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Strong Acids/ Bases Strong Acids more readily release H+ into water, they more fully dissociate –H 2 SO 4  2 H + + SO 4 2- Strong Bases more readily release.

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Presentation on theme: "Strong Acids/ Bases Strong Acids more readily release H+ into water, they more fully dissociate –H 2 SO 4  2 H + + SO 4 2- Strong Bases more readily release."— Presentation transcript:

1 Strong Acids/ Bases Strong Acids more readily release H+ into water, they more fully dissociate –H 2 SO 4  2 H + + SO 4 2- Strong Bases more readily release OH- into water, they more fully dissociate –NaOH  Na + + OH - Strength DOES NOT EQUAL Concentration!

2 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…) 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!

3 pK x ? Why were there more than one pK for those acids and bases?? H 3 PO 4  H + + H 2 PO 4 - pK 1 H 2 PO 4 -  H + + HPO 4 2- pK 2 HPO 4 1-  H+ + PO 4 3- pK 3

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

5 Dissociation of H 2 O H 2 O  H + + OH - K eq = [H + ][OH - ] log K eq = -14 = log K w pH = - log [H+] pOH = - log [OH-] pK = pOH + pH = 14 If pH =3, pOH = 11  [H + ]=10 -3, [OH - ]= Definition of pH

6 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 + ]=10 3 = 1000 molal? –But what’s   ?? Turns out to be quite small  or so…

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 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

9 Buffering example Let’s convince ourselves of what buffering can do… Take a base-generating reaction: – Albite + 2 H 2 O = 4 OH- + Na + + Al SiO 2(aq) –What happens to the pH of a solution containing 100 mM HCO3- which starts at pH 5?? –pK 1 for H 2 CO 3 = 6.35

10 Think of albite dissolution as titrating OH - into solution – dissolve 0.05 mol albite = 0.2 mol OH mol OH-  pOH = 0.7, pH = 13.3 ?? What about the buffer?? –Write the pH changes via the Henderson-Hasselbach equation 0.1 mol H 2 CO 3(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 - /H 2 CO 3 ) = 6.35+log(50/50) After 20 mmoles albite react (80 mmoles OH - ): –pH=6.35+log(80/20) = = 6.95

11 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

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

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

14 THE RELATIONSHIP BETWEEN H 2 CO 3 * AND HCO 3 - We can rearrange the expression for K 1 to obtain: This equation shows that, when pH = pK 1, the activities of carbonic acid and bicarbonate are equal. We can also rearrange the expression for K 2 to obtain: This equation shows that, when pH = pK 2, the activities of bicarbonate and carbonate ion are equal.

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

16 THE CO 2 -H 2 O SYSTEM - I Carbonic acid is a weak acid of great importance in natural waters. The first step in its formation is the dissolution of CO 2 (g) in water according to: CO 2 (g)  CO 2 (aq) At equilibrium we have: Once in solution, CO 2 (aq) reacts with water to form carbonic acid: CO 2 (aq) + H 2 O(l)  H 2 CO 3 0

17 THE CO 2 -H 2 O SYSTEM - II In practice, CO 2 (aq) and H 2 CO 3 0 are combined and this combination is denoted as H 2 CO 3 *. It’s formation is dictated by the reaction: CO 2 (g) + H 2 O(l)  H 2 CO 3 * For which the equilibrium constant at 25°C is: Most of the dissolved CO 2 is actually present as CO 2 (aq); only a small amount is actually present as true carbonic acid H 2 CO 3 0.

18 THE CO 2 -H 2 O SYSTEM - III Carbonic acid (H 2 CO 3 *) is a weak acid that dissociates according to: H 2 CO 3 *  HCO H + For which the dissociation constant at 25°C and 1 bar is: Bicarbonate then dissociates according to: HCO 3 -  CO H +

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

20 SPECIATION IN OPEN CO 2 -H 2 O SYSTEMS - I In an open system, the system is in contact with its surroundings and components such as CO 2 can migrate in and out of the system. Therefore, the total carbonate concentration will not be constant. Let us consider a natural water open to the atmosphere, for which p CO 2 = atm. We can calculate the concentration of H 2 CO 3 * directly from K CO 2 : Note that M H 2 CO 3 * is independent of pH!

21 SPECIATION IN OPEN CO 2 - H 2 O SYSTEMS - II The concentration of HCO 3 - as a function of pH is next calculated from K 1 : but we have already calculated M H 2 CO 3 * : so

22 SPECIATION IN OPEN CO 2 - H 2 O SYSTEMS - III The concentration of CO 3 2- as a function of pH is next calculated from K 2 : but we have already calculated M HCO 3 - so: and

23 SPECIATION IN OPEN CO 2 - H 2 O SYSTEMS - IV The total concentration of carbonate C T is obtained by summing:

24 Plot of log concentrations of inorganic carbon species H + and OH -, for open-system conditions with a fixed p CO 2 = atm.

25 Plot of log concentrations of inorganic carbon species H + and OH -, for open-system conditions with a fixed p CO 2 = atm.

26 Calcite Solubility? CaCO 3 -> Ca 2+ + CO 3 2- Log K=8.48 Ca 2+ in Ocean = m


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