1. GEOCHEMISTRY CLASS 4

3. : 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 K a. K a is often reported as pK a pK a = - log K a For example for HF pK a = 3.18 Hence a H + a F - = 10 -3.18 a HF

4. 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: a H + a F - = 10 -3.18 a HF 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 - ]

5. 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 a i ’s and [i]’s. a H + = γ H + [H + ] and so on Recall from Debye-Huckel Model: Log γ i = -Az i 2 I 0.5 1 + B a i I 0.5 And I = ½ Σ [i] z i 2 Note that in order to calculate γ i 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

6. Solution: Begin by assuming that the solution will be so dilute that I ~ 0, then : Log γ i ~ 0 or γ i ~ 1 and a i ~ [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 γ i ’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 γ i ’s and will redo the calculations. We then repeat and repeat Until I no longer changes significantly.

7. 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: a H + a F - = 10 -3.18 a HF 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 a H + = [H + ], a F - = [F - ], = a HF = [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

8. [H + ] 2 + 10 -3.18 [H + ] – 0.2 *10 -3.18 = 0 Using the quadratic equation X = - b +/- ( b 2 – 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] z i 2 = ½ ( 0.0111* 1 2 + 0.0111*1 2 ) = 0.0111 Should probably calculate activity coefficients and redo the problem, but we wont.

9. Acids that give off two protons are known as diprotic acids Carbonic acid (H 2 CO 3 ) is one of the geologically most important examples: H 2 CO 3 ↔ HCO 3 - + H + pK a1 = 6.35 Diprotic acids will have two pK a: pk a1 for the first dissociation and pK a2 for the second dissociation. HCO 3 - ↔ CO 3 -- + H + pk a2 = 10.33 Acids that give off three protons are triprotic and have 3 dissociation constants. Most important geological example is phosphoric acid H 3 PO 4

10. H 2 CO 3 ↔ HCO 3 - + H + pK a1 = 6.35 HCO 3 - ↔ CO 3 -- + H + pk a2 = 10.33 Note from this equation a HCO3- = 10 -6.35 /a H+ a H2CO3 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

12. An acid is a strong acid if it has a small pK a and hence undergoes extensive dissociation. Most important natural strong acid is sulfuric acid: H 2 SO 4 pK a1 = ~ -3, It forms by weathering of sulfides under oxidizing conditions: 4FeS 2 + 8H 2 O + 15O 2 ↔ 2Fe 2 O 3 + 8H 2 SO 4 Or by the dissolution of volcanic gasses in water. Less important naturally occurring strong acids are: Nitric acid HNO 3 pK a = 0 Hydrochloric acid HCl pK a ~ -3 Pyrite Hematite

13. Weak acids have relatively large pK a and hence dissociate to a relatively small degree. Most important naturally occurring weak acids are: 1. Carbonic acid 2. Silicic acid H 4 SiO 4 pK a1 = 9.83, pK a2 = 13.17 It forms through the weathering of silicate minerals: MgSiO 3 + 2 H + + H 2 O ↔ Mg ++ + H 4 SiO 4 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

14. Bases General 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 K b. K b is often reported as pK b pK b = - log K b

15. Example of a geologically important base Amorphous Al (OH) 3 pK b1 = 12.3 Al(OH) 3 ↔ Al(OH) 2 + + OH - Al(OH) 2 + OH - = 10 -12.3 Al (OH) 3