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Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

+ ] from added O + -1 1 M = 0.1 M = 10 - Suppose you have [H 3 Suppose you have [H 3 O ] from added HCl HCl = 0.1 M = 10 M - ] = 10 ][OH - -1 1 ][OH -14 14 Then [10 - ] = 10 - Then [10 - ] = 10 [OH - -13 13 ] = 10 - [OH - ] = [H + ] = 10 O, [OH - O + -7 7 , ] = 10 - In pure H 2 In pure H 2 O, [OH ] = [H 3 3 O , but but - ] < than in pure H in acid solution [OH - in acid solution [OH ] < than in pure H 2 2 O O Weak Acids and Bases Weak Acids and Bases Acetic Acid Acetic Acid Add acetic acid to H 2 Add acetic acid to H 2 O: O: + + CH → H → O → → → → → → O + COO - - Acetate ion Acetate ion CH 3 COOH + H 2 H 3 + CH 3 CH 3 COOH + H 2 O 3 O 3 COO K = [ H 3 O + ][ CH 3 COO − ] − 5 = 1.85 × 10 K=K K a K= [ CH 3 COOH ] a + ], [CH - ]? O + COO - What are [CH 3 What are [CH 3 COOH], [H COOH], [H 3 3 O ], [CH 3 3 COO ]?

- → → → → → → → → + from H → → H → → If we ignore H O → → → → If O + O + + + OH + OH - we ignore H 3 from H 2 O + H 2 H 3 3 O 2 O + H 2 O 3 O + + CH → → H O → → → → → → O + COO - - CH 3 COOH + H 2 H 3 + CH 3 CH 3 COOH + H 2 O 3 O 3 COO + ] + ] [H + ] O + O + O + C o C o - -[H [H 3 3 O ] [H 3 [H 3 O ] [H 3 3 O ] → have [H → → → → → → → O + + ] = [CH COO - - ] have [H 3 ] = [CH 3 ] (Stoichiometry ( Stoichiometry) ) 3 O 3 COO ≤ 10 ≤ O ≤ ≤ ≤ ≤ ≤ ≤ O + + ] from H 10 - -7 7 } {[H 3 ] from H 2 } {[H 3 O 2 O ≡ [ ≡ COOH] ≡ ≡ ≡ ≡ ≡ ≡ Let C o = initial [CH 3 [HOAc HOAc] ] Let C o = initial [CH 3 COOH] + ] (Simple mass balance) O + [HOAc [ HOAc]= C ]= C o o - -[H [H 3 3 O ] (Simple mass balance) + ] 2 Small K means Small K means [ H 3 O − 5 K = + ] = 1.85 × 10 K=K K a K= equilibrium is equilibrium is C O − [ H 3 O a far to left! far to left! ≅ C ≅ ≅ ≅ ] ≅ ≅ ≅ ≅ → → → [H → <<<1 ( → → → → + ]<<< C Guess, [HOAc HOAc] C o , since K=K K a [H 3 O + ]<<< C o ) Guess, [ o , since K= a <<<1 ( 3 O o ) K = [ H 3 O + ] 2 + ] = O + [H 3 ] = KC o [H 3 O C O

[ H 3 O + ] 2 − 5 K = + ] = 1.85 × 10 + ] = O + [H 3 ] = KC o [H 3 O C O − [ H 3 O 5 × × × × × 10 × 1 M = 18.5 × × × × 10 × × × 2 = 1.85 = 1.85 × × × × × × × 1 M = 18.5 × × × × + ] O + -5 -6 6 ] 2 10 - 10 - Example : C o = 1 M, [H 3 Example : C o = 1 M, [H 3 O × × × × 10 ] = 4.3 × × × × O + + ] + ] = 4.3 O + -3 3 But [CH 3 COOH]= C o - [H [H 3 ] 10 - But [CH 3 COOH]= C o - 3 O [H 3 [H 3 O ≅ ≅ ≅ ≅ C ] = 0.9957 M ≅ ≅ ≅ ≅ + ] = 0.9957 M O + C o - [H [H 3 C o = 1 M (0.43% error) C o - 3 O o = 1 M (0.43% error) → → → [H → × × × × 10 = 0.01 M → → → → ]= 4.3 × × × × + ]= 4.3 O + -4 4 10 - Example : C o [H 3 Example : C o = 0.01 M 3 O × × × × 10 × × × 10 × × 10 × × × ] = 100 × × × × 4 - 4.3 × × × × 4 = 95.7 = 95.7 × × × × + ] = 100 O + -4 -4 -4 4 10 - 10 - 10 - C o - [H [H 3 - 4.3 C o - 3 O × × × × 10 = 100 × × × × -4 4 (4.3% error!) vs C C o 10 - (4.3% error!) vs o = 100 → → [H → → × × 10 × × = .001 M → → → → 2 = 1.85 = 1.85 × × × × + ] O + -8 8 ] 2 10 - Example : C o [H 3 Example : C o = .001 M 3 O × 10 × × × ] = 1.36 × × × × + ] = 1.36 O + -4 4 [H 3 10 - [H 3 O × 10 × × × × × × 10 × × × × 10 × ] = 10 × × × × 4 - 1.36 × × × × 4 = 8.64 = 8.64 × × × × + ] = 10 O + -4 -4 -4 4 10 - 10 - 10 - C o - [H [H 3 - 1.36 C o - 3 O × 10 × × × = 10 × × × × -4 4 (13.6% error!!) vs C C o 10 - (13.6% error!!) vs o = 10

