Aqueous Equilibria: Acids and Bases

Transcription

1 6//4 What is an? What is a? There are actually multiple definitions Aqueous quilibria: Acids and Bases Ch. 6 Arrhenius: Dealt with species in aqueous solutions. Most basic definition of acis. Acid: increases in water. Base: increases O in water BrønstedLowrey: AcidBase need not be in aqueous solutions. Acid and s are part of a related conjugate pair. Acid: Proton ( ) donor Base: Proton ( ) acceptor Lewis: No need for hydrogen in definitions Acid: lectron (e ) acceptor Base: lectron (e ) donor Ag :N 3 [ 3 N:Ag:N 3 What does strong as in strong s and s mean? ighly Concentrated? Dangerous? Strong just mean % dissociation That s all. O in the gas phase in the aqueous phase (g) (aq) K s large is a strong =.3 6 Weak just means much less than % dissociation. Most F molecules are actually NOT dissociated O F F in the gas phase F in the aqueous phase F(g) F(aq) K s are very small F is a weak F = Just like with s, strong s are those with% dissociation. Weak just means much less than % dissociation! For N 3, it itself is not dissociated, however, it takes an from water to create O O O O O O O O in the solid phase O O O O O N 3 (l) O(l) N 4 (aq) O (aq) N N 3 in the liquid phase O Most N 3 molecules are actually NOT converted to N 4 O N O in the aqueous phase N in the aqueous phase O(s) O(aq) K s large O is a strong N 3 (l) N 4 (aq) K s are very small N 3 is a weak N 3 =.8 5

2 6//4 Strong Acids (aq) Br(aq) (aq) NO 3 O 3 O 4 SO 4 ** Strong vs. Weak s f you know the strong s/s, assume the rest are weak % ionization Strong Bases (soluble metal hydroides) LiO O KO Ca(O) (kindaish) Sr(O) Ba(O) amples of weak s F C 3 COO (acetic) COO (formic) N 4 (ammonium) 3 PO 4 O O (small, highlycharged metal ions) Al 3 amples of weak s (nitrogencontaining s) N R R = anything N 3 (ammonia) ( nsoluble hydroides) <% ionization Auto ionization of water K w At any time in a sample of pure water, some water molecules, very few, will dissociate by themselves Only.8 6 % of O molecules within a container of water do this!! K w = [ [O = 4 The amount (concentration) that dissociates is quantified by K w A common rule for oos: f there are at least more O s than s in the molecule, it is strong **Only the st removed is strong The p scale A simple way of epressing concentration of ions in solution. p and [ [ Common in M p solutions [O in M..... p = log[ [ = p p = log[ Determine the p of a solution with.55 6 M. Also, what will be the concentration of O? p = log(.55 6 M) = 5.8 Determine the p of a solution with.977 M O. Also, what will be the concentration of? 4. 4 [. M.977M Notice the relationship: [ [O =. 4 = K w (.55 6 M)[O =. 4 p = log(. 4 M) = [[O =. 4 = K w Also notice the logarithmic nature of p vs. [ For eample, what is the difference in [ between the p of 7 and 4? p of 4 has 3, (or,) more ions in solution than p [O 6.45 M 6.55 M Determine the and O concentrations for a solution with a p of 7. 4 [ = p [ 9.8 M [O. M M Notice that the closer to p = 7, the closer the [ and [O p and po Although most people know about p, we can also calculate something called po [ [O =. 4 = K So let s say we have a solution with a [ of.9 5 M. Determine w the concentration of [O, find p and po. take the log (.9 5 M)[O =. 4 or p of a strong /strong Remember that when a strong or a strong is placed in water, we assume that it is % dissociated. So, for, all of the becomes dissociated as. f we know the concentration of, we also know the concentration of. What is the p of a.5 solution of? And for O, all of the O becomes dissociated as O. f we know the concentration of O, we also know the concentration of O. What is the p of a.5 solution of O? p po = 4 4. [O = 5 = M M p = log(.9 5 M) = 4.54 Since is % dissociated, the [ is also.5 Since O is % dissociated, the [O is also.5 po = log(3.4 M) = 9.47 po = = 9.46 p = log[ p = log(.5) =.3 po = log[o po = log(.5) =.3 p = 4. po = 4.53 [O = 9.46 = 3.5 M p = 4 po Difference in last digit due to rounding (keep an etra sig fig in calculations) p = 4..3 =.7 Qualitatively: The larger the [, to lower the [O The smaller the p, the larger the po

3 6//4 Conjugate Acid/Base pairs Conjugate Acid/Base pairs Conjugate / pairs are s and s that differ by only an Conjugate / pair (aq) O(l) 3 O (aq) (aq) Reaction proceeds in direction that results in a and produced (aq) O(l) 3 O (aq) (aq) Conjugate / pair NO 3 (aq) O(l) O (aq) NO 3 (aq) The reaction will not proceed in the forward direction The the, the its conjugate. N 4 (aq) C 3 COO (aq) N 3 (aq) C 3 COO(aq) (aq) O(l) 3 O (aq) (aq) The Acid and Base quilibrium constants: K a and K b The strength of a or can be related to the an equilibrium constant Conjugate Acid/Base pair relationship K a and K b ow are conjugate / pairs related? A(aq) O(l) 3 O (aq) A (aq) B(aq) O(l) O (aq) B (aq) 3PO 4 K a = PO 4 K b =.3 [ 3 O [A [A pk a = log(k a ) K a = p pk b = log(k b ) [O [B Kb [B K b = pkb C 3COO K a =.6 5 C 3COO K b = 6. N 4 K a = 5.4 N 3 K b =.9 5 The larger the K a, and thus smaller the pk a, the the The larger the K b, and thus smaller the pk b, the the A certain has a K a of. 7. Calculate it s pk a. A certain has a pk a of 9.3. Calculate it s K a. 3PO 4 pk a C 3COO pk a PO 4 pk b C 3COO pk b N 4 pk a N 3 pk b Based on pk a or K a values, which is? For conjugate / pairs K b *K a = K w = 4 We can also take the negative log of K a, K b, and K w. pk b pk a = pk w = 4 Polyprotic Acids While binary s only have one to donate, some oos have multiple protons. They can be diprotic or triprotic. SO SO 4 pk 4 pk SO a 3 a =.99 4 % dissociation <% dissociation Only the first is considered strong. ach subsequent dissociates less. 3 Acidic, Basic, and Neutral Salts Remember: in chemistry, a salt is just an ionic compound containing an anion and cation, but is neutral overall. The anion and cation in this case can be viewed as conjugate /s of something else Rules to remember: The child of a strong or is negligible in strength. The child of weak or is also weak 3 PO 4 PO 4 PO 4 PO 3 4 pk a =. pk a = 7. pk a3 =.38 3

