How to Convert Solubility in g/L to Ksp for HSC Chemistry

A saturated solution of lead(II) iodide contains \(0.461\ \mathrm{g\,L^{-1}}\) of dissolved salt. Can you put \(0.461\) straight into a \(K_{sp}\) calculation?

No. Solubility in grams per litre measures the mass of salt dissolved, while \(K_{sp}\), the solubility product constant, depends on the equilibrium concentrations of its ions in moles per litre.

To convert solubility from \(\mathrm{g\,L^{-1}}\) into \(K_{sp}\), divide by the salt’s molar mass to find its molar solubility, use the balanced dissolution equation to calculate each ion concentration, then substitute those concentrations into the \(K_{sp}\) expression.

There’s one important catch: if the solution already contains a common ion, you must include that ion’s original concentration. Otherwise, even a perfectly calculated molar solubility can give a completely wrong \(K_{sp}\).

01Convert grams per litre into molar solubility

Suppose a salt has a mass solubility of \(0.461\ \mathrm{g\,L^{-1}}\).

That means \(0.461\ \mathrm{g}\) of the salt is dissolved in each litre of saturated solution. It does not tell us directly how many moles have dissolved.

To find that, we need the molar mass.

The conversion is:

\[ s=\frac{c_m}{M} \]

where:

• \(s\) is molar solubility, in \(\mathrm{mol\,L^{-1}}\).
• \(c_m\) is mass solubility, in \(\mathrm{g\,L^{-1}}\).
• \(M\) is the salt’s molar mass, in \(\mathrm{g\,mol^{-1}}\).

The gram units cancel, leaving moles per litre.

For lead(II) iodide, \(\ce{PbI2}\):

\[ \begin{aligned} M&=207.2+2(126.9)\\ &=461.0\ \mathrm{g\,mol^{-1}} \end{aligned} \]

If its mass solubility is \(0.461\ \mathrm{g\,L^{-1}}\):

\[ \begin{aligned} s&=\frac{0.461}{461.0}\\ &=1.00\times10^{-3}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

So each litre of saturated solution contains \(0.00100\) mol of dissolved lead(II) iodide.

This is the first conversion. We still need to turn that amount of dissolved salt into ion concentrations.

Why the dissolution equation changes the answer

Lead(II) iodide dissolves according to:

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

Every mole of \(\ce{PbI2}\) that dissolves produces:

• One mole of \(\mathrm{Pb}^{2+}\) ions.
• Two moles of \(\mathrm{I}^{-}\) ions.

Predict what happens if \(0.00100\) mol of \(\ce{PbI2}\) dissolves in one litre of pure water.

Would both ion concentrations equal \(0.00100\ \mathrm{mol\,L^{-1}}\)?

Only the lead concentration would. The iodide concentration would be twice as large because the solid releases two iodide ions per formula unit.

If the molar solubility is \(s\), and the ions come only from the dissolving solid:

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

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

Square brackets mean the equilibrium concentration of an aqueous species in \(\mathrm{mol\,L^{-1}}\).

The solubility product expression is:

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

The coefficient of 2 in the dissolution equation becomes the exponent on the iodide concentration.

Substitute the concentrations:

\[ \begin{aligned} K_{sp}&=s(2s)^2\\ &=4s^3 \end{aligned} \]

Notice that the coefficient of 2 does two jobs. It gives the iodide concentration \(2s\), and it makes that concentration squared in the \(K_{sp}\) expression.

The undissolved solid does not appear in this expression because the activity of a pure solid is constant.

The formula depends on the salt

You cannot use \(K_{sp}=4s^3\) for every ionic compound.

For example, barium sulfate dissolves as:

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

One mole of barium sulfate produces one mole of each ion.

Therefore, in pure water:

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

Lead(II) iodide gives \(4s^3\), while barium sulfate gives \(s^2\).

Both results come from the same method. The balanced equation determines which one applies.

02Where does the solubility measurement come from?

Before calculating \(K_{sp}\), we need to know whether the stated mass actually represents dissolved salt.

Imagine investigating the solubility of lead(II) iodide at a fixed temperature.

An initial idea might be to add excess crystals to water, stir the mixture, remove a sample, evaporate the water, and weigh whatever remains.

What should that tell us?

Ideally, the remaining dry mass would equal the amount of salt that had been dissolved in the sampled liquid.