≅ ≅ ≅ ≅ C ] ≅ ≅ ≅ ≅ * * * The approximation that [HOAc HOAc] C o is good * * * The approximation that [ o is good only for small small K K a , large C large C o . only for a , o . + ] >>10 O + -7 7 ] >>10 - Note: In all cases [H In all cases [H 3 Thus, OK to Note: 3 O Thus, OK to + from H + + OH - O + O → → H → → → → O + + OH - → → neglect H 3 from H 2 O + H 2 H 3 neglect H 3 O 2 O + H 2 O 3 O Could obtain an exact solution by solving a quadratic: Could obtain an exact solution by solving a quadratic: [ H 3 O + ] 2  [ H 3 O + ] 2 + K [ H 3 O + ] − KC O = 0 ( y 2 + by + c = 0 ) K = C O − [ H 3 O + ]  → Such equations are trivial to solve with a modern Such equations are trivial to solve with a modern (K=K K a ) (K= a ) graphing calculator. graphing calculator. A powerful, approximate alternative method: A powerful, approximate alternative method:

+ ][ OH − ] K b = [ CH 3 NH 3 − 4 Method of Successive Method of Successive = 5.0 × 10 [ CH 3 NH 2 ] Approximations Approximations → → 7 (ignore hydrolysis of H O) → → → → → → - ]>>10 Take [OH - -7 ]>>10 - (ignore hydrolysis of H 2 Take [OH 2 O) [OH - - ]=[CH + ] + ]=[CH 3 NH 3 ] [OH 3 NH 3 + + OH - (weak base) → CH → → → O → → → → + + OH - CH 3 NH 2 + H 2 CH 3 NH 3 (weak base) CH 3 NH 2 + H 2 O 3 NH 3 (Stoichiometry) C o C o - - x x i x i x x i x i i i Let initial concentration [CH 3 NH 2 ] = C o = 0.1 Let initial concentration [CH 3 NH 2 ] = C o = 0.1 First Approximation : First Approximation : 2 = 2 K b = x 1 → x 1 [OH - - ] = [CH + ] = x + ] = [CH 3 NH 3 ] = x 1 [OH 3 NH C o 3 1 × × × × 10 × × 10 × × (0.1)(5 × × × × ) = 50 × × × × [CH 3 NH 2 ] = C o - x x 1 , [CH 3 NH 2 ] = C o - 1 , -4 4 ) = 50 -6 6 10 - 10 - (0.1)(5 → → o → → → → → → Assume x 1 << C o Assume x 1 << C × 10 × = 7.1 × × × × × × 10 - -3 3 x 1 x 1 = 7.1

