Solutions, Solubility and Precipitation

Two clear solutions are poured together in the dark and something solid appears. That is not a laboratory curiosity; it is how every silver gelatin emulsion that has ever been made was made, and the properties of the solid that appears — how big its crystals are, what sits on their surfaces, how nearly impossible they are to redissolve — decide what the film can do. This page is about which substances dissolve, which fall out, and what governs the difference, because that difference is the manufacturing process of photography.

Dissolving, seen from the molecule

Three things have to happen for a solid to dissolve. The solid’s own particles have to be pulled apart from each other; the solvent’s particles have to be pushed aside to make room; and the solvent has to attach itself to the freed particles. The first two cost energy. The third releases it. Whether dissolution overall absorbs or releases energy is simply which of those is larger.

For an ionic solid in water the accounting is stark. Pulling the lattice apart means overcoming the full electrostatic attraction between the ions — the same attraction the previous page put at 769 kJ per mole for sodium chloride. Paying it back means surrounding each freed ion with water molecules, which grip it by ion-dipole attraction: oxygen ends inwards to a cation, hydrogen ends inwards to an anion. That payback is called hydration, and it is the only thing standing between a salt and permanent insolubility.

OpenStax states the two outcomes and names an example of each. Where the lattice attraction greatly exceeds what hydration can pay back, dissolution is strongly endothermic and the compound may barely dissolve at all — calcium carbonate, the material of a coral reef, is the case it gives. Where hydration dominates, dissolution is strongly exothermic and the compound is highly soluble — sodium hydroxide is the case it gives, and CAMEO’s reactivity profile records that dissolving it liberates enough heat to make a solution steam and spatter.

Why some dissolution makes the beaker cold

Between those extremes sit the salts where the two terms nearly cancel, and a small residual can go either way. When the endothermic side wins by a little, the solution takes heat from its surroundings and the beaker goes cold in your hand. OpenStax uses ammonium nitrate, the instant cold pack, as its example.

Photography has its own. Ammonium chloride, the chloride of some salted-paper recipes, is recorded by HSDB as dissolving strongly endothermically: a salting bath made up in a jug noticeably chills it. The counterpart is sodium hydroxide, which is the reason a caustic developer is mixed into cold water and never the reverse.

!Endothermic does not mean reluctant

A cold beaker is not evidence that the salt is dissolving badly. Ammonium chloride dissolves freely while absorbing heat, because a dissolved solution is a far more disordered arrangement than a crystal plus separate water, and that gain in disorder can carry a process that is uphill in energy alone. This is the one place in this part where thermodynamics is doing work behind the curtain, and this course does not derive it here: entropy and free energy sit above the level Part III sets. Where a later part needs a thermodynamic argument it makes it locally, as Part XX does for the redox potentials of toning.

Dissolving, as a balance of two energies

1. 1Breaking the lattice — always costs energy: the ions are being pulled off each other
2. 2Hydrating the ions — always releases energy: water grips each freed ion
3. 3Exothermic: the beaker warms — hydration more than repays the lattice, as for sodium hydroxide
4. 4Endothermic: the beaker cools — hydration repays a little less, as for ammonium chloride

Drawn to show the direction of each term, not to scale: the two arrows in a real salt are both very large and their difference is small, which is why solubility is hard to predict from first principles.

Saturation is a dynamic equilibrium

Put more salt into water than the water can hold, stir, and after a while nothing more dissolves. It is tempting to picture that as the water being full, and the picture is wrong.

What has happened is that two processes are running at once. Ions are still leaving the surface of the undissolved crystals and entering the solution. Ions are still leaving the solution and rejoining the crystals. At the start, dissolving is faster. As the solution fills with ions, rejoining speeds up. When the two rates become equal, the concentration stops changing — and stays there, not because anything stopped, but because everything is going both ways at once. OpenStax puts the general point in one sentence: a reaction at equilibrium has not stopped, it is proceeding in both directions at the same rate.

A saturated solution is one at that balance. Its concentration is the substance’s solubility at that temperature. Nothing about it is static.

Saturation, drawn as two equal rates

1. 1Early — plenty of undissolved solid, few ions in solution: dissolving is much faster than crystallising
2. 2Approaching saturation — the solution is filling; crystallising is catching up
3. 3Saturated — both rates equal; the concentration stops changing and the traffic does not

The arrows are drawn to show the ratio of the two rates, not measured values.