But there’s a problem. If the sample contains even a few suspended crystals, those crystals will be weighed alongside the previously dissolved material. The result will overestimate solubility.

An improved experimental design

A more reliable design separates dissolved material from undissolved particles before measuring the mass.

The apparatus can be arranged in the following order.

First, establish saturation.

Place excess solid in water inside a vessel maintained at a constant temperature. Stir the mixture and allow enough time for equilibrium to be established.

Some undissolved solid should remain. Its presence, together with sufficient equilibration time, allows the liquid to reach saturation.

At equilibrium, dissolution and precipitation continue microscopically, but there is no net change in the dissolved concentrations.

Next, separate the solid.

Withdraw a sample and pass it through a filter while maintaining the equilibrium temperature.

The solid particles remain on the filter. The filtrate contains dissolved ions but no suspended crystals.

Finally, measure a known volume.

Transfer a measured volume of filtrate into a pre-weighed evaporation dish.

Remove the water and determine the mass of residue, provided the dissolved salt can be recovered unchanged by this method.

What should the measurements show?

For a saturated solution at the same temperature, the mass of recovered salt should be proportional to the volume of filtered solution.

Consider these illustrative results. They are constructed teaching data, not reported experimental measurements.

Both filtered samples give the same mass concentration.

The unfiltered sample gives a larger value, but this does not establish that more salt was dissolved. Suspended particles could account for the extra residue.

This design also has limitations.

It assumes the recovered dry material is the original salt, without decomposition, retained water of crystallisation, or additional dissolved substances. Changing the temperature during filtration can also disturb the equilibrium.

For some salts, a calibrated ion-analysis method is more reliable than evaporation and weighing.

The important distinction is that \(K_{sp}\) must be calculated from dissolved species at equilibrium, not from undissolved particles or contaminants.

03What changes when a common ion is already present?

Now consider the same lead(II) iodide dissolving in a potassium iodide solution rather than pure water.

Potassium iodide dissociates according to:

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

Suppose the initial KI concentration is \(0.0100\ \mathrm{mol\,L^{-1}}\).

That means the solution already contains \(0.0100\ \mathrm{mol\,L^{-1}}\) of iodide ions before any lead(II) iodide dissolves.

Would lead(II) iodide become more soluble or less soluble?

It becomes less soluble under the usual common-ion model. The existing iodide ions favour the reverse direction of the dissolution equilibrium, reducing the additional amount of lead(II) iodide that can dissolve.

At the same temperature, however, the solubility product constant does not change in the idealised model.

The amount of salt that dissolves changes because the initial ion concentrations are different.

The correct ion concentrations

Let \(s\) represent the molar solubility of \(\ce{PbI2}\) in the potassium iodide solution.

There are initially no lead ions from the potassium iodide.

Therefore:

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

But iodide comes from two sources:

1. The potassium iodide already in solution.
2. The additional iodide produced by dissolving \(\ce{PbI2}\).

So:

\[ [\mathrm{I}^{-}]=0.0100+2s \]

The correct equilibrium expression becomes:

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

Compare the two situations:

The value of \(s\) is generally different in the two rows.

The common mistake is to use \(2s\) for the iodide concentration in both situations.

That ignores the iodide already present and can underestimate \(K_{sp}\) by several orders of magnitude.

04A reliable decision process for grams per litre to Ksp

Before substituting any numbers, identify what the mass solubility actually represents.

There are three further assumptions to make explicit.

Temperature must be fixed. The value of \(K_{sp}\) depends on temperature. Comparing measurements at different temperatures can produce different equilibrium constants.

The dissolution equation must describe the important dissolved species. Hydrolysis, acid-base reactions, and complex-ion formation can change the relationship between the amount of solid dissolved and the free-ion concentrations.

The HSC concentration model is an approximation. In calculations, the ion concentrations are expressed in \(\mathrm{mol\,L^{-1}}\), and \(K_{sp}\) is usually reported as a numerical value. Strict thermodynamic equilibrium constants use dimensionless activities. The familiar concentration expressions approximate those constants, especially in dilute solutions.

If the question provides a molar solubility but does not tell you whether common ions are present, check the stated conditions before using a pure-water formula.

05Questions and solutions

The following questions use illustrative teaching data. Hypothetical salts are used where their stoichiometry is the important feature, rather than any particular real-world solubility value.

