Aqueous Equilibria: Acids and Bases

Transcription

1 /3/014 Aqueous Equilibria: Acids and Bases Ch. 16 What is an? What is a? There are actually multiple definitions 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: Electron (e ) acceptor Base: Electron (e ) donor Ag :N 3 [ 3 N:Ag:N 3 ] 1

2 /3/014 Weak vs. Strong Acids/Bases What does strong as in strong s and s mean? ighly Concentrated? Dangerous? Cl Strong just mean 100% dissociation That s all. Cl Cl Cl Cl Cl in the gas phase O Cl Cl Cl Cl in the aqueous phase Cl(g) Cl(aq) K s large Cl is a strong Cl = Weak vs. Strong Acids/Bases Weak just means much less than 100% dissociation. Most F molecules are actually NOT dissociated F F F O F F F F F in the gas phase F F F F in the aqueous phase F(g) F(aq) K s are very small F is a weak F =

3 /3/014 Weak vs. Strong Acids/Bases Just like with s, strong s are those with100% dissociation. O Na O Na O Na O Na O Na O Na O O Na O Na Na O NaO in the solid phase Na O NaO in the aqueous phase NaO(s) NaO(aq) K s large NaO is a strong Weak vs. Strong Acids/Bases Weak just means much less than 100% dissociation! For N 3, it itself is not dissociated, however, it takes an from water to create O N 3 (l) O(l) N 4 (aq) O (aq) N O Most N 3 molecules are actually NOT converted to N 4 N O N 3 in the liquid phase N in the aqueous phase N 3 (l) N 4 (aq) K s are very small N 3 is a weak N 3 =

4 /3/014 Strong vs. Weak s If you know the strong s/s, assume the rest are weak Strong Acids Cl(aq) Br(aq) I(aq) NO 3 ClO 3 ClO 4 SO 4 ** 100% ionization Strong Bases (soluble metal hydroides) LiO NaO KO Ca(O) (kindaish) Sr(O) Ba(O) Eamples of weak s F C 3 COO (acetic) COO (formic) N 4 (ammonium) 3 PO 4 ClO ClO (small, highlycharged metal ions) Al 3 Eamples of weak s (nitrogencontaining s) N R R = anything N 3 (ammonia) ( Insoluble hydroides) <100% ionization A common rule for oos: If there are at least more O s than s in the molecule, it is strong **Only the 1 st removed is strong Auto ionization of water K w At any time in a sample of pure water, some water molecules, very few, will dissociate by themselves K w = [ ][O ]= Only % of O molecules within a container of water do this!! The amount (concentration) that dissociates is quantified by K w 4

5 /3/014 The p scale A simple way of epressing concentration of ions in solution. Common solutions [ ] in M p [O ] in M p = log[ ] [ ] = 10 p Notice the relationship: [][O] = = 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 10 3, (or 1,000) more ions in solution than p 7. p and [ ] p = log[ ] Determine the p of a solution with M. Also, what will be the concentration of O? p = log( M) = [ ][O ] = = K w Determine the p of a solution with M O. Also, what will be the concentration of? [ ] M 14 M p = log( M) = ( M)[O ] = [O ] M 9 M Determine the and O concentrations for a solution with a p of 7.0 [ ] = 10 p [ ] M [O ] M 7 M Notice that the closer to p = 7, the closer the [ ] and [O ] 5

