Aqueous Equilibria. Acids and Bases

What is an acid? What is a base? There are actually multiple definitions Arrhenius: Dealt with species in aqueous solutions. Most basic definition of acis-base. •Acid: increases H+ in water. •Base: increases OH- in water Aqueous Equilibria: Acids and Bases Brønsted-Lowrey: Acid-Base need not be in aqueous solutions . Acid and bases are part of a related conjugate pair. •Acid: Proton (H+) donor •Base: Proton (H+) acceptor Ch. 16 Lewis: No need for hydrogen in definitions •Acid: Electron (e- ) acceptor •Base: Electron (e- ) donor Weak vs. Strong Acids/Bases Ag+ + 2 :NH3 → [H3N:Ag:NH3]+ Weak vs. Strong Acids/Bases What does “strong” as in strong acids and bases mean? •Highly Concentrated? Weak just means much less than 100% dissociation. •Dangerous? Strong just mean 100% dissociation That’s all. + H H Most HF molecules are actually NOT dissociated H H Cl Cl F H2O F F H + H H Cl Cl - H F H F F H Cl H HCl in the gas phase - + Cl H H H - Cl H2O H F - Cl H F H F H HF in the gas phase + F + HF in the aqueous phase HCl in the aqueous phase HF(g) HCl(g) HCl(aq) K’s large HF(aq) K’s are very small HCl is a strong acid Weak vs. Strong Acids/Bases Weak vs. Strong Acids/Bases Weak just means much less than 100% dissociation! For NH3, it itself is not dissociated, however, it takes an H+ from water to create OH- Just like with acids, strong bases are those with100% dissociation. + Na H O Na H O H Na O Na H O Na H H O Na H O O Most NH3 molecules are actually NOT converted to NH4+ NH3(l)+H2O(l) → NH4+(aq) + OH-(aq) - N N H + H2O H O Na - Na H + Na NaOH in the solid phase H2O + O NaOH(aq) + Na - H O - H O NH3 in the liquid phase NH+ in the aqueous phase NaOH in the aqueous phase NaOH(s) HF is a weak acid HF = 6.6x10-4 HCl = 1.3x106 K’s large NaOH is a strong base NH3(l) NH 4 +(aq) K’s are very small NH3 is a weak base NH3 = 1.8x10-5 1 Strong vs. Weak acids Auto ionization of water – Kw If you know the strong acids/bases, assume the rest are weak At any time in a sample of pure water, some water molecules, very few, will dissociate by themselves Strong Acids • • • • • • • Strong Bases HCl(aq) HBr(aq) HI(aq) HNO3 HClO3 HClO4 H2SO4** Examples of weak bases Examples of weak acids • (soluble metal hydroxides) • • • • • • LiOH NaOH KOH Ca(OH)2 (kinda-ish) Sr(OH)2 Ba(OH)2 100% ionization • • • • • • • HF CH3COOH (acetic) HCOOH (formic) NH4+ (ammonium) H3PO4 HClO2 HClO • (small, highly-charged metal ions) • Al3+ • • • • Kw = [H+][OH-]= 1x10-14 (nitrogen-containing bases) NH2R R = anything NH3 (ammonia) Only 1.8x10-6% of H2O molecules within a container of water do this!! The amount (concentration) that dissociates is quantified by Kw (“Insoluble” hydroxides) <100% ionization A common rule for oxo-acids: If there are at least 2 more O’s than H+’s in the molecule, it is strong **Only the 1st H+ removed is “strong” The pH scale pH and [H+] A simple way of expressing concentration of H + ions in solution. [H+] in M pH Common solutions 0.00000000000001 1 0.0000000000001 0.1 0.000000000001 0.01 0.00000000001 0.001 0.0000000001 0.0001 0.000000001 0.00001 0.000001 (1.55x10-6M)[OH-] = 1.0x10-14 [H+][OH-] = 1.0x10-14 = Kw 0.00000001 0.000000001 0.0001 0.0000000001 0.001 0.00000000001 0.01 0.1 0.0000000000001 1 0.00000000000001 For example, what is the difference in [H+] between the pH of 7 and 4? 