Supersaturation is the state in which a solution holds more than its saturation concentration. It is prepared by dissolving to saturation while hot and cooling without disturbance, and it is fragile: OpenStax notes that a seed crystal or mechanical agitation makes the excess precipitate. Every photographer who has found a bottle of stock solution grown solid overnight has met it, and so has every photographer who has knocked the bench and seen a clear solution go cloudy in front of them.

Equilibrium, stated generally

Since this is the course’s first equilibrium, take the general form once, so that the buffer, redox, kinetics and complex-formation pages can use it without restating it.

For a reaction that runs both ways, the reaction quotient Q is the concentrations of the products multiplied together, divided by the concentrations of the reactants multiplied together, each raised to the power of its coefficient in the balanced equation. When the system reaches equilibrium, Q stops changing, and the value it settles at is the equilibrium constant K. A large K means the products are strongly favoured; a small K means the reactants are.

Two conventions matter here. A pure solid or a pure liquid does not appear in the expression, which is exactly why a solubility product contains only the dissolved ions. And K depends on temperature: it is a constant for a reaction at a stated temperature, not a constant of the universe.

Le Chatelier’s principle says that if a system at equilibrium is disturbed, it shifts in the direction that relieves the disturbance. Add a product and it shifts back towards reactants; remove a product and it shifts forwards. OpenStax explains it mechanically rather than as a slogan: the disturbance changes the forward and reverse rates unequally, so for a while one outruns the other, and the composition moves until they are equal again.

✕Le Chatelier says nothing about how fast

A shift that the principle predicts may take a second or a century. Position and rate are separate questions with separate answers, and confusing them is the commonest error in applied chemistry. An unexposed silver halide grain would eventually be reduced by a developer — the position of that equilibrium says so — and the reason a photograph exists at all is that it does not get round to it in ten minutes. Rate belongs to the kinetics page later in this part, and it is a different subject.

Which salts dissolve, in the narrow world of photography

The general solubility guidelines fill a page in a textbook. Photography needs three lines of them.

• Nitrates all dissolve. Every one. That is why nitrate is the anion silver is supplied with: silver nitrate is soluble in anything, so it can be weighed, dissolved and delivered.
• Sodium, potassium and ammonium salts all dissolve. That is why those three are the counter-ions of almost every photographic formula: whatever anion you want, one of them will make it into a storable powder that dissolves cleanly.
• The silver halides do not dissolve. Silver chloride, silver bromide and silver iodide are the exceptions the whole medium is built on.

Put the first and the third together and you have the reaction that makes an emulsion. Two soluble salts meet; the silver and the halide find each other; the solid falls out and the leftovers stay in solution.

AgNO₃ + KBr → AgBr + KNO₃

Precipitation, molecular form

Ag⁺ + Br⁻ → AgBr

Precipitation, net ionic form: the only ions that did anything

Kodak’s 1928 primer calls this “double decomposition” and adds a physical observation worth having: if the solutions are at all concentrated, the silver bromide comes down as a “thick, curdy precipitate”. Curdy is what a fast, uncontrolled precipitation looks like, and controlling it is most of what emulsion making is.

Some real solubilities, so that “soluble” and “insoluble” stop being adjectives:

Substance	Solubility	Source
Silver nitrate	245 g per 100 g of water, temperature not stated; 122 g per 100 mL at 0 °C	HSDB via PubChem
Potassium bromide	67.8 g per 100 g at 25 °C	HSDB via PubChem
Sodium chloride	36.0 g per 100 g of water at 25 °C	HSDB via PubChem
Sodium carbonate	30.7 g per 100 g at 25 °C	HSDB via PubChem
Sodium sulfite	30.7 g per 100 g at 25 °C; 22 g per 100 mL at 20 °C	HSDB and the ILO-WHO card
Borax	3.17 g per 100 g at 25 °C; 5.1 g per 100 mL at 20 °C	HSDB and the ILO-WHO card
Silver bromide	about 0.00013 g per litre at 25 °C	calculated below from the solubility product

Notice two things. The first is that the range covers seven orders of magnitude, which is why one word cannot cover it. The second is that two of the rows disagree with themselves: published compilations give sodium sulfite as 30.7 and as 22, and borax as 3.17 and as 5.1, at temperatures five degrees apart. That is normal, and it is why this course names the source beside every number. Where a figure decides something, measure it or cite it; do not average two tables.