4 6//4 Acidic, Basic, and Neutral Salts Cation Anion (conjugate ) Salt From strong of a strong Neutral salts Conjugate of weak of a strong Acidic salts Whether a salt is neutral, ic, or basic will depend on the specific anion and anion N 4 from from O from from N 3 Metal ions as s Certain metal ions act as s, increasing the concentration in water. When these ions dissolve in water, water ions coordinate with the ions creating a comple. Small/highlycharged metal on From strong of a strong of a weak Acidic salts Basic salts Al 3 O from from O from O The metal ions pulls enough electron density from one water molecule outside the comple to allow it to dissociate into 3O while an O stays bound to the comple. Thus, increasing 3O concentration. Conjugate weak or small/highlycharged metal ion of a weak K a K b?????? Neutral salts N 4 C 3 COO from C 3COO from N 3 Al 3 Al 3 K a > K b Acidic salts N 4 NO from NO from N 3 O outside the comple 3O outside the comple K a < K b Basic salts N 4 CN from CN Acidity increases with higher charge and smaller size of ion from N 3 Determining the p of a Weak Acid Solution Determine the p of a.3m aqueous solution of O. K a = Write the balanced equation for the in O producing 3O. Just like any other equilibrium problem, plug in what we know [ 3 O [O [O ()() = 6. 5 [ 3O = = 6. 5 M p = log(6. 5 M) p = 4. Since O is a pure liquid (and does not c appreciably over time), we don t need to deal with it O(aq) O(l) 3O (aq) O (aq).3m.3m %.3 Less than 5%, good assumption 5 (6. ) Determining the p of a Weak Base Solution Determine the p of a.555m aqueous solution of N 3. pk b = Write the balanced equation for the in O producing O and the conjugate, N = First, we need to deal with pk b Kb.78 [O [N4 Kb [N 3 ()().555 [O = = M po = log(3.4 3 M) po =.53 p = =.497 N 3(aq) O(l) O (aq) N 4 (aq).555m.555m %.555 Less than 5%, good assumption 3 (3.4 ) Calculating K a from p f.55m solution of an has a p of 5.5, determine the K a for the (assume it is monoprotic). Since we know p, we also know [ at equilibrium [ = 5.5 = 7. 6 M [ 3 O [A [A 6 6 K (7. )(7. ) a 6 (.55.7 ) M = A(aq) O(l) 3O (aq) A (aq).55m.55m 7. 6 M.55M 7. 6 M 7. 6 M [weakor ShortCuts f you complete enough ice tables, you will start noticing trends. For eample, placing a weak or in water will give you the following epression using an ice table: This assumes that K c is small enough where will also be small. Therefore we leave it out. This also assumes that there are no products of ionization in the water before the weak or is added. f you are ever unsure of a short cut always do it the long way!! 4

5 Bond Length p ydronium concentration at equilibrium (M) % ionization 6//4 [3O %ionizationof an [A Percent ionization Percent ionization is the amount of a weak or that has become 3 O or O. Determine the percent ionization if.76m aqueous solution of an unknown mied with water resulted in a an [ 3 O of M %ionization M.76M.% Determine the percent ionization if 5. aqueous solution of an unknown mied with water resulted in a p of 3. [ 3 O = p = 3. = M %ionization M 5..5% equilibrium [O %ionizationof a [B equilibrium Determine the percent ionization of a. aqueous solution of N 3. pk b = K b = 4.75 =.8 5 Let s use the shortcut =.3 3 = [O %ionization 3.3 M. [weakor.3% Percent onization and Concentration Weak s and s do not ionize to the same etent in every solution. The amount to which it ionizes depends on, among other things, the concentration of the weak or. Determine the % ionization of a weak with Let s use the shortcut a of. 5 with the concentrations of [weakor.8m,.4, and M.4.63M.8.36M [3O.3M % ionization 3.%..63M % ionization.6%.4.36m % ionization.8%.8m The more concentrated a weak solution, the lower the % ionization nitial weak concentration (M).5.5 nitial concentration (M) Binary s Oo Strengths nductive effects n oos, The the O bond, the the O F pk a =5.74 pk a = 3.7 S pk a = 6.9 pk a 6 Se pk a = 3.9 Br pk a 8 Te pk a =.6 pk a F O S Bond Length (pm) Se Br Te What can make the O bond? ighly electronegative atoms withdraw electron density from the O bond, making it Shift in electron density away from O makes the bond (more ic) lectronegativity Moving down a group, the bond length gets longer. Longer bond length = X bond Less electron density at the Carboylic s Oo Strengths nductive effects O lectronegativity of central atom Across each row, electron density of being shifted away from the O bond Stabilizes negative charge after proton loss plain the following trend in strength: