1. ACIDS and BASES

2. DEFINITIONS of Acids and Bases: Arrhenius Theory Acid: A molecular substance that ionizes in aqueous solution to form hydrogen ions (H + ). Base: A substance that produces hydroxide ions (OH - ) in aqueous solution.

3. An alternative definition: NH 3 (g) + H 2 O( l ) NH 4 + (aq) + OH - (aq) Base: H + acceptor Acid: H + donor Brønsted-Lowry Acid Brønsted-Lowry Acid= Proton donor Brønsted-Lowry Base Brønsted-Lowry Base= Proton acceptor Works for non-aqueous solutions and explains why NH 3 is basic: Brønsted-Lowry Concept

4. Weak Weak acids and bases do not fully ionize. Strong Strong acids and bases almost completely ionize. HNO 3 (aq) + H 2 O( l ) H 3 O + (aq) + NO 3 - (aq) HF(aq) + H 2 O( l )  H 3 O + (aq) + F - (aq) H 2 O( l ) + NH 3 (aq)  NH 4 + (aq) + OH - (aq) Base: H + acceptor Acid: H + donor Conj.Base: H + acceptor Conj.Acid: H + donor Brønsted-Lowry Concept Conjugate acid-base pairs

5. HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) acid ionization constant The acid ionization constant is used to report the degree of ionization: [A - ][H 3 O + ] [HA] Ka = Ka = (Water omitted, as usual) Strong acids have large K a values Weak acid have small K a values When an acid ionizes in water: Acid Ionization Constants, K a

6. Dissociation Constants: Strong Acids

7. Dissociation Constants: Weak Acids

8. For a base in water: base ionization constant The base ionization constant, K b, is: K b = [BH + ][OH - ] [B:] A - (aq) + H 2 O( l ) HA(aq) + OH - (aq) K b = [HA][OH - ] [A - ] B:(aq) + H 2 O( l ) BH + (aq) + OH - (aq) If the base is an anion: Base Ionization Constants, K b

9. Dissociation of Strong Bases  Strong bases are metallic hydroxides (MOH)  Group I hydroxides (NaOH, KOH) are very soluble  Group II hydroxides (Mg, Ca, Ba, Sr) are less soluble MOH(s)  M + (aq) + OH - (aq)

10. K b for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

11. Strong acids are better H + donors than weak acids Strong bases are better H + acceptors than weak bases Stronger acids have weaker conjugate bases. Weaker acids have stronger conjugate bases. HCl(aq) + H 2 O(ℓ) → H 3 O + (aq) + Cl - (aq) Relative Strength of Acids & Bases strong acid (100 % ionized) strong acid (100 % ionized) HF(aq) + H 2 O(ℓ) H 3 O + (aq) + F - (aq) weak acid (many HF are un-ionized) weak acid (many HF are un-ionized) F - readily forms HF very very weak base (no tendency to form HCl) very very weak base (no tendency to form HCl)

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13. Water acts as a base when an acid dissolves in water: HBr(aq) + H 2 O (l) H 3 O + (aq) + Br - (aq) acid base amphiprotic/amphoteric Water is amphiprotic/amphoteric - it can donate or accept a proton (act as acid or base). But water also acts as an acid for some bases: H 2 O (l) + NH 3 (aq) NH 4 + (aq) + OH - (aq) acid base acid base Water can be Acid or Base

14. Two water molecules can react to form ions. Autoionization Autoionization occurs: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Heavily reactant favored. Only a very small fraction is ionized: K w = [ H 3 O + ] [ OH - ] ionization constant for water [H 2 O] 2 is omitted… = (1.0 x 10 -7 )(1.0 x 10 -7 ) = 1.0 x 10 -14 (at 25°C) Base Acid Base Autoionization of Water

15. K w, like all equilibrium constants, is T-dependent. T = 25°C (77°F) is usually used as the standard T. T (°C) K w 100.29 x 10 -14 150.45 x 10 -14 200.68 x 10 -14 251.01 x 10 -14 301.47 x 10 -14 505.48 x 10 -14 Ionization Constant for Water

16. The pH Scale pH is a means of expressing the acidity or basicity of a solution.

17. Acid concentration can vary over a very large range. A logarithmic scale is more convenient: pH = −log 10 [H 3 O + ] At 25°C a neutral aqueous solution has: pH = −log 10 [1.0 x 10 -7 ] = −(−7.00) = 7.00 Acidic solutions: pH < 7.00 Basic solutions: pH > 7.00 The pH scale

