General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Contents 16-1 The Arrhenius Theory: A Brief Review 16-2 Brønsted-Lowry Theory of Acids and Bases 16-3 The Self-Ionization of Water and the pH Scale 16-4 Strong Acids and Strong Bases 16-5 Weak Acids and Weak Bases 16-6 Polyprotic Acids 16-7 Ions as Acids and Bases 16-8 Molecular Structure and Acid-Base Behavior 16-8 Lewis Acids and Bases Focus On Acid Rain General Chemistry: Chapter 16

5-3 Acid-Base Reactions Latin acidus (sour) Sour taste Arabic al-qali (ashes of certain plants) Bitter taste Svante Arrhenius 1884 Acid-Base theory.

Some Definitions Arrhenius An acid is a substance that, when dissolved in water, ionizes and increases the concentration of hydrogen ions, H+. HCl → H+ + Cl- A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions, OH-. But, Not all bases contain OH- NaOH → Na+ + OH-

16-1 The Arrhenius Theory: A Brief Review Chemistry 140 Fall 2002 16-1 The Arrhenius Theory: A Brief Review H2O HCl(g) → H+(aq) + Cl-(aq) NaOH(s) → Na+(aq) + OH-(aq) H2O Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) H+(aq) + OH-(aq) → H2O(l) Arrhenius theory did not handle non OH- bases such as ammonia very well. General Chemistry: Chapter 16

Brønsted-Lowry An acid is a proton donor. A base is a proton acceptor. e.g. ammonia, NH3, is a base.

A Brønsted-Lowry acid… …must have an ionizable or removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons.

In an acid – base reaction, the acid donates a proton (H+) to the base.

What Happens When an Acid Dissolves in Water? Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed.

If a substance can be either… …it is amphiprotic. H2O HCO3-

Conjugate Acids and Bases The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids.

What is the conjugate base of each of the following acids? HClO4 H2S HCO3- PH4+ All the above substances are acid; simply remove H+ to obtain the conjugate base.

What is the conjugate acid of each of the following? CN- SO22- H2O HCO3- CN- + H+ → SO22- + H+ → H2O + H+ → HCO3- + H+ →

Acid and Base Strength Strong acids are completely dissociated in water. Weak acids only dissociate partially in water.

General Chemistry: Chapter 16

General Chemistry: Chapter 16 The Solvated Proton General Chemistry: Chapter 16

General Chemistry: Chapter 16 In an acid-base reaction, the favored direction of the reaction is from the stronger to the weaker member of a conjugate acid-base pair. General Chemistry: Chapter 16

General Chemistry: Chapter 16 A Strong Acid General Chemistry: Chapter 16

Predict the favored direction of the following reactions: HCl + OH- H2O + I- H2O + H2O CH3COOH + H2O NH3 + H2O

General Chemistry: Chapter 16

A Weak Acid

Acid Ionization Constant conjugate base conjugate acid acid base CH3CO2H + H2O CH3CO2- + H3O+ Kc= [CH3CO2H][H2O] [CH3CO2-][H3O+] Ka= Kc[H2O] = = 1.810-5 [CH3CO2H] [CH3CO2-][H3O+] General Chemistry: Chapter 16

General Chemistry: Chapter 16 A Weak Base General Chemistry: Chapter 16

Base Ionization Constant conjugate acid conjugate base base acid NH3 + H2O NH4+ + OH- Kc= [NH3][H2O] [NH4+][OH-] Kb= Kc[H2O] = [NH3] [NH4+][OH-] = 1.810-5 General Chemistry: Chapter 16

Autoionization of Water As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

16-3 The Self-Ionization of Water and the pH Scale General Chemistry: Chapter 16

General Chemistry: Chapter 16 The equilibrium condition of the self-ionization of water is called the Ion Product of Water conjugate acid conjugate base base acid H2O + H2O H3O+ + OH- Kc= [H2O][H2O] [H3O+][OH-] KW= Kc[H2O][H2O] = = 1.010-14 [H3O+][OH-] General Chemistry: Chapter 16

For ALL aqueous solutions: Kc = [H3O+] [OH-] At 25C, Kw = [H3O+]·[OH-] = 1.0·10-14 = Kw (at 25 ºC) [H3O+] = [OH-] the solution is…. [H3O+] > [OH-] the solution is…. [H3O+] < [OH-] the solution is….

pH pH is defined as the negative base-10 logarithm of the concentration of hydronium ion. pH = -log [H3O+]

pH In pure water, Kw = [H3O+] [OH-] = 1.0  10-14 Since in pure water [H3O+] = [OH-], [H3O+] = 1.0  10-14 = 1.0  10-7

pH Therefore, in pure water, pH = -log (1.0  10-7) = 7.00 An acid has a higher [H3O+] than pure water, so its pH is <7. A base has a lower [H3O+] than pure water, so its pH is >7.

pH These are the pH values for several common substances.

