How to Calculate Solubility with a Common Ion (HSC Chemistry)

Add excess calcium hydroxide to a solution containing 0.100 mol L⁻¹ sodium hydroxide. Some solid dissolves, but most remains at the bottom of the beaker. How much actually dissolves?

The tempting approach is to treat the solution as though it started with pure water. That would give you the wrong hydroxide concentration because the sodium hydroxide has already supplied OH⁻ ions.

The key is to account for both sources of ions: the soluble compound already in the solution and the sparingly soluble solid that dissolves.

For example, calcium hydroxide dissolves according to:

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

If its molar solubility is \(s\), then:

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

\[ [\mathrm{OH}^{-}]=0.100+2s \]

Using the illustrative value \(K_{sp}=5.5\times10^{-6}\) at 25 °C:

\[ K_{sp}=s(0.100+2s)^2 \]

Solving gives approximately:

\[ \boxed{s=5.4\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

That’s the essential method: add the initial common-ion concentration to the amount produced by dissolution, then substitute the resulting equilibrium concentrations into the \(K_{sp}\) expression.

01An experiment: measuring solubility when a common ion is already present

There is an interesting measurement problem here. If calcium hydroxide dissolves in sodium hydroxide, how could you experimentally determine how much has dissolved?

You cannot simply measure the total hydroxide concentration and assume it all came from calcium hydroxide. Most of it was already there.

A useful investigation compares two solutions.

• A control containing 0.100 mol L⁻¹ NaOH.
• A test solution containing the same NaOH concentration, with excess solid Ca(OH)₂ added until equilibrium is reached.

What should we predict?

If calcium hydroxide dissolves, every mole produces two moles of hydroxide ions.

Therefore, the saturated test solution should contain more OH⁻ than the original NaOH solution.

How could we measure that difference?

One option is an acid-base titration.

Following the apparatus

First, prepare the two solutions and maintain both at 25 °C.

The test solution is stirred with excess Ca(OH)₂ until equilibrium is established. Some undissolved solid must remain, confirming that the solution can become saturated.

Next, filter the test solution at the same temperature to remove the remaining solid. Collect the clear filtrate in a stoppered flask.

Use a pipette to transfer 50.00 mL of the control solution into one conical flask and 50.00 mL of the saturated filtrate into another.

Finally, titrate both samples with standardised 0.1000 mol L⁻¹ hydrochloric acid.

The hydrochloric acid neutralises hydroxide ions:

\[ \ce{H+ + OH- -> H2O} \]

If the common-ion calculation is correct, the saturated sample should require more acid than the control.

Illustrative predicted results

The following values are calculated from the equilibrium model. They are not reported experimental measurements.

The additional 0.54 mL of hydrochloric acid accounts for the extra hydroxide produced by dissolving calcium hydroxide.

Step 1: Calculate the extra moles of hydrochloric acid.

\[ \begin{aligned} n(\ce{HCl})&=CV\\ &=0.1000\times0.00054\\ &=5.4\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Step 2: Calculate the additional hydroxide concentration.

The acid reacts with hydroxide in a 1:1 mole ratio.

\[ \begin{aligned} \Delta[\ce{OH-}] &=\frac{5.4\times10^{-5}}{0.05000}\\ &=1.08\times10^{-3}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 3: Use the dissolution stoichiometry.

Each mole of calcium hydroxide produces two moles of hydroxide ions.

\[ s=\frac{1.08\times10^{-3}}{2} \]

\[ \boxed{s=5.4\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

The experimental approach and the \(K_{sp}\) calculation give consistent results.

Why the experimental design matters

A first attempt might use a pH probe to compare the two solutions.

The problem is that pH is logarithmic. The additional hydroxide concentration is small compared with the 0.100 M background, so the change in pH is tiny.

A more useful design uses a matched control and an acid titration. It measures the extra hydroxide rather than trying to distinguish a very small pH change.