The solubility product, before the algebra and then with it

Silver bromide does dissolve. Just barely. A saturated solution of it exists, and in it a very small number of silver ions and an equal number of bromide ions are wandering about, in equilibrium with the solid.

AgBr ⇌ Ag⁺ + Br⁻

The dissolution equilibrium of silver bromide

Apply the general rule. The solid does not appear in the expression, so the equilibrium constant is just the two ion concentrations multiplied together. That product is the solubility product, Ksp.

K_sp = [Ag⁺][Br⁻]

Solubility product of a 1:1 salt

The square brackets mean “the concentration of, in moles per litre”. The statement Ksp makes is this: you may have as much silver ion as you like or as much bromide as you like, but the product of the two cannot exceed this number while solid is present. Push one up and the other must come down. That single sentence is the whole of emulsion chemistry, developer restraint and the first half of fixing.

OpenStax Appendix J gives the three that matter, all at 25 °C:

Salt	Ksp at 25 °C	Saturated [Ag⁺]	As a mass
Silver chloride, AgCl	1.6 × 10⁻¹⁰	1.3 × 10⁻⁵ mol/L	1.8 mg/L
Silver bromide, AgBr	5.0 × 10⁻¹³	7.1 × 10⁻⁷ mol/L	0.13 mg/L
Silver iodide, AgI	1.5 × 10⁻¹⁶	1.2 × 10⁻⁸ mol/L	0.0029 mg/L

The middle column is obtained by noticing that in pure water the two ions arrive in equal numbers, so Ksp is the concentration squared, and the concentration is the square root of Ksp. The right-hand column multiplies that by the molar mass — 143.32, 187.77 and 234.77 respectively.

🧮Working the numbers yourself

For silver bromide: the square root of 5.0 × 10⁻¹³ is 7.07 × 10⁻⁷ mol/L. Multiply by the molar mass 187.77 g/mol and you get 1.33 × 10⁻⁴ g/L, which is 0.13 milligrams in a whole litre.

For silver chloride: the square root of 1.6 × 10⁻¹⁰ is 1.26 × 10⁻⁵ mol/L, which is 1.8 mg/L — about fourteen times as much mass and eighteen times as many silver ions.

Between silver chloride and silver iodide the ratio of dissolved silver ion is about one thousand to one. These are not three similar substances.

⁇Compilations disagree, and this course picks one and says so

OpenStax Appendix J gives the solubility product of silver bromide as 5.0 × 10⁻¹³ at 25 °C. The LibreTexts general-chemistry treatment of the same reaction gives 5.35 × 10⁻¹³, and quotes the resulting dissolved-ion concentration as 7.31 × 10⁻⁷ mol/L against the 7.07 × 10⁻⁷ that follows from the OpenStax figure.

Neither is wrong. Solubility products are measured quantities, they depend on ionic strength and on how the measurement was made, and the standard compilations differ in the last figure and sometimes more. This course uses the OpenStax table throughout, so that every comparison it makes is internally consistent, and the 7 per cent disagreement changes no conclusion on this page. If you take a number from a different table, label it with its source and do not mix the two in one calculation.

What those numbers predict about fixing

Silver bromide’s dissolved-silver concentration of 0.13 mg per litre is the reason fixing needs chemistry rather than patience. The LibreTexts treatment does the sum: removing the unexposed silver bromide from a single roll of film using pure water would take tens of thousands of litres, and a great deal of time.

And yet the same solid dissolves in a tray of thiosulfate in three minutes. Hold that contradiction; it is the last section of this page.

The three-way comparison also predicts the fixing difficulty of the three halides, and it is borne out by practice: chloride emulsions fix quickly, bromide emulsions more slowly, and an iodide-containing film slowest of all. That ordering is a thousand-to-one ratio in dissolved silver ion, and Part XI turns it into clearing times.

The common-ion effect

Ksp is a product, and a product can be pushed around. Add bromide ion to a saturated solution of silver bromide, from any source at all, and the product [Ag⁺][Br⁻] momentarily exceeds Ksp. The system does what Le Chatelier says: it precipitates silver bromide until the product is back down to Ksp, which means the silver ion concentration has fallen. That is the common-ion effect — a salt is less soluble in a solution that already contains one of its own ions.

The arithmetic is a single division. If bromide is present at some concentration you control, then

[Ag⁺] = K_sp ÷ [Br⁻]

Silver ion under bromide control

Take a real formula. Kodak’s D-71 negative motion picture developer, from the 1928 primer, carries 3.0 g of potassium bromide in 4.0 litres — 0.75 g per litre. Potassium bromide’s molar mass is 119.00, so that is 0.75 ÷ 119.00 = 6.3 × 10⁻³ mol/L of bromide ion. Then

[Ag⁺] = 5.0 × 10⁻¹³ ÷ 6.3 × 10⁻³ = 7.9 × 10⁻¹¹ mol/L

Worked: silver ion in a developer carrying 0.75 g/L of potassium bromide

Against 7.07 × 10⁻⁷ mol/L in pure water, that is a suppression of about 8,900-fold, from three quarters of a gram of a cheap salt.

Silver ion concentration against added potassium bromide

• Silver ion in equilibrium with solid silver bromide

Show the numbers behind this plot

A steeply falling curve. With 0.1 g of potassium bromide per litre the silver ion concentration is about 595 picomol per litre; at 0.25 g it is 238; at 0.5 g it is 119; at 0.75 g, the amount in Kodak's D-71 developer, it is 79; at 1 g it is 60; at 2 g it is 30; and at 3 g it is 20. The curve is a reciprocal: doubling the bromide halves the silver ion concentration. A note records that in pure water, with no bromide added, the figure would be 707,000 picomol per litre, about nine thousand times the value at 0.75 g per litre and far above the top of the plotted scale.

Series	Potassium bromide added, g per litre	Silver ion concentration, picomol per litre
Silver ion in equilibrium with solid silver bromide	0.10	595.00
Silver ion in equilibrium with solid silver bromide	0.15	397.00
Silver ion in equilibrium with solid silver bromide	0.20	298.00
Silver ion in equilibrium with solid silver bromide	0.25	238.00
Silver ion in equilibrium with solid silver bromide	0.35	170.00
Silver ion in equilibrium with solid silver bromide	0.50	119.00
Silver ion in equilibrium with solid silver bromide	0.75	79.30
Silver ion in equilibrium with solid silver bromide	1.00	59.50
Silver ion in equilibrium with solid silver bromide	1.50	39.70
Silver ion in equilibrium with solid silver bromide	2.00	29.80
Silver ion in equilibrium with solid silver bromide	2.50	23.80
Silver ion in equilibrium with solid silver bromide	3.00	19.80

The line is arithmetic, not measurement: it is the solubility product 5.0 × 10⁻¹³ at 25 °C divided by the bromide concentration, and it assumes solid silver bromide is present and nothing else competes for the silver. In pure water the value would be 707,000 on this axis, off the top of the scale by three orders of magnitude. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

The two places photography uses it

In making an emulsion. Kodak’s 1928 primer states the rule flatly: a material that is to be developed must contain no excess of soluble silver, and the emulsion must be made so that there is always an excess of bromide or chloride, “since any excess of soluble silver will produce a heavy fog over the whole of the surface as soon as the material is placed in the developer.” A deliberate excess of halide drives the free silver ion concentration down to nothing, and the fog with it.

The same primer supplies the mirror image. Solio, a printing-out paper, is made from a chloride emulsion with an excess of silver nitrate, precisely because that excess makes it darken rapidly in light. Mike Ware describes the microscopic consequence: a precipitate made with silver in excess carries adsorbed silver ions on its crystal surfaces, and one made with halide in excess carries adsorbed halide ions. Same two compounds, opposite excess, opposite material.

In restraining a developer. Potassium bromide in a developer is a restrainer, and the plot above is why. It holds the free silver ion concentration down, so that development of a grain that has no latent image — which needs free silver to get going — is suppressed far more than development of one that has. The cost is that too much of it slows the wanted development as well, because the same suppression applies everywhere. Part VIII turns that trade into a working practice; the mechanism is this equation.

📷What you see

Too little bromide in a print developer: grey, veiled whites, because unexposed grains have developed. Too much: long development times, and highlights that will not come up. The dial you are turning is the free silver ion concentration, and the reason it is a dial rather than a switch is that the relationship is a smooth reciprocal, as the curve shows.

Temperature, and the trap in the obvious rule

For most solids, solubility rises with temperature. OpenStax states that as a general trend and then immediately names an exception, cerium sulfate, whose solubility falls. The rule is a trend, not a law, and treating it as a law is how stock solutions get ruined.

