Concentration and Dilution: The Arithmetic Every Formula Assumes

Every formula in the rest of this course assumes you can do the arithmetic on this page, and every formula in the last two hundred years has assumed the same thing about its own readers — which is why the historical literature is littered with arguments about what a percentage means. This is the page the other pages link to instead of deriving it again.

By the end you should be able to read any concentration statement and say exactly what it asks you to weigh and into what; convert between per cent, grams per litre and moles per litre; correct for a hydrate; make a working solution from a stock in either direction; and recognise, before you pour, the two mistakes that account for most bad batches.

What a concentration statement has to say

Three questions, and a statement that leaves any of them open is not an instruction.

1. How much of what? A mass, or a volume, of a named substance — named precisely enough to settle its hydrate and its cation. “Sodium thiosulfate” does not do this; “sodium thiosulfate pentahydrate, CAS 10102-17-7” does.
2. In how much of what? A mass of solution, a volume of solution, or a volume of solvent. These are three different denominators and they give three different answers.
3. By mass or by volume, on both sides? This is what the letters after the per cent sign are for, and it is why this course never writes a bare percentage.

A general chemistry text gives the three definitions cleanly, and photography uses all three.

Notation	Definition	Reads as
% w/w (mass percentage)	mass of solute ÷ mass of solution × 100	grams per 100 g of solution
% v/v (volume percentage)	volume of solute ÷ volume of solution × 100	millilitres per 100 ml of solution
% w/v (mass-volume percentage)	mass of solute ÷ volume of solution × 100	grams per 100 ml of solution
g/L	mass of solute per litre of solution	grams per 1000 ml of solution

The text’s own example of the third is physiological saline at 0.9 % (m/v), which is 0.9 g of sodium chloride per 100 ml of solution. Note the mixed units: per cent w/v is not a ratio of like to like, and the “per cent” in it is a convention rather than a fraction. That is exactly why it needs its letters.

🧮The two conversions you will use constantly

% w/v to g/L: multiply by 10. 100 ml is a tenth of a litre, so grams per 100 ml is a tenth of grams per litre.

• 10 % w/v = 100 g/L
• 1 % w/v = 10 g/L
• 0.1 % w/v = 1 g/L

That last line is the one that catches people: 1 g/L and 0.1 % w/v are the same concentration. The trap is a factor of ten in either direction, and it is worth ten times your developer.

% w/w to % w/v: you need the density. They coincide only when the solution’s density is 1.00 g/ml, which is true for very dilute aqueous solutions and for nothing else. A 10 % w/w solution of density 1.08 g/ml is 10.8 % w/v.

Which unit photographic formulas actually use

Modern manufacturer sheets mostly avoid percentages altogether and give the whole formula as masses made up to a stated volume: so many grams of this, so many of that, water to 1 litre. That is effectively grams per litre, and it is unambiguous.

The historical literature is another matter, and it knew it. Wall’s Photographic Facts and Formulas of 1924 opens its section on percentage solutions by admitting that “some dispute has prevailed as to the exact meaning of ‘x per cent solution’”, settles it for photographic practice as x parts in 100 parts of the total bulk of solution — the make-up-to-volume convention — and then explains why the dispute existed: chemicals were sold by the avoirdupois ounce of 437½ grains and liquids measured by the fluid ounce of 480 minims.

His two tables show what that costs. For a 1 per cent solution he gives 4.375 grains to be dissolved to make 100 parts, and 4.8 grains to be dissolved to make one fluid ounce. Both are labelled “1 per cent”. They differ by 480 ÷ 437.5, which is 9.7 per cent — a tenth of your developing agent, sitting inside a single word.

Making up to volume, and adding to a volume

These are different operations and they give different concentrations. It is the most common mistake on this page and the easiest to avoid.

The same weighed mass, two procedures, two concentrations

1. 1Made up to volume — correct — dissolve in part of the water, then top up to the mark; final volume exactly 100 ml
2. 2Added to a volume — wrong — solid tipped into a measured 100 ml; final volume exceeds 100 ml, so the solution is weaker than stated
3. 3The typographic tell — "water to 1 litre" or "water to 20 ounces" means make up to; "in 1 litre of water" would mean add to

The discrepancy grows with concentration: unnoticeable in a 0.1 per cent solution, real in a 10 per cent stock. The habit costs nothing, so it is used at every strength.

