Chemistry 121 Lectures 20 & 21: Brønstead-Lowry Acid/Base Theory Revisited; Acid & Base Strength - Acids & Bases in Aqueous Solution; Acid Dissociation Constants and the Autoionization of Water; ph or the Acidity/Basicity of Aqueous Solutions; Determining Acidity; Buffers; Biological Buffer Systems; Common Acid/Base Reactions; Titration; Acid/Base Characteristics of Salt Solutions Chapter 10 in McMurry, Ballantine, et. al. 7 th edition HW #10: 10.38, 10.40, 10.46, 10.52, 10.58, 10.66, 10.68, 10.74, 10.78, 10.84, 10.88, 10.90, 10.102, 10.110 Learning Objectives 1. Use Brønsted-Lowry theory to define acids and bases 2. Define the concept of conjugate acids and bases 3. Utilize conjugate acid-base theory to interpret the acid-base behavior of a given compound 4. Define strong and weak acids, strong and weak bases 5. Rationalize why a compound is an acid (strong or weak) or base (strong or weak) 6. Describe the behavior of polyprotic acids 7. Know which acid-base reactions go to completion (or which case must be considered on an individual basis) 8. Define the hydronium and hydroxide ions in terms of the dissociation of water 9. Identify the major properties of the ph scale 10. Make simple ph calculations 11. Qualitatively describe how indicators and ph meters function 12. Define buffer and buffer capacity 13. Develop the Henderson-Hasselbalch equation a. Define Ka and pka 14. Calculate the ph of a simple buffer system 1
15. Prepare a simple buffer system of given ph 16. Given a system of known ph, calculate the % ionization of an AH acid 17. Define titration, end point, and equivalence point 18. Describe the features of the curve for titrating a strong acid with a strong base 19. Describe the features of the curve for titrating a strong base with a strong acid 20. Calculate the amount of strong acid in an unknown sample by titrating against a base solution of known concentration 21. Calculate the amount of strong base in an unknown sample by titrating against an acid solution of known concentration 22. Appreciate the ability of some salts to influence ph 10.3: The BrØnsted-Lowry Definition of Acids and Bases An acid is a proton donor The above is a general definition, useful in all environments. As a practical matter, we call a compound an acid if it generates [at least some] H3O + in water. Examples o Sulfuric acid strong, 100 % H3O + in water o Acetic acid weak, 0.5 % H3O + in water A base is a proton acceptor A base must have a pair of non-bonded electrons in order to accept a proton Examples o NaOH (s) dissolves well in water to give essentially 100 % OH -. As such, lye is a strong base o NH3 (g) reacts with water to give 5 % OH -. As such, ammonia is a weak base 2
Acids react with bases to give [ionic] salts. If, and only if, the base is hydroxide you get water as well. The acid-base reaction when HCl gas is bubbled into water (water as reluctant base): The acid-base reaction when HCl gas is bubbled into a solution of sodium hydroxide (hydroxide as Oh yeah base) The acid-base reaction when HCl gas is mixed with NH3 gas (ammonia as shy base) Conjugate Acid-Base Pairs When HCl gas and NH3 gas were mixed together above, NH4Cl formed. However, ammonium chloride is mildly acidic and may be used to acidify the blood of individuals suffering metabolic alkalosis. Let s consider the reaction in equilibrium terms: HCl (g) + NH3 (g) NH4Cl (s) In the forward direction HCl acted as the acid, leaving behind Cl -. Since the Cl - had bonded with the H + in the first place, Cl - must be able to act as a base (it does have pairs of non-bonded electrons). Cl - is the conjugate base of HCl. In the forward direction NH3 acted as the base, becoming NH4 +. If NH3 was stable in the first place (all atoms do have a noble gas electronic configuration) NH4 + should be able to give up H + and revert back to NH3. NH4 + is the conjugate acid of NH3. 3
