frac{0.0100\times100.0}{20.0}\\ &=0.0500\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The original saturated solution therefore contained \(0.0500\ \mathrm{mol\,L^{-1}}\) of silver ions.

Step 2: Write the dissolution equation.

\[ \ce{Ag2SO4(s) <=> 2Ag+(aq) + SO4^2-(aq)} \]

Two moles of silver ions are produced for every mole of silver sulfate that dissolves.

Therefore:

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

Rearrange:

\[ s=\frac{[\mathrm{Ag}^{+}]}{2} \]

\[ \begin{aligned} s&=\frac{0.0500}{2}\\ &=0.0250\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 3: Calculate the sulfate concentration.

The dissolution equation produces one mole of sulfate for every mole of silver sulfate.

Therefore:

\[ [\mathrm{SO_{4}}^{2-}]=s \]

\[ \boxed{[\mathrm{SO_{4}}^{2-}]=0.0250\ \mathrm{mol\,L^{-1}}} \]

Step 4: Explain what happens during dilution.

The original solution was filtered, so no excess silver sulfate solid was transferred into the volumetric flask.

Adding water therefore dilutes the dissolved ions. There is no solid available to dissolve further and restore saturation.

This distinction is important:

• Diluting a filtered solution simply reduces its concentrations.
• Diluting a saturated solution while excess solid remains can allow more solid to dissolve.

The two situations do not produce the same result.

Question 4

A clear solution at \(25.0^\circ\mathrm{C}\) initially contains:

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

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

A small silver chloride seed crystal is introduced.

Use:

\[ K_{sp}=1.8\times10^{-10} \]

Determine:

(a) Whether silver chloride dissolves or precipitates.

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

(c) The net amount of silver chloride formed per litre.

Assume precipitation is the only process changing the silver and chloride concentrations, and the initial solution is able to reach equilibrium.

Solution 4

Silver chloride precipitates.

The final equilibrium concentrations are:

\[ \boxed{[\mathrm{Ag}^{+}]=[\mathrm{Cl}^{-}]=1.34\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

The net amount of silver chloride formed is:

\[ \boxed{9.87\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

This question tests something slightly different: the system is not initially dissolving more solid. It is already supersaturated.

Step 1: Calculate the initial reaction quotient, \(Q_{sp}\).

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

Substitute the initial concentrations:

\[ \begin{aligned} Q_{sp} &=(1.00\times10^{-3})^2\\ &=1.00\times10^{-6} \end{aligned} \]

Compare with the solubility product:

\[ Q_{sp}=1.00\times10^{-6} \]

\[ K_{sp}=1.8\times10^{-10} \]

Therefore:

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

The solution is supersaturated, so the net reaction proceeds towards precipitation.

Step 2: Define the amount precipitated.

Let \(x\) represent the number of moles of silver chloride precipitated per litre.

We define \(x\) as positive for precipitation.

The reaction is:

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

Each mole of silver chloride formed removes one mole of silver ions and one mole of chloride ions.

Therefore:

\[ [\mathrm{Ag}^{+}]=0.00100-x \]

\[ [\mathrm{Cl}^{-}]=0.00100-x \]

Step 3: Apply the equilibrium condition.

At equilibrium:

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

Substitute:

\[ 1.8\times10^{-10}=(0.00100-x)^2 \]

Take the square root:

\[ 0.00100-x=\sqrt{1.8\times10^{-10}} \]

\[ 0.00100-x=1.34\times10^{-5} \]

Therefore, the final concentrations are:

\[ \boxed{[\mathrm{Ag}^{+}]=[\mathrm{Cl}^{-}]=1.34\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

Step 4: Calculate the amount precipitated.

Rearrange:

\[ \begin{aligned} x&=0.00100-1.34\times10^{-5}\\ &=9.87\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This is the net amount of silver chloride formed per litre.

Notice that we did not use a positive molar solubility \(s\) to represent this change.

We used \(x\) as a precipitation variable, so the dissolved ion concentrations decreased.

This is the same stoichiometric bookkeeping used in dissolution problems, but the direction of the reaction has changed.

