Why Comparing Ksp Values Can Misrank Salts with Different Formulas

A salt has a \(K_{sp}\) of \(1.0 \times 10^{-10}\). Another has a \(K_{sp}\) of \(4.0 \times 10^{-9}\). Which salt is less soluble?

The obvious answer is the first one. Its \(K_{sp}\) is smaller, so surely less of it dissolves.

Not necessarily. If the two salts have different chemical formulas, they produce different numbers of ions when they dissolve. This changes the relationship between their equilibrium concentrations and their \(K_{sp}\) values.

The key idea: a \(K_{sp}\) value measures an equilibrium relationship, not solubility directly. To compare how much of two different salts dissolves, you must account for their dissolution stoichiometry and calculate their molar solubilities.

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

01Why the chemical formula changes the comparison

Imagine dissolving two salts in separate beakers of pure water at the same temperature.

Salt A produces one positive ion and one negative ion per formula unit. Salt B produces one positive ion and two negative ions per formula unit.

Even if both salts dissolve to the same molar solubility, their equilibrium ion concentrations will be different.

Consider the two dissolution equations:

For a 1:1 salt:

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

For a 1:2 salt:

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

Notice the difference. One formula unit of MX produces two ions, while one formula unit of MX₂ produces three ions.

This affects the equilibrium expression.

Deriving the two Ksp expressions

For MX:

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

For MX₂:

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

The square on the concentration of \(\mathrm{X}^-\) is important. It comes directly from the coefficient of 2 in the balanced dissolution equation.

More generally, for a salt that dissolves according to:

\[ \ce{A_mB_n(s) <=> mA + nB} \]

the solubility product is:

\[ K_{sp}=[\mathrm{A}]^m[\mathrm{B}]^n \]

The exponents come from the stoichiometric coefficients, not from the charges on the ions.

This is why comparing two \(K_{sp}\) values without considering their formulas can produce the wrong ranking.

02The important distinction: Ksp versus molar solubility

We need a consistent quantity to compare.

The best choice is usually molar solubility, represented by \(s\).

Molar solubility is the number of moles of a salt that dissolve per litre of solution to form a saturated solution.

Its unit is:

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

Unlike \(K_{sp}\), molar solubility always has the same concentration unit, regardless of the salt’s formula.

Let’s see how we calculate it.

Case 1: A 1:1 salt

Suppose MX dissolves in pure water with molar solubility \(s\).

For every mole of MX that dissolves, one mole of M⁺ and one mole of X⁻ are produced.

Therefore:

Substituting into the equilibrium expression:

\[ K_{sp}=s\times s \]

Therefore:

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

And:

\[ \boxed{s=\sqrt{K_{sp}}} \]

Case 2: A 1:2 salt

Now consider MX₂.

For every mole of MX₂ that dissolves, one mole of M²⁺ and two moles of X⁻ are produced.

The equilibrium concentrations become:

Substitute these concentrations into the equilibrium expression:

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

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

\[ \boxed{K_{sp}=4s^3} \]

Rearranging:

\[ \boxed{s=\sqrt[3]{\frac{K_{sp}}{4}}} \]

Notice that the relationship is now cubic rather than quadratic.

A larger or smaller \(K_{sp}\) does not translate into the same change in solubility for every salt.

04Why the units can be misleading

There is another reason students sometimes make incorrect comparisons.

Using the conventional concentration-based HSC equilibrium expressions, the units of \(K_{sp}\) depend on the number of concentration terms and their exponents.

For MX:

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

Its concentration units are:

\[ (\mathrm{mol\,L^{-1}})^2 \]

For MX₂:

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

Its concentration units are:

\[ (\mathrm{mol\,L^{-1}})^3 \]

Strictly speaking, thermodynamic equilibrium constants are dimensionless because they use activities relative to standard states. However, HSC calculations conventionally use equilibrium concentrations in mol L⁻¹.

The practical point is that you should not treat concentration-based \(K_{sp}\) numbers for different stoichiometries as directly comparable measures of solubility.

Convert them to molar solubility first.

05What happens when the salts have more complicated formulas?

The same reasoning applies to salts with different ratios of positive and negative ions.

Consider three general dissolution patterns.

For the final example:

\[ K_{sp}=[\mathrm{A}]^2[\mathrm{B}]^3 \]

Since two A ions and three B ions are produced per formula unit:

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

\[ K_{sp}=4s^2\times27s^3 \]

\[ \boxed{K_{sp}=108s^5} \]

The coefficients in the balanced equation determine both the concentrations and the exponents.

You do not need to memorise every possible formula. You need to be able to derive the relationship from the balanced dissolution equation.

06How the experimental measurement connects to Ksp

There is a useful experimental reason to keep molar solubility separate from \(K_{sp}\).

Imagine that a chemist wants to determine the solubility of a sparingly soluble salt.

A possible approach is to prepare a saturated solution at a controlled temperature, allow equilibrium to establish, separate the undissolved solid, and analyse a measured volume of the clear solution.

One gravimetric approach is to evaporate a known volume of the filtered solution and measure the mass of the dissolved salt left behind. This method requires care because heating may cause decomposition, hydration changes, or loss of volatile products.

What should the experiment show?

If the solution is genuinely saturated and kept at constant temperature, its dissolved concentration should remain constant while excess solid is present.

However, a measurement of total dissolved mass alone is not enough to compare salts with different formulas. The chemist must convert the measured amount into moles of the salt.

For example, suppose a 250.0 mL sample of saturated solution contains 0.00300 g of a salt with molar mass 200.0 g mol⁻¹.

The amount of salt dissolved is:

\[ n=\frac{m}{M} \]

where \(n\) is the amount in moles, \(m\) is the mass in grams, and \(M\) is the molar mass in g mol⁻¹.

\[ n=\frac{0.00300}{200.0} \]

\[ n=1.50\times10^{-5}\ \mathrm{mol} \]

The volume is:

\[ V=0.2500\ \mathrm{L} \]

Therefore, the molar solubility is:

\[ s=\frac{n}{V} \]

\[ s=\frac{1.50\times10^{-5}}{0.2500} \]

\[ \boxed{s=6.00\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

The measured molar solubility can then be substituted into the correct \(K_{sp}\) expression for that salt.

The original measurement has limitations. Temperature changes, incomplete filtration, evaporation before measurement, and contamination can all affect the result.

A revised experimental design would control the temperature, use excess solid to establish saturation, minimise exposure to changing conditions, and use an appropriate analytical method to determine the dissolved ion concentrations.

This is the experimental logic behind the calculation: measure or establish equilibrium concentrations first, then apply the stoichiometric equilibrium expression.

07Common ions make the comparison more complicated

So far, we have assumed that each salt dissolves in pure water.

What happens if the water already contains one of the ions produced by the salt?

Consider:

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

Now imagine adding MX₂ to a solution that already contains X⁻ from another soluble compound.

The concentration of X⁻ is no longer just \(2s\).

This is the common ion effect.

A common ion is an ion that is already present in solution and is also produced by the dissolving salt.

According to Le Chatelier’s principle, increasing the concentration of a product shifts the equilibrium towards the solid. This reduces the amount of additional salt that dissolves.

Importantly, the \(K_{sp}\) does not change just because a common ion is added.

At constant temperature, the equilibrium constant remains the same. What changes is the equilibrium concentration of the dissolved salt.

08A reliable method for comparing Ksp values

When an exam asks which of two salts is more soluble, use this decision process.

1. Write the balanced dissolution equation for each salt.
2. Write the correct \(K_{sp}\) expression using the stoichiometric coefficients.
3. Define the molar solubility, \(s\), for each salt.
4. Express each equilibrium ion concentration in terms of \(s\).
5. Substitute into the equilibrium expression.
6. Solve for the molar solubility of each salt.
7. Check whether common ions or other equilibrium reactions affect the concentrations.
8. Compare the final values in mol L⁻¹.

One important condition: compare solubilities at the same temperature and under equivalent solution conditions. Comparing a salt in pure water with another salt in a concentrated common-ion solution does not tell you which solid is intrinsically more soluble under identical conditions.

For a broader explanation of why the smallest \(K_{sp}\) can be misleading, read Why the Smallest Ksp Does Not Always Mean the Least Soluble Salt.

09Questions and solutions

The following questions progress from straightforward calculations to more demanding equilibrium reasoning. All numerical equilibrium data are illustrative.

Question 1

Two salts dissolve in pure water at the same temperature.

Salt A:

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

\[ K_{sp}=9.0\times10^{-8} \]

Salt B:

\[ \ce{CD2(s) <=> C^2+(aq) + 2D-(aq)} \]

\[ K_{sp}=3.6\times10^{-7} \]

Calculate the molar solubility of each salt and identify which is more soluble.

Solution 1

Salt B is more soluble, with a molar solubility five times that of salt A.

Start by deriving the appropriate expression for each salt.

For salt A:

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

\[ s_A=\sqrt{9.0\times10^{-8}} \]

\[ \boxed{s_A=3.0\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

For salt B:

\[ K_{sp}=s_B(2s_B)^2 \]

\[ K_{sp}=4s_B^3 \]

Therefore:

\[ s_B=\sqrt[3]{\frac{3.6\times10^{-7}}{4}} \]

\[ s_B=\sqrt[3]{9.0\times10^{-8}} \]

\[ \boxed{s_B\approx4.48\times10^{-3}\ \mathrm{mol\,L^{-1}}} \]

Compare:

\[ \frac{s_B}{s_A} =\frac{4.48\times10^{-3}}{3.0\times10^{-4}} \]

\[ \boxed{\frac{s_B}{s_A}\approx14.9} \]

Salt B is approximately 14.9 times more soluble.

The tempting mistake is to compare the two \(K_{sp}\) values and stop there. Although salt B also has the larger \(K_{sp}\), the important result comes from using the correct stoichiometric relationship for each salt.

Question 2

Two salts have the following equilibrium constants in pure water at the same temperature.

(a) Derive the relationship between \(K_{sp}\) and molar solubility for each salt.

(b) Calculate the molar solubility of each salt.

(c) Explain why the larger numerical \(K_{sp}\) does not provide a reliable direct comparison of their solubilities.

Solution 2

Salt P is more soluble, even though its numerical \(K_{sp}\) is much smaller.

(a) Derive the expressions.

For P:

\[ K_{sp}=[\mathrm{P}^{3+}]^2 [\mathrm{Q}^{2-}]^3 \]

Using molar solubility \(s_P\):

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

\[ \boxed{K_{sp}=108s_P^5} \]

For R:

\[ K_{sp}=[\mathrm{R}^+]^2 [\mathrm{S}^{2-}] \]

Using molar solubility \(s_R\):

\[ K_{sp}=(2s_R)^2(s_R) \]

\[ \boxed{K_{sp}=4s_R^3} \]

(b) Calculate the solubilities.

For P:

\[ s_P=\sqrt[5]{\frac{3.456\times10^{-25}}{108}} \]

\[ s_P=\sqrt[5]{3.2\times10^{-27}} \]

\[ \boxed{s_P=5.0\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

For R:

\[ s_R=\sqrt[3]{\frac{6.4\times10^{-15}}{4}} \]

\[ s_R=\sqrt[3]{1.6\times10^{-15}} \]

\[ \boxed{s_R\approx1.17\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

Therefore, R is actually more soluble.

(c) Explain the comparison.

The numerical \(K_{sp}\) values cannot be directly compared as if they both represented the same concentration relationship.

Salt P produces five ions per formula unit, while salt R produces three. Their equilibrium expressions therefore have different powers of molar concentration.

The correct calculation shows that R has approximately 2.34 times the molar solubility of P.

Question 3

A sparingly soluble salt, \(\ce{XY2}\), has a \(K_{sp}\) of \(1.6\times10^{-11}\).

It is added to a solution that initially contains \(0.020\ \mathrm{mol\,L^{-1}}\) of \(\mathrm{Y}^{-}\) ions.

Assume the salt is at equilibrium, the common-ion source dissociates completely, and there are no other significant equilibria.

(a) Write the \(K_{sp}\) expression.

(b) Calculate the approximate molar solubility of \(\ce{XY2}\) in the solution.

(c) Explain why using \(K_{sp}=4s^3\) would produce an incorrect result.

Solution 3

The approximate molar solubility is \(4.0\times10^{-8}\ \mathrm{mol\,L^{-1}}\). The initial Y⁻ concentration must be included in the equilibrium expression.

(a) Write the equilibrium expression.

\[ \ce{XY2(s) <=> X^2+(aq) + 2Y-(aq)} \]

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

(b) Calculate the molar solubility.

Let the molar solubility be \(s\).

The equilibrium concentrations are:

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

\[ [\mathrm{Y}^-]=0.020+2s \]

Substitute:

\[ 1.6\times10^{-11} =s(0.020+2s)^2 \]

Since the expected solubility is small compared with \(0.020\ \mathrm{mol\,L^{-1}}\), use:

\[ 0.020+2s\approx0.020 \]

Then:

\[ 1.6\times10^{-11}=s(0.020)^2 \]

\[ s=\frac{1.6\times10^{-11}}{4.0\times10^{-4}} \]

\[ \boxed{s=4.0\times10^{-8}\ \mathrm{mol\,L^{-1}}} \]

Check the approximation:

\[ 2s=8.0\times10^{-8}\ \mathrm{mol\,L^{-1}} \]

This is negligible compared with the initial concentration of \(0.020\ \mathrm{mol\,L^{-1}}\).

(c) Explain the incorrect approach.

The expression \(K_{sp}=4s^3\) assumes the salt dissolves in pure water, where both ions originate entirely from the dissolving solid.

Here, the solution already contains Y⁻ ions.

The concentration of Y⁻ at equilibrium is therefore not \(2s\). It is \(0.020+2s\).

Ignoring the initial common-ion concentration would significantly overestimate the solubility.

Question 4

Two sparingly soluble salts, A and B, have the following properties at the same temperature.

Both salts are initially studied in pure water.

A student concludes that salt B must be less soluble because its \(K_{sp}\) is smaller.

(a) Calculate the molar solubility of both salts in pure water.

(b) A second experiment is performed by placing salt B into a solution already containing \(0.10\ \mathrm{mol\,L^{-1}}\) of chloride ions. Calculate its approximate molar solubility in this solution.

(c) Explain whether the common-ion solution changes the \(K_{sp}\) of salt B.

Solution 4

Salt A is more soluble in pure water, but adding chloride ions reduces the solubility of salt B without changing its \(K_{sp}\).

(a) Calculate the solubilities in pure water.

For salt A:

\[ \ce{ACl(s) <=> A+ + Cl-} \]

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

\[ s_A=\sqrt{1.0\times10^{-8}} \]

\[ \boxed{s_A=1.0\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

For salt B:

\[ \ce{BCl2(s) <=> B^2+ + 2Cl-} \]

\[ K_{sp}=4s_B^3 \]

\[ s_B=\sqrt[3]{\frac{4.0\times10^{-10}}{4}} \]

\[ s_B=\sqrt[3]{1.0\times10^{-10}} \]

\[ \boxed{s_B\approx4.64\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

Salt B is more soluble in pure water.

(b) Calculate the solubility of B in the chloride solution.

Let its new molar solubility be \(s\).

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

\[ [\mathrm{Cl}^-]=0.10+2s \]

Substitute into the equilibrium expression:

\[ 4.0\times10^{-10}=s(0.10+2s)^2 \]

Assume the additional chloride concentration is negligible compared with \(0.10\ \mathrm{mol\,L^{-1}}\).

\[ 4.0\times10^{-10}=s(0.10)^2 \]

\[ s=\frac{4.0\times10^{-10}}{0.010} \]

\[ \boxed{s=4.0\times10^{-8}\ \mathrm{mol\,L^{-1}}} \]

The new solubility is dramatically smaller.

(c) Explain the effect on Ksp.

The \(K_{sp}\) remains \(4.0\times10^{-10}\) because the temperature has not changed.

The added chloride ions increase the concentration of a product, shifting the dissolution equilibrium towards the solid.

The system reaches a new equilibrium with a much smaller concentration of dissolved B²⁺.

The trap is assuming that a change in solubility must mean a change in \(K_{sp}\). It does not. Solubility depends on the solution conditions, while \(K_{sp}\) is fixed at a given temperature for the relevant equilibrium.

Question 5

Two hypothetical salts, A and B, are studied at the same temperature.

Their dissolution equations are:

\[ \ce{A2X3(s) <=> 2A^3+(aq) + 3X^2-(aq)} \]

\[ \ce{B2Y(s) <=> 2B^+(aq) + Y^2-(aq)} \]

Their equilibrium constants are:

\[ K_{sp,A}=3.456\times10^{-25} \]

\[ K_{sp,B}=6.4\times10^{-15} \]

Both salts are initially placed separately in pure water.

A student calculates their molar solubilities and finds:

\[ s_A=5.0\times10^{-6}\ \mathrm{mol\,L^{-1}} \]

\[ s_B\approx1.17\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

The student concludes that B is more soluble.

Now consider a second experiment.

The salts are placed in separate solutions, each initially containing \(0.010\ \mathrm{mol\,L^{-1}}\) of the relevant negative ion:

• Solution A initially contains \(0.010\ \mathrm{mol\,L^{-1}}\) of X²⁻.
• Solution B initially contains \(0.010\ \mathrm{mol\,L^{-1}}\) of Y²⁻.

Assume both common-ion sources dissociate completely, the temperature remains constant, and there are no other significant equilibria.

(a) Verify the original molar solubilities in pure water.

(b) Calculate the approximate molar solubility of each salt in its respective common-ion solution.

(c) Determine whether the ranking changes.

(d) Explain why this result does not contradict the original comparison or indicate that either \(K_{sp}\) has changed.

Solution 5

Salt B is more soluble in pure water, but salt A becomes more soluble in the specified common-ion solutions. The common-ion concentrations affect the two dissolution equilibria differently because their stoichiometries are different.

This is the important distinction: a solubility ranking is meaningful only when the solution conditions are specified.

(a) Verify the pure-water solubilities.