There are also two important experimental controls:

• Keep the solutions at the same temperature. Solubility and \(K_{sp}\) depend on temperature.
• Minimise exposure to atmospheric carbon dioxide. CO₂ reacts with hydroxide and can form calcium carbonate, changing the measured concentrations.

This investigation illustrates an important distinction: the common-ion calculation predicts an equilibrium concentration, while the experiment needs a measurement that separates the contribution from the background electrolyte.

02How to set up a common-ion solubility calculation

The most reliable approach is to work from the balanced dissolution equation.

Step 1: Write the dissolution equation

Consider a general ionic solid:

\[ \ce{M_aX_b(s) <=> aM(aq) + bX(aq)} \]

The coefficients tell you how many moles of each ion are produced when one mole of solid dissolves.

For example:

\[ \ce{BaF2(s) <=> Ba^2+(aq) + 2F^-(aq)} \]

One mole of barium fluoride produces:

• One mole of \(\mathrm{Ba}^{2+}\).
• Two moles of \(\mathrm{F}^{-}\).

Step 2: Define the molar solubility

Let \(s\) represent the molar solubility of the solid.

Molar solubility is the number of moles of solid that dissolve per litre of saturated solution, measured in mol L⁻¹.

For barium fluoride:

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

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

These expressions would be correct if the solution initially contained no barium or fluoride ions.

A common-ion solution changes that starting point.

Step 3: Calculate the initial common-ion concentration

Suppose the solution already contains 0.100 mol L⁻¹ sodium fluoride.

Sodium fluoride dissociates:

\[ \ce{NaF(aq) -> Na+(aq) + F-(aq)} \]

It therefore supplies 0.100 mol L⁻¹ fluoride ions before any barium fluoride dissolves.

When barium fluoride dissolves, it adds another \(2s\) mol L⁻¹ fluoride.

The equilibrium concentrations become:

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

\[ [\mathrm{F}^{-}]=0.100+2s \]

This is the central adjustment in a common-ion problem.

Think of the background electrolyte as giving one type of ion a head start. Dissolving the solid adds more ions to an amount that is already present.

The analogy is only a way to track concentrations. Once dissolved, ions do not remain in separate groups according to where they came from.

Common-ion concentration bookkeeping

The following table shows how different soluble compounds contribute to a dissolution equilibrium.

Notice that the concentration of the common ion depends on the dissociation stoichiometry of the background electrolyte.

For example, 0.0200 M MgCl₂ produces:

\[ [\mathrm{Cl}^{-}]=2(0.0200) \]

\[ [\mathrm{Cl}^{-}]=0.0400\ \mathrm{mol\,L^{-1}} \]

Do not automatically use the concentration of the dissolved compound as the concentration of its ions.

Step 4: Write the \(K_{sp}\) expression

The solubility product expression comes directly from the balanced dissolution equation.

For barium fluoride:

\[ \ce{BaF2(s) <=> Ba^2+(aq) + 2F^-(aq)} \]

Therefore:

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

Substitute the equilibrium concentrations:

\[ \boxed{K_{sp}=s(C_0+2s)^2} \]

Here, \(C_0\) is the initial fluoride concentration supplied by the background solution, after any dilution or mixing.

The coefficient 2 has two jobs:

1. It means the solid produces \(2s\) fluoride ions.
2. It becomes the exponent of the fluoride concentration in the \(K_{sp}\) expression.

For a salt that produces one cation and two anions, the common-ion expression depends on which ion is already present.

The balanced equation determines which expression you need. Never choose an expression simply because it looks familiar.

Step 5: Decide whether you can use an approximation

Most HSC-style common-ion calculations become much easier when the amount of solid dissolving is small compared with the initial common-ion concentration.

If:

\[ 2s\ll C_0 \]

then:

\[ C_0+2s\approx C_0 \]

For a salt with one cation and two anions, when the anion is the common ion:

\[ K_{sp}\approx sC_0^2 \]

Therefore:

\[ \boxed{s\approx\frac{K_{sp}}{C_0^2}} \]

But this is an approximation, not a rule that works for every problem.