Question 5

An illustrative sparingly soluble ionic solid has the formula \(\ce{M2X}\) and dissolves according to:

\[ \ce{M2X(s) <=> 2M+(aq) + X^2-(aq)} \]

At a fixed temperature:

\[ K_{sp}=1.0\times10^{-10} \]

A solution is prepared containing \(0.100\ \mathrm{mol\,L^{-1}}\) of soluble \(\ce{MNO3}\), together with excess solid \(\ce{M2X}\).

The solution reaches saturation.

It is then diluted tenfold with pure water. Excess \(\ce{M2X}\) remains present throughout the dilution and subsequent equilibration.

Assume:

• \(\ce{MNO3}\) dissociates completely.
• \(\ce{NO3-}\) is a spectator ion.
• There are no other reactions or sources of \(\mathrm{M}^{+}\) or \(\mathrm{X}^{2-}\).
• Temperature remains constant, so \(K_{sp}\) does not change.
• At both saturated equilibria, \(2s\) is much smaller than the concentration of \(\mathrm{M}^{+}\) supplied by \(\ce{MNO3}\).

Determine:

(a) The initial molar solubility of \(\ce{M2X}\).

(b) The reaction quotient immediately after dilution but before additional solid dissolves.

(c) The new molar solubility after equilibrium is restored.

(d) The final concentrations of \(\mathrm{M}^{+}\) and \(\mathrm{X}^{2-}\).

(e) Explain why the new molar solubility is not simply one-tenth of the original value.

Solution 5

The new molar solubility is approximately:

\[ \boxed{s_1=1.0\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

This is 100 times the original molar solubility, not one-tenth of it.

The key is that dilution reduces the concentration of the common ion. Because that ion appears squared in the \(K_{sp}\) expression, the amount of solid that can dissolve changes substantially.

Step 1: Write the equilibrium expression.

The dissolution equation is:

\[ \ce{M2X(s) <=> 2M+(aq) + X^2-(aq)} \]

Therefore:

\[ K_{sp}=[\mathrm{M}^{+}]^2[\mathrm{X}^{2-}] \]

The soluble compound \(\ce{MNO3}\) supplies the initial common ion:

\[ \ce{MNO3(aq) -> M+(aq) + NO3-(aq)} \]

If the initial molar solubility of \(\ce{M2X}\) is \(s_0\), the equilibrium concentrations are:

\[ [\mathrm{M}^{+}]=0.100+2s_0 \]

\[ [\mathrm{X}^{2-}]=s_0 \]

Step 2: Calculate the initial molar solubility.

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

\[ 1.0\times10^{-10} =(0.100+2s_0)^2s_0 \]

The question tells us that \(2s_0\) is negligible compared with \(0.100\).

Therefore:

\[ 1.0\times10^{-10} \approx(0.100)^2s_0 \]

Rearrange:

\[ \begin{aligned} s_0&=\frac{1.0\times10^{-10}}{0.0100}\\ &=1.0\times10^{-8}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

So:

\[ \boxed{s_0=1.0\times10^{-8}\ \mathrm{mol\,L^{-1}}} \]

Initially, the solution contains approximately:

\[ [\mathrm{M}^{+}]=0.100\ \mathrm{mol\,L^{-1}} \]

\[ [\mathrm{X}^{2-}]=1.0\times10^{-8}\ \mathrm{mol\,L^{-1}} \]

Step 3: Calculate the concentrations immediately after dilution.

The solution is diluted tenfold.

Before any additional solid dissolves, every dissolved species has its concentration divided by ten.

Therefore:

\[ [\mathrm{M}^{+}]_{\text{immediate}} \approx0.0100\ \mathrm{mol\,L^{-1}} \]

And:

\[ [\mathrm{X}^{2-}]_{\text{immediate}} =1.0\times10^{-9}\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate the new reaction quotient.