Manufacturers write the instruction into their mixing directions. ILFORD’s powder developer sheet has you dissolve the powders in about three-quarters of the total solution volume of warm water, then add cold water to make up to the final volume. Its stop-bath sheet has you add the concentrate to the mixing vessel, measure the dilution water, use some of that water to rinse the measuring cylinder out into the mixing vessel, and then add the remainder to make up to the final working volume.

🕰The nineteenth century wrote the same instruction with a colon of its own

The British Journal Photographic Almanac explains its own house style: British formulae are given in grains or ounces per 20 ounces of solution and metric formulae in grammes per 1000 cubic centimetres, and “our formulae are mostly drawn up on the basis that the total bulk after the solution of the solids is that stated in the formula”. The tell, it says, is the way the water is written: “water to x ounces”. Kodak was still using exactly that typography in 1928 — “Water to make . . . 32 ounces / 1.0 liter”. When you meet a period formula, look at the water line first. It tells you which of the two operations on the diagram above the author had in mind.

Moles and molarity

A mole is a counted quantity of particles, and the molar mass is the mass of one mole in grams, obtained by adding up the atomic masses in the formula. Molarity is moles of solute per litre of solution, written mol/L or M.

c (mol/L) = concentration in g/L ÷ molar mass in g/mol

Grams per litre to moles per litre

🧮A 10 per cent w/v sulfite stock, in three units

Sodium sulfite, Na₂SO₃, has a relative molecular mass of 126.05 on PubChem, CAS 7757-83-7.

• 10 % w/v = 10 g per 100 ml
• = 100 g/L
• = 100 ÷ 126.05 = 0.793 mol/L

Three ways of writing one bottle. The first is what the label says, the second is what you weighed per litre, and the third is what the chemistry of Part III will want, because reactions count particles rather than grams.

Molarity earns its place wherever the number of particles is what matters rather than their mass: acid–base equilibria and pH, solubility products, and the stoichiometry of fixing, where two thiosulfate ions complex one silver ion. Those are Parts III, IV and XI. For weighing out a developer, grams per litre is the practical unit and molarity is an unnecessary detour.

Hydrates

Many photographic salts crystallise with water built into the lattice, and that water is part of the mass you weigh and none of the chemistry you want.

Substance	Formula	Relative molecular mass	CAS
Sodium thiosulfate, anhydrous	Na₂S₂O₃	158.11	7772-98-7
Sodium thiosulfate pentahydrate	Na₂S₂O₃·5H₂O	248.19	10102-17-7
Sodium carbonate, anhydrous	Na₂CO₃	105.99	497-19-8
Sodium carbonate decahydrate	Na₂CO₃·10H₂O	286.14	6132-02-1

The conversion is one ratio, in whichever direction you need it.

mass of form B = mass of form A × (M_B / M_A)

Substituting one form of a salt for another

M is the relative molecular mass of each form. The reasoning is that the number of thiosulfate ions or carbonate ions has to be the same, so the masses stand in the ratio of the molar masses.

🧮Two substitutions, and the volume nobody counts

Thiosulfate. A formula asks for 100 g of the anhydrous salt and you have the pentahydrate: 100 × (248.19 ÷ 158.11) = 157 g. Going the other way, 100 g of the pentahydrate is equivalent to 100 × (158.11 ÷ 248.19) = 63.7 g of the anhydrous salt.

Carbonate. A formula asks for 100 g of the decahydrate crystals and you have the anhydrous powder: 100 × (105.99 ÷ 286.14) = 37.0 g. Weigh 100 g of the anhydrous salt instead and you have put in 2.70 times the sodium carbonate the formula wanted.

The volume nobody counts. In the thiosulfate case, the extra 57 g is water — about 57 ml of it — and it goes into your solution. If you dissolve the pentahydrate and then add a litre of water you have made roughly 1,057 ml of solution. Making up to the final volume absorbs this automatically, which is one more reason the make-up-to-volume habit is not fussiness.

✕An unqualified name meant the opposite thing in 1906

The British Journal Photographic Almanac states its convention explicitly: “sodium carbonate” and “sodium sulphite” are used for the crystallised forms, and where the dry or anhydrous forms are meant, a qualifying term is added. A modern reader’s default is the opposite — an unqualified name now means the anhydrous salt.

So a period British formula calling for 100 g of “sodium carbonate”, worked with a modern jar, delivers 2.70 times the intended alkali. That is not a subtle error in a developer: pH governs developing-agent activity, and Part VIII will show you what it does to a negative. Whenever you use a formula older than about 1950, find the source’s own statement of its conventions before you weigh anything, and if it has none, say so in your notebook and treat your first batch as a test.

