What is check the approximation?

s' / 0.10 = 1.8 × 10^-8, far below 5%. Valid.

Write the Ksp expression for silver chromate, Ag_2CrO_4(s) 2Ag^+_(aq) + CrO_4^2-(aq), and state the units. [2 marks]

The Ksp of PbI_2 is 7.1 × 10^-9. Calculate the molar solubility of PbI_2 in pure water. [3 marks]

A solution contains [Ba^2+] = 1.0 × 10^-4 mol L^-1 and [SO_4^2-] = 5.0 × 10^-5 mol L^-1. Given that K_sp(BaSO_4) = 1.1 × 10^-10, (a) calculate Q, (b) predict whether a precipitate forms, (c) explain how adding sodium sulfate would change the answer. [2+1+2 marks]

§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 4: How does solubility relate to chemical equilibrium and the position of equilibrium?

Predict the solubility of an ionic substance by applying solubility equilibrium principles, and perform calculations involving the solubility product constant (Ksp) and the ionic product

A focused answer to the HSC Chemistry Module 5 dot point on the solubility product. Writing Ksp expressions, predicting precipitation using the ionic product Q vs Ksp, the common ion effect, and worked HSC calculations for solubility and precipitation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Worked example 2: the common ion effect
Examples in context
Try this

In plain English

Some salts barely dissolve at all - stir them into water and almost all of the solid just sits there, with only a tiny, fixed pinch of it actually going into solution. Ksp is a number that measures exactly how tiny that pinch is: a small Ksp means "basically insoluble", a bigger Ksp means "a bit more soluble". If you work out the same kind of number (called Q) for a solution someone has just mixed and it is bigger than Ksp, that solution is over-stuffed with dissolved salt, like adding one spoon of sugar too many to a cold drink, and the extra has to crash out as solid. Adding more of one of the ions that is already in the salt (say, more chloride to a silver chloride solution) is like adding more sugar to that already-saturated drink - it pushes even more solid out of solution. That is the common ion effect, and it is just Le Chatelier's principle doing its job.

What this dot point is asking

NESA wants you to write Ksp expressions for sparingly soluble salts, calculate molar solubility, predict whether a precipitate will form by comparing the ionic product Q to Ksp, and understand the common ion effect. This dot point is examined as a 3-5 mark calculation question and as part of extended response questions on environmental chemistry.

The answer

The solubility product Ksp

For a sparingly soluble ionic compound, dissolution is an equilibrium between the solid and its ions in solution:

\text{M}_a\text{X}_{b(s)} \rightleftharpoons a\text{M}^{n+}_{(aq)} + b\text{X}^{m-}_{(aq)}

The solubility product constant Ksp is the equilibrium constant for this dissolution. The pure solid is excluded, so:

K_{sp} = [\text{M}^{n+}]^a [\text{X}^{m-}]^b

Examples:

Compound	Dissolution	Ksp expression
$AgCl$	$AgCl_{(s)} \rightleftharpoons Ag^+ + Cl^-$	$[Ag^+][Cl^-]$
$PbCl_2$	$PbCl_{2(s)} \rightleftharpoons Pb^{2+} + 2Cl^-$	$[Pb^{2+}][Cl^-]^2$
$Ca_3(PO_4)_2$	$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}$	$[Ca^{2+}]^3[PO_4^{3-}]^2$

Ksp values are small (typically $10^{-5}$ to $10^{-50}$ ), reflecting the fact that the compound is only slightly soluble.

The ionic product Q

For any solution (not necessarily at equilibrium), the same expression evaluated using current concentrations is the ionic product Q.

Compare Q to Ksp to predict precipitation:

If $Q < K_{sp}$ , the solution is unsaturated. More solid can dissolve. No precipitate forms.
If $Q = K_{sp}$ , the solution is saturated. The system is at equilibrium.
If $Q > K_{sp}$ , the solution is supersaturated. A precipitate forms until Q falls back to Ksp.

The common ion effect

Adding a common ion to a saturated solution decreases solubility. For example, adding NaCl to a saturated solution of AgCl increases $[Cl^-]$ , so $[Ag^+]$ must decrease to keep $Q = K_{sp}$ . Some AgCl precipitates.