Always substitute your calculated \(s\) back into the original expression to check whether the omitted contribution was genuinely small.

Step 6: Convert the answer if necessary

Molar solubility and mass solubility are different quantities.

If the question asks for solubility in g L⁻¹:

\[ \text{Mass solubility} =s\times M \]

where:

• \(s\) is molar solubility in mol L⁻¹.
• \(M\) is molar mass in g mol⁻¹.

If the question asks for the mass dissolved in a particular volume:

\[ m=sVM \]

where \(V\) is the solution volume in litres and \(m\) is the dissolved mass in grams.

Do not put concentrations in g L⁻¹ directly into a \(K_{sp}\) expression. Convert to mol L⁻¹ first.

03Worked examples

04When the common-ion calculation needs a different approach

There is one more situation worth recognising.

So far, we have generally assumed that only one of the ions is initially present in the background solution.

What happens if the solution already contains both ions from the sparingly soluble salt?

Before calculating a new solubility, check the ion product, \(Q_{sp}\).

The ion product uses the same concentration expression as \(K_{sp}\), but it uses the concentrations that exist at that moment.

This prevents a major mistake: assuming every common-ion problem involves more solid dissolving.

Sometimes the solution already contains too many dissolved ions. In that case, the correct direction is precipitation, not dissolution.

For a broader explanation of the equilibrium calculation, see Calculate Solubility from Ksp for HSC Chemistry.

For a particle-level explanation of why adding chloride reduces silver chloride solubility, see Common-Ion Effect: Why Added Chloride Reduces AgCl Solubility.

05Questions and solutions

The following questions progress from a straightforward common-ion calculation to problems involving dilution, approximation limits, and precipitation. All numerical \(K_{sp}\) values are illustrative values supplied for these questions.

Question 1

At 25 °C, excess solid barium fluoride, BaF₂, is added to a solution containing 0.200 mol L⁻¹ NaF.

The illustrative solubility product is:

\[ K_{sp}=1.7\times10^{-6} \]

Calculate the molar solubility of BaF₂.

Assume complete dissociation, negligible complex formation, and that the concentration of fluoride supplied by the solid is negligible compared with the initial fluoride concentration.

Solution 1

The molar solubility is:

\[ \boxed{4.25\times10^{-5}\ \mathrm{mol\,L^{-1}}} \]

Step 1: Write the dissolution equation.

\[ \ce{BaF2(s) <=> Ba^2+(aq) + 2F^-(aq)} \]

Let the molar solubility be \(s\).

Then:

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

\[ [\mathrm{F}^{-}]=0.200+2s \]

Step 2: Write the \(K_{sp}\) expression.

\[ K_{sp}=s(0.200+2s)^2 \]

Step 3: Apply the approximation.

Since \(2s\) is negligible compared with 0.200:

\[ \begin{aligned} 1.7\times10^{-6}&\approx s(0.200)^2\\ s&=\frac{1.7\times10^{-6}}{0.0400}\\ &=4.25\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 4: Check the assumption.

\[ 2s=8.50\times10^{-5}\ \mathrm{M} \]

This is only 0.0425% of the initial fluoride concentration.

The approximation is justified.

The important detail is that NaF supplies the common ion. You must include that initial fluoride concentration before calculating the additional fluoride produced by BaF₂.

Question 2

A saturated solution of strontium fluoride, SrF₂, is prepared in 0.0500 mol L⁻¹ strontium nitrate, Sr(NO₃)₂, at 25 °C.

Use the illustrative value:

\[ K_{sp}=2.5\times10^{-9} \]

Calculate:

(a) The molar solubility of SrF₂.

(b) The mass of SrF₂ dissolved in 100.0 mL of the saturated solution.