Other “p” Scales The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are pOH: -log [OH-] pKw: -log Kw

-log [H3O+] + -log [OH-] = -log Kw = 14.00 Watch This! Because [H3O+] [OH-] = Kw = 1.0  10-14, we know that -log [H3O+] + -log [OH-] = -log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00

How Do We Measure pH? For less accurate measurements, one can use Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 Or an indicator.

How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H3O+] = [acid]. What is the pH of a solution of 0.3 M hydrochloric acid?

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 =

Dissociation Constants The greater the value of Ka, the stronger is the acid.

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.

Calculating pH from Ka [H3O+] [C2H3O2-] Ka = [HC2H3O2] [C2H3O2], M Initially Change At Equilibrium [H3O+] [C2H3O2-] [HC2H3O2] Ka = We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

General Chemistry: Chapter 16 EXAMPLE 16-5 Determining a Value of KA from the pH of a Solution of a Weak Acid. Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC4H7O2 is found to have a pH of 2.72. Determine Ka for butyric acid. HC4H7O2 + H2O C4H7O2 + H3O+ Ka = ? General Chemistry: Chapter 16

Chemistry 140 Fall 2002 EXAMPLE 16-5 Solution: For HC4H7O2 KA is likely to be much larger than KW. Therefore assume self-ionization of water is unimportant. HC4H7O2 + H2O C4H7O2 + H3O+ Initial conc. 0.250 M 0 0 Changes -x M +x M +x M Equilibrium (0.250-x) M x M x M Concentration

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 EXAMPLE 16-5 HC4H7O2 + H2O C4H7O2 + H3O+ Log[H3O+] = -pH = -2.72 [H3O+] = 10-2.72 = 1.910-3 = x [H3O+] [C4H7O2-] [HC4H7O2] Ka= 1.910-3 · 1.910-3 (0.250 – 1.910-3) = Ka= 1.510-5 Check assumption: Ka >> KW. General Chemistry: Chapter 16

General Chemistry: Chapter 16 Percent Ionization HA + H2O H3O+ + A- Degree of ionization = [H3O+] from HA [HA] originally Percent ionization = [H3O+] from HA [HA] originally  100% General Chemistry: Chapter 16

Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). Again, these substances dissociate completely in aqueous solution.

Bases react with water to produce hydroxide ion. Weak Bases Bases react with water to produce hydroxide ion.

General Chemistry: Chapter 16 A Weak Base General Chemistry: Chapter 16

Weak Bases [CH3NH3+][HO-] Kb= = 4.310-4 [CH3NH2] pKb= -log(4.210-4) = 3.37

Weak Bases [HB] [OH-] [B-] Kb = The equilibrium constant expression for this reaction is [HB] [OH-] [B-] Kb = where Kb is the base-dissociation constant.

Kb can be used to find [OH-] and, through it, pH. Weak Bases Kb can be used to find [OH-] and, through it, pH.

pH of Basic Solutions [NH4+] [OH-] [NH3] Kb = = 1.8  10-5 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

[NH3], M [NH4+], M [OH-], M Initially 0.15 At Equilibrium 0.15 - x  0.15 x

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.

16-7 Ions as Acids and Bases CH3CO2- + H2O CH3CO2H + OH- base acid [NH3] [H3O+] Ka= [NH4+] = ? NH4+ + H2O NH3 + H3O+ acid base [NH3] [H3O+] [OH-] Ka= [NH4+] [OH-] = KW Kb = 1.010-14 1.810-5 = 5.610-10 Ka Kb = Kw General Chemistry: Chapter 16

16-6 Polyprotic Acids Phosphoric acid: A triprotic acid. H3PO4 + H2O H3O+ + H2PO4- Ka = 7.110-3 H2PO4- + H2O H3O+ + HPO42- Ka = 6.310-8 HPO42- + H2O H3O+ + PO43- Ka = 4.210-13

Phosphoric Acid Ka1 >> Ka2 All H3O+ is formed in the first ionization step. H2PO4- essentially does not ionize further. Assume [H2PO4-] = [H3O+]. [HPO42-]  Ka2 regardless of solution molarity.

EXAMPLE 16-9 Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H3PO4 solution, calculate: (a) [H3O+]; (b) [H2PO4-]; (c) [HPO42-] (d) [PO43-] H3PO4 + H2O H2PO4- + H3O+ Initial conc. 3.0 M 0 0 Changes -x M +x M +x M Equilibrium (3.0-x) M x M x M Concentration

General Chemistry: Chapter 16 EXAMPLE 16-9 H3PO4 + H2O H2PO4- + H3O+ [H3O+] [H2PO4-] x · x Ka= = = 7.110-3 [H3PO4] (3.0 – x) Assume that x << 3.0 x2 = (3.0)(7.110-3) x = 0.14 M [H2PO4-] = [H3O+] = 0.14 M General Chemistry: Chapter 16