Stock and working solutions

A stock solution is a concentrated solution you keep; a working solution is what you dilute from it and actually use. Stocks exist for four reasons.

1. Shelf life. A concentrated solution keeps far longer than a dilute one, because oxidation depends on how much oxygen can reach how much reducing agent. ILFORD makes the point in its own directions: prepare 1+1 and 1+3 solutions from the stock directly before they are needed, do not reuse diluted developers, and do not keep them more than 24 hours.
2. Precision. Weighing 0.2 g of a developing agent is at the edge of a domestic balance; weighing 2 g into ten times the volume and measuring out a tenth moves the precision from your balance to your graduate. The measurement page works through why that helps.
3. Convenience and consistency. One careful weighing serves ten sessions, and all ten are the same.
4. Because some ingredients must be kept apart. A two-part developer is supplied as part A and part B because the two would react, or would not keep, if combined. ILFORD’s process-control guide gives a worked case: mixing the A and B concentrates of one of its developers together before adding water turns the mixture milky, because a chemical in part A comes out of solution in the conditions of the part B concentrate. Add water to part A first and the problem does not arise. The order in a multi-part instruction is part of the instruction.

Dilution ratios, and the ambiguity in them

This course writes 1+9 and means one volume of stock plus nine volumes of water, giving ten volumes of working solution. It says so the first time on every page that uses it, and it never writes 1:9 or 1:10.

The reason is not pedantry. Look at what two manufacturers actually publish.

• ILFORD writes 1+4, 1+9, 1+19, 1+29 and gives the volumes. Its ILFOTEC LC29 table says that a 600 ml tank at 1+9 takes 60 ml of concentrate and 540 ml of water. Sixty and five hundred and forty make six hundred. The notation is defined by the table.
• Kodak writes the same idea with a colon. Its HC-110 sheet labels a working solution made from 75 ml of stock and 225 ml of water as “1:3”, and its D-76 sheet speaks of the developer “diluted 1:1”, meaning equal volumes of stock and water.

So in this corpus, both notations mean parts of stock to parts of water. The problem is that a colon is also the ordinary way of writing a ratio to a total: “a 1 in 10 dilution” almost everywhere means one part made up to ten. This course could not find a photographic manufacturer using the colon that way, so it does not accuse anyone of doing so. It simply refuses a notation that can be read two ways when a notation that cannot exists.

One manufacturer is not even consistent with itself. ILFORD’s ancillary chemicals sheet gives its wetting agent as “diluted 1+200”; its own washing instructions, for the same product in the same role, write “(1:200)”. Nothing turns on it — both mean the same thing and the solution is far too dilute for the difference between 1 in 200 and 1 in 201 to matter — but if a manufacturer’s technical department can drift between two notations in two documents, a reader copying a formula from a forum at two in the morning certainly can.

🧮What the misreading costs

Take a stock at 100 g/L.

• 1+9 read correctly — 1 part in 10 total — gives 10.0 g/L.
• Read as “1 part in 9 total” gives 100 ÷ 9 = 11.1 g/L: 11 per cent too strong.
• A sheet meaning 1+10 (1 in 11), read as “1:10” meaning 1 in 10, gives 10.0 g/L where 9.09 g/L was intended: 10 per cent too strong.

Ten per cent of developer strength is not a rounding error. ILFORD’s own compensation chart makes 1 °C of temperature worth about 9 per cent of development time, and people take great trouble over one degree.

Three readings of the same instruction, and the three solutions they make

1. 11 part in 9 total — 11.1 % concentrate — what "1:9" means if the colon is read as a ratio to the whole
2. 21 part plus 9 parts water, 10 in all — 10.0 % — what this course writes 1+9, and what "1:10" would mean read as a ratio to the whole
3. 31 part plus 10 parts water, 11 in all — 9.09 % — what a sheet writing 1+10 intends

The plus sign says what to add. The colon says a ratio and leaves you to guess what it is a ratio to — and the guess is worth about ten per cent of your developer either way.

C₁V₁ = C₂V₂, derived rather than asserted

The equation is not a rule to memorise; it is a bookkeeping statement about a quantity that does not change. Diluting a solution adds solvent. It does not add or remove solute.