Unless stated otherwise, assume constant temperature, saturated solutions in equilibrium with excess solid, complete dissociation of soluble background salts, and negligible side reactions and activity effects.

Question 1

A saturated solution of barium sulfate, \(\ce{BaSO4}\), in pure water has a mass solubility of \(2.33\times10^{-3}\ \mathrm{g\,L^{-1}}\).

The molar mass of \(\ce{BaSO4}\) is \(233.4\ \mathrm{g\,mol^{-1}}\).

Calculate \(K_{sp}\).

Solution 1

The solubility product is \(K_{sp}=9.97\times10^{-11}\), using the given illustrative data. Barium sulfate has a 1:1 dissolution ratio, so the two ion concentrations equal its molar solubility.

The balanced dissolution equation is:

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

First convert the mass solubility:

\[ \begin{aligned} s&=\frac{2.33\times10^{-3}}{233.4}\\ &=9.98\times10^{-6}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

Therefore:

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

\[ [\mathrm{SO_4}^{2-}]=s \]

The solubility product expression is:

\[ K_{sp}=[\mathrm{Ba}^{2+}][\mathrm{SO_4}^{2-}] \]

Substitute:

\[ \begin{aligned} K_{sp}&=(9.98\times10^{-6})^2\\ &=9.97\times10^{-11} \end{aligned} \]

A tempting mistake is to use \(4s^3\) because that formula appeared in the lead(II) iodide examples. Barium sulfate releases one ion of each type, so the appropriate relationship is \(K_{sp}=s^2\).

Question 2

Two hypothetical salts, \(\ce{AX}\) and \(\ce{BX2}\), each have:

• A molar mass of \(150.0\ \mathrm{g\,mol^{-1}}\).
• A mass solubility of \(0.150\ \mathrm{g\,L^{-1}}\) in pure water.
• Saturated solutions at the same temperature.

Their ions are:

\[ \ce{AX(s) <=> A+(aq) + X^-(aq)} \]

\[ \ce{BX2(s) <=> B^2+(aq) + 2X^-(aq)} \]

(a) Calculate \(K_{sp}\) for each salt.

(b) Which saturated solution contains more moles of dissolved ions per litre?

(c) A student says, “BX₂ has a much smaller \(K_{sp}\), so it must be much less soluble.” Explain why this conclusion is incorrect.

Solution 2

Both salts have the same molar solubility, \(1.00\times10^{-3}\ \mathrm{mol\,L^{-1}}\), but different solubility products. The \(\ce{BX2}\) solution actually contains more individual dissolved ions per litre, despite having the smaller numerical \(K_{sp}\).

Part (a): Calculate both solubility products.

Because the salts have identical mass solubilities and molar masses:

\[ \begin{aligned} s&=\frac{0.150}{150.0}\\ &=1.00\times10^{-3}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

For \(\ce{AX}\):

\[ [\mathrm{A}^{+}]=[\mathrm{X}^{-}]=s \]

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

For \(\ce{BX2}\):

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

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

\[ \begin{aligned} K_{sp}&=s(2s)^2\\ &=4(1.00\times10^{-3})^3\\ &=4.00\times10^{-9} \end{aligned} \]

Part (b): Compare the total dissolved ion concentrations.

For \(\ce{AX}\), each dissolved formula unit produces two ions:

\[ 2s=2.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

For \(\ce{BX2}\), each dissolved formula unit produces three ions:

\[ 3s=3.00\times10^{-3}\ \mathrm{mol\,L^{-1}} \]

Therefore, \(\ce{BX2}\) produces the greater total concentration of dissolved ions.

Part (c): Evaluate the student’s claim.

The student is wrong because the numerical values of \(K_{sp}\) come from different mathematical expressions.

For \(\ce{AX}\), two concentrations are multiplied. For \(\ce{BX2}\), three concentration factors are involved because the iodide-like \(\mathrm{X}^{-}\) concentration is squared.

Those different powers can produce very different numerical \(K_{sp}\) values even when the salts have identical molar solubilities.

In this question, the mass solubilities are explicitly equal. The smaller \(K_{sp}\) of \(\ce{BX2}\) cannot overturn that measurement.

Question 3

A laboratory measures the solubility of a hypothetical salt, \(\ce{MX2}\), in a solution initially containing \(0.00600\ \mathrm{mol\,L^{-1}}\) of \(\ce{NaX}\).