Examples In the acid-base reaction below, indicate the conjugate base of the acid and the conjugate acid of the base by labeling and drawing a line connecting them. H2SO4 + 2NaOH Na2SO4 + 2H2O In the acid-base reaction below, indicate the conjugate base of the acid and the conjugate acid of the base by labeling and drawing a line connecting them. H2O + NH3 (NH4)OH In the acid-base reaction below, indicate the conjugate base of the acid and the conjugate acid of the base by labeling and drawing a line connecting them. HCl + CH3CO2Na NaCl + CH3CO2H In the examples above, notice Sulfuric acid (H2SO4) is a polyprotic (many protons) acid it gives up 2 protons in order to form sulfate We may view these reactions as a competition between base and conjugate base for a proton. Ammonia is our standard weak base, hydroxide is our standard strong base (NH4)OH produces a weak electrolyte solution because the equilibrium favors ammonia and water CH3CO2Na is the conjugate base of the weak acid CH3CO2H (acetic acid) and as such is weakly basic, since weak acids don t like to give up H + all that much (or they would be strong acids). Thus, when a strong acid is added to a solution containing CH3CO2Na, it takes the H + thereby dampening the effect of the strong acid this is the basis by which buffers work (more below). 4
10.4: Strengths of Acids and Bases Strong Acids and Strong Bases By definition, a strong acid completely dissociates in water to give an equal concentration of H3O +. Our primary strong acids are H2SO4, HNO3, and all the hydrogen halides barring HF By definition, a strong base completely dissociates in water to give an equal concentration of OH -. Our primary strong bases are the hydroxides of group 1 and also Ba(OH)2, Sr(OH)2. Mg(OH)2 is poorly soluble and as such used as an antacid 1. Ca(OH)2 is somewhere in between with a saturated solution (lime water) containing 17.3 g/l @ 20 o C (with a MM = 74.09 g/mol this is equivalent to a 0.233 M solution) Weak Acids and Weak Bases By definition, a weak acid dissociates incompletely in water but does produce some H3O +. Our primary weak acids are HF, NaHSO4, CH3CO2H (or other carboxylic acids), H3PO4, and H2CO3 By definition, a weak base dissociates incompletely in water but does produce some OH -. Our primary weak bases are NH3 or the organic derivatives the amines, Mg(OH)2, and Ca(OH)2 1 Tums is much more pleasant to ingest than Drano 5
Insight into acid strength the stability of the conjugate base Let s look at the structure of sulfuric acids and compare it to the weak acid acetic acid and phosphoric acid, as well as ethanol (which is neither acidic or basic): H2SO4 HSO4 - + H + HSO4 - resonance: H3PO4 H2PO4 - + H + H2PO4 - resonance: CH3CO2H CH3CO2 - + H + CH3CO2 - resonance: For comparison, let s look at ethanol as an acid and compare to hydroxide below CH3CH2OH CH3CH2O - + H + CH3CH2O - resonance: Insight into base strength [whether conjugate or not] electronegativity and concentration of charge Question: Which of the following ions in each pair represents the strongest base? OH - vs F - F - vs. Cl - OH - vs. H: - 6
More Equations for Acid-Base Reactions The single most important reaction between acids and bases is their reaction together to form water (if the base is hydroxide) and a salt: Strong Acid + Strong Base Completion Warning! HCl + NaOH Strong Acids + Weak Bases Completion HNO3 + NH3 Weak Acids + Strong bases Completion CH3CO2H + LiOH Weak acids + Weak Bases Equilibrium CH3CO2H + CH3NH2 As a reference for the equilibrium between a weak acid and weak base, note that amino acids exist as zwitterions: Problem: Write the conjugate acid-base pairs for the above reactions 7
Table 10.1 Summarizes Acid & Base Strength in Terms of Conjugate Acid/Base Pairs Notice that weak acids give rise to weak conjugate bases and vice versa this is very important to our understanding of buffer systems, since buffers are composed of a weak acid/conjugate base pair which allows them to resist changes in ph when strong acids or bases are added 10.5: Acid Dissociation Constants We can gain quantitative insight by examining the acid dissociation constant Ka 8
