Acids and Bases Operational definitions are based on observed properties. Compounds can be Classified as acid or base by observing these sets of properties.

Properties of Acids Taste sour (acere – Latin for sour) (Lemons, vinegar) Cause certain organic dyes to change colour (Turns blue litmus paper to red – BAR) Acid properties are destroyed by Bases (React with bases to form a salt and water) Acid solutions are Electrolytes (substance in solution that conduct an electric current – Acids can be strong or weak electrolytes) Acids react (corrode) with active metals (Group I and II as well as Zn and Aluminum) (Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)) Acids react with carbonates (CO32-) and hydrogen carbonates (HCO31-) to produce carbon dioxide gas {2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + H2O(l) + CO2(g)} Certain nonmetal oxides will dissolve to produce acid solutions. (SO3(g) + H2O → H2SO4(aq) (SO3(g) is the acid anhydride – without water)

Properties of Bases Bases taste bitter; mustard and soap Bases cause weak organic acids (dyes) to change colour (red litmus paper to blue {BB} Basic Blue Acids destroy base properties - react with acids to form salts and water Bases are electrolytes {strong or weak} Feel soapy, slippery Bases are formed when the oxide of some metals dissolve in water (CaO(s) + H2O → Ca(OH)2(aq) {CaO is the base anhydride}

Acid/Base definitions Definition #1: Arrhenius (traditional) Acids are compounds with ionizable hydrogen– produce H+ ions (or hydronium ions H3O+) in solution Bases are compounds that produce OH- ions in solution (problem: some bases don’t have hydroxide ions!) The reaction between an acid and a base: H+(aq) + OH-(aq) → H2O (l)

Arrhenius acid is a substance that produces H+ (H3O+) in water. The HCl molecule is ionized. (ionization) Arrhenius base is a substance that produces OH- in water. The ions are dissociated. (dissociation)

Some acids have more than one ionizable hydrogen H2SO4 → H+(aq) + HSO41-(aq HSO41- → H+(aq) + SO42-(aq) H2SO4 is diprotic H3PO4(aq) → H+(aq) + H2PO41-(aq) H2PO41-(aq) → H+(aq) + HPO42-(aq) HPO42-(aq) → H+(aq) + PO43-(aq) Phosphoric acid is a triprotic acid.

Water self-ionization H2O ↔ H+(aq) + OH-(aq) [H+] = [OH-] = 10-7M at SATP Keq = [H+][ OH-] [H2O(l)] Kw = [H+][ OH-] = 10-7 x 10-7 (at 25ºC) Kw = 10-14 at SATP

H2O ↔ H+(aq) + OH-(aq) What happens to this equilibrium if HCl(g) dissolves in the water? HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq) Increasing Decreasing H2O ↔ H3O+(aq) + OH-(aq) [H+] > [OH-] = acidic What happens when sodium hydroxide dissolves? NaOH(s) + H2O → Na+(aq) + OH-(aq) Decreasing Increasing [H+] < [OH-] = basic (alkaline solution) If [H+] = 10-7 then [OH-] = 10-7 solution is neutral (SATP)

pH and logs [H+] is important in the study of acid-base chemistry. pH is the widely used scale to show [H+]. pH = -log[H+] or pH = 1 . log[H+] [H+] = 10 – pH (the antilog) A logarithm is the power to which ten must be raised to get a number. log1000 = log(103) = 3

pH calculations For a neutral solution pH = -log[H+] pH = -log [10-7] pH = 7 at SATP Example: [H+] = 5 x 10-3 pH = -log [5 x 10-3] pH = -log [0.005] pH = - (-2.3) = 2.3

pH and pOH pOH = - log [OH-] or [OH-] = 10 - pOH Kw = [H+] x [OH-] = 1 x 10-14 (at 25ºC) pKw= pH + pOH 14= pH + pOH Example: If pH = (2.3) what is the [OH-]? pH + pOH = 14 pOH = 14 – pH pOH = 14 – 2.3 pOH = 11.7 pOH = -log [OH-] [OH-] = inverse log -11.7 or (10 - 11.7) [OH-] = 2.0 x 10-12

[H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10-3 M) pOH = - log 0.0010 pOH = 3 pH + pOH = 14 pH = 14 – 3 = 11 OR Kw = [H3O+] [OH-] 1.0 x10-14 = [H3O+] x 1.0 X 10-3 [H3O+] = 1.0 x 10-11 M pH = - log (1.0 x 10-11) = 11.00

Problem 1: The pH of rainwater collected in a certain region of the northeastern New Brunswick on a particular day was 4.82. What is the H+ ion concentration of the rainwater? [H+] = 1.51 x 10-5 Problem 2: The OH- ion concentration of a blood sample is 2.5 x 10-7M. What is the pH of the blood? pOH = 6.6 pH = 7.4 Problem 3: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C. a) [H3O+] = [3.0], pH = - 0.48, pOH = 14.48, [OH-] = 3.3 x 10-15 b) [H3O+] = [2.4x10-3], pH = 2.62, pOH = 11.38, [OH-] = 4.2 x 10-12 Problem 4: What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral? [H3O+] = 2.14 x10-4, pOH = 10.33, [OH-] = 4.68x 10-11 It is an acid. Problem 5: Problem #4 with pH = 8.05? [H3O+] = 8.92 x10-9, pOH = 5.95, [OH-] = 1.12x 10-6 It is an base.

Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – proton donor A “proton” is a hydrogen ion (the atom lost it’s electron) Bases – proton acceptor (accepts a hydrogen ion) No longer needs to contain the OH- ion

A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate acid conjugate base base

The Bronsted-Lowry concept + Cl H O base conjugate acid conjugate base acid conjugate acid-base pairs Acids and bases are identified based on whether they donate or accept H+. “Conjugate” acids and bases are found on the products side of the equation. A conjugate base is the same as the starting acid minus H+.

OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l) Practice problems Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs: CH3COOH(aq) + H2O(l)  CH3COO–(aq) +H3O+(aq) acid base conjugate base conjugate acid conjugate acid-base pairs OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l) base acid conjugate base conjugate acid conjugate acid-base pairs

Base Conjugate acid \ \ NH3(g) + H2O(l) ↔ NH4+(aq) + OH-(aq) / / Acid Conjugate Base HCl(aq) + H2O(l) ↔ H3O+(aq) + Cl-(aq) Acid Base Conjugate Conjugate Acid Base The water has acted as both an acid and a base, depending on what it is mixed with. Substances that can act as both an acid and a base are amphoteric (also called amphiproteric).

Strong acid and base : HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq) At equilibrium the ionic form is favored Weak acid and base : At equilibrium the molecular form is favored

CH3COOH(aq) + H2O(l) ↔ H+(aq) + CH3COO-(aq) Keq= [H+] [CH3COO-] . [CH3COOH] [H2O(l)] [H2O] is a constant, so collect the constants (Keq)[H2O(l)] = [H+] [CH3COO-] [CH3COOH]  (Keq)[H2O] is represented Ka(ionization constant for an acid) Ka = [H+]e [CH3COO-]e = 1.8 x 10-5 [CH3COOH]e Ka < 1 weak acid General Formula for the ionization constant of a weak acid.

a) What is the pH of an ethanoic acid solution with a concentration of 0.100M? Ka CH3COOH = 1.82 x 10-5 CH3COOH(aq) ↔ H+(aq) + CH3COO -(aq) [H+] = x [CH3COOH]i = 0.100M Ka = [H+]e [CH3COO-]e [H+]e= [CH3COO-]e 1:1 ratio [CH3COOH]e [H+]e = x = [CH3COOH]R ([CH3COOH]R – ionized) Ka = x2e [CH3COOH]e = [CH3COOH]i - [H+]e = 0.100M [CH3COOH]e Because it is a very weak acid [CH3COOH]R = 0 x2 = Ka x [CH3COOH]e x2 = 1.82 x 10-5 x 0.100 x2 = 1.82 x 10-6 [H+] = x = 0.00135M pH = -log[H+] pH = -log[0.00135] pH = 2.87

b) What is the percent ionization of this acetic (ethanoic) acid solution? % ionization = [H+] x 100 [CH3COOH] = 0.00135 x 100 = 1.35% 0.100 (very low degree of ionization)

The value of Ka for phosphoric acid, H3PO4(aq), is 7. 0 x 10-3 at 25C The value of Ka for phosphoric acid, H3PO4(aq), is 7.0 x 10-3 at 25C. a) Calculate the [H3O+] in a 0.10 M solution of H3PO4. H3PO4 ↔ H+(aq) + H2PO4-(aq) Ka= [H+] [H2PO4-] [H+] = [H2PO4-] = x [H3PO4]e = [H3PO4]i - x [H3PO4]  Ka= x2 . 0.10 – x  7.0 x 10-3 = x2 . 0.10 – x x2 = -7.0 x 10-3x + 7.0 x 10-4 x2 + 7.0 x 10-3 x – 7.0 x 10-4 =0 x = -b± √ b2 – 4ac 2a x = -7.0 x 10-3 ± √ (7.0 x 10-3)2 – 4 x 1 x – 7.0 x 10-4 2 x 1

x = -7.0 x 10-3 ± √ 4.9 x 10-5 – -2.8 x 10-3 2 x = -7.0 x 10-3 ± √ 2.85 x 10-3   x = -7.0 x 10-3 ± 5.34 x 10-2 x = 4.64 x 10-2 x = 2.32 x 10-2 = [H+] pH = - log[2.32x10-2] = 1.63