The photographic case where the trend is strong and matters is sodium carbonate. HSDB gives its solubility as 6, 8.5, 17 and 28 per cent by weight at 0, 10, 20 and 30 °C. From a cold night to a warm afternoon the solubility changes almost fivefold. A saturated carbonate stock mixed in a warm kitchen and stored in an unheated garage will grow crystals, and the solution left above them is no longer the concentration on the label.

Borax behaves the same way, more dramatically. HSDB gives 1 g dissolving in 16 mL of cold water and 1 g in 0.6 mL of boiling water — a factor of about 27. It is also the salt whose two published room-temperature figures disagree most, at 3.17 g and 5.1 g per 100, so treat any borax solubility number as approximate unless you measured it.

Why gentle warming is not always the cure

The instinct on finding crystals in a bottle is to warm it until they redissolve. Sometimes that is right. Three reasons it may not be.

The salt may not oblige. If the solubility does not rise steeply with temperature, warming achieves little. Sodium chloride is the standard example of a nearly flat curve, and sodium sulfite’s published figures — 30.7 g per 100 g at 25 °C from one compilation and 22 g per 100 mL at 20 °C from another, with an unstable heptahydrate in the picture — do not describe a simple rising line. This course has not verified the shape of the sulfite curve and does not assert one.

The crystals may not be the salt you think. A hydrate can change identity with temperature. Kodak’s 1928 primer records that sodium carbonate decahydrate loses nine of its ten waters simply by drying in air, becoming the monohydrate; and the tetrahydrate of sodium metaborate melts at 53.5 °C. What redissolves may not be what crystallised.

Warming costs you elsewhere. Warming a developer accelerates its oxidation as surely as it accelerates its dissolution, and that damage does not reverse when the bottle cools. The storage page has the practice; the aerial oxidation page later in this part has the mechanism.

Precipitation as a synthesis

Everything above has treated precipitation as something that happens. In emulsion making it is something you do, with the intention of producing crystals of a particular size, habit and surface composition, because those decide the film’s speed and grain.

Kodak’s 1928 primer gives the outline. Gelatin is soaked in water until swollen and dissolved by gentle warming; the halide — potassium bromide, or sodium chloride for a chloride emulsion — is dissolved in it; a weighed quantity of silver nitrate is dissolved separately and then added slowly, in the darkroom, since what is being made is light-sensitive from the moment it forms. Without the gelatin the precipitate would settle to the bottom and there would be no emulsion; the gelatin holds each crystal where it forms, distributed through the liquid.

Four handles change the outcome, and the general chemistry says which way each pushes.

Handle	Fast, or high	Slow, or low
Rate of addition	Many nuclei form at once: a fog of small crystals	Few nuclei, each growing longer: fewer, larger crystals
Temperature	Faster growth and more redistribution of material between crystals	Less of both
Stirring	Even conditions everywhere; no local excess of either reagent	Local excesses, and a wider spread of crystal sizes
Which reagent is in excess	Halide in excess: adsorbed halide, low fog	Silver in excess: adsorbed silver, printing-out behaviour

Those four rows are general solution chemistry, not photographic measurement. This course has not yet checked them against the emulsion literature Part V will cite, so read them as the directions the chemistry of nucleation and growth implies, and expect Part V to qualify them: real emulsion making also uses double-jet addition, controlled silver ion concentration and a ripening step, none of which this page has described.

Kodak does add the fact that connects crystal history to film speed: the different degrees of sensitiveness are obtained by varying the temperature and the duration of heating the emulsion receives during manufacture, the most sensitive emulsions being heated hotter and longer. That heating step is ripening, and Part V is where it is properly explained. This page’s contribution is that the crystal you develop in Part IV is the crystal a precipitation reaction made, under conditions somebody chose.

Washing a precipitate, and what happens if you do not

Look again at the equation. Silver bromide is the product you want; potassium nitrate is the product you do not. It is soluble, so it stays in the liquid, dispersed through the same gelatin that is holding your crystals.

Abney’s 1885 manual divides emulsion processes into exactly two classes: those in which the soluble salts, present “owing to the double decomposition of the salts employed”, are extracted before use, and those in which they are not. Where they are extracted, the set emulsion is broken up small and washed in running water. Abney’s own timings are instructive: finely squeezed through canvas, the salts come out in about thirty-five minutes; cut in strips instead, twelve to twenty-four hours may be needed “in order to obtain a maximum sensitiveness.”