General Chemistry: Chapter 16 EXAMPLE 16-9 H2PO4- + H2O HPO42- + H3O+ Initial conc. 0.14 M 0 0.14 M Changes -y M +y M +y M Equilibrium (0.14 - y) M y M (0.14 +y) M Concentration [H3O+] [HPO42-] [H2PO4-] Ka= y · (0.14 + y) (0.14 - y) = = 6.310-8 y << 0.14 M y = [HPO42-] = 6.310-8 General Chemistry: Chapter 16

EXAMPLE 16-9 HPO4- + H2O PO43- + H3O+ [H3O+] [HPO42-] (0.14)[PO43-] Ka= = [H2PO4-] 6.310-8 [PO43-] = 1.910-19 M

Sulfuric Acid Sulfuric acid: A diprotic acid. H2SO4 + H2O H3O+ + HSO4- Ka = very large HSO4- + H2O H3O+ + SO42- pKa = 1.96

General Chemistry: Chapter 16 pH of Salt Solutions   Salt solutions can be neutral, acidic or basic in nature. The pH of a salt solution depends on the behavior of its corresponding Cations and Anions towards water. General Chemistry: Chapter 16

General Chemistry: Chapter 16 Hydrolysis The reaction between an ion and water is called hydrolysis. Na+ + H2O → Na+ + H2O No reaction Cl- + H2O → Cl- + H2O No reaction NH4+ + H2O → NH3 + H3O+ Hydrolysis General Chemistry: Chapter 16

General Chemistry: Chapter 16 In general: Cations of strong bases do not undergo hydrolysis; they don’t react with water.  Cations of weak bases undergo hydrolysis.  Anions of strong acids do not react with water.  Anions of weak acids undergo hydrolysis. General Chemistry: Chapter 16

16-8 Molecular Structure and Acid-Base Behavior Why is HCl a strong acid, but HF is a weak one? Why is CH3CO2H a stronger acid than CH3CH2OH? There is a relationship between molecular structure and acid strength. Bond dissociation energies are measured in the gas phase and not in solution. General Chemistry: Chapter 16

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.

Strengths of Binary Acids Chemistry 140 Fall 2002 Strengths of Binary Acids HI HBr HCl HF Bond length 160.9 > 141.4 > 127.4 > 91.7 pm Bond energy 297 < 368 < 431 < 569 kJ/mol Acid strength 109 > 108 > 1.3106 >> 6.610-4 HF + H2O → [F-·····H3O+] F- + H3O+ ion pair H-bonding free ions

General Chemistry: Chapter 16 Strengths of Oxoacids Factors promoting electron withdrawal from the OH bond to the oxygen atom: High electronegativity (EN) of the central atom. A large number of terminal O atoms in the molecule. H-O-Cl H-O-Br ENCl = 3.0 ENBr= 2.8 Ka = 2.910-8 Ka = 2.110-9 General Chemistry: Chapter 16

Factors Affecting Acid Strength In oxoacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

General Chemistry: Chapter 16 Strengths of Oxoacids S O H ·· S O H ·· Ka 103 Ka =1.310-2 S O H ·· - 2+ S O H ·· - + General Chemistry: Chapter 16

Strengths of Organic Acids ·· O H H H ·· ·· H C C O H H C C O H ·· ·· H H H acetic acid ethanol Ka = 1.810-5 Ka =1.310-16 General Chemistry: Chapter 16

Focus on the Anions Formed Ethoxide H H ·· - H C C O ·· ·· H H Acetate C O H - ·· C O H - ·· General Chemistry: Chapter 16

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Structural Effects ·· H O Ka = 1.810-5 Ka = 1.310-5 H C C ·· - O H ·· H C H C H C H C H C H C H C O - ·· H C H General Chemistry: Chapter 16

General Chemistry: Chapter 16 Chemistry 140 Fall 2002 Structural Effects ·· H O Ka = 1.810-5 Ka = 1.410-3 H C C ·· - O H ·· ·· Cl O H C C ·· - O H ·· General Chemistry: Chapter 16

Strengths of Amines as Bases ·· Br N ·· H H ammonia bromamine pKb = 4.74 pKb = 7.61 General Chemistry: Chapter 16

Strengths of Amines as Bases Chemistry 140 Fall 2002 Strengths of Amines as Bases H H H H H H H C NH2 H C C NH2 H C C C NH2 H H H H H H methylamine ethylamine propylamine pKb = 4.74 pKb = 3.38 pKb = 3.37 General Chemistry: Chapter 16

General Chemistry: Chapter 16 Resonance Effects General Chemistry: Chapter 16

General Chemistry: Chapter 16 16-9 Lewis Acids and Bases Lewis Acid A species (atom, ion or molecule) that is an electron pair acceptor. Lewis Base A species that is an electron pair donor. base acid adduct General Chemistry: Chapter 16

Showing Electron Movement