For A:

\[ K_{sp,A}=(2s_A)^2(3s_A)^3 \]

\[ K_{sp,A}=108s_A^5 \]

Therefore:

\[ s_A=\sqrt[5]{\frac{3.456\times10^{-25}}{108}} \]

\[ \boxed{s_A=5.0\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

For B:

\[ K_{sp,B}=(2s_B)^2(s_B) \]

\[ K_{sp,B}=4s_B^3 \]

Therefore:

\[ s_B=\sqrt[3]{\frac{6.4\times10^{-15}}{4}} \]

\[ \boxed{s_B\approx1.17\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

In pure water, B is approximately 2.34 times more soluble than A.

(b) Calculate the solubilities with common ions.

For salt A:

\[ \ce{A2X3(s) <=> 2A^3+ + 3X^2-} \]

Let its new molar solubility be \(s_A'\).

The equilibrium concentrations are:

\[ [\mathrm{A}^{3+}]=2s_A' \]

\[ [\mathrm{X}^{2-}]=0.010+3s_A' \]

Substitute:

\[ 3.456\times10^{-25} =(2s_A')^2(0.010+3s_A')^3 \]

Since the common-ion concentration is much greater than the additional X²⁻ concentration:

\[ 0.010+3s_A'\approx0.010 \]

Therefore:

\[ 3.456\times10^{-25} =4(s_A')^2(0.010)^3 \]

\[ 3.456\times10^{-25} =4\times10^{-6}(s_A')^2 \]

\[ (s_A')^2=8.64\times10^{-20} \]

\[ \boxed{s_A'=2.94\times10^{-10}\ \mathrm{mol\,L^{-1}}} \]

For salt B:

\[ \ce{B2Y(s) <=> 2B+ + Y^2-} \]

Let its new molar solubility be \(s_B'\).

The equilibrium concentrations are:

\[ [\mathrm{B}^+]=2s_B' \]

\[ [\mathrm{Y}^{2-}]=0.010+s_B' \]

Substitute:

\[ 6.4\times10^{-15} =(2s_B')^2(0.010+s_B') \]

Using the approximation:

\[ 0.010+s_B'\approx0.010 \]

\[ 6.4\times10^{-15} =4(s_B')^2(0.010) \]

\[ 6.4\times10^{-15}=0.040(s_B')^2 \]

\[ (s_B')^2=1.6\times10^{-13} \]

\[ \boxed{s_B'=4.0\times10^{-7}\ \mathrm{mol\,L^{-1}}} \]

(c) Determine the new ranking.

Compare the results:

Salt B remains more soluble in the common-ion solutions.

So the ranking does not actually reverse. Both salts become less soluble, but salt A is suppressed much more strongly.

This is a useful correction to a tempting conclusion: different stoichiometries do not guarantee that a common-ion effect will reverse a ranking.

(d) Explain why the result is consistent with equilibrium theory.

The two salts have different equilibrium expressions.

For A:

\[ K_{sp,A}=108s^5 \]

in pure water, but in the common-ion solution:

\[ K_{sp,A}=(2s)^2(0.010+3s)^3 \]

For B:

\[ K_{sp,B}=4s^3 \]

in pure water, but in the common-ion solution:

\[ K_{sp,B}=(2s)^2(0.010+s) \]

The initial common-ion concentration changes the mathematical relationship between the solubility and the equilibrium constant.

It does not change the equilibrium constant itself.

The key is that the two salts have different powers of the ion concentrations in their expressions. A common-ion concentration therefore affects each solubility calculation according to a different relationship.

The student must not assume that the pure-water ranking automatically predicts the ranking in every other solution.

A final check: the calculated additional common-ion concentrations are tiny compared with \(0.010\ \mathrm{mol\,L^{-1}}\), so the approximations used above are valid.

10The main idea to take into your HSC exam

A \(K_{sp}\) value is not a direct measurement of how soluble a salt is.

The formula of the salt determines how many ions are produced, and the stoichiometric coefficients determine the powers in the equilibrium expression.

When comparing salts with different formulas:

• Do not rank them using their numerical \(K_{sp}\) values alone.
• Write each balanced dissolution equation.
• Derive the equilibrium concentrations in terms of molar solubility.
• Account for any common ions already present.
• Compare the final molar solubilities in mol L⁻¹.

For more calculation practice, the next useful skill is learning to move in both directions between \(K_{sp}\) and molar solubility. See Calculate Solubility from Ksp for HSC Chemistry and How to Calculate Ksp from Calcium Fluoride Solubility.

Once you can derive the correct concentration relationship from a dissolution equation, comparing salts becomes a calculation you can justify rather than a guess based on which number looks smaller.