The molar mass of SrF₂ is 125.62 g mol⁻¹.

Assume complete dissociation and negligible complex formation.

Solution 2

The molar solubility is approximately:

\[ \boxed{1.12\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]

The mass dissolved in 100.0 mL is:

\[ \boxed{1.41\times10^{-3}\ \mathrm{g}} \]

Step 1: Identify the common ion.

Strontium nitrate dissociates:

\[ \ce{Sr(NO3)2(aq) -> Sr^2+(aq) + 2NO3^-(aq)} \]

It supplies 0.0500 M strontium ions.

The nitrate ions are spectators for the SrF₂ dissolution equilibrium.

Step 2: Write the dissolution equation.

\[ \ce{SrF2(s) <=> Sr^2+(aq) + 2F^-(aq)} \]

If \(s\) is the molar solubility:

\[ [\mathrm{Sr}^{2+}]=0.0500+s \]

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

Step 3: Substitute into the \(K_{sp}\) expression.

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

\[ 2.5\times10^{-9}=(0.0500+s)(2s)^2 \]

Step 4: Apply the approximation.

Since \(s\) is small compared with 0.0500:

\[ \begin{aligned} 2.5\times10^{-9} &\approx0.0500(2s)^2\\ &=0.200s^2 \end{aligned} \]

Therefore:

\[ \begin{aligned} s^2&=\frac{2.5\times10^{-9}}{0.200}\\ s&=\sqrt{1.25\times10^{-8}}\\ &=1.12\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 5: Check the approximation.

\[ \frac{s}{0.0500}=0.00224 \]

The contribution from dissolved SrF₂ is only 0.224% of the initial strontium concentration.

The approximation is reasonable.

Step 6: Calculate the mass dissolved.

Convert the volume into litres:

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

Calculate the moles:

\[ \begin{aligned} n&=sV\\ &=(1.12\times10^{-4})(0.1000)\\ &=1.12\times10^{-5}\ \mathrm{mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m&=nM\\ &=(1.12\times10^{-5})(125.62)\\ &=1.41\times10^{-3}\ \mathrm{g} \end{aligned} \]

The final mass is approximately 1.41 mg.

Notice that the common ion is the cation this time. The method is unchanged, but the concentration that receives the initial addition is different.

Question 3

A saturated solution of calcium sulfate, CaSO₄, is prepared in 0.100 mol L⁻¹ sodium sulfate, Na₂SO₄, at 25 °C.

Use:

\[ K_{sp}=2.4\times10^{-5} \]

The solution is filtered at 25 °C to remove all undissolved CaSO₄. The clear filtrate is then diluted tenfold with pure water.

Answer the following:

(a) Calculate the original molar solubility of CaSO₄, using an appropriate approximation.

(b) Calculate \(Q_{sp}\) immediately after dilution.

(c) Is the diluted solution saturated, unsaturated, or supersaturated?

(d) Will more CaSO₄ dissolve immediately after dilution if no solid remains? What happens if fresh CaSO₄ is added?

Assume additive volumes, no significant complex formation, and constant temperature.

Solution 3

The diluted solution is unsaturated:

\[ \boxed{Q_{sp}\approx2.4\times10^{-7}<K_{sp}} \]

No additional CaSO₄ dissolves while there is no solid present. If fresh CaSO₄ is added, it can dissolve until saturation is restored.

Step 1: Calculate the original molar solubility.

The dissolution equation is:

\[ \ce{CaSO4(s) <=> Ca^2+(aq) + SO4^2-(aq)} \]

Let the molar solubility be \(s\).

The equilibrium concentrations are:

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

\[ [\mathrm{SO_4}^{2-}]=0.100+s \]

Therefore:

\[ 2.4\times10^{-5}=s(0.100+s) \]

Since \(s\) is much smaller than 0.100:

\[ \begin{aligned} s&\approx\frac{2.4\times10^{-5}}{0.100}\\ &=2.4\times10^{-4}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 2: Establish the concentrations before dilution.