18. Base concentrations: pOH = −log 10 [OH - ] A neutral solution (25°C) has: pOH = −log 10 [1.0 x 10 -7 ] = −(−7.00) = 7.00 Since K w = [ H 3 O + ][ OH - ] = 1.0 x 10 -14 −log(K W )= −log[H 3 O + ] + (−log[OH - ]) = −log(1.0 x 10 -14 ) all (Valid in all aqueous solns. at 25°C: acidic, neutral or basic) pK w = pH + pOH = 14.00 The pOH scale

19. pH of common solns:

20. A. Given two aqueous solutions (25°C). Solution A: [OH - ] = 4.3 x 10 -4 M, Solution B: [H 3 O + ] = 7.5 x 10 -9 M. Which has the higher pH? Which is more acidic? B. Calculate the hydronium and hydroxide ion concentration of an aqueous solution with pH 4.21

21. H 3 O + concentrations can be measured with an: pH meterElectronic pH meter:  fast and accurate  preferred method. IndicatorAcid-base Indicator :  substance that changes color within a narrow pH range  may have multiple color change (e.g. bromthymol blue)  one “color” may be colorless (e.g. phenolphthalein)  cheap and convenient. Measuring pH

22. Relationship between K a and K b values For an acid-base conjugate pair: HA and A - Phenol, C 6 H 5 OH, is a weak acid, K a = 1.3 x 10 -10 at 25°C. Calculate K b for the phenolate ion C 6 H 5 O - [HA][OH - ] [A - ] K a x K b = [H 3 O + ][A - ] [HA] = [H 3 O + ][OH]= K w

23. Reactions of Acids, Bases Strong acid strong basepH = 7 Strong acid + strong base→ salt solution, pH = 7 Strong acid weak base pH < 7 Strong acid + weak base → salt solution, pH < 7 Weak acid strong base pH > 7 Weak acid + strong base → salt solution, pH > 7 Weak acid weak base pH = ? Weak acid + weak base → salt solution, pH = ? Need K’s “strongest wins”

24. The problem-solving approach Make a habit of applying the following steps:  Write the balanced equation and K a expression.  Define x as the unknown change in concentration that occurs during the reaction. Frequently, x = [HA] dissoc, the concentration of HA that dissociates, which, through the use of certain assumptions, also equals [H 3 O + ] and [A - ] at equilibrium.  Construct a reaction table (ICE) that incorporates the unknown.  Make assumptions that simplify the calculations: usually that x is very small relative to the initial concentration.  Substitute the values into the K a expression, and solve for x.  Check that the assumptions are justified. (Apply the 5% rule) If they are not justified, use the quadratic formula to find x.

25. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #1: Write the dissociation equation HC 2 H 3 O 2  C 2 H 3 O 2 - + H +

26. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #2: ICE it! HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 0 0 - x +x 0.50 - x xx

27. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #3: Set up the law of mass action HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

28. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #4: Solve for x, which is also [H + ] HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E [H + ] = 3.0 x 10 -3 M

29. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #5: Convert [H + ] to pH HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

30. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #1: Write the equation for the reaction NH 3 + H 2 O  NH 4 + + OH -

31. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #2: ICE it! 0.50 0 0 - x +x 0.50 - x xx NH 3 + H 2 O  NH 4 + + OH -

32. A Weak Base Equilibrium Problem Step #3: Set up the law of mass action 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

33. A Weak Base Equilibrium Problem Step #4: Solve for x, which is also [OH - ] 0.50 - xxx E [OH - ] = 3.0 x 10 -3 M NH 3 + H 2 O  NH 4 + + OH - What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ?

34. A Weak Base Equilibrium Problem Step #5: Convert [OH - ] to pH 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

35. 35 Acid-base Titrations 35 Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color In a titration, a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

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37. 37 Acid-base Titrations  Primary standard – A chemical compound which can be used to accurately determine the concentration of another solution. Examples include KHP and sodium carbonate.  Standard solution – A solution whose concentration has been determined using a primary standard.  Standardization – The process in which the concentration of a solution is determined by accurately measuring the volume of the solution required to react with a known amount of a primary standard.

38. 38 Titrations can be used in the analysis of Acid-base reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 EXAMPLE: A sodium hydroxide solution was standardized and used to titrate 25.00 mL of a sulfuric acid solution. The titration requires 43.79 mL of the 0.1172 M NaOH solution to completely neutralize the acid. What is the concentration of the H 2 SO 4 solution?

39. Think! In a titration experiment, a student finds that 23.48 mL of a NaOH solution are needed to neutralize 0.5468 g of KHP. What is the concentration (in molarity) of the NaOH solution? Potassium hydrogen phthalate is a very good primary standard. –It is often given the acronym, KHP. –KHP has a molar mass of 204.2 g/mol.