Immediately after dilution:

\[ Q_{sp}=[\mathrm{M}^{+}]^2[\mathrm{X}^{2-}] \]

Substitute:

\[ \begin{aligned} Q_{sp} &=(0.0100)^2(1.0\times10^{-9})\\ &=1.0\times10^{-13} \end{aligned} \]

Compare with \(K_{sp}\):

\[ Q_{sp}=1.0\times10^{-13} \]

\[ K_{sp}=1.0\times10^{-10} \]

Therefore:

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

The solution is now unsaturated.

Because excess solid remains in contact with the solution, more \(\ce{M2X}\) must dissolve to restore equilibrium.

This is the important point: the concentrations immediately after dilution are not the final equilibrium concentrations.

Step 5: Calculate the new molar solubility.

Let \(s_1\) represent the total molar solubility of \(\ce{M2X}\) after the new equilibrium is established.

The diluted solution still contains approximately:

\[ 0.0100\ \mathrm{mol\,L^{-1}} \]

of \(\mathrm{M}^{+}\) from the soluble \(\ce{MNO3}\).

At the new equilibrium:

\[ [\mathrm{M}^{+}]=0.0100+2s_1 \]

\[ [\mathrm{X}^{2-}]=s_1 \]

The \(K_{sp}\) expression becomes:

\[ 1.0\times10^{-10} =(0.0100+2s_1)^2s_1 \]

Using the supplied approximation that \(2s_1\) is much smaller than \(0.0100\):

\[ 1.0\times10^{-10} \approx(0.0100)^2s_1 \]

Rearrange:

\[ \begin{aligned} s_1&=\frac{1.0\times10^{-10}}{(0.0100)^2}\\ &=\frac{1.0\times10^{-10}}{1.0\times10^{-4}}\\ &=1.0\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{s_1=1.0\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

The approximation is consistent because:

\[ \frac{2s_1}{0.0100} =\frac{2.0\times10^{-6}}{0.0100} =2.0\times10^{-4} \]

The additional contribution from dissolution is indeed very small compared with the common-ion concentration.

Step 6: Calculate the final ion concentrations.

For the common ion:

\[ \begin{aligned} [\mathrm{M}^{+}]_{\text{final}} &=0.0100+2(1.0\times10^{-6})\\ &=0.010002\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{[\mathrm{M}^{+}]_{\text{final}}\approx0.0100\ \mathrm{mol\,L^{-1}}} \]

For the other ion:

\[ [\mathrm{X}^{2-}]_{\text{final}}=s_1 \]

\[ \boxed{[\mathrm{X}^{2-}]_{\text{final}}=1.0\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

Step 7: Explain the surprising result.

Compare the initial and final molar solubilities:

\[ \frac{s_1}{s_0} = \frac{1.0\times10^{-6}}{1.0\times10^{-8}} \]

\[ \boxed{\frac{s_1}{s_0}=100} \]

The molar solubility has increased by a factor of 100.

A tempting prediction is that dilution simply reduces every concentration by a factor of ten, so the solubility should also decrease.

That prediction ignores the excess solid.

Immediately after dilution, the ion concentrations produce a reaction quotient far below \(K_{sp}\). The system responds by dissolving more solid until the equilibrium condition is restored.

The common-ion concentration has fallen by a factor of ten. Because it is squared in the \(K_{sp}\) expression, the molar solubility increases by approximately a factor of \(10^2\).

This is the deeper reason you must distinguish between:

• The concentration immediately after dilution.
• The concentration after the system has reached a new equilibrium.
• The molar solubility of the solid under the new solution conditions.

They are not necessarily the same.

06What to do next

Before using \(K_{sp}\), always establish the actual equilibrium concentration of every ion.

The dissolution equation tells you how many ions the solid produces. The initial solution tells you which ions are already present. Equilibrium then determines how much additional solid can dissolve or how much precipitation occurs.

Once those concentrations are correct, you can confidently move on to calculating \(K_{sp}\), finding molar solubility from a given \(K_{sp}\), or predicting precipitation using \(Q_{sp}\).

Two useful next steps are:

• How to Calculate Ksp from Calcium Fluoride Solubility
• How to Calculate Solubility with a Common Ion (HSC Chemistry)