× 10 × × 10 × = 7.1 × × × × 7.1) × × × × × × × × 10 - -3 3 10 - -3 3 x 1 C o - x x 1 = (100 - - 7.1) x 1 = 7.1 C o - 1 = (100 + + OH - (weak base) → → CH O → → → → → → + + OH - CH 3 NH 2 + H 2 CH 3 NH 3 (weak base) CH 3 NH 2 + H 2 O 3 NH 3 ≅ ≅ ≅ C ≅ 1 ≅ ≅ ≅ ≅ C o - x x 1 C o x 1 x 1 C o - x x o 1 1   7.1 × × × × 100 = 7.1 % is thus × × × ×   % error in taking C o - x x 1 = C o 100 = 7.1 % % error in taking C o - 1 = C o is thus 100 Second Approximation : Second Approximation : [OH - - ] = [CH + ] = x + ] = [CH 3 NH 3 ] = x 2 (unknown) [OH 3 NH 2 (unknown) 3 + + OH - (weak base) → CH → O → → → → → → + + OH - CH 3 NH 2 + H 2 CH 3 NH 3 (weak base) CH 3 NH 2 + H 2 O 3 NH 3 C o - x x 1 x 2 x 2 C o - x x 1 2 2 ( x 2 ) 2 ( x 2 ) 2 ] ≅ ≅ ≅ C ≅ ≅ ≅ ≅ ≅ [CH 3 NH 2 C o - x x 1 = [CH 3 NH 2 ] o - 1 = K b = ( C o − x 1 ) = − 3 92.9 × 10 × × 10 × × × × × × 10 7.1) × × × × 3 = 92.9 = 92.9 × × × × -3 -3 3 10 - 10 - (100 - - 7.1) (100 2 = (92.9 = (92.9 × × 10 × × × × )(5.0 × × 10 × × × × ) = 46.45 × × 10 × × × × × × × × × × ) 2 10 - -3 3 )(5.0 10 - -4 4 ) = 46.45 10 - -6 6 (x 2 (x 2 ) × 10 × × × 10 = 6.8 × × × × 3 (Note x = 7.1 × × × × 3 see above) × × × × 10 - -3 10 - -3 x 2 (Note x 1 see above) x 2 = 6.8 1 = 7.1

≅ C ≅ ] must equal ≅ ≅ ≅ ≅ ≅ ≅ = 93.2x10 - -3 3 Thus [CH 3 NH 2 C o -x x 2 Thus [CH 3 NH 2 ] must equal o - 2 = 93.2x10 -3 3 We approximated [CH 3 NH 2 ] as C o -x x 1 = 92.9x10 - We approximated [CH 3 NH 2 ] as C o - 1 = 92.9x10 Value we chose was only off by 0.3 out of 93.2 (about 0.3%) Value we chose was only off by 0.3 out of 93.2 (about 0.3%) Try further iterations (x 3 ) and will find no change in values to three significant figures.

Hydrolysis Hydrolysis + + → → H → → × × × × 10 O → → → → = 1.85 × × × × - O + OAc - -5 5 HOAc + H + H 2 H 3 + OAc K a 10 - HOAc 2 O 3 O K a = 1.85 - =CH OAc - COO - - HOAc=CH =CH 3 COOH, OAc =CH 3 HOAc 3 COOH, 3 COO - is a (conjugate) base: OAc - HOAc is an acid, therefore is an acid, therefore OAc is a (conjugate) base: HOAc OAc - - Na + + , Add NaOAc NaOAc to H to H 2 O, which dissociates completely to Na , OAc Add 2 O, which dissociates completely to - + H OAc - is a proton acceptor. → → HOAc → → O → → → → - OAc - + OH - + H 2 HOAc + OH OAc 2 O K h = [ HOAc ][ OH − ] (“ “hydrolysis hydrolysis” ” constant) constant) ( − ] [ OAc

by 1 = [ H 3 O + ] Multiply K Multiply K h h by 1 = + ] [ H 3 O − ][ H 3 O + ] [ HOAc ][ OH - ][H + ] K h = , but K K w = [OH - ][H 3 O + ] K h = , but w = [OH 3 O − ][ H 3 O + ] [ OAc [ HOAc ] + ] = K w K h = K K w , where K h = − ][ H 3 O , where w [ OAc K a - ][H + ]/[ OAc - O + K a =[OAc ][H 3 ]/[HOAc HOAc] is the ionization constant for ] is the ionization constant for K a =[ 3 O the reaction: the reaction: + + O + OAc - - HOAc + H + H 2 O = H 3 + OAc HOAc 2 O = H 3 O 5 → → → K → → → → → =10 - -14 14 , =1.85x10 - -5 =5.4x10 - -10 10 K w , K K a K h K w =10 a =1.85x10 h =5.4x10 - combines OAc - Note: K K h is very small, indicating that little OAc combines Note: h is very small, indicating that little With H 2 With H 2 O to form O to form HOAc HOAc Since K K h =K K w /K K a , the smaller K K a (weaker the acid) the Since h = w / a , the smaller a (weaker the acid) the - is more extensively hydrolyzed) � - OAc � Larger K K h (or OAc is more extensively hydrolyzed) Larger h (or