The molar mass of \(\ce{MX2}\) is \(200.0\ \mathrm{g\,mol^{-1}}\).

A \(250.0\ \mathrm{mL}\) sample of the original NaX solution is evaporated. Its dry residue has a mass of \(0.0900\ \mathrm{g}\).

A separate \(250.0\ \mathrm{mL}\) sample of saturated, filtered \(\ce{MX2}\) solution prepared in the same NaX solution leaves \(0.0950\ \mathrm{g}\) of dry residue.

Assume that evaporation recovers both salts unchanged, and that the NaX contribution per litre remains the same in both samples.

(a) Calculate the mass solubility of \(\ce{MX2}\).

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

(c) Explain why using \(0.0950\ \mathrm{g}\) as the mass of dissolved \(\ce{MX2}\) would give an incorrect answer.

Solution 3

The mass solubility of \(\ce{MX2}\) is \(0.0200\ \mathrm{g\,L^{-1}}\), and \(K_{sp}=3.84\times10^{-9}\). The crucial adjustment is subtracting the background salt’s residue before calculating the molar solubility.

Part (a): Find the mass contributed by MX₂.

The NaX solution alone leaves \(0.0900\ \mathrm{g}\).

The saturated mixture leaves \(0.0950\ \mathrm{g}\).

Therefore, the additional mass attributable to \(\ce{MX2}\) is:

\[ \begin{aligned} m&=0.0950-0.0900\\ &=0.0050\ \mathrm{g} \end{aligned} \]

Convert the sample volume:

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

Then:

\[ \begin{aligned} c_m&=\frac{0.0050}{0.2500}\\ &=0.0200\ \mathrm{g\,L^{-1}} \end{aligned} \]

Part (b): Calculate Ksp.

First convert to molar solubility:

\[ \begin{aligned} s&=\frac{0.0200}{200.0}\\ &=1.00\times10^{-4}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

The dissolution equation is:

\[ \ce{MX2(s) <=> M^2+(aq) + 2X^-(aq)} \]

The metal-ion concentration is:

\[ [\mathrm{M}^{2+}]=1.00\times10^{-4} \]

The NaX already supplied \(0.00600\ \mathrm{mol\,L^{-1}}\) of \(\mathrm{X}^{-}\).

Therefore:

\[ \begin{aligned} [\mathrm{X}^{-}]&=0.00600+2s\\ &=0.00600+0.000200\\ &=0.00620 \end{aligned} \]

Both equilibrium concentrations are in \(\mathrm{mol\,L^{-1}}\).

Now:

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

\[ \begin{aligned} K_{sp}&=(1.00\times10^{-4})\\ &\quad\times(0.00620)^2\\ &=3.84\times10^{-9} \end{aligned} \]

Part (c): Explain the measurement error.

The \(0.0950\ \mathrm{g}\) residue includes the original NaX as well as the dissolved \(\ce{MX2}\).

Using that entire mass would falsely attribute the background electrolyte to \(\ce{MX2}\).

There are actually two separate errors to avoid here:

• The mass measurement needs correction for the original dissolved salt.
• The \(K_{sp}\) calculation needs the original common-ion concentration.

Subtracting the residue of the control solves the first problem. Adding \(0.00600\ \mathrm{mol\,L^{-1}}\) to the \(\mathrm{X}^{-}\) concentration solves the second.

Both corrections are necessary.

Question 4

A hypothetical salt, \(\ce{A2B}\), dissolves as follows:

\[ \ce{A2B(s) <=> 2A+(aq) + B^2-(aq)} \]

Its molar mass is \(250.0\ \mathrm{g\,mol^{-1}}\).

At a particular temperature, its measured mass solubility in pure water is \(0.0250\ \mathrm{g\,L^{-1}}\).

A second saturated solution at the same temperature has a much lower measured mass solubility of \(0.00100\ \mathrm{g\,L^{-1}}\).

The second solution was prepared using exactly one of these two background solutions:

• Preparation A: \(9.92\times10^{-4}\ \mathrm{mol\,L^{-1}}\) of a soluble salt \(\ce{AZ}\), supplying \(\mathrm{A}^{+}\).
• Preparation B: \(9.92\times10^{-4}\ \mathrm{mol\,L^{-1}}\) of \(\ce{Na2B}\), supplying \(\mathrm{B}^{2-}\).

