Zumdahl’s Chapter 14 AcidsBasesAcidsBases

Chapter Contents Acid-Base Models Acid-Base Models Acidity and K a Acidity and K a pH, pOH, and pK a pH, pOH, and pK a Calculating pH Calculating pH Dominant Sources Dominant Sources Polyprotic Acids Polyprotic Acids % dissociation % dissociation Bases and K b Bases and K b Conjugate Acids and Bases Structure and Strength Oxoacids Lewis Acids Acid-Base Solution Al Gore Rhythms

Acid-Base Models Definition of Acid Example of Acid Definition of Base Example of Base SvanteArrhenius Donates H + HCl Donates OH – NaOH Brøsted- Lowry Donates H + HCl Accepts H + NH 3 or H 2 O G.N.Lewis Accepts electron BF 3 Donates electron : NH 3

Caustic Characteristics Greek: kaustikos  kaiein “to burn” Greek: kaustikos  kaiein “to burn” Acids & bases corrode nearly everything Acids & bases corrode nearly everything HA(aq) + H 2 O( l )  H 3 O + (aq) + A – (aq) HA(aq) + H 2 O( l )  H 3 O + (aq) + A – (aq) HA “acid” and A – “conjugate base” HA “acid” and A – “conjugate base” H 2 O “base” and H 3 O + “conjugate acid” H 2 O “base” and H 3 O + “conjugate acid” H 3 O + “hydronium ion” no free protons! H 3 O + “hydronium ion” no free protons! K a = [H + ] [A – ] / [HA] “acid dissociation contant” K a = [H + ] [A – ] / [HA] “acid dissociation contant”

Acidity and K a Strong acid: K a   means [HA] eq = 0 Strong acid: K a   means [HA] eq = 0 Weak acid: K a  0 or [HA] eq  [HA] 0 Weak acid: K a  0 or [HA] eq  [HA] 0 A – (aq) + H 3 O + (aq)  HA(aq) + H 2 O( l ) A – (aq) + H 3 O + (aq)  HA(aq) + H 2 O( l ) K=K a –1 or weak acid is associated with a strong conjugate base and vice versa. K=K a –1 or weak acid is associated with a strong conjugate base and vice versa. AcidsHSO 4 – HC 2 H 3 O 2 HCN AcidsHSO 4 – HC 2 H 3 O 2 HCN K a 1.2  10 –2 1.8  10 –5 6.2  10 –10 K a 1.2  10 –2 1.8  10 –5 6.2  10 –10

pH, pOH, and pK a pX  – log 10 (X) pX  – log 10 (X) Reduces K w =  10 –14 to pK w = to a manageable entity for comparison Reduces K w =  10 –14 to pK w = to a manageable entity for comparison Actually pK w = 14 is usually memorized. Actually pK w = 14 is usually memorized. 2 H 2 O( l )  H 3 O + (aq) + OH – (aq) K = K w 2 H 2 O( l )  H 3 O + (aq) + OH – (aq) K = K w pH+pOH=pK w because K w = [H + ][OH – ] pH+pOH=pK w because K w = [H + ][OH – ] Acid (pH 7) Acid (pH 7) E.g., pH(0.1 M acetic acid) = 2.87 E.g., pH(0.1 M acetic acid) = 2.87

Calculating pH Strong acids Strong acids [H + ] eq = [HA] 0 [H + ] eq = [HA] 0 pH = p[HA] 0 pH = p[HA] 0 Strong bases Strong bases [OH – ] eq = [BOH] 0 [OH – ] eq = [BOH] 0 pOH = p[BOH] 0 pOH = p[BOH] 0 pH = 14 – pOH pH = 14 – pOH Don’t be fooled by [HA] 0 < 10 –7 Don’t be fooled by [HA] 0 < 10 –7 Weak Acids K a = [H + ][A – ] / [HA] K a = x 2 / ( [HA] 0 – x) K a  x 2 / [HA] 0 x  ( K a [HA] 0 ) ½ pH = – log 10 (x) Weak Bases x  ( K b [MOH] 0 ) ½ pH = 14 - px

