Geology 560, Prof. Thomas Johnson, Fall 2009 Class Notes: Unit III Part 2. Weak Acids Speciation Reading: White 6.2; Walther Ch. 6 Sections on Acid/Base equilibria, buffers, and carbonate solubility We start with a simple question: How (quantitatively) does the solubility of calcite vary as a function of ph, provided all other things are not varied? As you saw in our look at calcite stability calculations, carbonate activity is linked to bicarbonate activity by: HCO 3 H + 2 + CO 3 (1) This reaction is a deprotonation reaction (loss of H + = loss of a proton, essentially). The reverse reaction is a protonation reaction. You can write this in log form: logk a = loga H+ + loga CO32 loga HCO3 logk a = ph + loga CO32 loga HCO3 logk a + ph = log(a CO32 /a HCO3 ) In english, this says, The relative abundances of carbonate and bicarbonate are controlled by the equilibrium constant and ph. As ph increases, carbonate increases relative to bicarbonate as a linear function of ph. Looking at this, we see that if we decrease ph by one ph unit, we decrease the activity of carbonate, relative to bicarbonate, by a factor of ten (one log unit). AND because carbonate, and not bicarbonate, is needed to stabilize calcite: CaCO 3(s) Ca 2+ + CO 3 2 Calcite is greatly destabilized by a decrease in ph. Digression: You may ask, Why is only carbonate considered and not bicarbonate? Why can t a bicarbonate attach to a calcite crystal and help it grow? The answer is that it can, but before it can truly become part of the calcite crystal and have more crystal grow on top of it, it needs to shed the H+ to become carbonate. The equilibrium would then look like: Ca 2+ + HCO 3 CaCO 3(s) +H + But this reaction is really just the sum of two reactions: Ca 2+ + CO 3 2 CaCO 3(s) HCO 3 H + + CO 3 2 But these two equilibria are what we were studying already. If you look into this you see that there is no need to consider the reaction where calcite grows via addition of bicarbonate. Just considering just one of the equilibria captures them all. Proof of this: If you did use Ca 2+ + HCO 3 CaCO 3(s) +H + to calculate the equilibrium, you would still have the same ph dependence because H + is on the right side. The Main Point: Protonation/deprotonation of an anion greatly affects the stability of any solids formed by that anion. More generally: Speciation of a component involved in an equilibrium greatly influences the equilibrium. So now we will explore protonation/deprotonation reactions. 1. Definitions: a. Acid: A substance that, when added to water, results in release of H + into the water. b. Strong acid: Readily releases H + c. Weak acid: Releases H + less readily; a large fraction of the substance may be protonated; speciation depends on concentration of H + in the solution d. pk = logk eq (same idea as ph) 2. Most (all?) anions can be paired up with H + to form an acid a. Elements to the right side of the periodic table form anions b. Halides all can be made into acids: HF, HCl, HBr, etc. (HCl and HBr are strong acids).
c. Recall the elements that form oxyanions in water. We can add H + to make neutral compounds: i. e.g., HNO 3, H 2 SO 4, H 3 AsO 4, many more examples ii. These compounds release H + ions when they dissolve into water d. Then think about the oxygens in those oxyanions. Some will have hydrogen ions attached to them just like the other oxygens in the surrounding water. But to some extent the bonding tendencies of those oxygens are satisfied by the central atom, and there may be very little ability to attract H +. i. NO 3, SO 4 2, SeO 4 2, do not attract H + ions strongly. e.g., put HNO 3 in water and it loses the H + very quickly. This leaves the 3 oxygens with no attached protons at all. Thus, we call it a strong acid (same for H 2 SO 4, H 2 SeO 4 ) ii. H 3 PO 4, H 2 CO 3, H 3 AsO 4, H 2 CrO 4 and others (including organic acids) are weak acids e. H 2 O can be thought of as a very weak acid also see below. 3. Let s start with a simple one: Acetic acid, CH 3 COOH, as an example of a weak organic acid. Digression: Many other organic acids have carboxyl groups that behave in similar fashion. a. If we dissolve one acetic acid molecule in pure water, how likely is it that a proton (H + ) will escape from it? (only moderately likely it s a weak acid) b. What is the name for the fragment left after the proton leaves the acetic acid? i. Acetate; CH 3 COO an anion. Note that sulfate nitrate, perchlorate, carbonate, borate, chromate, etc. have acid forms (sulfuric, nitric, perchloric, carbonic, boric, chromic acids) a. If we put one acetate ion into a solution with a high H + concentration, the likelihood that a proton will replace the missing one (i.e., the acetate becomes protonated) is high. b. So consider a population of 1000 acetate ions in water. At any given instant, some will be protonated and some won t. Look again, say 10 4 seconds later, and the number will be similar, but the H + ions will be in different