Acid Strength: STRONG ACIDS REVIEW: Strong acids dissociate completely in water (strong electrolytes) HA(aq) + H2O(l) <-> H3O+(aq) + A-(aq) Acid Base Conjugate A. C.B. Equilibrium lies far to the RIGHT (almost all of the HA dissociates at equilibrium) A STRONG acid has a WEAK conjugate base (a much weaker base than water) Water wins the competition for H+ ions

ACID STRENGTH: WEAK ACIDS HA(aq) + H2O(l) <-> H3O+(aq) + A-(aq) Acid Base Conjugate A. C.B. Very little dissociation of HA Equilibrium lies far to the LEFT Conjugate base is a much stronger base than water Water loses the competition for H+ ions

Common Strong Acids: MEMORIZE!! Sulfuric Acid: H2SO4(aq) Hydrochloric Acid: HCl(aq) Nitric Acid: HNO3(aq) Perchloric Acid: HClO4(aq) Diprotic acids have two acidic protons (H2SO4) H2SO4(aq) -> H+(aq) + HSO4-(aq) strong HSO4-(aq) <-> H+(aq) + SO42-(aq) weak Monoprotic acids have one acidic proton

Acids… Oxyacids: acidic proton is attached to an oxygen atom (weak and strong) Organic Acids: acids with a carbon atom backbone, commonly contain the carboxyl group (usually weak acids)

Ka For monoprotic acids, Ka can be used to determine strength Larger Ka = stronger acid = equilibrium lies farther RIGHT BASE STRENGTH: opposite of acid strength…larger Ka = stronger acid = weaker conjugate base Water acts stronger than weak, but weaker than strong :)

Monoprotic Ka Values

H2O(l) + H2O(l) -> H3O+ + OH- Water Amphoteric substances can act as either an acid or a base (ex: water) H2O(l) + H2O(l) -> H3O+ + OH- Equilibrium: Kw = [H3O+][OH-] Kw = ion-product constant (dissociation constant for water) At 25°C in pure water [H3O+] = [OH-] = 1.0 X 10-7 M, so Kw = 1.0 X 10-14

About Kw In any aqueous solution at 25°C, no matter what it contains, [H+]*[OH-] must always equal 1.0 X 10-14 Neutral solutions, [H+] = [OH-] Acidic solutions, [H+] > [OH-] Basic solutions, [OH-] > [H+] NO MATTER WHAT, at 25°C, Kw = [H+][OH-] = 1.0 X 10-14

Examples Calculate [H+] or [OH-] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic. 1.0 X 10-5 M OH- 1.0 X 10-7 M OH- 10.0 M H+

Temperature The equilibrium constant Kw varies with temperature… If Kw increases with temperature, energy is a reactant (endothermic) I will not ask you questions about these

2H2O(l) <-> H3O+(aq) + OH-(aq) More Complex Example At 60°C, the value of Kw is 1 X 10-13 Using Le Chatelier’s principle, predict whether the following reaction is exothermic or endothermic 2H2O(l) <-> H3O+(aq) + OH-(aq) Calculate [H+] and [OH-] in a neutral solution at 60°C

pH Scale Ranges from 0-14 and represents the small concentrations used in the Kw expression for [OH-] and [H+] pH = -log[H+] pOH = -log[OH-] pK = -log K Log scale is based on 10, so the pH changes by 1 for every power of 10 change in [H+] ACIDS: low pHs, BASES: high pH, NEUTRAL: pH = 7

Example Fill in the following table: Significant figures: The # of decimal places in the log = # SF in the original # pH pOH [H+] [OH-] Acid, base, neutral Sol’n A 6.88 Sol’n B 8.4 X 10-14 Sol’n C 3.11 Sol’n D 1.0 X 10-7

Polyprotic Acids Acids can break up into more than one proton (ex. H2SO4 = diprotic or H3PO4 = triprotic) Ka describes the first proton, Ka2 describes the second, etc. For a typical weak polyprotic acid, Ka1 > Ka2 > Ka3…the acid will get successfully weaker *For a typical polyprotic acid in water, only the first dissociation step is important in pH calculation

Example Using table 14.4 in your book, calculate the pH of a 1.40 M H2C2O4 (oxalic acid) solution and the equilibrium concentrations of H2C2O4, HC2O4-, C2O42-, and OH-. ICE table necessary…

Example #2 Using data from Table 14.4 in your textbook, calculate the pH, [PO43-], and [OH-] in a 6.0 M phosphoric acid (H3PO4) solution.