Write down the amount of solute before and after. Amount equals concentration times volume, in whatever consistent units you like — grams and litres, grams per 100 ml and millilitres, moles and litres.

n₁ = C₁V₁ and n₂ = C₂V₂

Amount of solute, before and after

Since nothing was added or taken away, n₁ = n₂, and therefore:

C₁V₁ = C₂V₂

The dilution equation

C₁ and V₁ are the concentration and volume of the stock you take; C₂ and V₂ are the concentration and volume of the working solution you end up with. A general chemistry text derives it in exactly this way and notes that although it is usually written in molarity and litres, any units may be used as long as they cancel.

Rearranged for the two things you actually want:

V₁ = C₂V₂ / C₁

Forwards: how much stock do I take?

C₂ = C₁V₁ / V₂

Backwards: what have I ended up with?

What the equation is a statement about: the solute that does not change

1. 1Before: C₁ in V₁ — the same solute, crowded into a small volume
2. 2The operation: add solvent — nothing is added to or taken from the solute
3. 3After: C₂ in V₂ — the same solute, spread through a larger volume

C₁V₁ = C₂V₂ is not a rule about dilution; it is the observation that dilution does not change the amount of solute, written down twice.

🧮Worked forwards: 500 ml of 1.5 % w/v sulfite from a 10 % w/v stock

C₁ = 10 % w/v, C₂ = 1.5 % w/v, V₂ = 500 ml.

V₁ = (1.5 × 500) ÷ 10 = 75 ml of stock.

Then make up to 500 ml — put the 75 ml in the flask and add water to the mark. Do not add 425 ml of water to it: that is the operation on the right of the diagram above, and the make-up route absorbs any volume change for nothing.

Check by a second route, always. The 75 ml of stock contains 75 × 0.10 = 7.5 g of sulfite. In 500 ml, 7.5 g is 7.5 ÷ 5 = 1.5 g per 100 ml, which is 1.5 % w/v. The check took ten seconds and it uses different arithmetic from the first calculation, which is what makes it a check rather than a repetition.

🧮Worked backwards: what did I actually make?

You meant to take 75 ml but the graduate read 78 ml, and you made up to 500 ml.

C₂ = (10 × 78) ÷ 500 = 1.56 % w/v, not 1.50.

That is 4 per cent strong. Whether it matters depends on the solution — 4 per cent on a sulfite preservative is nothing, 4 per cent on a restrainer is visible — but the number belongs in the notebook either way, because a result you cannot repeat is usually a result whose inputs you never wrote down.

The limits of the equation

C₁V₁ = C₂V₂ is exact as bookkeeping. Two things can still go wrong.

Volumes do not necessarily add. When two liquids mix, the total volume is not always the sum of the parts, because the molecules pack differently together than apart. For dilute aqueous solutions the discrepancy is small; for concentrated ones it is not. Notice, though, that the equation is not what fails — the procedure is. If you calculate V₁ = 75 ml and then measure out 425 ml of water and combine them, you are assuming additivity. If you calculate V₁ = 75 ml and then make up to 500 ml, you are not assuming anything, and the question never arises. This is the third time the same habit has solved a different problem, which is a good sign that it is the right habit.

The concentration of the stock has to be true. The equation propagates whatever error is in C₁. If your stock is 4 per cent strong, every working solution made from it is 4 per cent strong, and no amount of careful dilution will find that out. This is why the stock bottle carries the date it was made and, if you checked it, its measured specific gravity.

✦Both manufacturers do it both ways, and the difference is instructive

ILFORD’s LC29 table gives additive volumes: 60 ml plus 540 ml for a 600 ml tank. Its powder developer sheet gives make-up-to-volume: dissolve in three-quarters of the volume, then top up. Both are correct for what they describe. At 1+9 with a liquid concentrate the volume change is far below anything a tank cares about, and additive volumes are quicker at the sink. When solids go into solution, the volume change is real and the same sheet switches to make-up-to-volume. Read which one your instruction is giving you, and follow that one.

Serial dilution

Sometimes the single dilution needs a volume you cannot measure. ILFORD says where the floor is, for its own product: do not use amounts of concentrate less than 10 ml when mixing working-strength solutions, because such small quantities are difficult to measure accurately with a measuring cylinder — and if very small quantities are necessary, use a graduated pipette.

🧮When two steps beat one

You want 100 ml of a 0.1 % w/v solution and you have a 10 % w/v stock. A single step needs V₁ = (0.1 × 100) ÷ 10 = 1 ml.

