ph scale [H 3 O + ] varies over a very wide range. The ph scale simplifies expressing acidity/basicity. ph = log[h 3 O + ] = log 1 + [H3O ] (really an approximation; just as we did for K expressions we will use molar concentrations instead of activities. The true definition uses the activity of H 3 O +, not molarity.) ph frequently ranges between 0 and 14, but can be negative (a very acidic solution) or greater than 14 (a very basic solution). Regarding sig figs and logs: the number of sig figs appears after the decimal point in the log. e.g., a solution that has [H 3 O + ] = 0.013 M has ph = log(0.013) = 1.89 p function negative log 10 poh = log[oh ] TOT. So, using the properties of logs, [H 3 O + ][OH ] = K W. Strong enough for a man; ph balanced for a woman log([h 3 O + ][OH ]) = log K W log[h 3 O + ] + log[oh ] = log K W log[h 3 O + ] log[oh ] = log K W ph + poh = 14.00 (at 25 C) at 25 C, If ph = 7.00, the solution is NEUTRAL If ph < 7.00, the solution is ACIDIC If ph > 7.00, the solution is BASIC Strength of Acids and Bases For any acid HA in water, an equilibrium is established: + [H3O ][A ] HA(aq) + H 2 O(l) º H 3 O + (aq) + A (aq) K a = [HA] acid base conj. acid conj. base By convention, the acids considered strong have K a > 1 while weak acids have K a < 1. Recall that a strong acid is 100% dissociated in solution (Chapter 4). For very strong acids, the equilibrium position lies far enough to the right that the reaction is essentially complete and the equation is usually written with a single arrow (). e.g., HCl (a selected strong acid in Table 4.3), K a 1 H 10 6 HCl(aq) + H 2 O(l) º H 3 O + (aq) + Cl (aq) + [H3O ][Cl ] K a = 1 H 10 6. Ratio of products to reactants is very large. [HCl] Therefore, we usually write HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) with a single arrow. Similarly for bases, an equilibrium is established: B(aq) + H 2 O(l) º BH + (aq) + OH + [BH ][OH ] (aq) K b = base acid conj. acid conj. base [B] same arguments as above for acids (strong vs. weak, single vs. equilibrium arrows) Big Question #1 11 3150:153 002/801
Molecular Properties and Acid Strength (Big Question #3 Alert!) Recall (again) that in Chapter 4 you learned several selected strong/weak acids/bases. Now we can consider these general trends: Strong Acids HCl, HBr, HI Oxoacids where the number of O atoms exceeds the number of ionizable H atoms by 2 or more, e.g., o HNO 3 o H 2 SO 4 o HClO 4 Strong Bases Soluble metal oxides & hydroxides (consider solubility rules, Chapter 4), e.g., o Na 2 O o Ba(OH) 2 o CsOH o SrO Explanations for a couple of the above Nonmetal Hydrides Nonmetal hydride acid strength increases with increasing EN across a period and decreasing bond strength down a group. Weak Acids HF Oxoacids where the number of O atoms exceeds the number of ionizable H atoms by 1 or fewer, e.g., o HNO 2 o H 2 SO 3 o HOCl Acids with H not bonded to O or halogen (group 7A), e.g., o H 2 S o HCN Carboxylic acids (R COOH), e.g., [15.4] o CH 3 COOH Weak Bases NH 3 Amines (R 3 N, R 2 NH, RNH 2 ), e.g., [15.4] o (CH 3 ) 3 N o CH 3 CH 2 NH 2 Oxoacids Same number of O atoms, different central atom o Acid strength increases with increasing electronegativity of central atom (O H bond polarized) HOI: K a = 2.3 H 10 11 HOBr: K a = 2.3 H 10 9 HOCl: K a = 2.9 H 10 8 Different number of O atoms, same central atom o Acid strength increases with increasing number of O atoms (more electronegative O atoms pull electron density to themselves, polarizing O H bond) HOCl: K a = 2.9 H 10 8 HOClO: K a = 1.1 H 10 2 (aka HClO 2 ) HOClO 2 : K a = ~ 5.0 H 10 2 HOClO 3 : K a = 1 H 10 8 Presence of Electron-Withdrawing Groups/Atoms Electron-withdrawing groups/atoms frequently increase acidity for the same reason as additional O atoms increase acidity of oxoacids with the same central atom increased polarization of the O H bond. CH 3 COOH: K a = 1.8 H 10 5 CCl 3 COOH: K a = 2.2 H 10 1 Big Question #1 12 3150:153 002/801
