1 / 15

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 -.

edalene
Download Presentation

GEOCHEMISTRY CLASS 4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. GEOCHEMISTRY CLASS 4

  2. 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

  3. 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: aH+ aF- = 10-3.18 aHF Equation 1: Equilibrium relationship: 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-]

  4. 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.

  5. 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.

  6. What is the pH of a solution in which 0.2 moles of HF is dissolved in 1 liter of “pure” water. aH+ aF- = 10-3.18 aHF Equation 1: Equilibrium relationship: 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] [H+]2 = 10-3.18 0.2 – [H] Plugging these relationships in we get Rearranging we get the quadratic polynomia [H+]2 + 10-3.18 [H+] – 0.2 10-3.18 = 0

  7. [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.

  8. 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+ pKa1= 6.35 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

  9. 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

  10. 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

  11. 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

  12. 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 Kb. Kb is often reported as pKb pKb = - log Kb

  13. 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

More Related