6 /3/014 p and po Although most people know about p, we can also calculate something called po [ ][O ] = = K w So let s say we have a solution with a [ ] of M. Determine the concentration of [O], find p and po. take the log ( M)[O ] = or p po = [O ] = 5 = M 10 M p = log( M) = 4.54 po = log( M) = 9.47 po = = 9.46 p = po = 4.53 [O ] = = M Difference in last digit due to rounding (keep an etra sig fig in calculations) Qualitatively: The larger the [], to lower the [O] The smaller the p, the larger the po p of a strong /strong Remember that when a strong or a strong is placed in water, we assume that it is 100% dissociated. So, for Cl, all of the becomes dissociated as. If we know the concentration of Cl, we also know the concentration of. And for NaO, all of the O becomes dissociated as O. If we know the concentration of NaO, we also know the concentration of O. What is the p of a 0.050M solution of Cl? Since Cl is 100% dissociated, the [ ] is also 0.050M What is the p of a 0.050M solution of NaO? Since NaO is 100% dissociated, the [O ] is also 0.050M p = log[ ] p = log(0.050) = 1.30 po = log[o ] po = log(0.050) = 1.30 p = 14 po p = =

7 /3/014 Conjugate Acid/Base pairs Conjugate / pairs are s and s that differ by only an Conjugate / pair Cl(aq) O(l) 3 O (aq) Cl (aq) Conjugate / pair The stronger the, the weaker its conjugate. Cl(aq) O(l) 3 O (aq) Cl (aq) stronger stronger weaker weaker Conjugate Acid/Base pairs Cl(aq) O(l) 3 O (aq) Cl (aq) stronger stronger weaker weaker Reaction proceeds in direction that results in a weaker and weaker produced NO 3 (aq) O(l) O (aq) NO 3 (aq) weaker weaker stronger stronger The reaction will not proceed in the forward direction N 4 (aq) C 3 COO (aq) N 3 (aq) C 3 COO(aq) weaker weaker stronger stronger 7

8 /3/014 The Acid and Base Equilibrium constants: K a and K b The strength of a or can be related to the an equilibrium constant A(aq) O(l) 3 O (aq) A (aq) B(aq) O(l) O (aq) B (aq) [ 3 O ][A ] Ka [A] [O ][B ] Kb [B] pk a = log(k a ) K a = 10 pka pk b = log(k b ) K b = 10 pkb The larger the K a, and thus smaller the pk a, the stronger the The larger the K b, and thus smaller the pk b, the stronger the A certain has a K a of Calculate it s pk a. A certain has a pk a of 9.3. Calculate it s K a. Based on pk a or K a values, which is stronger? Conjugate Acid/Base pair relationship K a and K b ow are conjugate / pairs related? 3 PO 4 K a = PO 4 K b = C 3 COO K a = C 3 COO K b = N 4 K a = N 3 K b = PO 4 pk a C 3 COO pk a N 4 pk a PO 4 pk b C 3 COO pk b N 3 pk b For conjugate / pairs K b *K a = K w = We can also take the negative log of K a, K b, and K w. pk b pk a = pk w = 14 8

9 /3/014 Polyprotic Acids While binary s only have one to donate, some oos have multiple protons. They can be diprotic or triprotic. Only the first is considered strong. SO SO 4 pk 4 pk SO a1 3 a = % dissociation <100% dissociation Each subsequent dissociates less. 3 3 PO 4 PO 4 PO 4 PO 3 4 pk a1 =.1 pk a = 7.0 pk a3 = 1.38 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 9

10 /3/014 Acidic, Basic, and Neutral Salts Cation From strong Anion (conjugate ) of a strong Salt Neutral salts Whether a salt is neutral, ic, or basic will depend on the specific anion and anion NaCl from Cl from NaO Conjugate of weak of a strong Acidic salts N 4 Cl from Cl from N 3 Small/highlycharged metal on of a strong Acidic salts AlCl 3 from Cl From strong of a weak Basic salts NaClO from ClO from NaO Conjugate weak or small/highlycharged metal ion of a weak K a K b?????? Neutral salts N 4 C 3 COO from C 3 COO from N 3 K a > K b Acidic salts N 4 NO from N 3 from NO K a < K b Basic salts N 4 CN from CN 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. The metal ions pulls enough electron density from one water molecule outside the comple to allow it to dissociate into 3 O while an O stays bound to the comple. Thus, increasing 3 O concentration. Al 3 Al 3 O outside the comple 3 O outside the comple Acidity increases with higher charge and smaller size of ion 10

