119_3_Acids&Bases_07F

# 119_3_Acids&amp;Bases_07F - III. ACIDS and BASES Harris:...

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1 III. ACIDS and BASES Harris: Chap 6: 105 – 111; Ex. 6 H; Chap 9: Ex. 9 B-G

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2 Weak Mono-protic Acids General Problem: Calculate the pH of a solution of a weak acid, HA, of known formality, F, and ionization constant K A . Necessary definitions: pH = -log[H 3 O + ] [X] = mol X per L of solution pK A = -log K A F = formality or analytical concentration = C T = total moles of compound added per L (dm 3 ) of solution, regardless of any dissociation.
3 Weak Mono-protic Acids: Fundamental Equations 1. Acid ionization reaction: HA + H 2 O H 3 O + + A K A =[H 3 O + ][A ]/[HA] 2. Water ionization reaction: 2H 2 O H 3 O + + OH K W =[H 3 O + ][OH ] 3. Charge balance: [H 3 O + ] = [A ] + [OH ] 4. Species balance: F = [HA] + [A ]

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4 Weak Mono-protic Acids Standard approximations: Chemical: Since the ionization of the acid is the dominant reaction forming H 3 O + , assume it is the only reaction producing H 3 O + . If K A *F >> K W , then the [H 3 O + ] from the ionization of water (~1*10 -7 M) is much less than the [H 3 O + ] from the ionization of the weak acid and can be neglected (= 0). or Rxn (1) >> Rxn (2) Rxn(1) ONLY Mathematical: The solution must be acidic and [H 3 O + ] >> [HO ] Therefore, [H 3 O + ] 2245 [A ] >> [OH ]
5 Weak Mono-protic Acids, Standard Approximation Rxn: HA + H 2 O H 3 O + + A Equations: K A =[H 3 O + ][A ]/[HA] F = [HA] + [A ] [H 3 O + ] 2245 [A ] F, K A will be known Results: [ ][ ] [ ] - - + - = A F A O H A K 3 [ ] [ ] + + - = O H F O H K 3 2 3 A [ ] [ ] 0 F K O H K O H A 3 A 2 3 = - + + +

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6 Weak Mono-protic Acids: Aspirin Calculate the pH of a saturated solution of aspirin (HA) Aspirin (acetylsalicylic acid, C 9 H 8 O 4 , or 2-acetoxybenzoic acid) Data from Merck Index 1 g in 300 mL H 2 O K A = 3.27*10 -4 S 2245 { (1 g)/(180.15 g/mol)}/(0.3 L) = 0.019 M = F HA H 3 O + + A - K A = [H 3 O + ] 2 /(F - [H 3 O + ]) [H 3 O + ] 2 +K A [H 3 O + ] -K A *F = 0 a*x 2 + b*x+ c = 0 TI 82,TI 85:(Poly); TI 89 (Solve) HP 48: Solve poly a = 1, b = 3.27*10 –4 , c = -0.019*3.27*10 –4 [H 3 O + ] = 2.33*10 –3 M pH = 2.63 {[H 3 O + ] > 1*10 –7 M pH < 7}
7 Weak Mono-protic Acids: Aspirin K A = [H 3 O + ] 2 /(F-[H 3 O + ]) 3.27*10 -4 = x 2 /(0.019 - x) TI85 Solver HP48 Solve equation TI89 F2 Solve TI82 Math(Solve) x = [H 3 O + ] = 0.00233 pH = 2.63 The procedure gives only one solution with TI 82, TI85 and HP48. You need an initial guess (<0.019). TI89 gives both + and – solutions and no guess is needed. Check answer to ensure that the chosen solution is physically real. It’s easy to mis-punch numbers.

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8 Drug Delivery/Transport/Absorption and pH Ionic compounds generally do not penetrate/diffuse efficiently through biological membranes. The salts of many {large} organic
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## This note was uploaded on 04/13/2008 for the course CHEM 119H taught by Professor Munson during the Fall '07 term at University of Delaware.

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119_3_Acids&amp;Bases_07F - III. ACIDS and BASES Harris:...

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