0.000000000001 pH of 4 has 103, (or 1,000x) more H+ ions in solution than pH 7. Although most people know about pH, we can also calculate something called pOH So…let’s say we have a solution with a [H+] of 2.9x10-5 M. Determine the concentration of [OH], find pH and pOH. (2.9x10-5M)[OH-] = 1.00x10-14 1.0x10 14 [OH-] = = 3.4x10-10M 2.9x10 5 M pOH = -log(3.4x10-10M) = 9.47 pH = 14.00 – pOH = 4.53 1.00x10 14  6.452x10 9 M 1.55x10 -6 M Determine the H+ and OH- concentrations for a solution with a pH of 7.02 [H+] = 10-pH [H ]  107.01  9.8x10 8 M [OH- ]  1.00x10 14  1.0x10 7 M 9.8x10 -8 M Notice that the closer to pH = 7, the closer the [H +] and [OH-] pH of a strong acid/strong base pH and pOH pH + pOH = 14 [OH- ]  Also notice the logarithmic nature of pH vs. [H+] 0.00001 1.00x10 14  1.02x10 14 M 0.977M pH = -log(1.02x10-14M) = 13.991 [H+][OH-] = 1.0x10-14 = Kw Notice the relationship: 0.0000001 take the -log [H ]  pH = -log(1.55x10-6M) = 5.810 0.0000001 [H+][OH-] = 1.0x10-14 = Kw Determine the pH of a solution with 0.977 M OH-. Also, what will be the concentration of H+? pH = -log[H+] [H+] = 10-pH 0.00000001 0.000001 Determine the pH of a solution with 1.55x10-6 M H+. Also, what will be the concentration of OH-? pH = -log[H+] [OH-] in M or pH = -log(2.9x10-5M) = 4.54 pOH = 14 – 4.54 = 9.46 [OH-] = 10-4.54 = 3.5x10-10M Difference in last digit due to rounding (keep an extra sig fig in calculations) Remember that when a strong acid or a strong base is placed in water, we assume that it is 100% dissociated. So, for HCl, all of the “H” becomes dissociated as H+. If we know the concentration of HCl, we also know the concentration of H+. And for NaOH, all of the “OH” becomes dissociated as OH-. If we know the concentration of NaOH, we also know the concentration of OH-. What is the pH of a 0.050M solution of HCl? What is the pH of a 0.050M solution of NaOH? Since HCl is 100% dissociated, the [H+] is also 0.050M Since NaOH is 100% dissociated, the [OH-] is also 0.050M pH = -log[H+] pOH = -log[OH-] pH = -log(0.050) = 1.30 pOH = -log(0.050) = 1.30 pH = 14 - pOH pH = 14.00 -1.30 = 12.70 Qualitatively: •The larger the [H+], to lower the [OH-] •The smaller the pH, the larger the pOH 2 Conjugate Acid/Base pairs Conjugate Acid/Base pairs HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) Conjugate acid/base pairs are acids and bases that differ by only an H + stronger acid Conjugate acid/base pair base acid base weaker acid weaker base How are conjugate acid/base pairs related? B(aq) + H2O(l) → OH-(aq) + BH+(aq) HA(aq) + H2O(l) → H3O+(aq) + A-(aq) [H3O ][A ] [HA] Kb  Ka = 10-pKa H3PO4 Ka = [OH- ][BH ] [B] H2PO4- Kb = 1.3x10-12 7.6x10-3 CH3COOH Ka = 1.6x10-5 CH3COO- Kb = 6.2x10-10 NH4+ Ka = 5.4x10-10 Kb = 10-pKb pKb = -log(Kb) NH3 Kb = 1.9x10-5 The larger the Kb, and thus smaller the pKb, the stronger the base The larger the Ka, and thus smaller the pKa, the stronger the acid A certain acid has a Ka of 1.2x10-7. Calculate it’s pKa. A certain acid has a pKa of 9.23. Calculate it’s Ka. H3PO4 pKa 2.12 11.88 CH3COOH pKa 4.79 9.21 NH4+ pKa 9.27 4.73 For conjugate acid/base pairs Based on pKa or Ka values, which acid is stronger? Kb*Ka = Kw = 1x10-14 - pKa1 ≈ -3 HSO4- 100% dissociation 2- ⇌ ⇌ pKa1 = 2.12 SO42- Each subsequent H+ dissociates less. <100% dissociation - H3PO4 Only the first H+ is considered strong. pKa2 = 1.99 2- ⇌ H2PO4- pKa2 = 7.20 H2PO4- pKb CH3COO- pKb NH3 pKb We can also take the negative log of Ka, Kb, and Kw. pKb +pKa = pKw = 14 Acidic, Basic, and Neutral Salts Polyprotic Acids While binary acids only have one H+ to donate, some oxoacids have multiple protons. They can be diprotic or triprotic. H2SO4 stronger acid Conjugate Acid/Base pair relationship – Ka and Kb The strength of a acid or base can be related to the an equilibrium