Acids and Bases Chapter 16 Johannes N. Bronsted Thomas M. Lowry 1879-1947. 1874-1936. Both independently developed Bronsted-Lowry theory of acids and bases. 1 1 1 1

Acids and Bases: A Brief Review Classical Acids: Taste sour Donate H+ (called “H-plus” or “proton”) Turn litmus red Generally formed from H-Z, where Z = nonmetal Classical Bases: Taste bitter and feel soapy. Donate OH- (called “O-H-minus” or “hydroxide”) Turn litmus blue Generally formed from MOH, where M = metal Neutralization: Acid + Base  Salt + water H-Z + MOH  MZ + HOH H+ in water is actually in the form of H3O+, “hydronium”

Brønsted-Lowry Acids and Bases Proton Transfer Reactions Brønsted-Lowry acid/base definition: acid donates H+ base accepts H+. Brønsted-Lowry base does not need to contain OH-. Consider HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq): HCl donates a proton to H2O. Therefore, HCl is an acid. H2O accepts a proton from HCl. Therefore, H2O is a base. Water can behave as either an acid or a base. Amphoteric substances can behave as acids and bases.

Brønsted-Lowry Acids and Bases Conjugate Acid-Base Pairs Whatever is left of the acid after the proton is donated is called its conjugate base. Similarly, whatever remains of the base after it accepts a proton is called a conjugate acid. Consider After H2O (base) gains a proton it is converted into H3O+ (acid). Therefore, H2O and H3O+ are conjugate acid-base pairs. After HCl (acid) loses its proton it is converted into Cl- (base). Therefore HCl and Cl- are conjugate acid-base pairs. Conjugate acid-base pairs differ by only one proton.

Brønsted-Lowry Acids and Bases Conjugate Acid-Base Pairs In each of following Bronsted-Lowry Acid/Base reactions, Which is acid? base? conjugate acid? conjugate base? HCl + H2O  H3O+ + Cl- acid base conj. acid conj. base NH3 + H2O  NH4+ + OH- base acid conj. acid conj. base CH3NH2 + H2SO4  CH3NH3+ + HSO4- conj. base base acid conj. acid

Brønsted-Lowry Acids and Bases Relative Strengths of Acids and Bases The stronger the acid, the weaker the conjugate base. The stronger the base, the weaker the conjugate acid. H+ is the strongest acid that can exist in equilibrium in aqueous solution. OH- is the strongest base that can exist in equilibrium in aqueous solution.

Strong and Weak Acids and Bases Strong acids: completely ionized in water: HCl, HNO3, H2SO4 (also HBr, HI) Strong bases: completely ionized in water: MOH, where M = alkali M(OH)2, where M = alkaline earth Weak acids: incompletely ionized in water: any acid that is not strong - acetic acid, etc. Ka is finite. Weak bases: incompletely ionized in water: any base that is not strong – NH3, etc. Kb is finite.

The Autoionization of Water The Ion Product of Water In pure water the following equilibrium is established but [H2O]2 = constant (at 25oC) This is called the autoionization of water

The Autoionization of Water In pure water at 25oC, [H3O+][OH-] = 1 x 10-14 and also, [H3O+] = [OH-] = 1 x 10-7 (From now on, for simplification, let’s use the abbreviation: [H+] = [H3O+] which means [H]+ = [OH-] = 1 x 10-7 We define pH = -log [H+] and pOH = -log [OH-] In pure water at 25oC, pH = pOH = 7.00 pH + pOH = 14 pKw = 14 Acidic solutions have pH < 7.00 Basic solutions have pH > 7.00

The pH Scale Conc. Drano Battery acid

The pH Scale For [H+] = 3.4 x 10-5M, calculate pH, pOH and [OH-] pH = 4.47, pOH = 9.53, [OH-] = 2.95 x 10-10 (b) For [OH-] = 4.4 x 10-3M, calculate pH, pOH, and [H+] pOH = 2.36, pH = 11.64, [H+] = 2.27 x 10-12 (c) For pH= 8.9, calculate, pOH, [H+], [OH-] pOH = 5.1, [H+] = 1.26 x 10-9, [OH-] = 7.94 x 10-6 (d) For pOH= 3.2, calculate pH, [H+], [OH-] pH = 10.8, [H+] = 1.58 x 10-11, [OH-] = 6.3 x 10-4 The pH meter is the most accurate way to measure pH values of solutions.

