Hydrology Lecture Notes – Water Chemistry - Hofstra ... - People Page

Hydrology Lecture Notes – Water Chemistry
Contaminant Concentration
Groundwater is never pure, rather it contains small amounts of dissolved gases and
solids.
Chemical analysis of an aqueous solution yields the amount of solute in a specified
amount of water. There are several ways this is reported:
Weight per Weight units (dimensionless ratios): ppm, ppb, ppt (parts per million,
billion, trillion). These are no longer commonly used in water analysis, although they are
sometimes reported to the public because they are easy to interpret.
Weight per Volume units: milligrams per liter (mg/L) and micrograms per liter (ug/L).
These are more commonly used.
1 liter of pure water = 1X10 6 milligrams H2O at 3.89°C therefore 1 ppm = 1 mg/L
Note: density and weight of water will change with temperature and dissolved mineral
matter, but not enough to worry about within the range of normal groundwater
conditions.
Legal Limits on Concentrations:
MCL (Maximum Contaminant Level) – the highest level legally permitted in
groundwater.
MCLG (Maximum Containment Level Goal) – the level of contaminant below which
there is no known risk. The ideal level, lower than the MCL.
AL (Action Level) – The concentration of a contaminant at which action must be taken to
treat it.
ND – Non Detect – contaminant not detected given sensitivity of the test.
Chemistry Measures of Concentration
Molarity is the number of moles of solute dissolved in one liter of solution (moles/L).
The units, therefore are moles per liter, specifically it's moles of solute per liter of
solution.
One Mole of a chemical species is its formula weight in grams, calculated by adding up
the atomic weights of all the atoms in that species. E.g. NaCl = Na (at.wt. 22.991) + Cl
(at.wt. 35.457) = 58.448 gm.

Thus, a 1 molar solution of NaCl would be 58.5 grams of NaCl dissolved in 1 Liter of
water.
Sample Calculations:
1. What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to
make 125 ml of solution?
molarity = moles of solute / liters of solution
2.5 g NaCl x (1 mole NaCl / 58.5 g NaCl) = 0.0427 mole
molarity = 0.0427 mole NaCl / 0.125 L = 0.34 M NaCl
2. How many moles of salt are contained in 300. mL of a 0.40 M NaCl solution?
0.40 moles NaCl / 300 mL = 0.4 moles x .3 L = 0.12 moles NaCl
Molarity: 1 mole per liter of solvent = 1 molar solution
Molality: 1 mole per 1000 grams of solvent = 1 molal solution
molality (M) = moles solute / kg of solution
For water solutions, 1 kg of water has a volume close to that of 1 liter, so molality and
molarity are similar in dilute aqueous solutions – up to .01 molal.
Thus for dilute solutions, you can calculate Molarity or Molality as:
Molal / Molar = mg/L X 10 -3 / formula wt in grams
Example:
48 mg/L NaCl = ? molality
48 mg/L x 10 -3 / 58.5 = .00082 molal / molar
Equivalent Weight units incorporate the chemical behavior of a solute. The equivalent
weight or combining weight of a dissolved ionic chemical species is the formula weight
divided by the electrical charge.
Example: A divalent ion (Ca+2) combines with Cl- (a monovalent ion) to form calcium
chloride (CaCl2). The combining ratio of calcium to chlorine is one-to-two, meaning it
takes half a mole of Ca +2 to combine with one mole (or 1 eq) of Cl - . So, half a mole of
calcium equals 1 equivalent.

Another way to look at it is to say one equivalent is the amount of ion required to cancel
out the electrical charge of an oppositely charged monovalent ion. You can also say that
the valence charge of the ion is the number of equivalents there are in one mole of that
ion. For example, nitrogen (N -3 ) is a trivalent ion; therefore one mole of nitrogen equals 3
equivalents. Using the calcium example from above, we know that one mole of calcium
equals 2 equivalents.
The amount of a given substance in equivalents is equal to the amount of the
substance in moles multiplied by the valence of the substance.
The equivalent weight is the mass of one equivalent of a substance.
Example:
1 mol H2SO4 = 2 equivalents H2SO4
There are "TWO" H+ released when the acid dissociates
The acid releases two H+ ions and one SO4 2- ion.
Thus 1 equivalent weight of sulfuric acid is 1/2 the mass of a mol.
1 mol H2SO4 = 98 grams H2SO4; 1 equivalent H2SO4 = 49 grams H2SO4
In practice, the amount of a substance in equivalents often has a very small magnitude, so
it is frequently described in terms of milliequivalents (mEq or meq) -- the prefix milli
denoting that the measure is divided by 1000. Very often, the measure is used in terms of
milliequivalents of solute per liter of solvent (mEq/L).
Example:
1 mol H3PO4 = 3 equivalents H3PO4
The acid releases three H+ and one phosphate PO4 2- ion.
The equivalent weight of phosphoric acid is 1/3 the mass of a mol.
1 mol H3PO4= 98 grams H3PO4; 1 equivalent H3PO4= 32 grams H3PO4
The relation between equivalents and milliequivalents:
1 equivalent H3PO4= 1000 milliequivalent H3PO4
1 equivalent mol H3PO4= 32 grams H3PO4;
1 milliequivalent mol H3PO4= 32 milligrams H3PO4;
Shortcut:
Dividing the concentration in mg / L by the equivalent weight of the ion (one equivalent)
yields a concentration in milliequivalents per L (meq/L). Non ionic species such as
silica cannot be expressed in meq/L.
Example:
Ca 2+ = 40.08 gm/mole = 2 equivalents
1 equivalent = 20.04 gm
134 mg/L of Ca 2+ = 134 mg/L / 20.04 gm = 6.69 meq/L

Equivalents have advantage over moles in quantitative analysis of reactions. The best
feature of using equivalents is that there is no need to study much about the nature of
reaction, i.e. no need to analyse and balance chemical equations. Equivalents of reactants
react in equal numbers to yield the same equivalents of products.