Strong Acids Bases Strong Acids more readily release

Strong Acids/ Bases • Strong Acids more readily release H+ into water, they more fully dissociate – H 2 SO 4 2 H+ + SO 42 - • Strong Bases more readily release OHinto water, they more fully dissociate – Na. OH Na+ + OH- Strength DOES NOT EQUAL Concentration!

Acid-base Dissociation • For any acid, describe it’s reaction in water: – Hx. A + H 2 O x H+ + A- + H 2 O – 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! • p. K = -log K as before, lower p. K=stronger acid/base!

p. Kx? • Why were there more than one p. K for those acids and bases? ? • H 3 PO 4 H+ + H 2 PO 4 p. K 1 • H 2 PO 4 - H+ + HPO 42 p. K 2 • HPO 41 - H+ + PO 43 p. K 3

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

Dissociation of H 2 O • H 2 O H+ + OH • Keq = [H+][OH-] • log Keq = -14 = log Kw • p. H = - log [H+] • p. OH = - log [OH-] • p. K = p. OH + p. H = 14 Definition of p. H • If p. H =3, p. OH = 11 [H+]=10 -3, [OH-]=10 -11

p. H • Commonly represented as a range between 0 and 14, and most natural waters are between p. H 4 and 9 • Remember that p. H = - log [H+] – Can p. H be negative? – Of course! p. H -3 [H+]=103 = 1000 molal? – But what’s g. H+? ? Turns out to be quite small 0. 002 or so…

BUFFERING • When the p. H 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…

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

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 3+ + 3 Si. O 2(aq) – What happens to the p. H of a solution containing 100 m. M HCO 3 - which starts at p. H 5? ? – p. K 1 for H 2 CO 3 = 6. 35

• Think of albite dissolution as titrating OH- into solution – dissolve 0. 05 mol albite = 0. 2 mol OH • 0. 2 mol OH- p. OH = 0. 7, p. H = 13. 3 ? ? • What about the buffer? ? – Write the p. H changes via the Henderson-Hasselbach equation • 0. 1 mol H 2 CO 3(aq), as the p. H increases, some of this starts turning into HCO 3 • After 12. 5 mmoles albite react (50 mmoles OH-): – p. H=6. 35+log (HCO 3 -/H 2 CO 3) = 6. 35+log(50/50) • After 20 mmoles albite react (80 mmoles OH-): – p. H=6. 35+log(80/20) = 6. 35 + 0. 6 = 6. 95

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

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

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

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 p. H = p. K 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 p. H = p. K 2, the activities of bicarbonate and carbonate ion are equal.

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

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 30

THE CO 2 -H 2 O SYSTEM - II In practice, CO 2(aq) and H 2 CO 30 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 30.

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 3 - + H+ For which the dissociation constant at 25°C and 1 bar is: Bicarbonate then dissociates according to: HCO 3 - CO 32 - + H+

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

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 = 10 -3. 5 atm. We can calculate the concentration of H 2 CO 3* directly from KCO 2: Note that M H 2 CO 3* is independent of p. H!

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

SPECIATION IN OPEN CO 2 H 2 O SYSTEMS - III • The concentration of CO 32 - as a function of p. H is next calculated from K 2: but we have already calculated M HCO 3 - so: and