Neutralization H+1 + OH-1  HOH Acid + Base  Water + Salt (double replacement) HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq) H2SO4 (aq) + KOH (aq)  2 HOH (l) + K2SO4 (aq) HBr (aq) + LiOH (aq)  H2CrO4 (aq) + NaOH (aq)  HNO3 (aq) + Ca(OH)2 (aq)  H3PO4 (aq) + Mg(OH)2 (aq) 

pH A change of 1 in pH is a tenfold increase in acid or base strength. A pH of 4 is 10 times more acidic than a pH of 5. A pH of 12 is 100 times more basic than a pH of 10.

16.2: Dissociation of Water Autoionization of water: H2O (l) ↔ H+ (aq) + OH- (aq) KW = ion-product constant for water H3O+ (aq) or H+ (aq) = hydronium

Indicators At a pH of 2: Methyl Orange = red Bromthymol Blue = yellow Phenolphthalein = colorless Litmus = red Bromcresol Green = yellow Thymol Blue = yellow Methyl orange is red at a pH of 3.2 and below and yellow at a pH of 4.4 and higher. In between the two numbers, it is an intermediate color that is not listed on this table.

Alternate Theories Arrhenius Theory: acids and bases must be in aqueous solution. Alternate Theory: Not necessarily so! Acid: proton (H+1) donor…gives up H+1 in a reaction. Base: proton (H+1) acceptor…gains H+1 in a reaction. HNO3 + H2O  H3O+1 + NO3-1 Since HNO3 lost an H+1 during the reaction, it is an acid. Since H2O gained the H+1 that HNO3 lost, it is a base.

16.11: Lewis Acids & Bases Lewis acid: “e- pair acceptor” Brønsted-Lowry acid = H+ donor Arrhenius acid = produces H+ Lewis base: “e- pair donor” B-L base = H+ acceptor Arrhenius base = produces OH- Ex: NH3 + BF3 → NH3BF3 Lewis base Lewis acid Lewis salt 6 CN- + Fe3+ → Fe(CN)63- Lewis base Lewis acid Coordination compound Gilbert N. Lewis (1875 – 1946) Picture of Gilbert N Lewis (USA) who was the first to isolate D2O.

16.6: Weak Acids Weak acids partially ionize in water (equilibrium is somewhere between ions and molecules). HA (aq) ↔ A- (aq) + H+ (aq) Ka = acid-dissociation constant in water Weak acids generally have Ka < 10-3 See Appendix D for full listing of Ka values

Dissociation Constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, Ka. HA (aq) + H2O (l) A- (aq) + H3O+ (aq) [H3O+] [A-] [HA] Kc = © 2009, Prentice-Hall, Inc.

Dissociation Constants The greater the value of Ka, the stronger is the acid. © 2009, Prentice-Hall, Inc.

Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature. We know that [H3O+] [COO-] [HCOOH] Ka = © 2009, Prentice-Hall, Inc.

Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO-], from the pH. © 2009, Prentice-Hall, Inc.

Calculating Ka from the pH pH = -log [H3O+] 2.38 = -log [H3O+] -2.38 = log [H3O+] 10-2.38 = 10log [H3O+] = [H3O+] 4.2  10-3 = [H3O+] = [HCOO-] © 2009, Prentice-Hall, Inc.

Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M [HCOO-], M Initially 0.10 Change - 4.2  10-3 + 4.2  10-3 At Equilibrium 0.10 - 4.2  10-3 = 0.0958 = 0.10 4.2  10-3 © 2009, Prentice-Hall, Inc.

Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10] = 1.8  10-4 © 2009, Prentice-Hall, Inc.

Calculating Percent Ionization [H3O+]eq [HA]initial Percent Ionization =  100 In this example [H3O+]eq = 4.2  10-3 M [HCOOH]initial = 0.10 M 4.2  10-3 0.10 Percent Ionization =  100 = 4.2% © 2009, Prentice-Hall, Inc.

HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq) Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C. HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq) Ka for acetic acid at 25C is 1.8  10-5. © 2009, Prentice-Hall, Inc.

Calculating pH from Ka The equilibrium constant expression is [H3O+] [C2H3O2-] [HC2H3O2] Ka = © 2009, Prentice-Hall, Inc.