The stated background concentrations refer to the common ion itself. There are no other initial common ions.

The mass solubility measurements selectively determine how much \(\ce{A2B}\) dissolved, rather than the total amount of dissolved material.

Determine which preparation is consistent with both measurements.

Explain why the same initial concentration of a common ion does not necessarily produce the same solubility.

Solution 4

Preparation A, containing the initial \(\mathrm{A}^{+}\) ions, is consistent with both measurements. Preparation B is not, because the two common ions appear with different powers in the \(K_{sp}\) expression.

First calculate \(K_{sp}\) using the pure-water measurement.

\[ \begin{aligned} s&=\frac{0.0250}{250.0}\\ &=1.00\times10^{-4}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

For \(\ce{A2B}\) in pure water:

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

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

Therefore:

\[ \begin{aligned} K_{sp}&=(2s)^2s\\ &=4(1.00\times10^{-4})^3\\ &=4.00\times10^{-12} \end{aligned} \]

That equilibrium constant must also describe the second saturated solution at the same temperature under the stated assumptions.

Now calculate its molar solubility:

\[ \begin{aligned} s_2&=\frac{0.00100}{250.0}\\ &=4.00\times10^{-6}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

Test Preparation A: Initial A⁺ ions

The equilibrium concentration of \(\mathrm{A}^{+}\) is:

\[ \begin{aligned} [\mathrm{A}^{+}]&=9.92\times10^{-4}+2s_2\\ &=1.00\times10^{-3} \end{aligned} \]

The equilibrium concentration of \(\mathrm{B}^{2-}\) is:

\[ [\mathrm{B}^{2-}]=4.00\times10^{-6} \]

Substitute:

\[ \begin{aligned} K_{sp}&=[\mathrm{A}^{+}]^2[\mathrm{B}^{2-}]\\ &=(1.00\times10^{-3})^2\\ &\quad\times(4.00\times10^{-6})\\ &=4.00\times10^{-12} \end{aligned} \]

This matches the pure-water result exactly at the stated precision.

Test Preparation B: Initial B²⁻ ions

There are no initial \(\mathrm{A}^{+}\) ions, so:

\[ \begin{aligned} [\mathrm{A}^{+}]&=2s_2\\ &=8.00\times10^{-6} \end{aligned} \]

The \(\mathrm{B}^{2-}\) concentration is:

\[ \begin{aligned} [\mathrm{B}^{2-}]&=9.92\times10^{-4}+s_2\\ &=9.96\times10^{-4} \end{aligned} \]

Therefore:

\[ \begin{aligned} K_{sp}&=(8.00\times10^{-6})^2\\ &\quad\times(9.96\times10^{-4})\\ &=6.37\times10^{-14} \end{aligned} \]

This is inconsistent with the pure-water equilibrium constant of \(4.00\times10^{-12}\).

Therefore, Preparation A is the only consistent explanation.

The trap is assuming that equal initial concentrations of either common ion should have equivalent effects.

They do not. The \(\mathrm{A}^{+}\) concentration is squared in the solubility product expression, while the \(\mathrm{B}^{2-}\) concentration is not.

Which ion is initially present can strongly affect the measured mass solubility, even at the same nominal common-ion concentration.

Question 5

A hypothetical salt, \(\ce{MX}\), has molar mass \(100.0\ \mathrm{g\,mol^{-1}}\) and dissolves according to:

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

Its mass solubility in pure water at a fixed temperature is \(0.100\ \mathrm{g\,L^{-1}}\).

A beaker contains \(100.0\ \mathrm{mL}\) of this saturated solution, with additional undissolved \(\ce{MX}\) at the bottom.

A chemist adds \(100.0\ \mathrm{mL}\) of \(9.00\times10^{-4}\ \mathrm{mol\,L^{-1}}\) \(\ce{NaX}\) solution without removing any of the original solid.

The temperature stays constant. Assume the liquid volumes are additive, and that enough undissolved \(\ce{MX}\) remains to establish a new saturated equilibrium.

(a) Immediately after mixing, before any further dissolution or precipitation, will the original \(\ce{MX}\) crystals tend to dissolve or grow? Justify your prediction quantitatively.

(b) Calculate the additional mass of \(\ce{MX}\) that dissolves or precipitates before the new equilibrium is established. Determine the final mass concentration of dissolved \(\ce{MX}\).

