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Complex ion formation and solubility

Key idea: A ligand can increase solubility by binding a dissolved metal ion into a complex. Lower free-ion concentration drives more solid to dissolve, while Ksp remains unchanged.

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  • GCE A-Level H2 Chemistry 2027
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In the free-ion model, adding a common ion lowers solubility. A ligand can do the opposite: ammonia dissolves silver chloride, because it locks silver ions up in a complex ion. This lesson explains how complex formation raises solubility, and uses it to tell silver chloride, bromide and iodide apart.

Silver halides and ammonia

Adding aqueous silver nitrate to a solution of halide ions gives a precipitate of the silver halide:

Ag⁺(aq) + X⁻(aq) → AgX(s)

For the usual qualitative test with small precipitate samples, adding aqueous ammonia helps distinguish the halides:

Silver halideColourKₛₚ / mol² dm⁻⁶Dilute ammoniaConcentrated ammonia
AgClwhiteabout 1.8 × 10⁻¹⁰dissolvesdissolves
AgBrcreamabout 5 × 10⁻¹³partly dissolvesdissolves
AgIyellowabout 8 × 10⁻¹⁷does not dissolvedoes not dissolve

Check your understanding 1

Ammonia binds free Ag⁺. Before following the linked equilibria, predict the direction of the AgCl dissolution equilibrium and whether Ksp changes at fixed temperature.

Dissolution equilibrium
Ksp
Show answer

The dissolution equilibrium shifts right; Ksp stays the same. Follow how the two equilibria share free silver, rather than treating the silver in the complex as a free ion.

How a complex raises solubility

Two equilibria are linked through the free silver ions:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq)

Ammonia is a ligand. It bonds to free Ag⁺ ions to form the diamminesilver(I) complex, which lowers [Ag⁺]. The ionic product [Ag⁺][Cl⁻] falls below Kₛₚ, so the first equilibrium shifts to the right and more AgCl dissolves. With enough ammonia, all of the precipitate dissolves.

Kₛₚ itself does not change. The silver in the complex is still dissolved, but it is no longer a free Ag⁺ ion, so it does not count in Kₛₚ.

Whether a silver halide dissolves depends on how low ammonia can push [Ag⁺], compared with how low it must be. AgCl has the largest Kₛₚ, so dilute ammonia is enough. AgBr needs the larger effect of concentrated ammonia. Under these qualitative-test conditions, AgI has so small a Kₛₚ that concentrated ammonia does not lower free silver enough to dissolve the visible precipitate.

Exam-style question 1

Why ammonia dissolves silver chloride

4 marks

Problem

Explain why solid silver chloride dissolves when excess aqueous ammonia is added. [4 marks]

Try this before viewing the solution

Show solution step by step

  1. Show the complex forming

    1 mark

    Method

    Ammonia bonds to silver ions.

    Reason

    Ammonia is a ligand.

    Working

    Answer

    Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq)

  2. Free silver falls

    1 mark

    Reason

    Silver in the complex is not free Ag⁺.

    Working

    Answer

    [Ag⁺] falls, so [Ag⁺][Cl⁻] < Kₛₚ.

  3. The dissolving equilibrium shifts

    1 mark

    Reason

    The equilibrium replaces the silver ions that were removed.

    Working

    Answer

    AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) shifts to the right.

  4. Conclude

    1 mark

    Method

    More AgCl dissolves.

    Reason

    The shift to the right uses up the solid.

    Working

    Answer

    The solubility of AgCl increases, and the precipitate dissolves.

Check your understanding 2

A precipitate is either AgBr or AgI. In a particular ammonia solution, complex formation holds free [Ag⁺] at 1.0 × 10⁻¹⁰ mol dm⁻³. If the whole precipitate dissolved, the halide concentration would be 1.0 × 10⁻³ mol dm⁻³. Use the Kₛₚ values in the table to decide which precipitate would dissolve completely.

Unit: mol² dm⁻⁶
Which can dissolve completely in this model?
Show answer

If all of it dissolved, Q = [Ag⁺][X⁻] = (1.0 × 10⁻¹⁰)(1.0 × 10⁻³) = 1.0 × 10⁻¹³ mol² dm⁻⁶. For AgBr, Q is below Kₛₚ ≈ 5 × 10⁻¹³, so the solution can hold all of it: AgBr dissolves completely. For AgI, Q is far above Kₛₚ ≈ 8 × 10⁻¹⁷, so most of the AgI stays as solid. The much smaller Kₛₚ of AgI means free [Ag⁺] would have to be pushed far lower before it dissolved. Kₛₚ itself does not change.