11 /3/014 Determining the p of a Weak Acid Solution Determine the p of a 0.13M aqueous solution of ClO. K a = Write the balanced equation for the in O producing 3 O. Since O is a pure liquid (and does not change appreciably over time), we don t need to deal with it Just like any other equilibrium problem, plug in what we know. [ 3 O ][ClO ] Ka [ClO] ()() I C E ClO(aq) O(l) 3 O (aq) ClO (aq) nitial 0.13M 0M 0M hange quilibirum 0.13M = [ 3 O ] = = M p = log( M) p = % 0.13 Less than 5%, good assumption 5 (6.010 ) Determining the p of a Weak Base Solution Determine the p of a 0.555M 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. K b [O ][N4 ] Kb [N ] ()() I C E nitial hange quilibirum N 3 (aq) O(l) O (aq) N 4 (aq) 0.555M 0M 0M 0.555M [O ] = = M po = log( M) po =.503 p = = % Less than 5%, good assumption 3 ( ) =

12 /3/014 Calculating K a from p If 1.55M solution of an has a p of 5.15, determine the K a for the (assume it is monoprotic). Since we know p, we also know [ ] at equilibrium [] = = M [ 3 O ][A ] Ka [A] 6 6 K (7.110 )(7.110 ) a 6 ( ) Ka I C E nitial hange quilibirum M = A(aq) O(l) 3 O (aq) A (aq) 1.55M 0M 0M 1.55M 1.55M M M M ShortCuts If 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: Kc [weakor] initial 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. If you are ever unsure of a short cut always do it the long way!! 1

13 ydronium concentration at equilibrium (M) % ionization /3/014 [3O ] %ionizationof an [A] Percent ionization Percent ionization is the amount of a weak or that has become 3 O or O. equilibrium initial 100 [O ] % ionizationof a [B] equilibrium initial 100 Determine the percent ionization if 0.76M aqueous solution of an unknown mied with water resulted in a an [ 3 O ] of M M %ionization % 0.76M Determine the percent ionization if 5.00M aqueous solution of an unknown mied with water resulted in a p of 3.1 [ 3 O ] = 10 p = = M M %ionization % 5.00M Determine the percent ionization of a 0.100M aqueous solution of N 3. pk b = K b = = Let s use the shortcut = = [O ] M %ionization % 0.100M Kc [weakor] initial Percent Ionization 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 initial concentration of the weak or. Determine the % ionization of a weak with a Ka of with the initial concentrations of 1.8M, 0.040M, and 0.010M. Let s use the shortcut Kc [weakor] initial M M M [ O ] M % ionization % 0.010M M % ionization % 0.040M M % ionization % 1.8M The more concentrated a weak solution, the lower the % ionization Initial weak concentration (M) Initial concentration (M) 13

14 Bond Length pka /3/014 Binary Acid Strengths O F pk a =15.74 pk a = 3.17 S Cl pk a = 6.9 pk a 6 Br Se pk a = 3.9 pk a 8 I Te Acid Strength 0 O S F 5 Se Te Br Cl I 10 pk a =.6 pk a Bond Length (pm) Acid Strength Electronegativity Moving down a group, the bond length gets longer. Longer bond length = weaker X bond Carboylic Acid Strengths

15 /3/014 Oo Strengths Inductive effects In oos, The weaker the O bond, the stronger the What can make the O bond weaker? ighly electronegative atoms withdraw electron density from the O bond, making it weaker Shift in electron density away from O makes the bond weaker (more ic) Less electron density at the Carboylic Acid Strengths NaO 4.74 Eplain the following trend in strength:

16 /3/014 Oo Strengths Inductive effects Electronegativity of central atom Acid Strength Across each row, electron density of being shifted away from the O bond. Stabilizes negative charge after proton loss