Chemistry 201 Lecture 16 Polyprotic acids and bases Applications NC State University Titration curve summary [OH-]/[HA]0 use rxn table goes 100% calc pH or pOH use rxn table goes 100% buffer : H-H base : pOH Strong Acid Weak Acid Strong Base Weak Base use rxn table goes 100% buffer : H-H acid : pH Examples: Weak acid and weak base For a reaction of a weak acid and a weak base we need to calculate the equilibrium constant from the known Ka’s. We take the example of ammonium acetate. We see that the overall reaction is composed of two acid-base equilibria for acetate for ammonia Examples: Weak acid and weak base Therefore, the overall equilibrium constant for the reaction is Now, that we can see how to calculate the Equilibrium constant, we can solve any acid-base Reaction problem using the standard methods That we have used. Step. 1. determine dilutions Step. 2. set up the reaction table Step. 3. solve for the unknown and then calculate pH Examples: Weak acid and weak base Therefore, the overall equilibrium constant for the reaction is In this case we can look up the pKas pKa for acetic acid/acetate is 4.74 pKa for ammonium ammonia is 9.25 pK = pKa1 – pKa2 = 4.74 – 9.25 = -4.51 K = 10-pK = 104.51 = 3.23 x 104 Examples: Weak acid and weak base 200.0 mL of 0.30 M NaH2PO4 was added to 200.0 mL of 0.30 M NaHS. What are the concentrations of all species at equilibrium? Examples: Weak acid and weak base 200.0 mL of 0.30 M NaH2PO4 was added to 200.0 mL of 0.30 M NaHS. What are the concentrations of all species at equilibrium? Solution: Look up the Ka for each reaction involved in this acid-base equilibrium. Examples: Weak acid and weak base Species Initial Difference Final 0.15 -x 0.15-x 0.15 -x 0.15-x 0.0 x x 0.0 x x Examples: Weak acid and weak base Species Initial Difference Final 0.15 -x 0.15-x 0.15 -x 0.15-x 0.0 x x 0.0 x x Shortcut: realize that you can take the square of both sides Strong Acid Strong Base Weak Acid Weak Base if conjugates use H-H calc pH if not conjugates calc. K calc concs. Polyprotic acids: Phosphoric acid Note that each K value is over 1,000 times smaller than the preceding value. This is the case for most polyprotic acids, and it is why polyprotic acids can be treated in a relatively simple manner. The large decrease in K means that a negligible amount of each acid reacts in each step, so the concentration of any species is the concentration determined in the first step in which it is produced. Polyprotic acids: Phosphoric acid The point is that we can calculate the pH for each equilibrium separately. Determine [H3O1+], [H2PO41-], and [H3PO4] from the first acid dissociation. These concentrations are not affected by subsequent steps because the subsequent equilibrium constants are so small. Set up the reaction table for the second reaction using the [H3O1+] and [H2PO41-] determined in Step 1 as the initial conditions. Solve the equilibrium problem for [HPO42-], which will be shown to equal K2. Set up the reaction table for the third reaction using [H3O1+] and [HPO42-] determined in Steps 1 and 2 as initial concentrations. Solve the equilibrium problem for [PO43-] Polyprotic bases: Sulfide We can treat polyprotic bases in the same way as acids. The equilibrium constants are separated by orders of magnitude. They can be solved sequentially. Figure 7.7 Figure 7.7 Amphiprotic acid Amphiprotic Acids Amphiprotic acids have two equilibria. In theory this can require solution of two variables. This is possible because each equilibrium constant gives a constraint (two equations and two unknowns). However, there is a simpler solution. Mass balance requires Amphiprotic Acids The trick to solving these equations for the pH is to write The equation in terms of HA-. Amphiprotic Acids Finally we observe that the maximum value of HA- will be formed when which gives the following simple result and Buffers formed from polyprotic acids The buffer region of a polyprotic acid is within 1 pH unit of each pKa of the acid. The example on the next page shows a diprotic acid with two different buffer regions. There are also two equivalence points. Figure 7.4 Carbonic acid: major determinant of ocean chemistry Carbonic acid has two acidity constants. When CO2 is taken up by water it reacts to form hydrogen carbonate. HCO3- is the major form of this acid. The balance of carbonate and hydrogen carbonate is very important for life in the ocean. Coral reefs are highly colored collections of CaCO3 Diatoms are microorganisms that fix CO2 and produce O2 Amphiprotic carbonic acid The pKa’s of carbonic acid are pKa1 = 6.35 and pKa2 = 10.35. We see an interesting situation where: 1. Neither pKa is in the buffer range for the current ocean pH (pH ~ 8.1) 2. The amphiprotic pH of carbonic acid is the average of the two values, which is 8.35. Once we understand these two facts we can see the following. The ocean despite its large size is not particularly strongly buffered. Therefore, as CO2 dissolves it pulls the equilibrium of the CO32towards HCO3-. Thinking about ocean acidification The two equilibria are: The current ocean pH = 8.35 is outside the buffer range for both. However, the addition of CO2 will clearly shift the pH lower according to H-H. Thus, the equilibrium of the second reaction will shift accordingly. The net result is an increase in [HCO3-] at the expense of [CO32-]. Acid catalysis: General description Acid catalysis involves the acceleration of a reaction in rate by protonation. For example, in solvent S, the reaction rate may be proportional to the concentration of protonated solvent molecules [SH+]. The acid catalyst (AH) contributes to the rate acceleration by shifting the chemical equilibrium between solvent S and AH in favor of the SH+species. S + AH → SH+ + AFor example in an aqueous buffer solution the reaction rate for reactants R depends on the pH of the system but not on the concentrations of different acids. Zeolites are SiO2 structures that can have Al-OH doping to create acidic sites that can help to catalyze reactions. Acid catalysis: zeolites Goals • Calculate pH change when acid or base is added to a buffer • Find solution composition at a given pH • Review methods for acid-base problems • Consider implications for polyprotic acids and bases
© Copyright 2024 Paperzz