The saturated solution contains:

\[ [\mathrm{Ca}^{2+}]=2.4\times10^{-4}\ \mathrm{M} \]

\[ [\mathrm{SO_4}^{2-}]=0.10024\ \mathrm{M} \]

Filtration removes the undissolved solid, not the dissolved ions.

Therefore, these concentrations remain unchanged immediately after filtration.

Step 3: Calculate the concentrations after tenfold dilution.

Every dissolved ion concentration decreases by a factor of ten.

\[ [\mathrm{Ca}^{2+}]_{\text{new}} =\frac{2.4\times10^{-4}}{10} \]

\[ =2.4\times10^{-5}\ \mathrm{M} \]

Similarly:

\[ [\mathrm{SO_4}^{2-}]_{\text{new}} =\frac{0.10024}{10} \]

\[ =0.010024\ \mathrm{M} \]

Step 4: Calculate the new ion product.

Use the same expression as the \(K_{sp}\) equation, but substitute the concentrations immediately after dilution.

\[ Q_{sp}=[\mathrm{Ca}^{2+}][\mathrm{SO_4}^{2-}] \]

\[ \begin{aligned} Q_{sp} &=(2.4\times10^{-5})(0.010024)\\ &=2.40576\times10^{-7} \end{aligned} \]

Therefore:

\[ Q_{sp}\approx2.4\times10^{-7} \]

Step 5: Compare \(Q_{sp}\) with \(K_{sp}\).

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

The solution is unsaturated.

The important distinction is between the amount of dissolved salt already present and the amount that could dissolve at equilibrium.

The diluted solution contains less dissolved CaSO₄ per litre than the original saturated solution.

However, filtration removed all the solid. There is nothing available to dissolve, so the solution simply remains unsaturated.

If fresh CaSO₄ is added, some will dissolve, increasing both ion concentrations until the solution reaches saturation again.

This is why filtration, dilution, and the availability of solid must be considered separately.

Question 4

At 25 °C, barium sulfate, BaSO₄, has an illustrative solubility product:

\[ K_{sp}=1.1\times10^{-10} \]

Excess BaSO₄ is added to a solution containing \(1.00\times10^{-5}\) mol L⁻¹ Na₂SO₄.

(a) Calculate the exact molar solubility of BaSO₄ by solving the resulting quadratic equation.

(b) Calculate the solubility that would be obtained using the approximation \(s\approx K_{sp}/C_0\).

(c) Calculate the molar solubility in pure water.

(d) Explain why the approximation in part (b) is unreliable.

Assume complete dissociation and no significant side reactions.

Solution 4

The exact molar solubility is:

\[ \boxed{6.62\times10^{-6}\ \mathrm{mol\,L^{-1}}} \]

The approximation is unreliable because the amount of sulfate produced by dissolution is comparable with the initial sulfate concentration.

Step 1: Write the dissolution equation.

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

Let the molar solubility be \(s\).

Then:

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

\[ [\mathrm{SO_4}^{2-}]=1.00\times10^{-5}+s \]

Step 2: Write the \(K_{sp}\) expression.

\[ 1.1\times10^{-10} =s(1.00\times10^{-5}+s) \]

Expand:

\[ s^2+1.00\times10^{-5}s-1.1\times10^{-10}=0 \]

Step 3: Solve using the quadratic formula.

\[ s=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

Substitute:

\[ s= \frac{-1.00\times10^{-5} \pm\sqrt{(1.00\times10^{-5})^2+ 4(1.1\times10^{-10})}}{2} \]

The positive root gives:

\[ \begin{aligned} s&=\frac{-1.00\times10^{-5} +\sqrt{5.40\times10^{-10}}}{2}\\ &=6.62\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The negative root is physically impossible because a solubility cannot be negative.

Step 4: Calculate the approximate answer.

