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In a message dated 10/18/2001 10:07:33 PM Eastern Daylight Time,
[email protected] writes:
> Subj:Re: CS>Bubble or Stir?
> Date:10/18/2001 10:07:33 PM Eastern Daylight Time
> From: [email protected] (Kevin Nolan)
> To: [email protected]
>
>
>
>
> OK Roger, please send that material. I had assumed your classes were just
> about establishing solubility product equations for beginners
>
Ag+ Solubility in LVDC CS Assuming Aqueous Chemical Equilibria Exists
LESSON I
Introduction
Chemical reactions fall into two general types, those that react so
completely that the reactants disappear, and those that react partially so
that the reactants and products coexist.
An extreme example of a reaction in which the reactants disappear completely
is the explosion of gun powder,
(1) 2KNO3 + 3C + S -----> K2S + 3CO2 + N2
An example in which the reactants and products coexist is observed when solid
silver hydroxide is mixed with distilled water (DW),
(2) AgOH(s) = [Ag+] + [OH-]
Note that the "=" signifies that all species coexist, and the "(s)" indicates
that the AgOH is in a solid, crystalline form. The "+" and "-" attached to
"Ag" and "OH" represent the formation of ions which are single atoms or a
combination of atoms that have gained or lost one or more electrons. Of
course, a reaction like the one above can be written for practically any
salt, but reactions involving insoluble or slightly soluble salts have
characteristics that allow chemists to make accurate solubility predictions
by taking advantage of the unique properties of dilute aqueous solutions.
Here are some of these unique properties.
First, it can be shown that the product of Ag+ concentration and OH-
concentration equals a constant called the solubility product. Although the
value of this constant changes with temperature, it does not change with
variations in the concentration of Ag+ or OH-. I guess that's why it's
referred to as a 'constant'. So if we know the temperature of the DW we can
look up the solubility product of any similar salt.
Second, the solubility product of, say, AgOH, will not be effected by the
presence of other dilute ionic, or molecular species in the DW such as
dissolved CO2.
The concentration of OH- is important because it is associated with the pH of
the DW. pH is based on the concentration of H+ which, together with OH- are
produced from the ionization of water molecules as shown in reaction (3),
(3) H2O = [H+] + [OH-]
Notice that the form of this reaction is similar to that of reaction (2). The
"ionization constant" of water works like the solubility product just
mentioned, but instead of relating ion formation to a salt, it relates it to
water molecules shown as "H2O".
Water is practically all molecules. In fact, pure water has only 10^-7
moles/liter of hydrogen [H+] and hydroxyl [OH-] ions. The logarithm of the
hydrogen ion concentration is the pH. So, as you can see, pure water has a pH
of 7.
When pure DW is saturated with AgOH, two equations can be written,
[Ag+] * [OH] = K(sp) = 1.52 x 10^-8
[H+] * [OH-] = K(w) = 10^-14
Note that without some knowledge of the pH or Ag+ concentration,
it is not possible to calculate the concentrations of the remaining species.
In the example above, it was assumed that we have pure water in equilibrium
with silver hydroxide. In actuality, the DW that exists after a LVDC CS brew
has been prepared contains dissolved CO2, and it may not contain solid silver
hydroxide. So in order to determine the maximum solubility of silver ion in a
LVDC CS brew, we need to start out itemizing the things we know, or can
reasonably assume, so that the relevant solubility equations can be written.
More about that in the next lesson.
LESSON II
Review
OK. Let's review my closing remarks from the last class:
When pure DW is saturated with AgOH, two equations can be written,
[Ag+] * [OH] = K(sp) = 1.52 x 10^-8
[H+] * [OH-] = K(w) = 10^-14
Note that without some knowledge of the pH or Ag+ concentration, it is not
possible to calculate the concentrations of the remaining species. In the
example above, it was assumed that we have pure water in equilibrium with
silver hydroxide. In actuality, the DW that exists after a LVDC CS brew has
been prepared contains dissolved CO2, and it may not contain solid silver
hydroxide. So in order to determine the maximum solubility of silver ion in a
LVDC CS brew, we need to start out itemizing the things we know, or can
reasonably assume, so that the relevant solubility equations can be written.
