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Acid-base Dissociation

Acid-base Dissociation. For any acid, describe it’s reaction in water: H x A + H 2 O  x H + + A - + H 2 O Describe this as an equilibrium expression, K (often denotes K A or K B for acids or bases…)

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Acid-base Dissociation

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  1. Acid-base Dissociation • For any acid, describe it’s reaction in water: • HxA + H2O  x H+ + A- + H2O • Describe this as an equilibrium expression, K (often denotes KA or KB for acids or bases…) • Strength of an acid or base is then related to the dissociation constant  Big K, strong acid/base! • pK = -log K  as before, lower pK=stronger acid/base!

  2. pKx? • Why were there more than one pK for those acids and bases?? • H3PO4 H+ + H2PO4- pK1 • H2PO4-  H+ + HPO42- pK2 • HPO41-  H+ + PO43- pK3

  3. Geochemical Relevance? • LOTS of reactions are acid-base rxns in the environment!! • HUGE effect on solubility due to this, most other processes

  4. Dissociation of H2O • H2O  H+ + OH- • Keq = [H+][OH-] • log Keq = -14 = log Kw • pH = - log [H+] • pOH = - log [OH-] • pK = pOH + pH = 14 • If pH =3, pOH = 11  [H+]=10-3, [OH-]=10-11 Definition of pH

  5. pH • Commonly represented as a range between 0 and 14, and most natural waters are between pH 4 and 9 • Remember that pH = - log [H+] • Can pH be negative? • Of course!  pH -3  [H+]=103 = 1000 molal? • But what’s gH+?? Turns out to be quite small  0.002 or so…

  6. Henderson-Hasselbach Equation: • When acid or base added to buffered system with a pH near pK (remember that when pH=pK HA and A- are equal), the pH will not change much • When the pH is further from the pK, additions of acid or base will change the pH a lot

  7. BUFFERING • When the pH is held ‘steady’ because of the presence of a conjugate acid/base pair, the system is said to be buffered • In the environment, we must think about more than just one conjugate acid/base pairings in solution • Many different acid/base pairs in solution, minerals, gases, can act as buffers…

  8. Buffering example • Let’s convince ourselves of what buffering can do… • Take a base-generating reaction: • Albite + 2 H2O = 4 OH- + Na+ + Al3+ + 3 SiO2(aq) • What happens to the pH of a solution containing 100 mM HCO3- which starts at pH 5?? • pK1 for H2CO3 = 6.35

  9. Think of albite dissolution as titrating OH- into solution – dissolve 0.05 mol albite = 0.2 mol OH- • 0.2 mol OH-  pOH = 0.7, pH = 13.3 ?? • What about the buffer?? • Write the pH changes via the Henderson-Hasselbach equation • 0.1 mol H2CO3(aq), as the pH increases, some of this starts turning into HCO3- • After 12.5 mmoles albite react (50 mmoles OH-): • pH=6.35+log (HCO3-/H2CO3) = 6.35+log(50/50) • After 20 mmoles albite react (80 mmoles OH-): • pH=6.35+log(80/20) = 6.35 + 0.6 = 6.95

  10. Bjerrum Plots • 2 D plots of species activity (y axis) and pH (x axis) • Useful to look at how conjugate acid-base pairs for many different species behave as pH changes • At pH=pK the activity of the conjugate acid and base are equal

  11. Bjerrum plot showing the activities of reduced sulfur species as a function of pH for a value of total reduced sulfur of 10-3 mol L-1.

  12. Bjerrum plot showing the activities of inorganic carbon species as a function of pH for a value of total inorganic carbon of 10-3 mol L-1. In most natural waters, bicarbonate is the dominant carbonate species!

  13. Carbonate System Titration • From low pH to high pH

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