Check your understanding 3

Explain both trial-Q comparisons above and why the same given free silver allows complete dissolution of one halide but not the other.

Show answer

The trial Q = 1.0 × 10⁻¹³ is below AgBr’s Kₛₚ ≈ 5 × 10⁻¹³: with the given free silver held constant, all the sample can dissolve and leave an unsaturated solution. It is above AgI’s Kₛₚ ≈ 8 × 10⁻¹⁷, so that much dissolved iodide cannot coexist with the given free silver at equilibrium; some AgI remains solid. The dissolution constants differ and neither changes on adding ammonia.

Free silver and total dissolved silver

With a complex present, dissolved silver exists in two forms:

total dissolved silver = [Ag⁺] + [[Ag(NH₃)₂]⁺]

The graph follows both for AgCl at increasing specified free ammonia concentration, with excess AgCl solid remaining. Its two-species model includes only free silver and the diamminesilver complex, at fixed temperature and with no added chloride.

Check your understanding 4

Before reading the graph, predict the trend in free silver and total dissolved silver as free ammonia increases. Use the two linked equilibria, not just the falling concentration of one ion.

Free silver
Total dissolved silver
Show answer

Free silver falls while total dissolved silver rises. Inspect both series to test the distinction between a free ion and all dissolved forms of an element.

Free silver falls while dissolved silver rises

Concentrations of total dissolved silver and free silver ions against specified free ammonia concentration. Both begin at 13.4 μmol dm⁻³ without ammonia; increasing free ammonia produces opposing trends.

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Concentrations of total dissolved silver and free silver ions against specified free ammonia concentration. Both begin at 13.4 μmol dm⁻³ without ammonia; increasing free ammonia produces opposing trends.Concentrations of total dissolved silver and free silver ions against specified free ammonia concentration. Both begin at 13.4 μmol dm⁻³ without ammonia; increasing free ammonia produces opposing trends.
A calculated AgCl–ammonia teaching model at 25°C: total dissolved silver rises even though free Ag⁺ falls. Excess AgCl solid remains. Silver values in the graph and data table are in μmol dm⁻³.
Open full-size graph
What to noticeTwo-species concentration model: only free Ag⁺ and [Ag(NH₃)₂]⁺ are included. Kₛₚ = 1.8×10⁻¹⁰ and Kf = 1.7×10⁷; no added chloride, other silver complexes or activity effects. Free NH₃ is specified, not total ammonia added. Model values are not measurements.
View figure data
Figure data: Free silver falls while dissolved silver rises
Free NH₃ concentration (mmol dm⁻³)Total dissolved silverFree Ag⁺
0.00013.4213.416
0.02513.4913.346
0.05013.7013.140
0.07514.0412.818
0.10014.5112.403
0.12515.0911.926
0.15015.7711.410
0.17516.5410.880
0.20017.3910.351
0.22518.309.836
0.25019.279.342
0.27520.288.874
0.30021.348.435
0.32522.438.024
0.35023.567.642
0.37524.707.286
0.40025.886.956
0.42527.076.650
0.45028.286.365
0.47529.506.101
0.50030.745.855
0.52531.995.627
0.55033.255.413
0.57534.525.214
0.60035.805.028
0.62537.094.854
0.65038.384.690
0.67539.684.537
0.70040.984.392
0.72542.294.256
0.75043.604.128
0.77544.924.007
0.80046.243.892
0.82547.573.784
0.85048.903.681
0.87550.233.584
0.90051.563.491
0.92552.903.403
0.95054.243.319
0.97555.583.239
1.00056.923.162

Check your understanding 5

Use the two graph trends to judge the claim “less free silver means less AgCl has dissolved”. Identify which dissolved form accounts for the widening gap between the curves.

Show answer

The graph shows falling free Ag⁺ but rising total dissolved silver. The difference is silver in [Ag(NH₃)₂]⁺. The falling free-ion concentration drives more salt dissolution while complex formation holds much of that dissolved silver. Only free silver belongs in Ksp; both forms count towards solubility.

