The Physics of Washing

Two darkroom workers wash a film. The first fills a 300 millilitre tank three times, inverting it each time, and spends 900 millilitres. The second leaves the tap running into the same tank, and to reach the same point must pass about three and a half times as much water through it. That factor is not a matter of opinion or of equipment quality, and it does not depend on any number that has to be measured to more than an order of magnitude. It falls straight out of the arithmetic of successive dilution, and this page is mostly about doing that arithmetic, because once you have done it every published wash sequence in the course becomes legible — including the ones that look arbitrary.

What is leaving, and what is not

Part XI established what is in the gelatin when the fixer comes off, and this page does not re-derive any of it. Three things, and only two of them can be washed out.

Free thiosulfate, the bath itself, soaked into every part of the material that holds solution. ILFORD names it as the reason washing exists: washing “removes residual chemicals from the photographic process; in particular thiosulphate from the fix process can cause long-term image degradation if not effectively removed”.

The soluble silver complex, the bis(thiosulfato) ion that Part III numbered as the second step. It is soluble, so it goes with the water, and Reilly names it explicitly among the things a wash is for: “the purpose of washing is to remove the sodium thiosulfate and silver thiosulfate complexes that remain in the print after fixation”.

The insoluble first complex, which washing does not remove at all. Kodak’s 1924 primer says so in a sentence and adds the detail that makes it dangerous — the compound “is invisible”. If the bath was exhausted or the material left it too early, that compound is in the paper and no wash schedule on this page will touch it. That is a fixing failure, and the Part XI criterion is what prevents it. The rest of this page assumes fixing was done properly, because washing cannot rescue it.

Fick’s law, as an accountant reads it

Part III owns diffusion and derived everything below; what follows is the reading of it that a wash schedule needs.

flux = D × area × (concentration gradient)

Fick's first law, in words

Flux is how much substance crosses a boundary per second. D is the diffusion coefficient, which is a property of the ion, the gelatin and the temperature and which you do not control. Area is the surface the material presents to the water, which is fixed once you have chosen the format. Gradient is the difference in concentration divided by the distance over which it falls — and it is the only one of the four you can do anything about.

That single observation is the whole of wash design. You cannot make thiosulfate diffuse faster. You can only keep the concentration outside the material as close to zero as possible, so the gradient across the boundary stays as steep as it can be. Kodak’s 1928 primer put the consequence bluntly: the rate of washing “has nothing to do with solubility”, and “the process will then stop unless the water in the vessel is changed”. A material sitting in water it has already equilibrated with is not washing. It is soaking.

Part III also supplies the rate: the amount remaining is halved in a fixed interval, about 15 seconds under a running tap and about 30 seconds in most trays and tanks. Hold both facts together, because they answer different questions. The half-time tells you how fast the inside empties into the water around it. The change of water tells you how far it can empty before it stops.

The arithmetic of a change of water

Here is the model, stated so you can check it and so its failures are visible.

A washed material holds a volume v of solution inside it — in the swollen gelatin, and, for fibre-base paper, in the paper as well. When fixing ends that solution is at the fixer’s own concentration. Drop the material into a bath of clean water of volume V, agitate until inside and outside are the same, and the thiosulfate that was in v is now spread through v + V. What is left inside is:

fraction remaining = v ÷ (v + V)

One complete change of water

Write ρ = V ÷ v for the size of the bath in units of the material’s own held volume. Then one change leaves 1 ÷ (1 + ρ), and because each change starts from wherever the last one finished, n changes leave

f = (1 + ρ)⁻ⁿ

After n complete changes

Now compare that against the same total water used in one long soak, a bath of volume nV:

f = 1 ÷ (1 + nρ)

The same water, all at once

The first is exponential in n. The second is barely better than linear. That is the entire argument, and it is the same argument that makes a chemist extract three times with a third of the solvent rather than once with all of it.

🧮The numbers, with one assumption stated openly

Take a 135-36 film on a spiral in a 300 mL tank and suppose it carries v = 10 mL of fixer into the wash, so ρ = 30. That number is an assumption: no measured carry-over volume for any photographic material appears in any source this course holds, and the paragraph after this one shows why it does not matter as much as it looks as though it should.

