Coagulant dose is not calculated from first principles — it is found, on a bench, in six beakers, and then scaled. The scaling is the arithmetic.

$D = \dfrac{V_{st}\,C_{st}}{V_s}$ — D equals V-stock C-stock over V-sample. $D$ is the equivalent dose on the raw water in mg/L. $V_{st}$ is the volume of stock solution pipetted into the beaker, in millilitres; $C_{st}$ is the strength of that stock in grams per litre; and $V_s$ is the sample volume in the beaker, in litres. Subscript st is always the stock you made up; subscript s is always the raw water being dosed.

The units line up more kindly than they look. Millilitres of stock times grams per litre gives milligrams of chemical directly, and dividing by litres of sample gives mg/L. So 3 mL of a 10 g/L stock into a 1 L jar is 30 mg of alum in a litre — a 30 mg/L dose, with no conversion needed anywhere.

Take that dose to the plant with the mass-loading relation: kg/d = m³/d × mg/L ÷ 1000. A 30 mg/L dose at 8,000 m³/d is 240 kg of alum a day, which is a chemical order, a storage volume and a pump setting all at once.

Then the second question, and it is the one that catches people. Alum does not simply dissolve — it reacts, and it consumes alkalinity to do it: $A_f = A_0 - 0.45\,D$. $A_0$ is the raw water alkalinity as CaCO₃ in mg/L, $A_f$ is what is left after coagulation, and $D$ is the same alum dose in mg/L. The 0.45 is not empirical — it falls out of the molar arithmetic of aluminium sulphate reacting to aluminium hydroxide floc, and it is fixed.

Why it matters more than the chemistry suggests. Alkalinity is the water's buffer. While there is buffer, the pH holds and the floc forms. When the buffer runs out the pH does not drift, it falls off a cliff — and aluminium hydroxide is only insoluble across a narrow pH window, so below it the alum simply stays dissolved and leaves in the finished water. A soft, low-alkalinity supply is exactly where this bites, and the answer is lime or soda ash fed ahead of the coagulant, never after.