1. 1 Acids and Bases Operational definitions are based on observed properties. Compounds can be Classified as acid or base by observing these sets of properties.

2. 2 Properties of Acids þ Taste sour (acere – Latin for sour) (Lemons, vinegar) þ Cause certain organic dyes to change colour (Turns blue litmus paper to red – BAR) þ Acid properties are destroyed by Bases (React with bases to form a salt and water) þ 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) → ZnCl 2 (aq) + H 2 (g))  Acids react with carbonates (CO 3 2- ) and hydrogen carbonates (HCO 3 1- ) to produce carbon dioxide gas {2HCl(aq) + Na 2 CO 3 (s) → 2NaCl(aq) + H 2 O(l) + CO 2 (g)} þ Certain nonmetal oxides will dissolve to produce acid solutions. (SO 3 (g) + H 2 O → H 2 SO 4 (aq) (SO 3 (g) is the acid anhydride – without water)

3. 3 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) + H 2 O → Ca(OH) 2 (aq) {CaO is the base anhydride}

4. 4 Acid/Base definitions Definition #1: Arrhenius (traditional) –Acids are compounds with ionizable hydrogen– produce H + ions (or hydronium ions H 3 O + ) 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) → H 2 O (l)

5. 5 Arrhenius acid is a substance that produces H + (H 3 O + ) in water. The HCl molecule is ionized. (ionization) Arrhenius base is a substance that produces OH - in water. The ions are dissociated. (dissociation)

6. 6 Some acids have more than one ionizable hydrogen H 2 SO 4 → H + (aq) + HSO 4 1- (aq HSO 4 1- → H + (aq) + SO 4 2- (aq) H 2 SO 4 is diprotic H 3 PO 4 (aq) → H + (aq) + H 2 PO 4 1- (aq) H 2 PO 4 1- (aq) → H + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) → H + (aq) + PO 4 3- (aq) Phosphoric acid is a triprotic acid.

7. 7 Water self-ionization H 2 O ↔ H + (aq) + OH - (aq) [H + ] = [OH - ] = 10 -7 M at SATP Keq = [H + ][ OH - ] [H 2 O(l)] Kw = [H + ][ OH - ] = 10 -7 x 10 -7 (at 25ºC) Kw = 10 -14 at SATP

8. 8 H 2 O ↔ H + (aq) + OH - (aq) What happens to this equilibrium if HCl(g) dissolves in the water? HCl(g) + H 2 O(l) → H 3 O + (aq) + Cl - (aq) Increasing Decreasing H 2 O ↔ H 3 O + (aq) + OH - (aq) [H + ] > [OH - ] = acidic What happens when sodium hydroxide dissolves? NaOH(s) + H 2 O → Na + (aq) + OH - (aq) Decreasing Increasing H 2 O ↔ H 3 O + (aq) + OH - (aq) [H + ] < [OH - ] = basic (alkaline solution) If [H + ] = 10 -7 then [OH - ] = 10 -7 solution is neutral (SATP)

9. 9 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(10 3 ) = 3

10. 10 pH calculations For a neutral solution pH = -log[H + ] pH = -log [10 -7 ] pH = - [-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

11. 11 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

12. 12 [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution?What is the pH of the 0.0010 M NaOH solution? [OH - ] = 0.0010 (or 1.0 X 10 -3 M)[OH - ] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010pOH = - log 0.0010 pOH = 3pOH = 3 pH + pOH = 14pH + pOH = 14 pH = 14 – 3 = 11pH = 14 – 3 = 11 OR Kw = [H 3 O + ] [OH - ]OR Kw = [H 3 O + ] [OH - ] 1.0 x10 -14 = [HO + ] x 1.0 X 10 -31.0 x10 -14 = [H 3 O + ] x 1.0 X 10 -3 [HO + ] = 1.0 x 10 -11 M[H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00pH = - log (1.0 x 10 -11 ) = 11.00

13. 13 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? Problem 2: The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? Problem 3: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H 3 O + ], pH, [OH - ], and pOH of the two solutions at 25°C. [H + ] = 1.51 x 10 -5 pOH = 6.6 pH = 7.4 Problem 4: What is the [H 3 O + ], [OH - ], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral? Problem 5: Problem #4 with pH = 8.05? a) [H 3 O + ] = [3.0], pH = - 0.48, pOH = 14.48, [OH-] = 3.3 x 10 -15 b) [H 3 O + ] = [2.4x10 -3 ], pH = 2.62, pOH = 11.38, [OH-] = 4.2 x 10 -12 [H 3 O + ] = 2.14 x10 -4, pOH = 10.33, [OH - ] = 4.68x 10 -11 It is an acid. [H 3 O + ] = 8.92 x10 -9, pOH = 5.95, [OH - ] = 1.12x 10 -6 It is an acid.

14. 14 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

15. 15 A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acidconjugate base base conjugate acid

16. 16 The Bronsted-Lowry concept 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 +. + ClH H H O + H H HO + acid base conjugate acidconjugate base conjugate acid-base pairs

17. 17 Practice problems Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs: acidbase conjugate acidconjugate base CH 3 OOH (aq) + H 2 O (l)  CH 3 COO – (aq) + H 3 O + (aq) conjugate acid-base pairs acidbase conjugate acidconjugate base OH – (aq) + HCO 3 – (aq)  CO 3 2– (aq) + H 2 O (l) conjugate acid-base pairs

18. 18 Base Conjugate acid \ \ NH 3 (g) + H 2 O(l) ↔ NH 4 + (aq) + OH - (aq) / / Acid Conjugate Base HCl(aq) + H 2 O(l) ↔ H 3 O + (aq) + Cl - (aq) Acid BaseConjugate 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).

19. 19 Strong acid and base : HA(aq) + H 2 O(l) ↔ H 3 O + (aq) + A - (aq) B(aq) + H 2 O(l) ↔ BH + (aq) + OH - (aq) At equilibrium the ionic form is favored Weak acid and base : HA(aq) + H 2 O(l) ↔ H 3 O + (aq) + A - (aq) B(aq) + H 2 O(l) ↔ BH + (aq) + OH - (aq) At equilibrium the molecular form is favored

20. 20 CH 3 COOH(aq) + H 2 O(l) ↔ H + (aq) + CH 3 COO - (aq) K eq = [H + ] [CH 3 COO - ]. [CH 3 COOH] [H 2 O(l)] [H 2 O] is a constant, so collect the constants (K eq )[H 2 O(l)] = [H + ] [CH 3 COO - ] [CH 3 COOH] (K eq )[H 2 O] is represented K a (ionization constant for an acid) K a = [H + ] [CH 3 COO - ] = 1.8 x 10-5 [CH 3 COOH] K a < 1 weak acid General Formula for the ionization constant of a weak acid.

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