THE BASIC CHEMISTRY OF SYNTHETIC SEAWATER
The major salts used in formulation correspond approximately to the major constituents of ocean
water shown on Table 1. To ascertain how large a quantity of salt may be required it is usual practice to convert the natural readings into gram-equivalents, from their original grams per kilo quantities , as shown on the Table. This process is best described by giving
an example of the technique.
Column (a } of Table 1 shows the weight
of nine final
elements required. These g/kg weights are then divided. by the atomic
weight of each element concerned to convert them into gram-equivalent weights, eg;
Weight of element in seawater divided by the
atomic weight of the same element = Gram-Equivalent weight
Each manufacturer of laboratory chemicals states the molecular
weight of a reagent or salt on its packaging and labels, if not also in their catalogues. A simple multiplication
will then provide a working result as;-
'
Gram-Equivalent weight X Molecular weight =Final weight of salt needed.
(The mathematics involved also of course work in reverse, and formulae stating
weights of salt can be converted back into weights of elements per kilo and then compared to Table 1.}
I t is imperative that, if a formula is being 'worked out , one
works in reverse order dealing first with the minor constituents, then on up into the major constituents to the penultimate figure
;the one for sodium chloride.
The salt containing -bicarbonate is the last to be calculated, enough HCO ions (theoretically) being added to use any spare sodium ions as yet unpaired ,remembering that any ion carrying a positive {+) indicator has to be paired
with a similar ion carrying a {-) negative indicator and that where an ion carries
two symbols of one type, it requires double the quantity of an opposite type carrying only one symbol e.g. Na+ plus C1- = NaCl, which is common salt but Na+ plus
S04-- = Na2S04,or sodium .sulphate.
Thus only ten gram equivalents of chlorine (Cl}
are
needed to neutralise ten gm-equiv. of sodium in the first
example but twenty gram-equivalents of sodium would be used
in neutralising ten -gram-equivalents of sulphate in the second.
Let us assume that we wish to produce an extremely simple and not too accurate formula, and follow the complete process:
***
The final solution is to contain the following elements the
weights of which are given in grams per kilo.= g/kg
Sodium. Na+)10•9 Chloride(C1-)19•4 Magnesium Mg++)1•25
Calcium Ca++)0•5 Bromide (Br-) 0•1 Potassium K+) 0•4
Sulphate(S04--)2.5 Boric Acid (H3BO3) which is neutral,0•5
and lastly bicarbonate of, as yet, unknown quantity.
2.We intend using the following commonly available salts.
Common salt, Sodium chloride NaCl Molecular weight= 58•44
Anhydrous Magnesium chloride MgC12. MW = 95•22
Potassium chloride KC1. MW= 74•56 Magnesium
sulphate(crystallino)MgS047H20 MW = 246•48
Calcium chloride. Dry. CaCl 2H20 MW = 110•99
Boric Acid.H3803,which is already a salt and needs no MW.
calculations, Potassium bromide. KBr MW = 119•01
lastly Sodium bicarbonate. Na HC03 MW = 84•01
3e The elements and radicals concerned have the following
atomic weights according to the IUPAC 1973 list.
Sodium = 22•989 Chlorine = 35•453 Magnesium = 24•305
Calcium= 40•08 Bromine = 79•904 Potassium = 39.098 Sulphate = (S04 = Sulphur
(32906)plus 4xOxygen(15•999))or 96•057
Boric acid, which is a salt requiring no further
calculations.
Bicarbonate =(HCO3=1xHydrogen,1xCarbon & 3 x Oxygen ) or
61•017
***The measurements given in this section are as examples
and are not intended for use.
****Scientifically, elements have atomic weights whereas radicals or combinations (eg:HCO3 ) have molecular
weights, which as shown are the totals of the atomic weights of all component elements.This
section has intentionally been very much simplified,as has much of this chapter.(See preface)