Calculating pH from Ka We next set up a table… [C2H3O2], M [H3O+], M Initially 0.30 Change -x +x At Equilibrium 0.30 - x  0.30 x We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. © 2009, Prentice-Hall, Inc.

Calculating pH from Ka (x)2 (0.30) 1.8  10-5 = Now, (x)2 (0.30) 1.8  10-5 = (1.8  10-5) (0.30) = x2 5.4  10-6 = x2 2.3  10-3 = x © 2009, Prentice-Hall, Inc.

Calculating pH from Ka pH = -log [H3O+] pH = -log (2.3  10-3) © 2009, Prentice-Hall, Inc.

Polyprotic Acids… …have more than one acidic proton If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation. © 2009, Prentice-Hall, Inc.

Bases react with water to produce hydroxide ion. Weak Bases Bases react with water to produce hydroxide ion. © 2009, Prentice-Hall, Inc.

Weak Bases The equilibrium constant expression for this reaction is [HB] [OH-] [B-] Kb = where Kb is the base-dissociation constant. © 2009, Prentice-Hall, Inc.

Kb can be used to find [OH-] and, through it, pH. Weak Bases Kb can be used to find [OH-] and, through it, pH. © 2009, Prentice-Hall, Inc.

pH of Basic Solutions What is the pH of a 0.15 M solution of NH3? NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+] [OH-] [NH3] Kb = = 1.8  10-5 © 2009, Prentice-Hall, Inc.

pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH-], M Initially 0.15 At Equilibrium 0.15 - x  0.15 x © 2009, Prentice-Hall, Inc.

pH of Basic Solutions (x)2 (0.15) 1.8  10-5 = © 2009, Prentice-Hall, Inc.

pH of Basic Solutions Therefore, [OH-] = 1.6  10-3 M pOH = -log (1.6  10-3) pOH = 2.80 pH = 14.00 - 2.80 pH = 11.20 © 2009, Prentice-Hall, Inc.

Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw Therefore, if you know one of them, you can calculate the other. © 2009, Prentice-Hall, Inc.

Factors Affecting Acid Strength The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group. © 2009, Prentice-Hall, Inc.

Factors Affecting Acid Strength In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. © 2009, Prentice-Hall, Inc.

Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens. © 2009, Prentice-Hall, Inc.

Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic. © 2009, Prentice-Hall, Inc.

16.9: Salt Solutions as Acids & Bases Hydrolysis: acid/base reaction of ion with water to produce H+ or OH- Anion (A-) = a conjugate base A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq) Cation (B+) = a conjugate acid B+ (aq) + H2O (l) ↔ BOH (aq) + H+ (aq)

17.2: Buffers: Solutions that resist drastic changes in pH upon additions of small amounts of acid or base. Consist of a weak acid and its conjugate base (usually in salt form) Ex: acetic acid and sodium acetate: HC2H3O2 + NaC2H3O2 Or consist of a weak base and its conjugate acid (usually in salt form) Ex: ammonia and ammonium chloride: NH3 + NH4Cl

Buffers Buffers are solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added. © 2009, Prentice-Hall, Inc.

Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water. © 2009, Prentice-Hall, Inc.

Buffers Similarly, if acid is added, the F− reacts with it to form HF and water. © 2009, Prentice-Hall, Inc.

Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base). © 2009, Prentice-Hall, Inc.

Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. © 2009, Prentice-Hall, Inc.

Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly. © 2009, Prentice-Hall, Inc.

Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly. © 2009, Prentice-Hall, Inc.

Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. © 2009, Prentice-Hall, Inc.

Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off. © 2009, Prentice-Hall, Inc.

Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations. © 2009, Prentice-Hall, Inc.

Practice Problem on Titration: If 7.3 mL of 1.25 M HNO3 is required to neutralize 25.00 mL of a potassium hydroxide solution, what is the molarity of the potassium hydroxide? 0.044 M KOH

Titration of a Weak Base and Strong Acid 14 pH 7 0 30 Volume of HCl added (mL) Half Equivalence Point , pH= pKa pka or pkb of weak acid or base in a buffer should be clsoe to the desired pH of the buffer solution.