Dominant Sources If more than one H + source (multiple acids or a polyprotic acid), pK a will usually differ by a few units, and the largest will determine [H + ] eq. E.g., 0.01M H 3 PO 4 If more than one H + source (multiple acids or a polyprotic acid), pK a will usually differ by a few units, and the largest will determine [H + ] eq. E.g., 0.01M H 3 PO 4 pK a : 2.12, 7.21, and 12.32, respectively pK a : 2.12, 7.21, and 12.32, respectively Use K a for 1 st H only: pH = 2.24 (quadratic formula) Use K a for 1 st H only: pH = 2.24 (quadratic formula) Determines H 3 PO 4, H 2 PO 4 –, and H + Determines H 3 PO 4, H 2 PO 4 –, and H + Use other two pK a to find HPO 4 2– and PO 4 3– Use other two pK a to find HPO 4 2– and PO 4 3–

% Dissociation Weak electrolyte present in molecular form mostly with few ions & will have a small % dissociation unless diluted. Weak electrolyte present in molecular form mostly with few ions & will have a small % dissociation unless diluted. 100%  [ionic form] / [initial amount] 100%  [ionic form] / [initial amount] E.g., 0.1 M acetic acid = [initial amount] E.g., 0.1 M acetic acid = [initial amount] [acetate ion]  (0.10 K a ) ½ = M [acetate ion]  (0.10 K a ) ½ = M % dissoc.  100%  (K a /0.1) ½ = 1.3% % dissoc.  100%  (K a /0.1) ½ = 1.3% 0.01 M gives 4.2% shows more ionization upon dilution (but pushes the “5% approx.”) 0.01 M gives 4.2% shows more ionization upon dilution (but pushes the “5% approx.”)

Bases and K b Strong base: [OH – ] eq = [MOH] 0 Strong base: [OH – ] eq = [MOH] 0 where M=metal and pH = 14 - pOH where M=metal and pH = 14 - pOH Weak bases often amines, : NR 3 Weak bases often amines, : NR 3 where R = H and/or organic groups where R = H and/or organic groups Amine steals H + from H 2 O to leave OH – Amine steals H + from H 2 O to leave OH – K b = [NR 3 H + ] [OH – ] / [ : NR 3 ] K b = [NR 3 H + ] [OH – ] / [ : NR 3 ] For NH 3 (ammonia, not an amine), K b = 1.8  10 –5 For NH 3 (ammonia, not an amine), K b = 1.8  10 –5 Same pH rules & approximations apply. Same pH rules & approximations apply.

Conjugate Acids and Bases “Salt of a weak (acid/base) is itself a weak (base/acid).” “Salt of a weak (acid/base) is itself a weak (base/acid).” Example of a weak base, NH 3 Example of a weak base, NH 3 NH OH –  NH 3 + H 2 O K = K b –1 NH OH –  NH 3 + H 2 O K = K b –1 H 2 O  H + + OH – K = K w (add eqns) H 2 O  H + + OH – K = K w (add eqns) NH 4 +  NH 3 + H + K conjugate = K w / K b NH 4 +  NH 3 + H + K conjugate = K w / K b K acid (NH 4 + ) = 10 –14 /1.8  10 –5 = 5.5  10 –10 K acid (NH 4 + ) = 10 –14 /1.8  10 –5 = 5.5  10 –10 Weaker parent yields stronger conjugate! Weaker parent yields stronger conjugate!