One step. Measuring 1 ml in a 100 ml graduate is hopeless: an uncertainty of about 1 ml on 1 ml is 100 per cent. Even in a 10 ml graduate, ±0.2 ml on 1 ml is 20 per cent.

Two steps. Take 10 ml of the 10 % stock, make up to 100 ml: that is 1 % w/v, and the two volumes carry perhaps 2 per cent and 1 per cent. Then take 10 ml of the 1 %, make up to 100 ml: 0.1 % w/v, another 2 and 1 per cent. Relative uncertainties add through the chain, so the worst case is about 6 per cent against the single step’s 20 or 100.

And the warning. Errors multiply through a serial dilution: a step that is 10 per cent strong followed by a step that is 10 per cent strong gives a solution 21 per cent strong, not 20. Every extra step is another chance to make a mistake that nothing downstream can detect. Use the fewest steps that keep every measured volume inside a vessel that can measure it.

Do the maths: five photographic examples

1. The sulfite stock, and a working solution from it

Weigh 100.0 g of sodium sulfite, dissolve in about 700 ml of water, make up to 1 litre. That is 10 % w/v, 100 g/L, 0.793 mol/L. The 500 ml of 1.5 % w/v drawn from it needs 75 ml of stock, made up to 500 ml, as worked above. Sulfite is the preservative of Part VIII, and this is the stock that part will ask you to make.

The commissioning lab runs exactly this arithmetic one page later, with sodium chloride in place of the sulfite: 50.0 g made up to 500 ml for the stock, and 250 ml of a 2 % w/v working solution from it. It changes the substance and nothing else, so that the session’s whole attention goes on the measuring.

2. One liquid concentrate at three dilutions

ILFOTEC LC29 is published at 1+9, 1+19 and 1+29, with the choice offered for economy and for one-shot against reuse. Its table for a 600 ml tank:

Dilution	Concentrate	Water	Total
1+9	60 ml	540 ml	600 ml
1+19	30 ml	570 ml	600 ml
1+29	20 ml	580 ml	600 ml

Each row is V₂ ÷ (1 + n) for the concentrate and the remainder for the water, which is the arithmetic of the notation itself. The 600 ml at 1+9 is the fourth of the five worked examples asked of this page, and it takes one division.

A note on checking these by hydrometer. ILFORD publishes the specific gravity of LC29 at 1+9, 1+19 and 1+29 as 1.015, 1.006 and 1.002. Those are useful at 1+9 and useless at 1+29: two parts in a thousand is below what a hydrometer reading to 0.001 can resolve. The specific-gravity check that works so well for a stock developer runs out at high dilution, and knowing where a check stops working is part of knowing the check.

3. Glacial acetic acid down to a working stop bath

Kodak states the intermediate step in one line, in both its 1928 primer and its modern toning publication: to make approximately 28 per cent acetic acid from glacial acetic acid, add 3 parts of glacial acetic acid to 8 parts of water.

🧮Two arithmetics, and what 'approximately' costs

The intermediate. Three parts in eleven is 3 ÷ 11 = 27.3 % by volume. The 1928 primer says glacial acid may contain as much as 99.5 per cent acetic acid, so the product is about 0.995 × 27.3 = 27.1 % v/v. Kodak’s word is “approximately”, and it is doing about 3 per cent of work.

The working bath. Kodak’s formula SB-1, an acid rinse bath for prints, is 48.0 ml of 28 per cent acetic acid made up with water to 1 litre. So the working bath contains 48 × 0.28 = 13.4 ml of glacial-equivalent acid per litre — about 1.3 % v/v acetic acid.

Straight from the glacial acid. Skip the intermediate and take C₁V₁ = C₂V₂ with volumes: you need 48 ml of a solution that is 3 ÷ 11 glacial by volume, so 48 × (3 ÷ 11) = 13.1 ml of glacial acetic acid, made up to 1 litre. Using Kodak’s nominal 28 per cent instead of the 27.3 per cent the recipe actually produces gives 13.4 ml. The 2 per cent between those two answers is the price of a rounded label, and it is smaller than the error in measuring 13 ml with a 100 ml graduate — which is the real lesson: know which of your uncertainties dominates before you argue about the others.

The order of addition is not negotiable. Princeton’s guidance for photographic work states it as a rule: always add acid to the water when diluting. Water goes in the vessel first, acid goes in afterwards and slowly, with the vessel in a tray and eye protection on. Diluting a concentrated acid releases heat; adding water to acid concentrates that heat in the small volume of water that arrives first, which can spit. This course states the rule and its source; it does not quantify the heat of dilution of acetic acid, because it has no source that does.

⚠What glacial acetic acid is, before you dilute it

Glacial acetic acid carries the harmonised classification H226, flammable liquid and vapour, and H314, causes severe skin burns and eye damage. It is a Level B reagent in this course’s classification: splash goggles, gloves, a tray, and dilution done where the air moves. The protective equipment page gives the workplace exposure limits and the glove data. Do this calculation on paper long before you do it at the bench.

4. A historical formula, in grains per fluid ounce

Take a period formula calling for 90 grains of metol per 32 fluid ounces of solution. There are two routes, and the interesting part is where they disagree.

🧮Which grain, which ounce, and why only one of the two is a question

Which grain? Wall’s 1912 dictionary lays out both British systems in the same unit: apothecaries’ weight, 480 grains to the ounce; avoirdupois weight, 437½ grains to the ounce and 7,000 to the pound. Both are counted in the same grain, so “90 grains” is unambiguous. Wall’s metric table gives 1 gramme as 15.432 grains, so 1 grain = 64.8 mg, and 90 grains = 5.83 g.

Which ounce? This is the real question, and it is a fluid ounce here. Wall gives 1 litre as 35 ounces and 94 minims; at 480 minims to the fluid ounce that is 35.196 fluid ounces per litre, so the imperial fluid ounce is 28.4 ml. ILFORD’s own modern conversion table, printed beside its dilution figures, gives the US fluid ounce as 29.6 ml.

• Imperial: 32 × 28.4 = 909 ml. 5.83 g ÷ 0.909 L = 6.4 g/L.
• US: 32 × 29.6 = 947 ml. 5.83 g ÷ 0.947 L = 6.2 g/L.

Four per cent apart, decided entirely by whether the author was in London or Rochester.

Cross-check against the period’s own factor. The British Journal Photographic Almanac converted grains per ounce into grammes per litre by multiplying by 2.3. Ninety grains in 32 fluid ounces is 2.8125 grains per fluid ounce; times 2.3 gives 6.5 g/L, within 1 per cent of the imperial answer — which it should be, since 64.8 mg ÷ 28.4 ml = 2.28 g/L per grain per ounce, and the almanac rounded up.

How you tell which system a formula used. Look for drachms. The almanac’s footnote states that it is now customary in formulae to employ the avoirdupois ounce of 437½ grains, but that where drachms appear the apothecaries’ drachm of 60 grains is the unit. That footnote is partly illegible in the scan this course read, and is reported with that caveat. Look also at the water line: “water to 20 ounces” is a British pint made up to volume, “water to 32 ounces” an American quart.

✕A period metric column is not a conversion

Kodak’s 1928 formulas print an avoirdupois column beside a metric one, and it is tempting to read across. Do not. Its developer D-1 gives pyro as 2 ounces and 60.0 grams; 2 avoirdupois ounces is 2 × 437.5 × 64.8 mg = 56.7 g, not 60.0. The same formula gives water to make 32 ounces and 1.0 litre; 32 US fluid ounces is 947 ml, not 1000.

The metric column is the same formula rescaled so that the batch is a round litre — by 1000 ÷ 947 = 1.056, which is very nearly the factor between 56.7 and 60.0. Check it on another line: sodium carbonate at 2½ ounces is 70.9 g, times 1.056 is 74.9, and the metric column says 75.0.

So the two columns describe the same solution at different batch sizes, and reading a mass across without the water line is how you end up about 6 per cent out. If you want to convert a Kodak formula given per 32 fluid ounces into grams per litre, use Kodak’s own paired figures and the rescaling is already in them.

5. Dilution as a decision about the picture

Dilution is not only economy. ILFORD publishes different development times for the same film at different dilutions of the same concentrate: DELTA 100 Professional at EI 100 is given 6 minutes in LC29 at 1+19 and 7 minutes 30 seconds at 1+29, at the same 20 °C. It also states that only 1+9 and 1+19 are suitable when the developer is to be reused, and that for the highest image quality the developer should be used one-shot.

Those are the manufacturer’s statements about time and capacity, and they are the ones this page can make. What a higher dilution does to the look of a negative — the argument about compensating development, adjacency effects and sharpness — is a claim about the image, and this course will make it in Parts VIII and IX, where it can be measured on the course’s own densitometer rather than asserted. What belongs here is the reason the arithmetic matters: at 1+29 a 3 ml error in the concentrate for a 600 ml tank is a 15 per cent error in developer strength, where the same 3 ml at 1+9 is 5 per cent. The more dilute the working solution, the more of your result rides on a small measured volume — which is why ILFORD sets a floor of 10 ml of concentrate and tells you to reach for a pipette below it.

Recap

• A concentration statement must say how much of what, in how much of what, and whether each side is a mass or a volume. Per cent w/v is grams per 100 ml; multiply by ten for grams per litre.
• Make up to the final volume; do not add solid to a measured volume of water. Manufacturers’ own mixing instructions say so, and so did the almanacs, in the phrase “water to 20 ounces”.
• Molarity is grams per litre divided by the molar mass. It earns its place in equilibria and stoichiometry, not at the balance.
• For a hydrate, masses stand in the ratio of the molar masses: 100 g of anhydrous thiosulfate is 157 g of the pentahydrate; 100 g of decahydrate carbonate is 37.0 g of the anhydrous salt, and getting that one backwards multiplies your alkali by 2.7.
• Stocks exist for shelf life, precision, consistency, and because some parts must be kept apart. The order in a multi-part instruction is part of the instruction.
• This course writes 1+9 and defines it on first use. Manufacturers use both notations and both mean parts of stock to parts of water; the colon is refused because it can be read two ways.
• C₁V₁ = C₂V₂ follows in one line from the fact that dilution does not change the amount of solute. Work it forwards for how much stock to take, backwards for what you actually made, and check every answer by a second route.
• Serial dilution beats a single step when the single step needs a volume too small to measure, at the cost of adding the steps’ errors together — and multiplying their mistakes.

Check your understanding

Question 1. How much 10 % w/v sodium sulfite stock, and how much water, make 500 ml of a 1.5 % w/v solution?

• 75 ml of stock and 425 ml of water, combined
• 75 ml of stock, made up to 500 ml with water
• 150 ml of stock, made up to 500 ml with water
• 7.5 ml of stock, made up to 500 ml with water

Show the answer and why

Answer: 75 ml of stock, made up to 500 ml with water

V1 = C2V2 / C1 = (1.5 x 500) / 10 = 75 ml. The check by a second route: 75 ml of a 10 % w/v stock holds 7.5 g of sulfite, and 7.5 g in 500 ml is 1.5 g per 100 ml. The distinction between the first two options is the whole point of the make-up-to-volume rule - combining 75 and 425 assumes the volumes add, while topping up to the mark assumes nothing and costs nothing. The 150 ml answer is the factor-of-two slip you get by dividing the wrong way round, and 7.5 ml is the factor-of-ten slip between per cent w/v and grams per litre.

Question 2. A datasheet specifies 1+10. A student reads it as "1:10" and takes one part of concentrate made up to ten parts of working solution. By how much is the developer wrong, and in which direction?

• It is correct; the two notations mean the same thing
• About 10 per cent too strong, because 1 in 10 is 0.1000 of the concentrate where 1+10 means 1 in 11, or 0.0909
• About 10 per cent too weak, for the same reason reversed
• About 100 per cent too strong

Show the answer and why

Answer: About 10 per cent too strong, because 1 in 10 is 0.1000 of the concentrate where 1+10 means 1 in 11, or 0.0909

1+10 means one volume of concentrate plus ten of water, so eleven volumes in total and a concentrate fraction of 1/11 = 0.0909. Making one part up to ten total gives 1/10 = 0.1000. The ratio is 0.1000/0.0909 = 1.10, so the working solution carries ten per cent more concentrate than intended. Ten per cent of developer strength is comparable to a whole degree of temperature error on ILFORD's own compensation chart, which is why this course refuses the colon notation rather than trusting the reader to guess right.

Question 3. A period British formula from 1906 calls for "sodium carbonate 100 g" and you weigh 100 g from a modern jar of anhydrous sodium carbonate. What have you done?