places. c. Rapid, random exchanges of highly mobile H + occur between H 3 O + ions and acetate ions. Equilibrium occurs quickly. CH 3 COOH CH 3 COO + H + d. The activity of H + controls the speciation (acetic acid vs. acetate). e. Some questions we might want to answer: i. At what value of H + activity is acetate almost completely (e.g., > 90%) protonated? ii. At what value of H + activity is acetate 50% protonated? iii. What is H + activity if we dissolve acetic acid in water (1 mol/l)? iv. What is the speciation of acetic acid as a function of H + activity? f. Quantitative calculation: We have two equations in two unknowns: i. Mass action expression K a = a CH3COOH 1 x a CH3COO x a H+ ii. Mass balance: The two species concentrations must add up to the total of the component present: [CH 3 COOH] + [CH 3 COO ] = C total iii. These two equations can be solved in several ways. 1. One example approach: Solve the mass action equation for the concentration of one species, as a function of ph, the concentration of the other species, and the activity coefficients. Substitute this expression for the species concentration in the mass balance equation. Then solve this equation one unknown for the species concentration. g. More importantly, you should develop an intuitive sense for protonation as a function of ph: i. We can rearrange the mass action experession in log form to get: log(a CH3COO /a CH3COOH )= logk loga H+ = ph pk a ii. When pk a = ph, activities of acetate and acetic acid are equal. (the equivalence point or crossover point in the plot below). iii. Consider the speciation when H + activity is 10 times greater/less than this value (i.e. ph is shifted by one ph unit). The equilbrium ratio will also shift by a factor of 10, so the two species activities will be in a 10:1 or 1:10 ratio. If this is not obvious, work with the equation until it is. This approach can give you quick estimates of speciation for various ph values.
1.2 Fractional concentration 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ph Quantitatively: CH 3 COOH <==> CH 3 COO + H + a. G 0 R = 88.3+ 0.00 (94.8) = 6.5 kcal/mole 1. IMPORTANT: G 0 f for H + is 0.00 (by convention; this is an arbitrary zero point for the energy scale for aqueous ions) b. K = exp( G 0 R/RT) (where exp(x) means e x ) 1. At 25 C, RT = 0.5924 kcal/mol 2. Also, note that lnk = 2.303logK (because 10 = e 2.303 ) c. So. LogK = G 0 R /1.364 kcal/mole (this applies for all reactions at std. T + P) d. For this reaction, logk a = 4.77; pk a = 4.77 e. So the speciation is revealed in this rearranged mass action equation: log(a CH3COO /a CH3COOH ) = ph pk a = ph 4.77 f. Qualitatively, this shows us the speciation: 1. the equivalence point is at about ph = 4.77 a. This is an approximation, based on an assumption that the two activity coefficients are equal they actually differ a little 2. at ph = 5.77, about 90% is acetate, 10% is acetic acid (protonated) 3. at ph = 3.77, about 90% is acetic acid, 10% acetate g. Quantitatively: Example for 1 mm total (acetate + acetic acid) iv. Mass action: K a = γ CH3COOH [CH 3 COOH] 1 x γ CH3COO [CH 3 COO ] x a H+ v. Mass Balance: [CH 3 COOH] + [CH 3 COO ] = 1 mm vi. Solve mass action eqn. for [CH 3 COOH]: [CH 3 COOH] = K a 1 x γ CH3COOH 1 x γ CH3COO [CH 3 COO ] x a H+ Note that you need ionic strength to calculate gammas vii. Substitute into mass balance equation and do simple algebra to solve viii. See graph above for result when activity coefficients are assumed equal. 4. H 2 O itself can lose or gain protons in the chaos of the microscopic world (this is basic chemistry; work with this until you know the relationship between H + activity and OH activity without thinking about it much). a. Picture H + hopping from one water molecule and sticking to another temporarily b. 2 H 2 O OH + H 3 O + i. We usually write H +, realizing that it really means H 3 O + c. Mass action expression: a OH * a H+ = 10 14 (assume activity of water is 1) d. In pure water, what is a H+? What is a OH? e. In pure water, what are their concentrations, roughly (mol/l)? f. In a solution with a high [H + ], is [OH ] high or low? Quantitatively, how does a H+ relate to a OH? g. Review: Why do we care about H + ions so much anyway? i. Determines speciation of many elements / compounds 1. many elements form weak acids protonation/deprotonation is critical 2. Many organic molecules are weak acids or weak bases