If we ignore \(s\) in the sulfate concentration:

\[ K_{sp}\approx sC_0 \]

Therefore:

\[ \begin{aligned} s&\approx\frac{1.1\times10^{-10}}{1.00\times10^{-5}}\\ &=1.10\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

This is approximately 66% higher than the exact result.

Step 5: Calculate the solubility in pure water.

In pure water, neither ion is initially supplied by another electrolyte.

Therefore:

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

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

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

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

Hence:

\[ \begin{aligned} s&=\sqrt{1.1\times10^{-10}}\\ &=1.05\times10^{-5}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Step 6: Interpret the results.

The common-ion effect is still present: BaSO₄ is less soluble in the sodium sulfate solution.

However, the initial sulfate concentration is not large enough to dominate the sulfate contributed by dissolution.

This is the key test:

\[ \frac{s}{C_0} =\frac{6.62\times10^{-6}}{1.00\times10^{-5}} =0.662 \]

The contribution from dissolution is about 66% of the initial concentration.

You cannot reasonably ignore it.

Question 5

A chemist prepares two solutions at 25 °C:

• 100.0 mL of 0.0100 mol L⁻¹ strontium nitrate, Sr(NO₃)₂.
• 100.0 mL of 0.0200 mol L⁻¹ sodium sulfate, Na₂SO₄.

The solutions are mixed in a single beaker.

The illustrative solubility product of strontium sulfate, SrSO₄, is:

\[ K_{sp}=3.2\times10^{-7} \]

Assume the volumes are additive, both soluble compounds dissociate completely, and any SrSO₄ precipitate reaches equilibrium.

(a) Calculate the initial concentrations of Sr²⁺ and SO₄²⁻ immediately after mixing.

(b) Calculate \(Q_{sp}\) and predict whether precipitation occurs.

(c) A student suggests calculating the solubility by writing:

\[ K_{sp}=(0.00500+s)(0.0100+s) \]

Explain why this approach is incorrect.

(d) Calculate the final equilibrium concentrations of Sr²⁺ and SO₄²⁻.

(e) Calculate the mass of SrSO₄ precipitated. The molar mass of SrSO₄ is 183.68 g mol⁻¹.

Solution 5

Strontium sulfate precipitates because the initial mixture is supersaturated.

The final equilibrium concentrations are:

\[ \boxed{[\mathrm{Sr}^{2+}]=6.32\times10^{-5}\ \mathrm{M}} \]

\[ \boxed{[\mathrm{SO_4}^{2-}]=5.063\times10^{-3}\ \mathrm{M}} \]

The mass of SrSO₄ precipitated is approximately:

\[ \boxed{0.181\ \mathrm{g}} \]

Step 1: Calculate the initial moles of each ion.

For strontium nitrate:

\[ \begin{aligned} n(\mathrm{Sr}^{2+}) &=CV\\ &=0.0100\times0.1000\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

One mole of Sr(NO₃)₂ produces one mole of Sr²⁺.

For sodium sulfate:

\[ \begin{aligned} n(\mathrm{SO_4}^{2-}) &=CV\\ &=0.0200\times0.1000\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

One mole of Na₂SO₄ produces one mole of sulfate ions.

Step 2: Calculate the concentrations after mixing.

The total volume is:

\[ V=100.0+100.0=200.0\ \mathrm{mL} \]

Convert to litres:

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

Therefore:

\[ [\mathrm{Sr}^{2+}] =\frac{0.00100}{0.2000} =0.00500\ \mathrm{M} \]

And:

\[ [\mathrm{SO_4}^{2-}] =\frac{0.00200}{0.2000} =0.0100\ \mathrm{M} \]

Step 3: Calculate the initial ion product.

For strontium sulfate:

\[ Q_{sp}=[\mathrm{Sr}^{2+}][\mathrm{SO_4}^{2-}] \]

Substitute:

\[ \begin{aligned} Q_{sp}&=(0.00500)(0.0100)\\ &=5.00\times10^{-5} \end{aligned} \]

Compare with:

\[ K_{sp}=3.2\times10^{-7} \]

Clearly:

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

The solution is supersaturated, so SrSO₄ precipitation is favoured.