This is a direct application of Le Chatelier's principle. The added ion shifts the dissolution equilibrium to the left (toward the solid).

Worked example

Calculate the molar solubility of lead(II) chloride, $PbCl_2$ , in pure water at 25°C. Ksp = $1.7 \times 10^{-5}$ .

Step 1: Dissolution equation.

PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}

Step 2: Ksp expression.

K_{sp} = [Pb^{2+}][Cl^-]^2

Step 3: Set up with molar solubility $s$ .

Each mole of $PbCl_2$ that dissolves gives 1 mole of $Pb^{2+}$ and 2 moles of $Cl^-$ .

$[Pb^{2+}] = s$ , $[Cl^-] = 2s$ .

Step 4: Substitute and solve.

K_{sp} = (s)(2s)^2 = 4s^3 = 1.7 \times 10^{-5}

s^3 = 4.25 \times 10^{-6}

s = (4.25 \times 10^{-6})^{1/3} = 1.62 \times 10^{-2} \text{ mol/L}

The molar solubility is $1.62 \times 10^{-2}$ mol/L (about 0.016 mol/L).

Compare to AgCl. Note that $PbCl_2$ has a larger Ksp than AgCl but the relationship to solubility is not direct (different stoichiometries). For $AB$ salts, $s = \sqrt{K_{sp}}$ . For $AB_2$ salts, $s = \sqrt[3]{K_{sp}/4}$ .

Worked example 2: the common ion effect

Calculate the molar solubility of $AgCl$ ( $K_{sp} = 1.8 \times 10^{-10}$ ) in 0.10 mol/L $NaCl$ , and compare with its solubility in pure water.

Step 1: Account for the common ion. $NaCl$ fully dissociates, so $[Cl^-]$ from the salt is 0.10 mol/L before any AgCl dissolves. Let $s'$ be the molar solubility of AgCl in this solution.

	$Ag^+$	$Cl^-$
From $NaCl$	0	0.10
From dissolved $AgCl$	$+s'$	$+s'$
Equilibrium	$s'$	$0.10 + s'$

Step 2: Substitute into Ksp.

K_{sp} = [Ag^+][Cl^-] = s'(0.10 + s') = 1.8 \times 10^{-10}

Step 3: Approximate. Because Ksp is tiny and $[Cl^-]$ from NaCl is much larger than $s'$ , $0.10 + s' \approx 0.10$ .

s' \approx \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ mol/L}

Step 4: Compare. In pure water $s = 1.34 \times 10^{-5}$ mol/L; in 0.10 mol/L NaCl, $s' = 1.8 \times 10^{-9}$ mol/L. The solubility falls by a factor of ~7400. Adding the common ion drives the equilibrium to the left (toward solid), exactly as Le Chatelier's principle predicts.

Check the approximation. $s' / 0.10 = 1.8 \times 10^{-8}$ , far below 5%. Valid.

Examples in context

Example 1. Cadia gold mine cyanide leach circuit. At Newmont's Cadia operation near Orange, gold is dissolved as the soluble dicyanoaurate complex, but trace metals must be precipitated out before tailings discharge. Engineers control the silver content of the spent solution by manipulating the equilibrium $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$ , with $K_{sp} = 1.8 \times 10^{-10}$ . Adding excess chloride to the leach residue drives the common-ion effect, dropping $[Ag^+]$ to well below regulatory limits. The same Ksp framework students apply to lead chloride precipitation problems in HSC Section II underpins the real-time process control at the tailings dam.

Example 2. Hard-water scaling in Hunter Valley irrigation pipes. Bore water along the Hunter Valley vineyard belt contains calcium and carbonate above the saturation threshold for $CaCO_3$ , with $K_{sp} = 3.4 \times 10^{-9}$ . Pipe inspections after summer reveal a hard white scale on the inside of irrigation lines. Calculating the ionic product $Q = [Ca^{2+}][CO_3^{2-}]$ from a typical bore reading of $[Ca^{2+}] = 2.0 \times 10^{-3}$ and $[CO_3^{2-}] = 3.0 \times 10^{-5}$ mol L $^{-1}$ gives $Q = 6.0 \times 10^{-8}$ , well above Ksp, so calcite precipitates. The same calculation tells maintenance crews when to dose with acid or replace pipe runs.