Water used	Three changes of 300 mL	One soak in the same total
300 mL	10 ÷ 310 = 3.2 %	10 ÷ 310 = 3.2 %
600 mL	3.2 % of 3.2 % = 0.10 %	10 ÷ 610 = 1.6 %
900 mL	0.0034 %	10 ÷ 910 = 1.1 %

Three changes leave about one three-hundredth of what the same 900 mL leaves in a single soak.

Against a running tap, the comparison is cleaner still. A tank of volume V fed continuously and mixed perfectly loses its contents exponentially with the volume W of water passed through it, so f = e^(−W ÷ V). Set that equal to n complete changes and the two ways of spending water can be priced against each other:

one change ≡ ln(1 + ρ) tank-volumes of running water

What one complete change of water is worth

At ρ = 30 that is ln 31 = 3.4. Every time you dump the tank and refill it you have done what three and a half tankfuls of running water would have done. Ten changes, which is exactly what Kodak prescribes for a small tank, are worth about thirty-four tankfuls.

And now the assumption pays for itself. The answer depends on v only through a logarithm. If the film really carries 30 mL rather than 10, ρ falls to 10 and the value of a change falls from 3.4 to ln 11 = 2.4. A factor of three in the quantity nobody has measured moves the conclusion by less than a factor of one and a half. The arithmetic is robust in exactly the place where the evidence is thin, which is the best reason to trust it.

Three ways of spending the same water

• Complete changes of water
• Continuous flow through the tank
• One soak in all the water at once

Show the numbers behind this plot

Three curves showing how much thiosulfate is left in a material against how much water has been spent, on a logarithmic vertical scale where each step of one unit downward is a factor of ten. Complete changes of water fall in a straight line: after one tank-volume the fraction is ten to the minus 1.5, after three tank-volumes ten to the minus 4.5, and after five tank-volumes ten to the minus 7.5. Continuous flow through the same tank falls far more slowly, reaching only ten to the minus 0.4 after one tank-volume and ten to the minus 2.2 after five. A single long soak in all the water at once is worst of all after the first tank-volume, flattening almost immediately and reaching only ten to the minus 2.2 after five tank-volumes because a bath five times larger is only about five times more dilute. The gap between complete changes and either alternative widens without limit as more water is spent, which is the whole argument for changes of water over a running tap.

Series	Total water used, in tank-volumes	log₁₀ of the fraction of thiosulfate still in the material
Complete changes of water	0.00	0.00
Complete changes of water	1.00	-1.49
Complete changes of water	2.00	-2.98
Complete changes of water	3.00	-4.47
Complete changes of water	4.00	-5.97
Complete changes of water	5.00	-7.46
Continuous flow through the tank	0.00	0.00
Continuous flow through the tank	1.00	-0.43
Continuous flow through the tank	2.00	-0.87
Continuous flow through the tank	3.00	-1.30
Continuous flow through the tank	4.00	-1.74
Continuous flow through the tank	5.00	-2.17
One soak in all the water at once	0.00	0.00
One soak in all the water at once	1.00	-1.49
One soak in all the water at once	2.00	-1.78
One soak in all the water at once	3.00	-1.96
One soak in all the water at once	4.00	-2.08
One soak in all the water at once	5.00	-2.18

Arithmetic from the model stated above, at ρ = 30, not a measurement of any material. The straight line is the point: a complete change multiplies what is left by a constant factor, so the logarithm falls by a constant step, while flow and soaking both suffer from washing the material in water that already contains what came out of it. 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.

!Where this model breaks, and what breaks it

Two assumptions carry all of the arithmetic above, and both fail in ways worth knowing.

It assumes each change reaches equilibrium. If the material is dropped into fresh water and left still, the layer of water touching it saturates and the change is worth a fraction of ln(1 + ρ). This is why every published sequence specifies agitation rather than time alone — ILFORD’s five, ten and twenty inversions, Kodak’s “fill the tank to overflowing with fresh water and then dump it all out”, Reilly’s flat statement that with imperfect apparatus “hand agitation of the prints is absolutely necessary”.

It assumes nothing holds on to the thiosulfate. For film that is nearly true. For paper it is not, and Reilly says so in the strongest terms available: “with prints the rate of washing slows down tremendously at the lower levels of thiosulfate concentration, and in practice it is impossible to remove every trace of thiosulfate simply by washing in water”. Kodak Limited’s handbook says the same thing about the same material — traces of hypo are “tenaciously held by the paper fibres”.