(c) Determine the concentration of \(\ce{NaX}\) that could be added in the same \(100.0\ \mathrm{mL}\) volume to produce no immediate net dissolution or precipitation.

Explain why the result in part (a) does not contradict the common-ion effect.

Solution 5

More \(\ce{MX}\) dissolves, even though the added solution contains a common ion. An additional \(0.00600\ \mathrm{g}\) dissolves, the final dissolved mass concentration becomes \(0.0800\ \mathrm{g\,L^{-1}}\), and the no-change concentration for the added NaX solution is \(0.00300\ \mathrm{mol\,L^{-1}}\).

The key is that adding NaX also doubles the liquid volume. Dilution reduces the original ion concentrations, and the combined effect must be considered before predicting dissolution or precipitation.

Part (a): Determine the direction of the initial change.

First calculate the original molar solubility:

\[ \begin{aligned} s&=\frac{0.100}{100.0}\\ &=1.00\times10^{-3}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

Since the salt is 1:1:

\[ \begin{aligned} K_{sp}&=s^2\\ &=1.00\times10^{-6} \end{aligned} \]

Before mixing, the \(100.0\ \mathrm{mL}\) saturated solution contains:

\[ \begin{aligned} n(\mathrm{M}^{+})&=cV\\ &=(1.00\times10^{-3})(0.1000)\\ &=1.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]

It contains the same number of moles of \(\mathrm{X}^{-}\).

The added NaX contributes:

\[ \begin{aligned} n(\mathrm{X}^{-})&=(9.00\times10^{-4})\\ &\quad\times0.1000\\ &=9.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]

The total solution volume is now:

\[ V=0.2000\ \mathrm{L} \]

Immediately after mixing, before the solid responds, the metal-ion concentration is:

\[ \begin{aligned} [\mathrm{M}^{+}]&=\frac{1.00\times10^{-4}}{0.2000}\\ &=5.00\times10^{-4}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

The total iodide-like \(\mathrm{X}^{-}\) concentration comes from both sources:

\[ \begin{aligned} [\mathrm{X}^{-}]&=\frac{0.000100+0.0000900}{0.2000}\\ &=9.50\times10^{-4}\\ &\quad\mathrm{mol\,L^{-1}} \end{aligned} \]

Now calculate the ion product, \(Q_{sp}\), which uses the concentrations immediately after mixing:

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

\[ \begin{aligned} Q_{sp}&=(5.00\times10^{-4})\\ &\quad\times(9.50\times10^{-4})\\ &=4.75\times10^{-7} \end{aligned} \]

Compare:

\[ Q_{sp}=4.75\times10^{-7} \]

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

Since \(Q_{sp}<K_{sp}\), the mixed solution is initially unsaturated with respect to \(\ce{MX}\).

The crystals therefore tend to dissolve.

This is the first surprise. Adding a common ion does not guarantee precipitation if the operation also dilutes the existing ions.

Part (b): Calculate the additional amount dissolved.

Let \(x\) be the additional molar concentration of \(\ce{MX}\) produced by dissolution after mixing, in \(\mathrm{mol\,L^{-1}}\).

Each mole of \(\ce{MX}\) produces one mole of each ion.

At the new equilibrium:

\[ [\mathrm{M}^{+}]=0.000500+x \]

\[ [\mathrm{X}^{-}]=0.000950+x \]

The equilibrium expression is:

\[ \begin{aligned} K_{sp}&=(0.000500+x)\\ &\quad\times(0.000950+x)\\ &=1.00\times10^{-6} \end{aligned} \]

Solving for the physically meaningful positive value gives:

\[ x=3.00\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

We can verify it without relying on the algebra alone.

The final metal concentration becomes:

\[ [\mathrm{M}^{+}]=8.00\times10^{-4} \]

The final \(\mathrm{X}^{-}\) concentration becomes:

\[ [\mathrm{X}^{-}]=1.25\times10^{-3} \]

Their product is:

\[ \begin{aligned} K_{sp}&=(8.00\times10^{-4})\\ &\quad\times(1.25\times10^{-3})\\ &=1.00\times10^{-6} \end{aligned} \]

So the proposed value satisfies equilibrium.