Leveling Effect Acids stronger than H 3 O + react completely with water to produce a stoichiometric amount of H 3 O +. Bases stronger than OH react completely with water to produce a stoichiometric amount of OH. It s not possible to compare the strengths of strong acids in aqueous solution. A non-aqueous solution must be used. Equilibria of Acids and Bases in Water!! We will discuss several (four to be exact) different cases of equilibria of acids and bases. 1. Strong, monoprotic acids in water (1) HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) K a big (2) 2H 2 O(l) º H 3 O + (aq) + OH (aq) K W small [H 3 O + ] TOT = [H 3 O + ] HA + [H 3 O + ] W [H 3 O + ] W small compared to [H 3 O + ] HA. Can we neglect [H 3 O + ] W? Perhaps If [H 3 O + ] W 0.05[H 3 O + ] TOT, then we can neglect [H 3 O + ] W. Strong Acid Rules Formality (F) [Formal/analytical concentration] 1. If F acid > 4.5 H 10 7 M, then concentration of solution as prepared a. water dissociation unimportant (number of formula masses of initially b. [H 3 O + ] TOT F acid undissociated solute per liter of solution, 2. If F acid < 4.5 H 10 7 M, then regardless of subsequent dissociation) a. water dissociation important b. solve using: [H 3 O + ] 2 TOT F acid [H 3 O + ] TOT K W = 0 Similar analysis applies for strong monobasic bases in water (e.g., NaOH): Strong Base Rules 1. If F base > 4.5 H 10 7 M, then a. water dissociation unimportant b. [OH ] TOT F base 2. If F base < 4.5 H 10 7 M, then a. water dissociation important b. solve using: [OH ] 2 TOT F base [OH ] TOT K W = 0 2. Weak, monoprotic acids in water (1) HA(aq) + H 2 O(l) º H 3 O + (aq) + A (aq) K a (2) 2H 2 O(l) º H 3 O + (aq) + OH (aq) K W [H 3 O + ] TOT = [H 3 O + ] HA + [H 3 O + ] W Both equations involve equilibria that favor reactants. Will [H 3 O + ] W be small compared to [H 3 O + ] HA? i.e., can we neglect [H 3 O + ] W? Just as for strong acids, it depends on the relative amounts produced by the two reactions Big Question #1 13 3150:153 002/801
What are the four equations and four unknowns in a monoprotic acid solution equilibrium problem? Four Unknowns: [HA], [A ], [H 3 O + ], [OH ] Four Equations: (1) HA(aq) + H 2 O(l) º H 3 O + (aq) + A (aq) K a = + [H3O ][A ] [HA] (2) 2H 2 O(l) º H 3 O + (aq) + OH (aq) K W = [H 3 O + ][OH ] (3) Charge balance: total positive charge equals total negative charge [H 3 O + ] = [OH ] + [A ] (4) Mass balance: initial HA (F acid ) equals total HA and A at equilibrium F acid = [HA] + [A ] Being a physical system, there MUST be a unique solution even though the four equations are not all linear equations. Solving this system of equations is a mathematical nightmare, but a few chemically reasonable approximations simplify the math significantly and do not cost us accuracy in the sig figs that we can measure. A diprotic acid problem would have five equations and five unknowns ([H 2 A], [HA ], [A 2 ], [H 3 O + ], [OH ]), and is equally easy to solve using reasonable approximations. There is almost never a reason to solve one of these systems of equations rigorously.