Two lessons, and both go forward. Soluble by-products left in a coating do not go away — they crystallise as the layer dries, they attract water afterwards, and they alter the ionic environment around every grain. And the rate of removal is a diffusion problem, set by how far a dissolved ion has to travel through gelatin: Abney’s factor of twenty between finely divided emulsion and strips is the same physics that decides how long a fibre-base print takes to wash. The diffusion page at the end of this part owns it, and Part XII turns it into wash times.

The paradox this page ends on

Silver bromide is one of the most insoluble substances a photographer handles. A litre of water holds 0.13 mg of it. Removing one roll of film’s worth by washing would take tens of thousands of litres.

And a tray of sodium thiosulfate solution clears a film in a few minutes.

Both statements are true, and the second does not contradict the first, because solubility is not a property of a compound. It is a property of a compound in a particular solution. HSDB’s entry for silver iodide — the least soluble of the three — lists it as soluble in solutions of potassium iodide, sodium chloride, potassium cyanide, ammonium hydroxide and sodium thiosulfate. Something in those solutions is taking the silver ion out of the picture, and by Le Chatelier’s principle, removing a product drives the dissolution forward.

What that something is, how strongly it binds, why a fixer that is too dilute fixes worse rather than more slowly, and why an exhausted one leaves behind compounds that will not wash out, is the complex formation page later in this part.

Recap

Dissolving is a contest between the energy holding a lattice together and the energy water gives back by hydrating the freed ions; when the second wins the beaker warms, when it loses by a little the beaker cools, and when it loses badly the compound is what we call insoluble. A saturated solution is not full: it is at a dynamic balance where dissolving and crystallising run at equal rates, and that is this course’s first equilibrium. For a sparingly soluble salt the equilibrium constant is the solubility product, the product of the dissolved ion concentrations, and it cannot be exceeded while solid is present — so raising one ion forces the other down, which is the common-ion effect and the reason a developer contains bromide. The silver halides span a thousandfold range in dissolved silver ion, which orders their fixing times. Precipitation, run deliberately with control of addition rate, temperature, stirring and which reagent is in excess, is how an emulsion is manufactured. And the soluble by-product has to be washed out, at a rate set by how far it has to diffuse.

Next: what makes a solution acidic or alkaline, why the scale for it is a logarithm, and why almost every photographic bath specifies one.

Check your understanding

Question 1. Using the solubility products in this page, how many times more silver ion is present above saturated silver chloride than above saturated silver bromide, and what does that predict?

• About 3 times; the two fix at almost the same speed
• About 18 times; a chloride emulsion fixes faster than a bromide one
• About 320 times; silver chloride cannot be fixed at all
• None; the two are equally soluble

Show the answer and why

Answer: About 18 times; a chloride emulsion fixes faster than a bromide one

The saturated silver ion concentration is the square root of Ksp. For AgCl that is the square root of 1.6 × 10⁻¹⁰ = 1.26 × 10⁻⁵ mol/L; for AgBr it is the square root of 5.0 × 10⁻¹³ = 7.07 × 10⁻⁷ mol/L. The ratio is 1.26 × 10⁻⁵ ÷ 7.07 × 10⁻⁷ = 17.9. More dissolved silver means the fixing equilibrium has less work to do, and chloride papers do clear faster than bromide films. Note the ratio of Ksp values, 320, is the square of that: it is the ion concentration you compare, not the constant.

Question 2. A developer is to carry 1.5 g of potassium bromide per litre (molar mass 119.00). What silver ion concentration does that set above solid silver bromide, taking Ksp as 5.0 × 10⁻¹³?

• 4.0 × 10⁻¹¹ mol/L
• 7.5 × 10⁻¹³ mol/L
• 3.3 × 10⁻⁹ mol/L
• 7.1 × 10⁻⁷ mol/L

Show the answer and why

Answer: 4.0 × 10⁻¹¹ mol/L

Bromide concentration is 1.5 ÷ 119.00 = 1.26 × 10⁻² mol/L. Then [Ag⁺] = Ksp ÷ [Br⁻] = 5.0 × 10⁻¹³ ÷ 1.26 × 10⁻² = 4.0 × 10⁻¹¹ mol/L. That is about 18,000 times lower than the 7.1 × 10⁻⁷ mol/L of a saturated solution in pure water, and twice the suppression that 0.75 g/L gives — the relationship is a simple reciprocal, so doubling the restrainer halves the free silver.