10.7, 10.8: The Autoionization of Water and the ph Scale Water Reacting with Water Recall from our previous equilibrium discussion the autoionization of water produces the hydronium ion and the hydroxide ion, albeit at very low concentrations as indicated by the value for the equilibrium constant: H2O + H2O H3O + + OH - K = Since this is such a crummy reaction we can include the concentration of water with K (since its concentration is constant) and write a new constant, KW = K[H2O] 2, the value for which = 10-14 KW = 10-14 = [H3O + ][OH - ] The autoionization of water gives rise to the ph scale. Bear in mind The autoionization of water is a pretty crummy reaction; water is very stable, and the reverse reaction the mixing of a strong acid and strong base can be highly hazardous because it is so favorable Because of the crummy extent of reaction, the ph scale is a log scale; that is, low ph values mean a greater molar H3O + concentration and there is a 10x change in concentration between ph units o Basic solutions have [H3O + ] < 10-7 M or ph > 7 o Acidic solutions have [H3O + ] > 10-7 M or ph < 7 From the above you can see the autoionization of water is an equilibrium process; as such ph + poh = 14 always 9
The ph scale and ph Calculations Please indicate the ph of the following solutions 3.65 g HCl dissolved in 1 L water 3.65 mg HCl dissolved in 10 L water 4.0 g NaOH dissolved in 1L water 10.10: Laboratory determination of acidity Acidic and basic environments have the ability to change the ionization state of organic (carboxylic) acids and bases (amines) There are a number of organic compounds that change their ability to absorb light as a function of their ionization status. An example is phenolphthalein HO OH O O Phenolphthalein ph meters use the reaction between H + and a reactive metal such as Zn to calculate ph Zn (s) + 2H + (aq) Zn +2 (aq) + H2 (g) Since there is a voltage associated with such spontaneous reactions that is a function of [H + ] one can measure the voltage and relate it directly to ph 10
10.11: Buffers and the Blood Buffer System Question: What is a buffer? Answer: Question: How do buffers work? Answer: Question: What is the difference between neutralization and buffering? Answer: Question: What is buffer capacity? Answer: In order to understand the last question and buffers fully we must employ the dreaded (and not in the Jamaican fashion) Henderson-Hasselbalch equation The Henderson-Hasselbalch equation relates the relative concentrations of acid and conjugate base components to the strength of the acid component (pka value) and the resulting ph of the solution The Henderson-Hasselbalch equation is typically written in 1 of 2 forms Where log([a-]/[ah]) = ph - pka or [A-]/[AH] = 10 ph-pka [AH] = molar concentration of acid [A - ] = molar concentration of conjugate base of that acid pka is the log of the equilibrium constant for acid dissociation (Ka) The Henderson-Hasselbalch equation is derived on the last page refer to it if you dare! 11
Let s see how the Henderson-Hasselbalch equation allows us to determine the ph of a buffered system based on the relative concentration of the acid and base components log([a - ]/[AH]) = ph pka or [A - ph pka ]/[AH] = 10 If we know the pka of the acid component of the buffer and the ratio of acid to conjugate base, we can determine the ph of the buffer Example: What ratio of acetic acid (pka = 4.76) to sodium acetate would be required to make a buffer of ph 4.76? Example: What ratio of acetic acid (pka = 4.76) to sodium acetate would be required to make a buffer of ph 3.76? Notice a 10x shift in the ratio of acid to base causes a 1 ph unit shift in the buffer ph Example: What ratio of acetic acid (pka = 4.76) to sodium acetate would be required to make a buffer of ph 7.76? Answer: Question: Does the ratio of acetic acid to sodium acetate tell us anything about the strength of the buffer? Answer: Question: Would acetic acid be a good choice for a ph 7.76 buffer Answer: 12
Common conjugate acid-base pairs for buffer systems Acetic acid(pka = 4.76)/sodium acetate Dihydrogen phosphate(pka = 7.20)/hydrogen phosphate Carbonic acid(pka = 6.35)/bicarbonate buffer 10.12: Buffers in the Body Dihydrogen phosphate(pka = 7.20)/hydrogen phosphate (intracellular) Carbonic acid(pka = 6.35)/bicarbonate buffer (extracellular) Proteins, owing to the presence of weak acid (aspartic and glutamic acid) and weak base (lysine, arginine, histidine) amino acid side chains The Blood Buffering System While there are several systems at play to regulate body ph overall, including phosphate and protein, our primary means of regulating blood ph is via the carbonic acid/bicarbonate buffer 13