Use this “decision tree” to calculate pH values of solutions of specific solutions. Is it pure water? If yes, pH = 7.00. Is it a strong acid? If yes, pH = -log[HZ] Is it a strong base? If yes, pOH = -log[MOH] or pOH = -log (2 x [M(OH)2]) Is it a weak acid? If yes, use the relationship Ka = x2/(HZ – x), where x = [H+] Is it a weak base? If yes, use the relationship Kb = x2/(base – x), where x = [OH-] Is it a salt (MZ)? If yes, then decide if it is neutral, acid, or base; calculate its K value by the relationship KaKb = Kw, where Ka and Kb are for a conjugate system; then treat it as a weak acid or base. Is it a mixture of a weak acid and its weak conjugate base? It is a buffer; see next chapter.

2. Strong Acids 3. Strong Bases Calculate the pH of 0.2 M HCl . pH = -log[H+] = -log[0.2] = 0.70 3. Strong Bases Calculate the pH of 0.2 M NaOH . pOH = -log[OH-] = -log[0.2] = 0.70 pH = 14 – 0.70 = 13.30 Calculate the pH of 0.2 M Ba(OH)2. pOH = -log[OH-] = -log[0.4] = 0.40 pH = 13.60

4. Weak Acids Weak acids are only partially ionized in solution, and are in equilibrium: (shorthand): OR: (shorthand) Ka is the acid dissociation constant.

4. Weak Acids

4. Weak Acids Calculate pH of 0.2 M solution of acetic acid, HC2H3O2 From previous slide, Ka= 1.8 x 10-5 From now on, we’ll use the shorthand notation for HC2H3O2 = HAc Denote the acetate ion, C2H3O2- as “Ac-” Equilibrium is then: HAc  H+ + Ac- 0.2 0 0 Initial conc (M): -x +x +x change: at equilibrium: 0.2-x x x Ka= But, this is a quadratic equation!! We can assume x is small if Ka < 10-4. Then, x2/0.2= 1.8 x 10-5 x = 1.90 x 10-3 = [H+] = [Ac-] pH = -log (1.90 x 10-3) = 2.72

4. Weak Acids Polyprotic Acids Polyprotic acids have more than one ionizable proton. The protons are removed in steps not all at once: It is always easier to remove the first proton in a polyprotic acid than the second. Therefore, Ka1 > Ka2 > Ka3 etc. Most H+(aq) at equilibrium usually comes from the first ionization (i.e. the Ka1 equilibrium).

4. Weak Acids Polyprotic Acids

Carbonic acid, H2CO3, Ka1=4.3 x 10-7 H2CO3  H+ + HCO3- Ka1 = 4.3 x 10-7 HCO3-  H+ + CO32- Ka2 = 5.6 x 10-11 Since Ka1>>Ka2, nearly all H+ ions come from 1st equilibrium. Therefore, the 1st equilibrium determines the pH. Calculate pH of 0.4 M H2CO3 solution. Ka = x2/HZ-x 4.3 x 10-7 = x2/0.4 x = [H+] = 4.15 x 10-4 pH = -log (4.1 x 10-4) = 3.38

5. Weak Bases Weak bases remove protons from substances. There is an equilibrium between the base and the resulting ions: Example: The base dissociation constant, Kb is defined as:

5. Weak Bases The larger Kb the stronger the base.

5. Weak Bases x = 8.12 x 10 = [OH ] For convenience, denote it as “B” What is pH of 0.15 M solution of methylamine, NH2CH3, a weak base? For convenience, denote it as “B” Then: B + H2O  BH+ + OH- Kb =4.4 x 10-4 Initially: 0.15 0 0 change: -x +x +x At equil: 0.15-x x x Then: x = 8.12 x 10 - 3 = [OH - ]

Relationship Between Ka and Kb For a conjugate acid-base pair Ka  Kb = Kw (constant) Therefore, the larger the Ka, the smaller the Kb. That is, the stronger the acid, the weaker the conjugate base. Taking negative logarithms: -log Ka- log Kb= -log Kw pKa + pKb = pKw For HAc (acetic acid), the pKa = -log (1.8 x 10-5) = 4.74. Thus, the pKb for Ac- (acetate 0 is 14 – 4.74 = 9.26

6. Salts Salts may be acidic, basic or neutral. Salts made by a strong acid and a strong base are neutral, e.g., NaCl, KNO3. Salts made by a weak acid and a strong base are weakly basic, e.g., sodium acetate, NaAc, NaHCO3. Salts made by a strong acid and a weak base are weakly acidic, e.g., NH4Cl. Calculate the pH of a 0.35 M solution of sodium acetate. Since Ka Kb = Kw, where Ka = 1.8 x 10-5 for acetic acid, Then Kb for NaAc (sodium acetate) is Kb = Kw/Ka = 1.00 x 10-14/1.8 x 10-5 = 5.56 x 10-10. Now treat this as a weak base problem, Kb = x2/base = 5.56 x 10-10 = x2/0.35 x = [OH-] = 1.39 x 10-5 pOH = 4.85 and pH = 9.14