Try this

Q1. Write the Ksp expression for silver chromate, $Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO_4^{2-(aq)}$ , and state the units. [2 marks]

Cue. $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$ , with units mol $^3$ L $^{-3}$ for the unsymmetrical 2:1 salt.

Q2. The Ksp of $PbI_2$ is $7.1 \times 10^{-9}$ . Calculate the molar solubility of $PbI_2$ in pure water. [3 marks]

Cue. Let $s$ = molar solubility, then $[Pb^{2+}] = s$ and $[I^-] = 2s$ , so $K_{sp} = s(2s)^2 = 4s^3$ ; $s = \sqrt[3]{7.1 \times 10^{-9} / 4} = 1.21 \times 10^{-3}$ mol L $^{-1}$ .

Q3. A solution contains $[Ba^{2+}] = 1.0 \times 10^{-4}$ mol L $^{-1}$ and $[SO_4^{2-}] = 5.0 \times 10^{-5}$ mol L $^{-1}$ . Given that $K_{sp}(BaSO_4) = 1.1 \times 10^{-10}$ , (a) calculate $Q$ , (b) predict whether a precipitate forms, (c) explain how adding sodium sulfate would change the answer. [2+1+2 marks]

Cue. (a) $Q = 1.0 \times 10^{-4} \times 5.0 \times 10^{-5} = 5.0 \times 10^{-9}$ . (b) $Q > K_{sp}$ , so $BaSO_4$ precipitates. (c) Common-ion effect raises $[SO_4^{2-}]$ , $Q$ increases further, more precipitate forms.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2020 HSC4 marksThe Ksp of silver chloride (AgCl) at 25°C is 1.8 × 10⁻¹⁰. Calculate the molar solubility of silver chloride in pure water at 25°C.

Show worked answer →

A 4 mark answer needs the dissolution equation, the Ksp expression, the algebraic solution, and the final value with units.

Step 1: Dissolution equation.

AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}

Step 2: Ksp expression. Pure solid AgCl is excluded.

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

Step 3: Set up using molar solubility $s$ .

Each mole of AgCl that dissolves gives 1 mole of $Ag^+$ and 1 mole of $Cl^-$ , so $[Ag^+] = [Cl^-] = s$ .

K_{sp} = s \times s = s^2 = 1.8 \times 10^{-10}

Step 4: Solve.

s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ mol/L}

Markers reward (1) the balanced dissolution equation with state symbols, (2) the correct Ksp expression (excluding the solid), (3) clear use of molar solubility $s$ with stoichiometry, (4) the answer in mol/L to appropriate significant figures.

2019 HSC3 marksEqual volumes of 0.0010 mol/L silver nitrate and 0.0020 mol/L sodium chloride are mixed. The Ksp of AgCl is 1.8 × 10⁻¹⁰. Determine, with calculation, whether a precipitate will form.

Show worked answer →

Mixing equal volumes halves each concentration.

After mixing:
$[Ag^+] = 0.0010 / 2 = 5.0 \times 10^{-4}$ mol/L.
$[Cl^-] = 0.0020 / 2 = 1.0 \times 10^{-3}$ mol/L.

Calculate the ionic product Q.

Q = [Ag^+][Cl^-] = (5.0 \times 10^{-4})(1.0 \times 10^{-3}) = 5.0 \times 10^{-7}

Compare to Ksp.

$Q = 5.0 \times 10^{-7}$ is much greater than $K_{sp} = 1.8 \times 10^{-10}$ .

Since $Q > K_{sp}$ , the solution is supersaturated and a precipitate of AgCl will form.

Markers reward (1) the dilution step (halving concentrations after mixing), (2) calculating Q correctly, (3) the explicit Q vs Ksp comparison and the conclusion.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksWrite the Ksp expression for magnesium hydroxide,

Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^-_{(aq)}

, and state its units.

Show worked solution →

Ksp expression. The solid is excluded:

K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2

Units. Three ion terms are multiplied ( $[\text{Mg}^{2+}]$ to the first power, $[\text{OH}^-]$ squared), giving mol $^3$ L $^{-3}$ .