Now calculate the additional number of moles dissolved in the \(0.2000\ \mathrm{L}\) mixture:

\[ \begin{aligned} n&=xV\\ &=(3.00\times10^{-4})(0.2000)\\ &=6.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Convert this into mass:

\[ \begin{aligned} m&=nM\\ &=(6.00\times10^{-5})(100.0)\\ &=0.00600\ \mathrm{g} \end{aligned} \]

Therefore, \(0.00600\ \mathrm{g}\) of additional solid dissolves.

The final mass concentration of \(\ce{MX}\) is calculated from the metal-ion concentration, because all the metal ions come from dissolved \(\ce{MX}\):

\[ \begin{aligned} c_m&=s_{\mathrm{final}}M\\ &=(8.00\times10^{-4})(100.0)\\ &=0.0800\ \mathrm{g\,L^{-1}} \end{aligned} \]

This is smaller than the original \(0.100\ \mathrm{g\,L^{-1}}\).

Yet more solid has dissolved.

How can both be true?

The original \(100.0\ \mathrm{mL}\) contained:

\[ \begin{aligned} m_{\mathrm{initial}}&=0.100\times0.1000\\ &=0.0100\ \mathrm{g} \end{aligned} \]

The final \(200.0\ \mathrm{mL}\) contains:

\[ \begin{aligned} m_{\mathrm{final}}&=0.0800\times0.2000\\ &=0.0160\ \mathrm{g} \end{aligned} \]

So the total dissolved mass has increased by \(0.00600\ \mathrm{g}\), while the mass per litre has decreased.

The total volume doubled. That is why total dissolved mass and mass concentration tell different stories.

Part (c): Find the no-change NaX concentration.

Let \(c\) represent the concentration of NaX in the \(100.0\ \mathrm{mL}\) solution being added, in \(\mathrm{mol\,L^{-1}}\).

Immediately after mixing, the original metal concentration has halved:

\[ [\mathrm{M}^{+}]=5.00\times10^{-4} \]

The original \(\mathrm{X}^{-}\) concentration also halves, but the added NaX contributes an additional \(c/2\) because its volume doubles during mixing.

Therefore:

\[ [\mathrm{X}^{-}]=0.000500+\frac{c}{2} \]

For no immediate net change, the mixed solution must already satisfy:

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

So:

\[ \begin{aligned} 1.00\times10^{-6} &=(5.00\times10^{-4})\\ &\quad\times\left(0.000500+\frac{c}{2}\right) \end{aligned} \]

Dividing by \(5.00\times10^{-4}\):

\[ 0.00200=0.000500+\frac{c}{2} \]

Therefore:

\[ \begin{aligned} \frac{c}{2}&=0.00150\\ c&=0.00300\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This gives a useful boundary:

• If the added NaX concentration is below \(0.00300\ \mathrm{mol\,L^{-1}}\), the mixed solution initially favours further dissolution.
• If it equals \(0.00300\ \mathrm{mol\,L^{-1}}\), the mixture is already at equilibrium, with no net dissolution or precipitation required.
• If it exceeds \(0.00300\ \mathrm{mol\,L^{-1}}\), the initial mixture favours precipitation.

This result does not contradict the common-ion effect.

Adding a common ion favours lower solubility when other conditions, including volume, are held appropriately fixed. Here, adding the NaX solution simultaneously dilutes the original ions.

The direction of the initial change is determined by the resulting ion product, not simply by the fact that a common ion was added.

05What this conversion lets you do next

Converting grams per litre into \(K_{sp}\) is a way of translating a mass measurement into an equilibrium statement.

The conversion itself is straightforward:

\[ \mathrm{g\,L^{-1}} \longrightarrow \mathrm{mol\,L^{-1}} \longrightarrow K_{sp} \]

The chemistry lies in deciding which equilibrium ion concentrations belong in the final expression.

A balanced dissolution equation tells you how many ions are produced. The starting solution tells you which ions were already present. The experimental conditions tell you whether the measured amount genuinely represents a saturated equilibrium.

Once those details are correct, you can reverse the calculation to find mass solubility from a known \(K_{sp}\). That method is covered in Calculate Solubility from Ksp for HSC Chemistry.

If the ions come from two solutions being mixed, the next decision is different: calculate their concentrations after dilution, find \(Q_{sp}\), and compare it with \(K_{sp}\). See Predicting Precipitation with Ksp for HSC Chemistry.

In both cases, the same principle carries through: use the actual ion concentrations, not just the amount of salt you started with.