constant pKa = -log(Ka) stronger base weaker base The Acid and Base Equilibrium constants: Ka and Kb Ka  stronger acid NH4+ (aq) + CH3COO-(aq) ⇌ NH3(aq) + CH3COOH(aq) HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) weaker acid Reaction proceeds in direction that results in a weaker acid and weaker base produced The reaction will not proceed in the forward direction The stronger the acid, the weaker its conjugate base. stronger base weaker base stronger base weaker acid weaker base Conjugate acid/base pair stronger acid weaker acid NO3- (aq) + H2O(l) → OH-(aq) + HNO3(aq) HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) acid stronger base 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 acid/bases of something else… Rules to remember: •The child of a strong acid or base is negligible in strength. •The child of weak acid or base is also weak 3- ⇌ HPO42- pKa3 = 12.38 PO43- 3 Acidic, Basic, and Neutral Salts Cation Anion Whether a salt is neutral, acidic, or basic will depend on the specific anion and anion Salt (conjugate base) “from” HCl NaCl From strong base …of a strong acid “from” NaOH Neutral salts Conjugate acid of weak base …of a strong acid Acidic salts Small/highly-charged metal on …of a strong acid Acidic salts From strong base …of a weak acid Basic salts Conjugate acid weak base or small/highlycharged metal ion …of a weak acid ?????? Metal ions as acids Certain metal ions act as acids, increasing the H + concentration in water. “from” HCl NH4Cl When these ions dissolve in water, water ions “coordinate” with the ions creating a “complex”. “from” NH3 “from” HCl AlCl3 “from” HClO NaClO The metal ions pulls enough electron density from one water molecule outside the complex to allow it to dissociate into H3O+ while an OH- stays bound to the complex. Thus, increasing H3O+ concentration. “from“NaOH Ka Ka ≈ Kb > Ka Kb < NH4NO2 Acidic salts Kb “from” CH3COOH NH4CH3COO Neutral salts Basic salts “from” NH3 Determine the pH of a 0.123M aqueous solution of HClO. Ka = 2.95x10-8. Ka   HClO(aq)+ H2O(l)⇌ H3O+(aq) + ClO-(aq) I C E  [H3O ][ClO ] [HClO] nitial (x)(x) 2.95x10 8  0.123 - x 0.123M 0M -x hange quilibirum Determining the pH of a Weak Base Solution First, we need to deal with pKb. +x x Write the balanced equation for the base in H2O…producing OH- and the conjugate acid, NH4+ . [OH ][NH4  ] Kb  [NH3 ] x 1.78x10 5  2.95x10 8  3.63x10 9 x 0.123  x 6.02x10 x100  0.0487% 0.123 x2 0.555 x = 6.02x10-5 [H3O+] = x = 6.02x10-5M (6.02x10-5 )2 0.123 - 6.02x10 -5 Less than 5%, good assumption 9.88x10 6  x2 -x hange 0.555M - x 0M 0M +x +x x x pOH = 2.503 pH = 14.00-2.503 = 11.497 3.14x10 -3 x100  0.566% 0.555 Less than 5%, good assumption (3.14x10-3 )2 0.555 - 3.14x10 -4 pH = 4.220 “Short-Cuts” If you complete enough ice tables, you will start noticing trends. If 1.55M solution of an acid has a pH of 5.15, determine the Ka for the acid (assume it is monoprotic). Since we know pH, we also know [H+] at equilibrium [H+] = 10-5.15 = 7.1x10-6M Ka  0.555M x = 3.14x10-3 Calculating Ka from pH Ka  nitial [OH-] = x = 3.14x10-3M pOH = -log(3.14x10-3M) pH = -log(6.02x10-5M) 2 I C E quilibirum (x)(x) 1.78x10  0.555 - x -5 -5 2 NH3(aq)+ H2O(l)⇌ OH-(aq) + NH4+(aq) Kb  104.75  1.78x10 5 0M +x 0.123M - x Acidity increases with higher charge and smaller size of ion Determine the pH of a 0.555M aqueous solution of NH 3. pKb = 4.75. Since H2O 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. H3O+ outside the complex H2O outside the complex “from” NH3 Determining the pH of a Weak Acid Solution Write the balanced equation for the