• Nothing; the names refer to the same substance
• Under-dosed the alkali, because the crystals are more concentrated than the powder
• Over-dosed the alkali by a factor of about 2.7, because the period convention used the unqualified name for the crystallised decahydrate
• Over-dosed by about 10 per cent, the difference between the two hydrates

Show the answer and why

Answer: Over-dosed the alkali by a factor of about 2.7, because the period convention used the unqualified name for the crystallised decahydrate

The British Journal Photographic Almanac states its own convention: sodium carbonate and sodium sulphite are used for the crystallised forms, and the dry or anhydrous forms are named in qualification. The decahydrate has a relative molecular mass of 286.14 against 105.99 for the anhydrous salt, so 100 g of crystals contains only 37.0 g of the anhydrous material - and weighing 100 g of anhydrous instead delivers 286.14/105.99 = 2.70 times the intended carbonate. This is why the first thing to look for in an old formula is the source's own statement of its conventions, and why an undated, unattributed formula off the internet is not usable.

Question 4. You need 100 ml of a 0.1 % w/v solution from a 10 % w/v stock. Why does this course prefer two dilution steps of ten to one over a single step?

• Because a single step would need 1 ml, which no vessel a home laboratory owns can measure to better than about 20 per cent, whereas each of two steps measures 10 ml
• Because serial dilution is always more accurate than a single dilution
• Because the solute will not dissolve at 10 per cent
• Because the errors in a serial dilution cancel out

Show the answer and why

Answer: Because a single step would need 1 ml, which no vessel a home laboratory owns can measure to better than about 20 per cent, whereas each of two steps measures 10 ml

The single step needs V1 = (0.1 x 100)/10 = 1 ml, which is below ILFORD's own stated floor of 10 ml for measuring a concentrate with a cylinder. Two steps keep every measured volume at 10 ml or above, and the worst-case relative uncertainties add to perhaps six per cent against twenty or more for the single step. The last two wrong answers matter: errors do not cancel, they accumulate, and a mistake in an early step is multiplied by every step after it - so serial dilution is not generally better, only better when a single step needs a volume you cannot measure.

Question 5. Kodak instructs that approximately 28 per cent acetic acid is made by adding 3 parts of glacial acetic acid to 8 parts of water. What is the actual concentration by volume, and what does the gap tell you?

• 28.0 per cent exactly; "approximately" refers to the measuring, not the arithmetic
• About 27.1 per cent, since 3 in 11 parts is 27.3 per cent by volume and the glacial acid is itself up to 99.5 per cent acetic acid - so the label is rounded by about 3 per cent
• About 37.5 per cent, since 3 parts to 8 is 3/8
• It cannot be determined without the density of the glacial acid

Show the answer and why

Answer: About 27.1 per cent, since 3 in 11 parts is 27.3 per cent by volume and the glacial acid is itself up to 99.5 per cent acetic acid - so the label is rounded by about 3 per cent

Three parts in eleven total, not three in eight: the water is 8 of the 11, so the fraction is 3/11 = 27.3 per cent by volume, and Kodak's own 1928 primer notes that glacial acid may be as much as 99.5 per cent acetic acid, giving about 27.1 per cent. The gap to the nominal 28 is about three per cent, and the useful conclusion is a habit rather than a number: a rounded label is a source of error you can quantify, and it should be compared with your other errors before you spend effort on it. Density would be needed only to express the result by mass rather than by volume.

Question 6. Kodak's 1928 developer D-1 lists pyro as "2 ounces" in its avoirdupois column and "60.0 grams" in its metric column, and water to make "32 ounces" and "1.0 litre". Two avoirdupois ounces is 56.7 g. Why is the metric column not 56.7?

• Because Kodak rounded 56.7 up to a convenient number
• Because the metric column is the same formula rescaled to a round litre: 32 US fluid ounces is 947 ml, and 1000/947 = 1.056, which turns 56.7 g into 59.9 g
• Because the 1928 ounce was a different size from the modern one
• Because pyro was sold by the apothecaries' ounce of 480 grains

Show the answer and why

Answer: Because the metric column is the same formula rescaled to a round litre: 32 US fluid ounces is 947 ml, and 1000/947 = 1.056, which turns 56.7 g into 59.9 g

Test the explanation on a second line rather than accepting it: sodium carbonate at 2 and a half ounces is 70.9 g, and 70.9 x 1.056 = 74.9, against the printed 75.0. The scale factor is consistent across the formula, so the two columns are the same solution at two batch sizes, not a conversion of one into the other. The practical rule is that you convert a formula, not an ingredient: read the water line first, and carry the ratio of the two batch volumes through every row. Reading a single mass across the page is how a formula ends up about 6 per cent out with no visible mistake anywhere.

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