Notice what those two failures imply together. The model says ten changes should leave one part in 10¹⁵, which is absurd; the material is obviously not that clean. So the bath was never the bottleneck. Once you are past three or four changes, what is left is governed by how fast the inside of the material can empty and by how much of the thiosulfate is bound rather than dissolved. That is the diagnosis, and the wash aid is the treatment.

Where the thiosulfate hides

Three materials, three different journeys out, and the difference is not thickness in the sense of bulk. It is which layers hold solution and how far the solution has to travel back.

Three materials in section, with the layers that hold thiosulfate marked

1. 1Film — gelatin only; the base is impermeable, so the path out is the emulsion thickness
2. 2Resin-coated paper — polythene on both faces seals the paper base; ILFORD calls the base waterproofed
3. 3Fibre-base paper — the paper base absorbs solution and becomes the reservoir; ILFORD says fibre papers absorb processing solutions more readily and therefore need longer washing

Drawn to compare the layers, not to scale: the real thickness of a paper base against an emulsion is far greater than the drawing can show, and the course holds no measured base thickness from any source it has read.

Part III already did the distance argument and the course does not repeat it: because the time to diffuse a distance goes as the square of that distance, a path an order of magnitude longer is two orders of magnitude slower. What this page adds is that the fibre base is not merely a longer path but a reservoir, holding a quantity of thiosulfate that the emulsion of a film never had, and holding part of it bound to the cellulose rather than dissolved in it.

The published times follow exactly from that ranking. Resin-coated print: ILFORD, thirty seconds in vigorous running water; Kodak, four minutes. Film: ILFORD, five to ten minutes running, or ten one-minute changes; Kodak, five minutes with a clearing agent or twenty to thirty without. Fibre-base print: ILFORD, thirty to forty-five minutes; Kodak, one hour.

✕The one place where longer is worse

ILFORD prints a warning that reads oddly until you know the structure: do not over-wash resin-coated papers, because water entering the cut edge of the paper can eventually swell the base and damage the print. The polythene seals the faces; nothing seals the edge you cut. Reilly makes the matching point for hand-made papers, where prolonged washing can begin to dissolve the image layer itself. Wash long enough and no longer, and the number that says how long is on the maker’s sheet.

The published sequences, side by side

Four published wash sequences, drawn on one clock in minutes

Film, spiral tank

FixThree changes: invert 5, 10, 20

FB print, standard

FixFresh running water, 30 to 45 min

FB print, optimum permanence

FixFirst wash 5WASHAID 1+4, 10 minFinal wash 5

RC print

FixRunning water 30 s

015 min30 min45 min

All four rows are ILFORD’s own, from the reduced-wash-water sheet and the MULTIGRADE FB CLASSIC sheet, and the film row’s five minutes is this course’s estimate of how long three fills, three drains and thirty-five inversions actually take rather than a published time. Read the second and third rows against each other, because that pair is the argument for a wash aid: twenty minutes against thirty to forty-five, and perhaps a fifth of the water, for a sequence the maker labels optimum permanence rather than adequate.

Hardening, temperature, and two levers that are not what they seem

Hardening costs you wash time, and both manufacturers say so. ILFORD’s washing sheet notes that its instructions assume a non-hardening fixer and that “using hardening fixers can significantly increase the required wash time”; the FB CLASSIC sheet is blunter — a hardening fixer “is not recommended as it reduces washing efficiency”. The mechanism is Part III’s: a hardener cross-links the gelatin, the layer swells less, and a less swollen gel is a slower road. Everything an alum fixer buys in physical toughness it charges for in wash time, and Part XI costed the same trade from the fixing side.

Fixing longer than necessary costs you wash time too. ILFORD’s paper sheet puts two causes in one sentence: exceeding the capacity of the fixer and extending the fixing time “both make washing more difficult”. The first loads the material with complexes that are harder to shift; the second gives the bath longer to drive them into the base.

Warm water does not help. This is Part III’s most counter-intuitive result and it is Kodak’s, from 1928: warmth raises the diffusion coefficient and swells the gelatin in about the same proportion, “so that, as a matter of fact, washing goes on at about the same rate at all ordinary temperatures”. What temperature does govern is the material, not the rate: ILFORD asks for wash water within 5 °C of the developer, and ILFORD’s FB sheet sets a floor of 5 °C rather than a target. The one real temperature effect on rate in the corpus is second-hand and belongs to the clearing agent rather than the water: Kodak says Hypo Clearing Agent makes “washing at lower wash-water temperatures practical”, which is a claim about the aid rescuing cold water, not about warm water being faster.