This immediately tells us that the student’s proposed equation is wrong.

Step 4: Establish the correct concentration changes.

The dissolution equation is:

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

The student has treated the problem as though more solid were dissolving.

That would increase both ion concentrations.

But the initial ion product is already too large. The system needs to remove dissolved ions by forming solid.

Let \(x\) represent the concentration of SrSO₄ precipitated, in mol L⁻¹.

Each mole of precipitated SrSO₄ removes one mole of Sr²⁺ and one mole of SO₄²⁻.

Therefore:

\[ [\mathrm{Sr}^{2+}]=0.00500-x \]

\[ [\mathrm{SO_4}^{2-}]=0.0100-x \]

Step 5: Apply the equilibrium condition.

Once precipitation reaches equilibrium:

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

Therefore:

\[ (0.00500-x)(0.0100-x)=3.2\times10^{-7} \]

To solve more easily, let \(y\) represent the final strontium concentration:

\[ y=0.00500-x \]

Since the initial sulfate concentration was 0.00500 M greater than the initial strontium concentration, the final sulfate concentration is:

\[ [\mathrm{SO_4}^{2-}]=0.00500+y \]

Substitute into the equilibrium expression:

\[ y(0.00500+y)=3.2\times10^{-7} \]

Expand:

\[ y^2+0.00500y-3.2\times10^{-7}=0 \]

Step 6: Solve the quadratic equation.

\[ y= \frac{-0.00500+ \sqrt{(0.00500)^2+4(3.2\times10^{-7})}}{2} \]

\[ y=6.32\times10^{-5}\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ [\mathrm{Sr}^{2+}] =6.32\times10^{-5}\ \mathrm{M} \]

And:

\[ \begin{aligned} [\mathrm{SO_4}^{2-}] &=0.00500+0.0000632\\ &=0.0050632\ \mathrm{M} \end{aligned} \]

The second possible quadratic root would give a negative concentration of strontium ions. It is physically impossible and must be rejected.

Step 7: Calculate the amount of SrSO₄ precipitated.

First, calculate \(x\):

\[ \begin{aligned} x&=0.00500-0.0000632\\ &=0.0049368\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Now calculate the total moles precipitated in 0.2000 L:

\[ \begin{aligned} n&=xV\\ &=(0.0049368)(0.2000)\\ &=9.874\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m&=nM\\ &=(9.874\times10^{-4})(183.68)\\ &=0.181\ \mathrm{g} \end{aligned} \]

Therefore:

\[ \boxed{m=0.181\ \mathrm{g\ SrSO_4}} \]

The important conceptual point is that you cannot automatically assume a common-ion problem involves additional dissolution.

When both ions are initially present, calculate \(Q_{sp}\) first. Its relationship with \(K_{sp}\) determines whether the solid dissolves or precipitates.

06What to remember when calculating solubility with a common ion

The most reliable approach is to follow the chemistry rather than memorise a shortcut.

1. Write the balanced dissolution equation.
2. Define \(s\), the molar solubility.
3. Calculate the initial common-ion concentration, accounting for dissociation and dilution.
4. Add the contribution from the dissolving solid using the stoichiometric coefficients.
5. Substitute the equilibrium concentrations into the \(K_{sp}\) expression.
6. Check whether a small-solubility approximation is justified.
7. If both ions are initially present, calculate \(Q_{sp}\) before assuming dissolution.
8. Convert to mass or another concentration unit only after finding the molar solubility.

The next useful step is applying the same equilibrium reasoning to precipitation problems. Once you can calculate solubility in a common-ion solution, you can use \(Q_{sp}\) to predict what happens when two ionic solutions are mixed, including when a precipitate forms.