Distribution of Species in a System at Equilibrium, Carbonate Equilibria
Reading Assignment: Study the material in this lecture and review the information on equilibrium and acids and bases at the web sites linked to Lecture 8. Read chapter 3 in Manahan and read the material on carbonate equilibria at the web sites linked below.
Homework: HW-4 (Due Wednesday, February 13)
Distribution of species in solution
Likewise, it can be shown
Figure 9.1. Distribution of Carbonate species as a function of pH
How do we use this and what good is it?
Measured Field Parameters
Alkalinity and pH
Alkalinity is the ability of a natural water to neutralize
Determined from a titration:
Moles of H+ = (Volume need to reach end point) x (concentration of acid)
From the composition of the natural water, we also know:
Moles of H+ = Moles HCO3- + Moles OH- + 2 [Moles CO32-]
This is total alkalinity
We also have Bicarbonate Alkalinity (Moles HCO3-))
Hydroxide Alkalinity (Moles OH-)
Carbonate Alkalinity (2x Moles CO32-)
Show addition of acid to a natural water sample
Strongest Base neutralized first
Next strongest base neutralized second
Next strongest base neutralized third, etc.
Figure 9.2. Hypothetical titration curve for a solution containing hydroxide, carbonate and bicarbonate ions.
Molar means moles of substance per liter of solution.
One mole of a substance is also one gram formula weight (GFW) of that substance
For instance, one mole of CaCO3 would be 100.09 g of CaCO3. This is because the GFW of CaCO3 is : 40.08 + 12.011 + (3) (15.9994) = 100.09.
Likewise, a normal solution contains one equivalent weight of a substance per liter of solution.
In acid base chemistry, an equivalent weight of a substance is the GFW of the substance divided by the charge of the aqueous ion.
For example, calcium carbonate dissociates to form Ca2+ and CO32-. The equivalent weight of calcium carbonate is therefore: 100.09/2 = 50.045. Since several common water quality measurements are expressed as meq/L, it is necessary to learn to work with these units.
100 mL of a water sample required 20.0 mL of 0.0100 M HCl to reach an end point of pH = 4.2 Express the alkalinity of this samlple as mg CaCO3/L
Since 1 Mole of HCl = 1 Equivalent of HCl
Since there were 10.01 mg of CaCO3 in 100 mL of Sample, the amount per liter would be:
10.01mgof CaCO3/0.1 L = 100.1 mg CaCO3/L
|Environmental Chemistry 440
Last Updated: 02/06/2007