Things We Know (or Can Reasonably Assume)
about the Product of a Well Mixed LVDC CS Brew
First, we know that Ag+ is produced at the anode according the anodic
reaction,
(1) Ag ----------> [Ag+] + 1e
and OH- is produced at the cathode according to the cathodic reaction,
(2) H2O + 1e -------------> 1/2H2(g) + [OH-]
Since I'm stipulating the DW in the reaction vessel is well mixed and the
electrical potential is current limited, it is reasonable to assume that
reactions (1) and (2) proceed to the "right", i.e., that there are no "back"
reactions that consume either Ag+ and/or OH-. [I'm going to have to ask the
class to accept these conclusions since getting into the "whys & wherefores"
of electrochemical reactions are outside the scope of our chemical equilibria
mini-class.]
Second, since it is assumed that the DW in the CS open reactor is well mixed
there's a reasonable chance that the DW is air saturated, and, therefore, the
0.033% of CO2 in air produces some H+ (see below) and, therefore, reacts with
OH- generated by reaction (2) above to form water. Furthermore, since the
reactor is open to the air during and shortly after CS production, it is
reasonable to assume that the CO2 from the air remains in equilibrium with
the DW and that the small amount of OH- produced from reaction (2) reacts
with a replenishable supply of H+ because of the constant CO2 partial pressure
that remains above the DW surface.
Setting Up the Equilibrium and Mass Balance Equations
to Determine the Solubility of Ag+ in a LVDC CS Product
I used the Internet to search the chemical literature to determine the
relationship between the percentage of CO2 in air (partial pressure and
"percentage"/100 are equivalent at one "atmosphere" total pressure) and the
concentration of dissolved "CO2" in DW in equilibrium with it. Equation (3)
gives this relationship,
(3) [CO2]dis. = Solubility Coefficient * Partial Pressure of CO2 in air
where the CO2 Solubility Coefficient = 2*10^-3 @25C, &
the Partial Pressure of CO2 in air = 3.55*10^-4 ATM
Therefore, the concentration of carbon (in all ionic & molecular forms -- see
below) in water is 7.1*10^-7 moles/liter
However, not all of the dissolved CO2 forms CO3=. As will be shown in the
next lesson, at pH 7 and below, the vast bulk of dissolved "CO2" remains as
molecular H2CO3 as well as HCO3- according to the equilibrium reactions,
(4) [H2CO3](aq) = [H+] + [HCO3-] Ksp = 4.45 * 10^-7
(5) [HCO3-] = [H+] + [CO3=] Ksp = 4.69 * 10^-11
As mentioned, the distribution of these species is pH dependent. So the
Ionization Constant of water should be include with equations (4) & (5) to
help define the state of this system.
(6) H2O = [H+] + [OH-] Kw = 10^-14
Finally, since Ag+ is also present, wouldn't it make sense to include
reaction (7) so that the equilibria of all species present in a LVDC CS
product can be accounted for?
(7) AgOH(s) = [Ag+] + [OH-] Ksp = 1.52 * 10-8
The answer is no because we are interested in the MINIMUM equilibria that
define the state of a typical LVDC CS product, and we have no a priori
knowledge that silver hydroxide is present. In any case, let's look for an
explanation from another perspective.
When CO2 dissolves in DW, it distributes itself as molecular H2CO3, HCO3- and
CO3=. The solubility products for reactions (4), and (5) govern the
distribution of these species, as well as the ionization constant from
reaction (6). Since we know that, for dilute solutions, positively charged
ions cannot influence the concentration of one another (the same is true for
negatively charged ions -- remember, I mentioned this concept in the
Introduction), we can assume that the equilibrium pH of a typical air
saturated, LVDC CS product which contains ionic silver will be identical to
the pH found in pure DW, saturated with air with no silver ion present. This
value is readily available from the chemical literature, and at 25 C, the pH
was found to be 5.65. It should be noted that this pH is very close to the pH
range mentioned to me by Ole' Bob (pH = 5.0--->5.5) from his LVDV CS research.
Since this approach is consistent with dilute solution theory, we can now use
the water ionization constant from reaction (6) to obtain the hydroxyl ion
concentration. The Ksp from reaction (7) then gives us the POTENTIAL silver
ion solubility with regard to the formation of silver hydroxide. I say
POTENTIAL because the presence of a separate silver hydroxide phase was never
stipulated in setting up the equilibrium equations. However, I prefer to save
the details of this calculation for the next lesson because this single
estimate is not the whole story since silver carbonate can also form, and
THAT calculation is somewhat more complicated than those I've just covered.
