Outline:2/28/07 è Hand in Seminar Reports – to me è Pick up Quiz #6 – from me è Pick up CAPA 14 - outside è 5 more lectures until Exam 2… Today: è End Chapter 17 è Start Chapter 18 Ù Polyprotic acids Ù Buffers

Factors that affect acid strength: 1.Polarity of H  X bond 2.Bond strength of H  X bond 3.Stability of conjugate base X  Acid-Base Behavior and Chemical Structure

Which is a stronger acid? Acid-Base Behavior and Chemical Structure HCl or HF HClO or HBrO or HIO HBrO 2 or HBrO 3 more electroneg weaker bond more O’s = electroneg

Carboxylic Acids Carboxylic acids all contain COOH. All carboxylic acids are weak acids. When the carboxylic acid loses a proton, it generate the carboxylate anion, COO . Acid-Base Behavior and Chemical Structure

Carboxylic Acids HCH 3 H   Formic Acid Formate Ion  Acetic Acid Benzoic Acid Acetate Ion Benzoate Ion

Representative Polyprotic Acids

Polyprotic Acid Problem n Determine the concentration of all ions in M H 2 CO 3 n Step #1: Write down all reactions H 2 CO 3 + H 2 O  CO H 3 O + K a1 =4.5 x  HCO H 2 O  CO H 3 O + K a2 =4.7 x  H 2 O + H 2 O  H 3 O + + OH  K w =1.0 x Most of the H 3 O + will come from the first Dissociation step. Treat as a simple weak acid problem.

H 2 CO 3 + H 2 O  CO H 3 O + K a1 =4.5 x M 0 0 init  (0.050  x )M x M x M equil Assume x is small and we obtain: x= (0.050  4.5x10 -7 ) 1/2 = 1.5  M Hence:  [  CO 3 - ]= [H 3 O + ] = 1.5  M  H 2 CO 3 = M  1.5  = M Step 2. Set up ICE Table

What is the concentration of CO 3 2- ? 1.5   init   x +x +x change (1.5  10  4  x) x 1.5  x equil x(1.5  x)/(1.5  –x) = 4.7  Notice K a2. Assume x is small. x = [CO 3 2- ] = 4.7  HCO H 2 O  CO H 3 O + K a2 =4.7 x10 -11

Chapter 17 : No Common Ions è A typical weak acid dissociation: HOAc  H + + OAc  K a 1.8 × 10  5 K a = x  x x = 2.1 × 10  3 pH = M

Chapter 18 : Common Ions è Add ions to both sides of equation: HOAc  H + + OAc  K a 1.8 × 10  5 K a = x(0.25+x) 0.25  x x = 1.8 × 10  5 pH = 4.74 vs LeChâtelier again

Acid Base reactions: è Add 0.10 mol NaOH to 1L H 2 O : Strong Base = complete dissociation pOH =  log(OH  ) =  log(0.1) = 1.0 pH = 13.0 No common ions: pH goes from 7.0 to 13.0

Acid Base reactions: è Add 0.10 mol NaOH to 1L HOAc: HOAc + OH   H 2 O + OAc  What is the correct K eq to use? HOAc  H + + OAc  KaKa H + + OH   H 2 O 1/K w HOAc + OH   H 2 O + OAc  K a /K w K a /K w = 1.8 × 10  5 /1.0 × 10  = 1.8 × 10  Acid + Base reaction 0.25M buffer

Reaction with Common Ions è Add 0.10 moles of NaOH: HOAc + OH   H 2 O + OAc  K a 1.8 × 10  x +x -x K eq = (0.35  x) (0.15  x)x x = 1.3 × 10  9 pOH = 8.89 pH = 5.11

Acid Base reactions: è Add 0.10 mol NaOH to …. No common ions: pH goes from 7.00 to With common ions: pH goes from 4.74 to 5.11 change = change = A Buffer!

Chapter 18 : Buffers  Can add lots of HA or A  and pH doesn’t change much mathematically. è Easier to use Henderson-Hasselbach equation (p. 811)…. pH = pK a + log [base] [acid]

pH = log [OAc  ]/[HOAc] pH = pK a + log ([conj base] / [acid]) If 50.0 g of sodium acetate is added to 100 mL of M solution of acetic acid, what will be the pH of the resultant buffer? 50 g CH 3 COONa = 0.61 mol/0.1L = 6.1 M pH = log (6.1/0.1) = = 6.53

Where does this come from?  K a = [H + ][A  ] / [HA]  pK a = pH + p[A  ] / [HA]  pK a = pH  log[A  ] / [HA] pH = pK a + log ([conj base] / [acid])

Buffer calculations…. pH = pK a + log ([conj base]/[acid]) This one also exists: pOH = pK b + log ([conj acid]/[base]) Practice!