acid in H2O…producing H3O+. Al3+ “from” NH3 “from” HCN NH4CN Al3+ “from” HNO2 [H3O ][A ] [HA] (7.1x10-6 )(7.1x10-6 ) (1.55 - 1.7x10 -6 ) For example, placing a weak acid or base in water will give you the following expression using an ice table: HA(aq)+ H2O(l) ⇌ H3O+(aq) + A-(aq) I C E nitial hange quilibirum 1.55M -x 1.55M - x 1.55M – 7.1x10-6M 0M 0M +x +x x x 7.1x10-6M 7.1x10-6M Kc  x2 [weakacidor base]initial This assumes that Kc is small enough where x 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 base or acid is added. If you are ever unsure of a short cut…always do it the long way!! 7.1x10-6 M = x Ka  3.3x10 11 4 Percent Ionization and Concentration Percent ionization Percent ionization is the amount of a weak acid or base that has become H 3O+ or OH-. [HA]initial Determine the percent ionization if 0.76M aqueous solution of an unknown acid mixed with water resulted in a an [H3O+] of 9.05x10-4M % ionization % ionizationof a base x100 [B]initial Weak acids and bases do not ionize to the same extent in every solution. The amount to which it ionizes depends on, among other things, the initial concentration of the weak acid or base. x2 Let’s use the “short-cut” K c  [weakacidor base] initial x100 Determine the % ionization of a weak acid with a Ka of 1.0x10-5 with the initial concentrations of 1.28M, 0.040M, and 0.010M. Determine the percent ionization of a 0.100M aqueous solution of NH3. pKb = 4.75. Kb = 10-4.75 = 1.0x10 5  1.8x10-5 Kc  Let’s use the “short-cut” 9.05x10 4 M x100  0.12% 0.76M 1.8x10 5  Determine the percent ionization if 5.00M aqueous solution of an unknown acid mixed with water resulted in a pH of 3.12 1.3x10-3 1.0x10 5  x2 0.100 [H3O+] = 10-pH = 10-3.12 = 7.6x10-4 M = % ionization % ionization 0.00032M x100  3.2% 0.010M x2 0.040 % ionization 0.00063M x100  1.6% 0.040M 0.004 0.0035 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 0 x  0.00063M 1.0x10 5  [OH-] 7.6x10 4 M x100  0.015% 5.00M x2 0.010 x  0.00032M x2 [weakacidor base]initial x2  1.8x10 6 x= % ionization [OH- ]equilibrium Hydronium concentration at equilibrium (M) [H3O ]equilibrium x2 1.28 % ionization x  0.0036M  [H3O ] 1.3x10 3 M x100  1.3% 0.100M % ionization % ionizationof anacid 0.0036M x100  0.28% 1.28M The more concentrated a weak base solution, the lower the % ionization 0.2 0.4 0.6 0.8 1 1.2 Initial weak acid concentration (M) 1.4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 Initial acid concentration (M) Binary Acid Strengths Carboxylic Acid Strengths H2O HF pKa = 6.9 pKa ≈ -6 H2Se HBr pKa ≈ -8 10 H2S HF 5 H2Te H2Se 0 80 100 120 140 160 180 -5 HBr HCl HI H2Te pKa = 2.6 H2O 15 pKa HCl Acid Strength H2S pKa = 3.9 20 pKa = 3.17 Bond Length pKa =15.74 HI -10 pKa ≈ -10 -15 Acid Strength Bond Length (pm) Moving down a group, the bond length gets longer. Longer bond length = weaker X-H bond Electronegativity Oxoacid Strengths – Inductive effects 4.79 2.86 1.35 0.77 Carboxylic Acid Strengths In oxoacids, The weaker the O-H bond, the stronger the acid NaOH What can make the O-H bond weaker? Highly electronegative atoms withdraw electron density from the O-H bond, making it weaker. 7.53 1.96 ≈ -1 ≈ -8 Shift in electron density away from O-H makes the bond weaker (more acidic) Less electron density at the H 4.74 Explain the following trend in acid strength: 4.74 2.86 1.35 0.77 5 Oxoacid Strengths – Inductive effects Electronegativity of central atom Across each row, electron density of being shifted away from the O-H bond. Acid Strength Stabilizes negative charge after proton loss 9.24 6.37 -1.4 9.84 2.15 ≈ -3 ≈ -8 6