Transcript GEOCHEMISTRY CLASS 4

GEOCHEMISTRY
CLASS 4
:
An acid is a proton (H+) donor
An acid is monoprotic if it gives off 1 H:
Example hydrofloric Acid:
HF ↔ H+ + F-
The extent to which an acid dissociates is given by the
equilibrium constant for the dissociation which is known as
the acid dissociation constant designated as Ka.
Ka is often reported as pKa
pKa = - log Ka
For example for HF pKa = 3.18
Hence
aH+ aF- = 10-3.18
aHF
Sample problem:
What is the pH of a solution in which 0.2 moles of HF is
dissolved in 1 liter of “pure” water.
Step 1: Develop the same number of equations as we
have unknowns:
a
+ aF- = 10-3.18
H
Equation 1: Equilibrium relationship:
aHF
Equation 2: From mass balance:
Since all F-1 comes from dissociation of HF and the initial
[HF] = 0.2 then: [HF] = 0.2 – [F-1]
Equation 3: From mass balance:
If we assume that almost all H+ comes from dissociation of HF
And that there is no additional sink of H+ then:
[H+] = [F-]
Note equilibrium relationship is in terms of activities, mass
mass balance equations in terms of concentrations – to make
a system of equations we need a relationship between ai’s
and [i]’s.
aH+ = γH+[H+] and so on
Recall from Debye-Huckel Model:
Log γi =
-Azi2I0.5
1 + BaiI0.5
And I = ½ Σ [i] zi2
Where [i] = concentration of I in moles per liter
Note that in order to calculate γi we need to know the
concentrations of all of the charged dissolved species but it is
these concentrations we are trying to calculate in the first place.
Solution: Begin by assuming that the solution will be so dilute that
I ~ 0, then :
Log γi ~ 0 or γi ~ 1 and ai ~ [i]
After we calculate all of the [i] we will go back and calculate I. If I
is not approximately 0 we will use I to calculate γi’s .
We will then calculate new [i]’s and when we are finished
calculate a new I. We will compare the new and the old I. If they
are close we will quit. If not we will use the new I to calculate
new γi’s and will redo the calculations.
We then repeat
and repeat
and repeat
and repeat
and repeat
Until I no longer changes significantly.
What is the pH of a solution in which 0.2 moles of HF is
dissolved in 1 liter of “pure” water.
Equation 1: Equilibrium relationship:
aH+ aF- = 10-3.18
aHF
Equation 2: From mass balance:
[HF] = 0.2 then: [HF] = 0.2 – [F-1]
Equation 3: From mass balance:
[H+] = [F-]
And aH+ = [H+], aF- = [F-], = aHF = [HF]
Plugging these relationships in we get
[H+]2 = 10-3.18
0.2 – [H]
Rearranging we get the quadratic polynomia
[H+]2 + 10-3.18 [H+] – 0.2 10-3.18 = 0
[H+]2 + 10-3.18 [H+] – 0.2 *10-3.18 = 0
Using the quadratic equation X = - b +/- ( b2 – 4ac)0.5
2a
We get [H+] = -10-3.18 +/- ((10-3.18)2 -4*-0.2*10-3.18)0.5
2
[H+] = either 0.0111 or -0.01183
Only positive concentrations are possible so
[H+] = 0.0111
And pH = - log [H+] = - log (0.0111) = 1.95
I = ½ Σ [i] zi2 = ½ ( 0.0111* 12 + 0.0111*12) = 0.0111
Should probably calculate activity coefficients and redo the
problem, but we wont.
Acids that give off two protons are known as diprotic acids
Diprotic acids will have two pKa: pka1 for the first dissociation and
pKa2 for the second dissociation.
Carbonic acid (H2CO3) is one of the geologically most important
examples:
H2CO3 ↔ HCO3- + H+
HCO3- ↔ CO3-- + H+
pKa1= 6.35
pka2 = 10.33
Acids that give off three protons are triprotic and have 3
dissociation constants. Most important geological example is
phosphoric acid H3PO4
H2CO3 ↔ HCO3- + H+
pKa1= 6.35
Note from this equation aHCO3- = 10-6.35/aH+
aH2CO3
A similar equation can be derived from the equation
HCO3- ↔ CO3-- + H+
pka2 = 10.33
Thus it is relatively easy to calculate the relative proportion of
Carbonate species as a function of pH
An acid is a strong acid if it has a small pKa and hence
undergoes extensive dissociation.
Most important natural strong acid is sulfuric acid:
H2SO4
pKa1 = ~ -3,
It forms by weathering of sulfides under oxidizing conditions:
4FeS2 + 8H2O + 15O2 ↔ 2Fe2O3 + 8H2SO4
Pyrite
Hematite
Or by the dissolution of volcanic gasses in water.
Less important naturally occurring strong acids are:
Nitric acid HNO3 pKa = 0
Hydrochloric acid HCl pKa ~ -3
Weak acids have relatively large pKa and hence dissociate to a
relatively small degree.
Most important naturally occurring weak acids are:
1. Carbonic acid
2. Silicic acid
H4SiO4 pKa1 = 9.83, pKa2 = 13.17
It forms through the weathering of silicate minerals:
MgSiO3 + 2 H+ + H2O ↔ Mg++ + H4SiO4
enstatite
3. organic acids
Most important naturally occurring organic acids contain the
carboxylic group –COOH
Most concentrated in waters in contact with decaying organic
material.
In most environments poorly characterized
Bases
General definition a proton acceptor
More restricted definition: a substance that produces OHwhen it dissociates in water.
The extent to which an base dissociates is given by the
equilibrium constant for the dissociation which is known as
the base dissociation constant designated as Kb.
Kb is often reported as pKb
pKb = - log Kb
Example of a geologically important base
Amorphous Al (OH)3 pKb1 = 12.3
Al(OH)3 ↔ Al(OH)2+ + OHAl(OH)2+ OH- = 10-12.3
Al (OH)3