What a hypo clearing agent actually does

Not what the name suggests. A wash aid does not dissolve thiosulfate — thiosulfate is already about as soluble as a salt gets — and it does not destroy it. ILFORD’s own description names the mechanism in three words: WASHAID is “formulated to aid the efficient removal of the thiosulphate by-products of fixation by ion exchange”.

Reilly gives the same mechanism in full: these treatments “displace the absorbed thiosulfate ions and replace them with less harmful and more soluble ions of various salts”. The paper fibre is holding some fraction of the thiosulfate; flood it with a large excess of a small, cheap anion and the thiosulfate is released to diffuse out, while what stays behind is the substitute. Which is why the substitute matters: it must be something that will not attack silver if a trace of it is never removed. Sulfite is the answer both manufacturers and the conservation literature converge on, and the one per cent sulfite washing aid is that formula printed as a formula, with a source, an exhaustion figure and the function of its single ingredient.

Two consequences follow and both are practical.

A wash aid is not a substitute for the first wash. Reilly puts a two-to-four-minute running-water wash before the sulfite bath specifically to avoid “overloading the mechanism of ion-exchange”, and ILFORD’s sequence does the same thing with five minutes. An exchange bath has a finite number of ions to trade; spending them on the bulk of the thiosulfate, which plain water would have removed for nothing, is the one way to make the sequence worse than the long wash it replaced.

A wash aid has a capacity, and it is small. ILFORD publishes forty sheets of 20.3 × 25.4 cm, or forty 135-36 films, per litre. Kodak’s paper-chemistry sheet quantifies the same argument from the other side, giving its clearing agent a capacity of 80 without a pre-rinse and 200 with one. The mirror the course holds has lost the header of that capacity column, so the course does not state the unit; the ratio is what the sentence is for, and it says that a rinse costing nothing multiplies the working life of the bath by two and a half.

⚗Hypo eliminators, and why the course does not use one

A hypo eliminator is a different idea again: not exchange but oxidation, converting residual thiosulfate into something that is not thiosulfate. Kodak Limited published one, HE-1, and the course prints it in full as evidence.

The objection is a century old and it is photographic rather than toxicological. Wall in 1924: it is “an open question whether the use of chemicals to destroy the last traces of hyposulphite of soda and the hyposulphites of silver is justifiable if permanency of results be the aim”, their action being “probably in most cases to convert these salts into tetrathionates” — that is, into another sulfur compound sitting in the paper rather than into nothing at all. His conclusion is the one the course adopts: a negative or print “can be practically freed from hypo in half an hour by proper washing”, so an eliminator is “only allowable in cases of great pressure of time or shortage of fresh water”.

Read Kodak’s own directions and the argument closes itself. HE-1 is used after a thirty-minute wash, not instead of one. A bath that reaches the last traces of thiosulfate is a bath that reaches the image, and Kodak’s own list of side effects — a change of image colour, a yellowing of the whites — says as much.

Water, and the sequence that uses least of it

ILFORD wrote its reduced-wash sheet for people metering water or living under drought restrictions, and everything in it is the arithmetic above turned into instructions. The film method is three changes with escalating agitation — fill and invert five times, drain and refill and invert ten times, drain and refill and invert twenty times. Kodak’s equivalent for a small tank is ten fill-and-dump cycles in place of five minutes of running water.

The escalation is worth a sentence, because ILFORD does not explain it and the model does. Each change removes a smaller absolute quantity than the one before it, so each later change needs longer in contact to get its full value; more inversions is how you buy contact time without buying water. That is a reading of the sequence rather than ILFORD’s stated reason, and the course says so.

For fibre prints the water saving is larger still, because the alternative is thirty to forty-five minutes of running water. ILFORD’s own note is that an archival print washer with discrete slots raises efficiency further and allows a lower flow rate — a washer’s job being to guarantee that every surface meets water that has not just come off another print. Kodak’s 1928 primer solved the same problem with two trays at different heights and a rule that prints always enter the lower one, which is a cascade and costs nothing.

🧪Measure your own change ratio in ten minutes

The one number this page had to assume is measurable on a kitchen scale. Take a processed, dry sheet of the paper you use. Weigh it. Soak it in water for the length of your usual fixing time, lift it, let it drain for the count you normally give it, and weigh it again. The difference in grams is very nearly the volume in millilitres it carries into the wash.

Do it for a resin-coated sheet and a fibre-base sheet of the same size and you have measured, for your own materials, the quantity the whole of this page’s arithmetic turns on — and you will be able to say why one of them needs sixty times the wash of the other.

What the course could not establish

Sea-water or salt-water washing. The claim that a brief wash in sea water accelerates the removal of thiosulfate recurs in darkroom folklore. No source in this course’s corpus, at any tier, states it, tests it or explains it, so the course neither teaches it nor contradicts it. What can be said is that the mechanism usually offered for it is the ion-exchange mechanism above with chloride as the exchanging ion, and that if that is the mechanism then a salt bath is a wash aid whose substitute ion is chloride — which would have to be washed out in turn, from a material whose permanence depends on not retaining halide.

A measured carry-over volume, for any film or paper, which is the assumption the arithmetic rests on.

A residual-thiosulfate figure for any of the published sequences. Manufacturers publish times, not outcomes. Which is precisely why the next two pages exist: one to say what the residue does over thirty years, and one to measure it.

Recap

• Washing is diffusion outward, so the only lever you hold is the gradient: keep the water in contact with the material as close to clean as possible.
• A complete change of water is worth ln(1 + ρ) tank-volumes of running water, where ρ is the bath’s volume divided by the volume the material carries. At a plausible ρ of 30 that is a factor of about three and a half per change, and it compounds.
• The model’s own absurdity at ten changes is informative: past three or four changes the bath is no longer the bottleneck, and what is left is bound rather than dissolved.
• Where the thiosulfate hides sets the wash time: gelatin only for film and resin-coated paper, gelatin plus the whole paper base for fibre. Thirty seconds against an hour, from one structural difference.
• Hardening and over-fixing both lengthen the wash; warm water does not shorten it.
• A wash aid works by ion exchange, needs a first wash in front of it to protect its capacity, and turns a forty-five-minute fibre wash into ILFORD’s twenty-minute optimum permanence sequence.
• A hypo eliminator oxidises rather than exchanges, and the literature has objected to it since 1924 on the grounds that it puts another sulfur compound where the thiosulfate was.

Check your understanding

Question 1. A tank holds 300 mL and the film in it carries 10 mL of fixer into the wash. You have 1.2 litres of water and intend to use all of it. Using the model on this page, what fraction of the original thiosulfate is left after four complete changes of 300 mL, and how does that compare with pouring all 1.2 L in at once?

• About 1 part in 10⁶ for the changes; about 1 part in 120 for the single soak
• About 1 part in 900,000 for the changes; about 1 part in 120 for the single soak
• About 1 part in 1,200 for the changes; about 1 part in 120 for the single soak
• The two are the same, because the same total volume of water was used

Show the answer and why

Answer: About 1 part in 900,000 for the changes; about 1 part in 120 for the single soak

One change leaves 10 ÷ 310 = 1/31 of what went in, so four changes leave (1/31)⁴ = 1/923,521, about one part in nine hundred thousand. The single soak leaves 10 ÷ (10 + 1200) = 1/121. The changes are better by a factor of about 7,600 on identical water, because each change restarts the dilution from a clean bath while the single soak dilutes once and then stops. The answer of one part in a million is the right order of magnitude reached by guessing rather than by arithmetic; raise 31 to the fourth power properly and the figure is nine per cent smaller than that.

Question 2. Why does a fibre-base print need roughly sixty times the wash of a resin-coated one?

• The fibre-base emulsion is thicker and holds more thiosulfate
• The polythene layers of an RC print repel thiosulfate chemically
• The RC base is sealed on both faces, so only the gelatin ever holds solution, while the fibre base absorbs solution into the paper and holds part of it bound to the cellulose
• Fibre-base paper is fixed for longer, so it starts with more thiosulfate

Show the answer and why

Answer: The RC base is sealed on both faces, so only the gelatin ever holds solution, while the fibre base absorbs solution into the paper and holds part of it bound to the cellulose

The difference is structural, not chemical. ILFORD describes RC paper as waterproofed by its polythene coating, so the fixer only ever occupies the emulsion and the journey out is one thin layer. A fibre base has no such barrier: solution soaks into the matted cellulose, which becomes a reservoir, and Reilly records that part of it is then held by the fibres rather than dissolved in them, which is why the last of it will not wash out with water at all. The emulsions are comparable; the bases are not.

Question 3. You put a fibre print straight from the fixer into a ten-minute bath of wash aid, skipping the first wash. What has gone wrong?

• Nothing; the aid is stronger than plain water and does the first wash as well
• The aid works by ion exchange and has a finite capacity, so it is spent on the bulk of the thiosulfate that plain water would have removed for nothing
• The aid will react with the acid in the fixer and release sulfur dioxide
• The print will be over-washed and the emulsion will begin to dissolve

Show the answer and why

Answer: The aid works by ion exchange and has a finite capacity, so it is spent on the bulk of the thiosulfate that plain water would have removed for nothing

A wash aid trades ions: it offers the paper something it will hold in preference to thiosulfate. That trade is finite, which is why ILFORD publishes a capacity of forty 8 × 10 sheets per litre and why Reilly puts a two-to-four-minute running wash in front of the sulfite bath specifically to avoid overloading the exchange. Kodak quantifies the same effect from the other direction: eighty prints per gallon without a pre-rinse, two hundred with one. Loading the bath with fixer also carries silver into it, which is a second reason the first wash is not optional.

Question 4. Kodak states that washing proceeds at about the same rate at all ordinary temperatures. Why, given that diffusion coefficients rise with temperature?

• Because thiosulfate is already fully soluble, so temperature cannot help it dissolve faster
• Because warmer water holds less dissolved gas and so wets the gelatin less well
• Because warmth raises the diffusion coefficient but also swells the gelatin, lengthening the path by about the same proportion
• Because the half-time is set by the agitation, which does not change with temperature

Show the answer and why

Answer: Because warmth raises the diffusion coefficient but also swells the gelatin, lengthening the path by about the same proportion

Kodak's 1928 reasoning is Fick's law applied twice. Raising the temperature does raise D, which would shorten the wash; but it also swells the gelatin, and a thicker swollen layer is a longer path, and time to diffuse goes as the square of distance. The two effects very nearly cancel over the range a darkroom works in. Note what the answer is not: solubility plays no part, because as the same primer says, the rate of washing has nothing to do with solubility. The real temperature rule in the sequences is about matching baths to avoid stressing the gelatin, not about speed.

Question 5. The model on this page predicts that ten complete changes leave about one part in 10¹⁵. What should you conclude from a prediction that is obviously wrong?

• That the model is useless and published times should simply be followed
• That the bath stopped being the limiting factor after a few changes, and what remains is governed by transport inside the material and by thiosulfate that is bound rather than dissolved
• That ten changes are unnecessary and three are always enough for any material
• That the held volume v must be far larger than 10 mL

Show the answer and why

Answer: That the bath stopped being the limiting factor after a few changes, and what remains is governed by transport inside the material and by thiosulfate that is bound rather than dissolved

A model that fails informatively is doing its job. The prediction assumes the only obstacle is the concentration of the bath, so when the prediction runs away from reality it has identified which assumption broke: the bath is no longer the constraint. What is left is limited by how fast the interior can empty — the half-time Part III owns — and, in paper, by the fraction Reilly says water cannot remove at all. That diagnosis is what selects the treatment: more changes will not help, but an ion-exchange bath will, because it attacks the bound fraction rather than the dissolved one.

Question 6. Which pair of processing decisions, both made before the wash begins, will lengthen it?

• Using a rapid fixer, and fixing at a higher temperature
• Using a hardening fixer, and fixing for longer than the recommended time
• Using a two-bath fixing sequence, and using a stop bath
• Using a non-hardening fixer, and fixing for the shortest time that clears the film

Show the answer and why

Answer: Using a hardening fixer, and fixing for longer than the recommended time

Both are ILFORD's own statements. A hardening fixer "reduces washing efficiency", because the cross-linked gelatin swells less and a less swollen gel is a slower road out. And exceeding the capacity of the fixer or extending the fixing time "both make washing more difficult", because the first loads the material with complexes that are harder to shift and the second gives them longer to reach the base. The other options mostly shorten the wash or leave it unchanged: two-bath fixing and a stop bath both protect the last bath the material meets, which is exactly what keeps the residue low.

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