3. Many metals form hydroxyl complexes (see below) and are thus bases 4. Minerals are made partly of components that become weak acids or weak bases upon dissolution 5. H 3 PO 4 is also a weak acid, but it has three protons it can lose. a. Write the three deprotonation reactions i. Terminology: Phosphate = PO 4 3 ; Biphosphate = HPO 4 2, triphosphate= H 2 PO 4, Phosphoric acid = H 3 PO 4 ii. Actually, that terminology is somewhat archaic (though still used) hydrophosphate = HPO 4 2, dihydrophosphate= H 2 PO 4 b. Write the equilibrium constants as pk A values: 2.1, 7.2 and 12.4 c. Find the three equivalence ( crossover ) points...ph values where 2 species have equal activities 1.2 Fractional concentration 1 0.8 0.6 0.4 0.2 0 0.2 0 5 10 ph 6. ph buffers. Weak acids and bases can act to resist changes in ph a. ph CANNOT CHANGE without a change in proportions of the acid or base species. This because the ph is tied to the equilibrium and thus the activity ratios. b. Therefore, when you add acid to a solution, it may be buffered, i.e., the solution may resist strong change in ph as the various species are protonated/deprotonated and the ratios change c. Example 1: Unbuffered water. Consider one liter of water with a small amount of base added so ph = 9. It takes only 10 5 moles of H + to change the ph to about 5. d. Example 2: One liter solution at ph = 9, containing about 1 molal HPO 4 2. One must convert almost all of the HPO 4 2 to H 2 PO 4 2 in order to get from ph = 9 to ph = 5. i. Thus, it takes ABOUT 1 MOLE of H + to do this MUCH more than the unbuffered water e. Buffers work best when the ph is close to a crossover point. For example, in Example 2 and the figure above, one can see that HPO 4 2 would not hold ph at 10.0 very well, but it would provide a strong buffer at about ph = 7. 7. Solubility of silica a. Write reaction: SiO 2(s) = SiO 2(aq) b. More helpful: 2H 2 O + SiO 2(s) = H 4 SiO 4(aq) i. Note this reaction is NOT phdependent c. Silicic acid (H 4 SiO 4 ) is a weak acid: pk 1 = 9.7, pk 2 = 11.7 d. Speciation of silicic acid as a function of ph: Equivalence points at ph = 9.7, 11.7 i. Speciation far below ph = 9.7 is nearly 100% silicic acid ii. H 3 SiO 4 (aq) begins to appear around ph = 9 e. Semiquantitatively, the solubility of silica (i.e., the sum of all silica species): i. We expect the activity of H4SiO4 to remain constant in equilibrium with quartz ii. At ph much less than 9: Boring essentially all silica present as H 4 SiO 4 no phdependence iii. At ph = 9.7, we have equal activities of H3SiO4(aq) and H4SiO4(aq) 1. So then how much total silica is in solution? Double the low ph amount iv. At ph = 10.7, we have H3SiO4(aq) : H4SiO4(aq) activity ratio of 10:1 (concentrations are similar) v. So then how much silica is in solution? About 10x the low ph amount f. We ADD THE CONCENTRATIONS OF ALL THE DISSOLVED SPECIES to calculate the solubility of quartz.
g. NOTE: In the real world, quartz may be VERY slow to precipitate, in which case, the dissolved silica concentration may become great enough to allow precipitation of amorphous silica, a noncrystalline (and therefore higher energy) form of the solid. REAL WORLD APPLICATION: We expect that the solubility of amorphous silica defines an upper limit to the amount of silica that can be dissolved in natural waters. In real ground water systems, the dissolved silica concentrations rarely exceed 110 mg/l in circumneutral ph waters. In unusually high ph waters, dissolved silica can be greater. 8. H + as a component. If you are creating an aqueous geochemical model and need to choose components to build that model, usually you should pick H + as a system component. That way you can control ph by increasing or decreasing H +. 9. Alkalinity. The ability of a solution to neutralize acid is called its alkalinity. Because, in many cases, bicarbonate is the dominant acidneutralizing species, we often focus on its contribution to alkalinity. a. Simplified example: Imagine a solution at ph 8.0 containing 1 molal HCO 3 b. It takes 1 mole of acid per kg of water to neutralize this, so we say the alkalinity is 1 molal c. Alkalinity titration; Practical procedure: We measure ph constantly as we add acid (must be a strong acid like HCl) and keep track of the amount of acid added. i. At first, the ph changes rapidly because at ph 8.0 the solution is not strongly buffered ii. But as ph approaches the HCO 3 /H 2 CO 3 equivalence point, the ph change slows 1. i.e., the solution is buffered, and the added acid goes mostly into converting HCO 3 to H 2 CO 3 and not much into increasing [H + ] iii. Eventually, essentially all of the HCO 3 is converted to H 2 CO 3 iv. This occurs at ph about = 4.5 v. When you reach ph = 4.5, you stop and record moles acid added per kg water; this is your alkalinity 2 d. Higher ph cases: At ph > about 8.3, a significant fraction of the DIC is present as CO 3 i. This too must be neutralized ii. It takes 2 H + ions to convert it to H 2 CO 3 iii. No change in procedure this is taken into account iv. The amount of CO 2 3, if it is large enough, can be determine from the amount of acid needed to get to ph about = 8.0. e. Other protonation/deprotonation reactions can add to alkalinity i. H 2 PO 4 / HPO 2 3 4 /PO 4 ii. Organic acids can be abundant in some water (e.g., soils, swamps) iii. B(OH) 3 /B(OH) 4 equivalence point is about ph = 8.8 f. If these other reactions are minor or can be corrected for, alkalinity titration gives a good measure of HCO 3 (and perhaps CO 2 3 ) concentration.