GEOCHEMISTRYCLASS 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 = 3.18Hence

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:

Equation 1: Equilibrium relationship: aH+ aF- = 10-3.18

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 HFAnd 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 + BaiI0.5And I = ½ Σ [i] zi

2

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.

Where [i] = concentration of I in moles per liter

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 repeatand repeat

and repeatand 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-]

Plugging these relationships in we get

And aH+ = [H+], aF- = [F-], = aHF = [HF]

[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

Carbonic acid (H2CO3) is one of the geologically most important examples:

H2CO3 ↔ HCO3- + H+ pKa1= 6.35

Diprotic acids will have two pKa: pka1 for the first dissociation andpKa2 for the second dissociation.

HCO3- ↔ CO3

-- + H+ 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

HCO3- ↔ CO3

-- + H+ pka2 = 10.33

Note from this equation aHCO3- = 10-6.35/aH+

aH2CO3

A similar equation can be derived from the equation

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

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

Pyrite Hematite

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++ + H4SiO4enstatite

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

BasesGeneral definition a proton acceptor

More restricted definition: a substance that produces OH- when 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+ + OH-

Al(OH)2+ OH- = 10-12.3

Al (OH)3