Problem: Blood ph =7.4 and little variation is tolerated, while carbonic acid pka = 6.35 Fortunately, the carbonic acid/bicarbonate buffer system is effective at buffering our blood since We produce acidic byproducts in about a 10:1 ratio to basic products One of the fastest known enzymes, carbonic anhydrase, greatly enhances the decomposition of H2CO3 into CO2 and H2O, so breathing can regulate/remove CO2 thus removing H2CO3 by Le Châtelier s principal H2CO3 (aq) Carbonic Anhydrase H2O (l) + CO2 (g) Question: Does rapid breathing cause an increase or decrease in blood ph? How about holding one s breath? 10.13: Acid and Base Equivalents Normality vs. Molarity 2.0 M HCl is 2.0 N HCl 2.0 M H2SO4 is 4.0 M H2SO4 10.15: Titration What do titration curves look like? Strong Acids Titrated with Strong Bases: 14
Strong Bases Titrated with Strong Acids: Question: What is the ph of a solution if a 50 ml sample requires 5.0 ml 0.100 M HCl to reach the equivalence point? Question: What mass of acetic acid is contained in 2.0 L of an unknown vinegar solution if 20 ml of 6 M NaOH is required to reach the equivalence point of a 50 ml sample? 10.16: Acidity and Basicity of Salt Solutions H2SO4 + NaOH NaHSO4 + H2O NaHSO4 + NaOH Na2SO4 + H2O Overall, H2SO4 + 2NaOH Na2SO4 + 2H2O H2SO4 is a very strong acid (since very weak conjugate base HSO4 - has average charge of -1/3 on each O). NaHSO4 has an acidic proton which may be donated (weak conjugate base SO4-2 has average charge of -1/2 on each O) NaHSO4 dissolved in water produces a mildly acidic solution (like vinegar would) Na2SO4 dissolved in water produces a mildly basic solution (like sodium acetate). Recall a strong base will drive the reaction with a weak acid towards completion and NaOH is our standard for strong bases; H2O is a very weak conjugate acid; it is as if the hydroxide caused the NaHSO4 to dissociate further than it would have in water, so when placed in water it attempts to reestablish equilibrium by taking a few protons from water 15
The same holds true for phosphoric acid, which will consume 3 hydroxide anions on a molar basis: Notice that PO4-3 is significantly basic (has average charge of -3/4 on each O) and will generate some hydroxide when placed in water; thus Na3PO4 is a weak base o Of course it will not generate an equivalent amount of hydroxide, since the charge on hydroxide is -1 We can summarize with table 10.1 Question: Na3PO4 produces a mildly basic solution and NH4Cl produces a mildly acidic solution. Please explain by showing the appropriate chemical equation Na3PO4 + H2O NH4Cl + H2O 16
Appendix 1: The Relationship between ph, Ionization, and Solubility Or The Revenge of the Henderson-Hasselbach Equation Where we look at the ph as fixed and see what happens to the ratio of ionized to unionized compound for compounds that contain acidic or basic structural motifs o [A - ph pka ]/[AH] = 10 For compounds that are largely non-polar ionization can have a tremendous influence on solubility o Recall that an ion-dipole interaction, such as that between Na + or Cl - in water, is highly favored energetically, so the ionized [salt] forms are often much more water soluble Question: What is the % ionization of acetic acid if placed in a buffered ph 5.8 environment? Answer: Question: Why does aspirin, pka = 3.49, crystallize out of solution in the stomach? Answer: 17
Appendix 2: pka s, ph, and the Henderson-Hasselbach Equation Any compound with hydrogen attached, regardless of charge status (i.e. AH or BH + ) may be treated as an acid: AH A - + H + And an equilibrium ratio established: Ka = [A - ][H + ]/[AH] (1) Reorganizing (1) gives Ka = [H + ]([A - ]/[AH]) Taking the log of both sides gives -logka = -log[h + ] + -log([a - ]/[AH]) Applying the definition of p gives pka = ph - log([a - ]/[AH]) log([a - ]/[AH]) = ph pka An easier to interpret form, obtained by exponentiating, is [A - ph pka ]/[AH] = 10 Major Results: When ph = pka, the acid is 50% ionized 10 0 =1, [A - ] = [AH] For each unit ph and pka differ, ionization changes by an order of magnitude; e.g., ph = 10 & pka = 9, [A - ] = 10[AH]; ph = 11 & pka = 9, [A - ] = 100[AH] 18