I.   Units used in Water Quality Analysis -  w/v    

               mg/L     or    parts per million (ppm)   same as mg/ml

mg/L     or    parts per billion (ppb)   same as ng/ml

               Molar (M)    or    moles/liter  - usually too small a number

               mmol/L    =   moles/L x 106 mmole/mole

            Normal    =  Equivalent per liter (eq/L)  -  usually too small a number an equivalent of A reacts with an equivalent of B since this is often a function of charge, an equivalent is usually a mole of charge.

                 meq/L      =   eq/L x 106 meq/Eq  

        II.  Converting Units -  multiply by factors of one

              A.  Convert mg/L to mmol/L for Na+ (atomic wt. = 23)

                           

   For these problems you always put what you have (mg/L) on the left hand side of the equation, and what you want (mmol/L on the right.  Then you just multiply by a series of fractions which equal one, so that the numerator cancels out and you're in the right range to give you reasonably sized numbers for convenience (generally 0.1 to 1,000).  When converting between molar units and gravimetric units, you have to include the molecular or atomic weight (g/mol).   In this case you turn it upside down (it's still 1) so that the weight units will cancel out.  So for instance if your solution contains calcium, the second fraction would be 1mol/40g, which is also equal to one.  Demonstrate to yourself, by canceling out units appropriately, that the units do cancel.

  B.  Convert mmol/L to meq/L for SO42- (molecular wt. = 32+4x16 = 96)

                            Note that the second fraction converts from the number of molecules in the solution to the number of units of charge (equivalents), which is 2 for sulfate.  All of these

problems can be answered in different ways.  A shortcut for this problem would be to replace the second, third and fourth fractions with the single fraction 2meq/mmol, recognizing that this has the same absolute value as eq/mol.             

C.  Convert mg/L to mg/L   (ppm to ppb)

Remember that there are 1,000 mg in a gram, 1,000 mg in a mg, and 1,000 ng (nanograms) in a mg.  Major elements (Ca2+, Mg2+, K+, Na+, SO4

2-, NO3-, Cl-,

DOC, and Silica) are usually measured ppm (mg/L), whereas trace elements (Zn2+, Mn2+, Cu2+, Fe3+, Al3+, etc.) are usually measured in ppb (mg/L)

I. Dilutions

A. Given a solution of 1000mg/L, prepare 1 Liter of a solution of 5mg/L.

For all dilution problems you can use a simple formula:

C1V1 = C2V2

This just says that the volume I need (V1) of the first solution times its concentration (C1), which is the 1000mg/L solution, will be equal to the volume (V2) of the second solution times its concentration (5mg/L):

Solve for V1 by dividing the right hand side of the equation by:

C1:V1 = C2V2/C1 = 5 x 1000/1000 = 5ml

In other words, you will pipet 5ml of your 1,000 ppm stock into a 1,000 ml volumetric flask and make it up to volume with deionized water.

I.   Preparing solutions of a given concentration from salts.

A.  Prepare a solution of 100 mg P/L from KH2PO4

Formula weight of KH2PO4  =  39 + 2x1 + 32 + 4x16 = 137

Formula weight for P   =  32

this means that there are 32 grams of P in 137 g (1 mol) of KH2PO4

                                          

For these problems, where you're making up a solution of atoms from a larger molecule, you must always multiply by a fraction representing the ratio of the atomic weight to the formula weight, so that the units cancel out.  The answer to this problem means that you would weigh out 0.428 g of KH2PO4 and make it up to a volume of one liter to make a 100 ppm solution of P.  Since the salt would dissolve as phosphate, we could also denote this as PO4-P.