Question 3. A stock solution kept in an unheated garage has grown crystals overnight. Which two explanations should you consider, and what single test separates them?

• Evaporation or contamination; test by smelling it
• It was supersaturated and something triggered it, or the cold lowered the solubility below the concentration you mixed; warm it back to the temperature it was mixed at and see whether the crystals redissolve completely
• The salt has decomposed or the bottle has leaked; test by weighing the bottle
• It has oxidised or frozen; test by filtering

Show the answer and why

Answer: It was supersaturated and something triggered it, or the cold lowered the solubility below the concentration you mixed; warm it back to the temperature it was mixed at and see whether the crystals redissolve completely

Both are ordinary and they behave differently. If the solution was supersaturated at the storage temperature, it has now simply reached saturation and warming to the mixing temperature will redissolve everything and leave a solution that stays clear. If the concentration genuinely exceeds the cold solubility — likely with sodium carbonate, whose solubility HSDB puts at 6 per cent by weight at 0 °C against 28 per cent at 30 °C — warming redissolves the crystals and they come back on the next cold night. Either way the liquid above the crystals is not the concentration on the label, so the answer is to remix rather than to decant.

Question 4. Why does an emulsion intended for development contain a deliberate excess of bromide, while a printing-out paper such as Solio is made with an excess of silver nitrate?

• Excess bromide makes the emulsion cheaper; excess silver makes it more expensive but faster
• Excess bromide suppresses the free silver ion concentration by the common-ion effect, so unexposed grains do not develop and fog is avoided; excess silver leaves adsorbed silver ions on the crystal surfaces, which is what makes a paper darken directly in light
• The two have no chemical connection; it is a matter of manufacturing convenience
• Excess bromide is needed to harden the gelatin, and excess silver to soften it

Show the answer and why

Answer: Excess bromide suppresses the free silver ion concentration by the common-ion effect, so unexposed grains do not develop and fog is avoided; excess silver leaves adsorbed silver ions on the crystal surfaces, which is what makes a paper darken directly in light

Kodak stated the developing-out rule in 1928: any excess of soluble silver produces heavy fog as soon as the material meets a developer, so the emulsion is always made with excess halide. Solio is the deliberate opposite, a chloride emulsion made with excess silver nitrate, which causes rapid darkening in light so the print is toned and fixed rather than developed. Mike Ware describes the microscopic difference: the surface of the crystal carries adsorbed ions of whichever reagent was in excess.

Question 5. Which statement about Le Chatelier’s principle is correct?

• It predicts both the direction and the speed of the change
• It predicts the direction in which a disturbed equilibrium will shift, and says nothing at all about how long that will take
• It applies only to gases
• It means an equilibrium cannot be disturbed

Show the answer and why

Answer: It predicts the direction in which a disturbed equilibrium will shift, and says nothing at all about how long that will take

The principle is about position, not rate. The two are genuinely independent: an unexposed silver halide grain sits on the wrong side of an equilibrium that a developer could in principle move, and the whole medium depends on the fact that it does not get there in ten minutes. Every prediction you make from the principle should be followed by the separate question of whether it happens fast enough to matter.

Question 6. Silver bromide is described as insoluble in water, yet it dissolves in a fixing bath in minutes. What is the correct way to state that?

• The fixer heats the solution, and warm water dissolves more
• Solubility is a property of a compound in a particular solution, not of the compound alone: thiosulfate removes silver ion from the solution as it appears, so by Le Chatelier’s principle the dissolution equilibrium keeps moving forward
• The fixer chemically destroys the silver bromide crystal by oxidising it
• Silver bromide is not really insoluble; the tables are wrong

Show the answer and why

Answer: Solubility is a property of a compound in a particular solution, not of the compound alone: thiosulfate removes silver ion from the solution as it appears, so by Le Chatelier’s principle the dissolution equilibrium keeps moving forward

Nothing about the solubility product changes. What changes is that the dissolved silver ion is continually consumed, so the product [Ag⁺][Br⁻] never reaches Ksp and solid keeps dissolving. Removing a product drives a reaction forward, and that is the entire trick. HSDB lists silver iodide, the least soluble of the three halides, as soluble in solutions of iodide, chloride, cyanide, ammonia and thiosulfate for exactly this reason. The complex formation page explains what the silver ion is being turned into.

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