Markers reward (1) the correct expression with OH- squared, (2) the solid correctly excluded, (3) correct units mol $^3$ L $^{-3}$ .

foundation3 marksThe Ksp of silver bromide,

AgBr

, is

5.0 \times 10^{-13}

at 25°C. Calculate its molar solubility in pure water.

Show worked solution →

Step 1: Dissolution equation.

AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^-_{(aq)}

Step 2: Ksp expression and substitution. Let molar solubility $= s$ . Since the stoichiometry is 1:1, $[Ag^+] = [Br^-] = s$ .

K_{sp} = [Ag^+][Br^-] = s^2 = 5.0 \times 10^{-13}

Step 3: Solve.

s = \sqrt{5.0 \times 10^{-13}} = 7.07 \times 10^{-7} \text{ mol/L (3 s.f.)}

Markers reward (1) the dissolution equation, (2) correct Ksp expression, (3) the final value in mol/L to appropriate significant figures.

core4 marksA student mixes 50.0 mL of

2.0 \times 10^{-4}

mol/L lead(II) nitrate with 50.0 mL of

4.0 \times 10^{-4}

mol/L potassium iodide. Given

K_{sp}(PbI_2) = 7.1 \times 10^{-9}

, determine with a full calculation whether a precipitate of

PbI_2

forms.

Show worked solution →

Step 1: Correct for dilution. Equal volumes mixed, so each concentration halves.

[Pb^{2+}] = \frac{2.0 \times 10^{-4}}{2} = 1.0 \times 10^{-4} \text{ mol/L}

[I^-] = \frac{4.0 \times 10^{-4}}{2} = 2.0 \times 10^{-4} \text{ mol/L}

Step 2: Calculate the ionic product Q. $PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$ , so $Q = [Pb^{2+}][I^-]^2$ .

Q = (1.0 \times 10^{-4})(2.0 \times 10^{-4})^2 = (1.0 \times 10^{-4})(4.0 \times 10^{-8})

Q = 4.0 \times 10^{-12}

Step 3: Compare to Ksp.

$Q = 4.0 \times 10^{-12}$ is less than $K_{sp} = 7.1 \times 10^{-9}$ .

Step 4: Conclusion. Since $Q < K_{sp}$ , the mixed solution is unsaturated with respect to $PbI_2$ , so no precipitate forms.

Markers reward (1) the dilution correction on both concentrations, (2) the correctly squared $[I^-]$ term in Q, (3) the numerical value of Q to 2 s.f., (4) the explicit Q vs Ksp comparison and correct conclusion.

core5 marksThe graph below shows the solubility of potassium nitrate (

KNO_3

) and sodium chloride (

NaCl

) against temperature (owned illustrative data, not measured Ksp values). (a) Describe the trend for each salt. (b) Explain, in terms of the dissolution equilibrium, why

KNO_3

solubility is far more temperature-sensitive than

NaCl

. (c) A saturated

KNO_3

solution at 60°C (solubility about 110 g per 100 g water) is cooled to 20°C (solubility about 32 g per 100 g water). Calculate the mass of

KNO_3

that crystallises from 250 g of water.

Show worked solution →

(a) Trend: $KNO_3$ solubility rises steeply and continuously with temperature (from about 20 g per 100 g water at 0°C to over 240 g per 100 g water at 100°C on the figure). $NaCl$ solubility rises only gently and almost linearly across the same range (about 36 to 39 g per 100 g water).
(b) Explanation: Dissolving $KNO_3$ is strongly endothermic, so by Le Chatelier's principle, raising temperature (adding heat, a "reactant") shifts the dissolution equilibrium strongly to the right, sharply increasing solubility. Dissolving $NaCl$ is close to thermally neutral (only weakly endothermic), so temperature has little effect on the equilibrium position and solubility barely changes.
(c) Calculation

Mass dissolved at 60°C in 250 g water:

m_1 = \frac{110}{100} \times 250 = 275 \text{ g}

Mass that remains dissolved at 20°C in 250 g water:

m_2 = \frac{32}{100} \times 250 = 80 \text{ g}

Mass that crystallises out:

\Delta m = m_1 - m_2 = 275 - 80 = 195 \text{ g (3 s.f.)}

Markers reward (1) both trends described with approximate figures from the graph, (2) the endothermic dissolution / Le Chatelier explanation linking heat to equilibrium shift, (3) correct calculation of mass dissolved at each temperature, (4) correct final mass crystallised with units.

core4 marksCalculate the molar solubility of calcium fluoride,

CaF_2

(

K_{sp} = 3.9 \times 10^{-11}

), in a solution that is already

0.020

mol/L in

NaF

Show worked solution →

Step 1: Set up the common-ion table. $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$ . Let $s'$ be the molar solubility of $CaF_2$ in this solution. $NaF$ fully dissociates, giving $[F^-] = 0.020$ mol/L before any $CaF_2$ dissolves.

[Ca^{2+}] = s', \quad [F^-] = 0.020 + 2s'

Step 2: Approximate. Since $K_{sp}$ is tiny, $2s' \ll 0.020$ , so $[F^-] \approx 0.020$ mol/L.

Step 3: Substitute into Ksp.

K_{sp} = [Ca^{2+}][F^-]^2 = s'(0.020)^2 = 3.9 \times 10^{-11}

s' = \frac{3.9 \times 10^{-11}}{4.0 \times 10^{-4}} = 9.75 \times 10^{-8} \text{ mol/L}

Step 4: Check the approximation. $2s' = 1.95 \times 10^{-7}$ , which is about $9.8 \times 10^{-4}$ percent of $0.020$ , far below the 5 percent threshold, so the approximation is valid.

Markers reward (1) correct common-ion table with the $2s'$ stoichiometric term, (2) valid approximation stated and later checked, (3) correct algebra substituting into Ksp, (4) final answer in mol/L to appropriate significant figures.

exam6 marksAnalyse how solubility equilibrium principles, including Ksp and the common ion effect, are applied to control dissolved metal ion concentrations in the treatment of mine wastewater before discharge.

Show worked solution →

Band 6 plan. (1) State the equilibrium underlying the problem: a sparingly soluble metal salt in equilibrium with its ions. (2) Define Ksp and Q, and state the precipitation criterion Q > Ksp. (3) Explain how operators exploit the common ion effect to force precipitation. (4) Link to Le Chatelier's principle. (5) Note the real engineering constraint (adding excess reagent, monitoring against a discharge limit) and evaluate a limitation.

Model answer.

Mine wastewater often contains dissolved heavy metal ions (e.g. $Ag^+$ , $Pb^{2+}$ ) at concentrations that exceed environmental discharge limits. These metals exist in equilibrium with sparingly soluble salts, for example $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$ , governed by $K_{sp} = [Ag^+][Cl^-]$ .

To remove the metal, treatment plants deliberately raise the concentration of a common ion (commonly $Cl^-$ or $S^{2-}$ ) by dosing the wastewater with a soluble salt of that ion. This increases the ionic product $Q = [Ag^+][Cl^-]$ above $K_{sp}$ , so the solution becomes supersaturated and the equilibrium shifts left (Le Chatelier's principle) as solid precipitates out, consuming free $Ag^+$ until $Q$ falls back to $K_{sp}$ .

Because $K_{sp}$ is fixed at a given temperature, adding a large excess of the common ion forces the equilibrium concentration of the target metal ion to a very low value, since $[Ag^+] = K_{sp} / [Cl^-]$ falls as $[Cl^-]$ rises. This is exactly the common ion effect used in HSC calculations, scaled to an industrial process.

A limitation is that excess reagent itself becomes a discharge concern (e.g. excess chloride or sulfide) and must be independently monitored, and the precipitate (sludge) requires safe disposal. Operators therefore dose only enough common ion to bring the target metal below its regulatory limit, verified by sampling and comparing measured concentrations against the calculated equilibrium value.

Marker's note: Top bands require the explicit Ksp/Q relationship (not just "add a chemical to precipitate it"), the Le Chatelier link, a real named ion pair, AND an evaluative limitation. Mid-band responses often state the common ion effect correctly but never connect it back to the Ksp expression or fail to evaluate.