Check your understanding 6 Which silver goes into Ksp?

In a saturated AgCl solution containing ammonia, free [Ag⁺] = 3.16 × 10⁻⁶, [Cl⁻] = 5.69 × 10⁻⁵, and total dissolved silver is 5.69 × 10⁻⁵ mol dm⁻³. A learner puts the total dissolved silver into Kₛₚ and concludes that Kₛₚ has increased. Find the error, check Kₛₚ correctly and calculate the concentration of silver held in the complex.

Silver concentration used in Ksp
Unit: mol² dm⁻⁶
Unit: mol dm⁻³
Show answer

Kₛₚ uses the free Ag⁺ concentration, not the silver held in the complex. With free silver: (3.16 × 10⁻⁶)(5.69 × 10⁻⁵) = 1.8 × 10⁻¹⁰, the same Kₛₚ as in pure water. Total dissolved silver equals [Cl⁻], because every dissolved AgCl released one chloride ion, so the solubility has risen to 5.69 × 10⁻⁵ mol dm⁻³ even though free Ag⁺ has fallen.

Check your understanding 7

Explain why the correct free-ion product is unchanged while solubility has risen. Show the complex-silver subtraction and identify which concentration measures dissolved AgCl.

Show answer

Kₛₚ = (3.16 × 10⁻⁶)(5.69 × 10⁻⁵) = 1.8 × 10⁻¹⁰ mol² dm⁻⁶. Complex silver is (5.69 × 10⁻⁵)-(3.16 × 10⁻⁶) = 5.37 × 10⁻⁵ mol dm⁻³. Total silver, 5.69 × 10⁻⁵ mol dm⁻³, measures AgCl dissolved because each formula unit supplies one silver atom. With no added chloride, chloride also measures that total, whereas the smaller free-silver concentration measures only one dissolved form.

Common ions and complexes compared

Check your understanding 8

Two identical saturated AgCl samples contain undissolved solid. Add NaCl to A and excess aqueous ammonia to B. Choose each solubility direction and compare the final free-silver trends. Keep temperature fixed, ignore volume changes and exclude chloride complexes in sample A.

A: added NaCl, AgCl solubility
B: added excess NH₃, AgCl solubility
Final free Ag⁺
Show answer

Solubility falls in A and rises in B; free silver falls in both. In the next response, explain both directions with equations and compare the initial Q with the equilibrium free-ion product.

Try it yourself 1

Sodium chloride or ammonia?

Minimal support

Problem

Two identical samples of saturated silver chloride contain some undissolved solid. Keep temperature fixed, ignore volume changes and exclude chloride complexes in A. Sodium chloride is added to sample A, and excess aqueous ammonia to sample B. Compare what happens to the solubility of AgCl in each, and explain using the free ions.

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Hints

Hint 1

sample A

Sodium chloride adds one of the ions of the equilibrium.
Hint 2

sample B

What does ammonia do to the free silver ions?
Show solution step by step

  1. Sample A

    Method

    Solubility decreases.

    Reason

    Added Cl⁻ is a common ion; the equilibrium shifts to the left.

    Working

    Answer

    More AgCl precipitates.

  2. Sample B

    Method

    Solubility increases.

    Reason

    Ammonia removes Ag⁺ as [Ag(NH₃)₂]⁺; the equilibrium shifts to the right.

    Working

    Answer

    More AgCl dissolves.

  3. Compare

    Method

    The two changes go in opposite directions.

    Reason

    A common ion adds a product of the dissolving equilibrium; complex formation removes one.

    Working

    Answer

    In both samples free [Ag⁺] ends up lower, and Kₛₚ is unchanged.

Common mistakes

  • Saying ammonia increases Kₛₚ. It increases the solubility; Kₛₚ stays the same.
  • Counting silver in the complex as free Ag⁺ in the Kₛₚ expression.
  • Saying ammonia “reacts with the chloride”. It bonds to the silver ions.
  • Mixing up the halide results: AgCl dissolves in dilute ammonia, AgBr fully only in concentrated ammonia, AgI in neither.
  • Using free [Ag⁺] alone as a measure of how much salt has dissolved.

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