Roger
Lesson III
Review
OK. Let's review my closing remarks from the last class:
Since this approach is consistent with dilute solution theory, we can now use
the water ionization constant from reaction (6) to obtain the hydroxyl ion
concentration. The Ksp from reaction (7) then gives us the POTENTIAL silver
ion solubility with regard to the formation of silver hydroxide. I say
POTENTIAL because the presence of a separate silver hydroxide phase was never
stipulated in setting up the equilibrium equations. However, I prefer to save
the details of this calculation for the next lesson because this single
estimate is not the whole story since silver carbonate can also form and THAT
calculation is somewhat more complicated than those I've just covered.
Solving Equilibria Expressions
As mentioned in the previous lesson, carbon is distributed among H2CO3, HCO3-
and CO3=. Reactions (1) & (2) show the solubility products that help
establish this distribution.
(1) [H2CO3](aq) = [H+] + [HCO3-] Ksp = 4.45 * 10^-7
(2) [HCO3-] = [H+] + [CO3=] Ksp = 4.69 * 10^-11
Note that even though we can establish the final pH to be 5.65 (Lesson II),
it is impossible to calculate the concentration of the above species without
additional information because we have two equations and three unknowns.
However, we do have one other piece of information that I mentioned earlier,
and that is the molar concentration of dissolved carbon which is based on the
equilibrium between CO2 from the air and the "CO2" dissolved in DW. If you
remember,
(3) [CO2]dis. = Solubility Coefficient * Partial Pressure of CO2 in air
where the CO2 Solubility Coefficient = 2*10^-3 @25C, &
the Partial Pressure of CO2 in air = 3.55*10^-4 ATM
Therefore, the concentration of carbon (in all ionic & molecular forms) in
water is calculated to be 7.1*10^-7 moles/liter
A carbon mass balance is based on the fact that there is one mole (OK,
gram-atom for the purists) of carbon in each of the three forms of dissolved
carbon, and that their molar sum is known (molar concentration, 7.1*10^-7
moles/liter, in 1 liter of DW is 7.1*10^-7 moles). Equation (4) gives this
relationship for 1 liter of DW.
(4) [HCO3-] + [H2CO3] + [CO3=] = 7.1*10^-7 moles
Substituting the hydrogen concentration in moles/liter (pH = 5.65) into Eq.
(1) & (2) gives,
(5) [H2CO3] = 5.031*[HCO3-]
(6) [CO3=] = 2.095*10^-5*[HCO3-]
Substituting the equivalent [HCO3-] from equations (5) and (6) into equation
(4) allows the concentration of HCO3- to be determined. The remaining
concentration values are obtained by substituting the calculated HCO3-
concentration into equations (5) and (6) which yields the following results:
[HCO3-] = 1.177*10^-7 m/l
[H2CO3] = 5.923*10^-7 m/l
[CO3=] = 2.466*10^-12 m/l
Silver carbonate can precipitate from electrolytically prepared CS if the
product of the [Ag+] and [CO2=] concentrations reach the Solubility Product
given in reaction (7).
(7) Ag2CO3 = 2[Ag+] + CO3= Ksp = 8.1*10^-12
Therefore, substituting the [CO3=] concentration obtained above, yields an
ionic silver solubility of 193.9 grams/liter.
As already mentioned, another source of precipitated ionic silver is the
equilibrium reaction,
(8) AgOH = [Ag+] + [OH-] Ksp = 1.52 * 10-8
For pH = 5.65, the ionic silver concentration will be 367.5 grams/liter
Therefore, if equilibrium prevails, the ionic silver solubility in a LVDC CS
product is the lesser of these two values, or 193.9 grams/liter since there
is always an excess of CO2 from the air to allow for the continuous
precipitation of Ag2CO3 to hold the maximum ionic silver concentration at
193.9 grams/liter.
>From a practical point of view, one can make virtually an unlimited
concentration of ionic silver in DW. Making 500 or 1000 PPM "CS'' should not
produce a precipitate of any known silver compounds from species likely to be
present in a typical CS electrolytic process. Whether or not one would WANT
to produce anything beyond the 5-15 PPM "CS" that we use (with great results)
is another question. Since there will always be a silver particulate
component in your CS brew, particle size will likely increase as the
concentration of silver ion increases, and these larger particles WILL drop
out of suspension. So the whole issue I discussed above is essentially moot,
but I hope you enjoyed stretching your thinking and perhaps learned a little
about aqueous chemical equilibria.
Roger
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