Download presentation
Presentation is loading. Please wait.
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 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+ + Al 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
After 12.5 mmoles albite react (50 mmoles OH-):
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 = ?? 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) = = 6.95
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. In slide 8 we saw that, in the pH range of most natural waters, bicarbonate was the predominant species in the CO2-H2O system. In this slide, we see that the predominant species in the H2S-H2O system over the pH range of most natural waters is H2S0 (pH < 7.0) or HS- (pH > 7.0). This diagram can be constructed in exactly the same way as outlined for the previous diagram. Note that, as expected, the positions of the lines representing the concentrations of H+ and OH- have not changed.
12
In most natural waters, bicarbonate is the dominant carbonate species!
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. Although Bjerrum plots can be constructed rigorously by solving the combined mass-action and mass-balance expressions in the system for the concentrations of each of the species, there is a faster, approximate route to the construction of these diagrams. Once the total carbonate concentration (CT) is chosen and the pK values are known, the first step is to plot points with pH coordinates equal to the pK values, and concentration coordinates equal to log CT At pH = pK, the concentrations of two species are equal, and therefore equal to CT/2, the log of which is log CT For example, at pH = pK1 = 6.35, the concentrations of H2CO3* and HCO3- are equal to one another and to CT/2. Likewise, at pH = pK2 = 10.33, the concentrations of HCO3- and CO32- are equal to one another and to CT/2. The points where species concentrations are equal are called cross-over points. At pH < pK1 = 6.35, H2CO3* accounts for more than 99% of CT, so the concentration of H2CO3* plots as a horizontal line with a Y-intercept of log CT. As pH nears pK1, the line must bend down to intersect the HCO3- line at the first cross-over point. The HCO3- line extends from the first cross-over point towards lower pH with a slope of +1. At pK1 < pH < pK2, HCO3- accounts for the bulk of CT, so its concentration now plots as a horizontal line. In this pH range, the H2CO3* line descends away from the cross-over point towards higher pH with a slope of -1. As pH approaches pK2, the HCO3- line drops down to the second cross-over point. At pH > pK2, CO32- is the predominant species, so its concentration now plots as a horizontal line at log CT, and the HCO3- line descends from the second cross-over point towards higher pH with a slope of -1. In the range pH > pK2, the H2CO3* line now descends towards higher pH with a slope of -2. As the CO32- line passes through the second cross-over point towards lower pH into the region where pK1 < pH < pK2, it descends with a slope of +1. When this same line crosses under the first cross-over point into the region where pH < pK1, its slope changes to +2. In most natural waters, bicarbonate is the dominant carbonate species!
13
Carbonate System Titration
From low pH to high pH
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.