Conjugate pH Calculations Salts of strong parents are neutral; those of very weak bases, very acidic: Salts of strong parents are neutral; those of very weak bases, very acidic: Fe(H 2 O) 6 3+  Fe(H 2 O) 5 OH 2+ + H + Fe(H 2 O) 6 3+  Fe(H 2 O) 5 OH 2+ + H + pK a1 = 2.2 or K a1 = 6.3  10 –3 (pK b = 11.8) pK a1 = 2.2 or K a1 = 6.3  10 –3 (pK b = 11.8) pH of 3 M Fe(NO 3 ) 3 ? (HNO 3 strong; ignore!) pH of 3 M Fe(NO 3 ) 3 ? (HNO 3 strong; ignore!) K a1 = [FeOH 2+ ] [H + ] / [Fe 3+ ] = x 2 / (3-x) K a1 = [FeOH 2+ ] [H + ] / [Fe 3+ ] = x 2 / (3-x) x  [3K a1 ] ½ = 0.14 (5% OK)  pH = 0.86 x  [3K a1 ] ½ = 0.14 (5% OK)  pH = 0.86 pK a2 = 3.5, pK a3 = 6, pK a4 = 10 pK a2 = 3.5, pK a3 = 6, pK a4 = 10

NON-Dominant Situations In that Fe complex, pK a1 and pK a2 are very close (2.2 and 3.5), so the second conjugate may be important: In that Fe complex, pK a1 and pK a2 are very close (2.2 and 3.5), so the second conjugate may be important: Fe(H 2 O) 5 OH 2+  Fe(H 2 O) 4 (OH) H + Fe(H 2 O) 5 OH 2+  Fe(H 2 O) 4 (OH) H + This will certainly be the case at lower formal iron concentrations, F, since % dissociation is greater there. Try F=0.01 This will certainly be the case at lower formal iron concentrations, F, since % dissociation is greater there. Try F=0.01

Simplifing Approximations Even at F=0.01, we can safely ignore pK a3, pK a4, and pK w, since they are dominated by pK a1 and pK a2. Even at F=0.01, we can safely ignore pK a3, pK a4, and pK w, since they are dominated by pK a1 and pK a2. Abbreviate those equilibria as Abbreviate those equilibria as A 3+  B 2+ + H + K 1 = B H / A A 3+  B 2+ + H + K 1 = B H / A B 2+  C + + H + K 2 = C H / A B 2+  C + + H + K 2 = C H / A But those are 2 eqns in 4 unknowns! But those are 2 eqns in 4 unknowns!

2 Additional Conditions! Conservation of iron requires that Conservation of iron requires that F = A + B + C F = A + B + C Conservation of charge must include a spectator anion for A 3+ such as 3 Cl – Conservation of charge must include a spectator anion for A 3+ such as 3 Cl – 3 A B 2+ + C + + H + = 3 F (conc. Cl – ) 3 A B 2+ + C + + H + = 3 F (conc. Cl – ) Notice that higher charges get scaled by their charge for proper balance. Notice that higher charges get scaled by their charge for proper balance.

Solve 4 Nonlinear Equations Original equations: Original equations: F = A + B + C F = A + B + C 3F = 3A + 2B + C + H 3F = 3A + 2B + C + H B H = K 1 A B H = K 1 A C H = K 2 B C H = K 2 B Remove A by substituting from 1 st eqn Remove A by substituting from 1 st eqn A = F – B – C A = F – B – C

Solve 3 Nonlinear Equations Remaining after substitution Remaining after substitution 3F = 3F – 3B – 3C + 2B + C + H 3F = 3F – 3B – 3C + 2B + C + H B H = K 1 ( F – B – C ) B H = K 1 ( F – B – C ) C H = K 2 B C H = K 2 B Simplify Simplify H = B + 2C H = B + 2C B H = K 1 F – K 1 B – K 1 C B H = K 1 F – K 1 B – K 1 C C H = K 2 B  substitute C = K 2 B / H C H = K 2 B  substitute C = K 2 B / H

Solve 2 Nonlinear Equations H = B + 2 K 2 B / H H = B + 2 K 2 B / H B H = K 1 F – K 1 B – K 1 K 2 B / H B H = K 1 F – K 1 B – K 1 K 2 B / H Simplify Simplify H 2 = B H + 2 K 2 B  B = H 2 / (H + 2K 2 ) H 2 = B H + 2 K 2 B  B = H 2 / (H + 2K 2 ) B H 2 = K 1 F H – K 1 B H – K 1 K 2 B B H 2 = K 1 F H – K 1 B H – K 1 K 2 B Substitute for B Substitute for B H 3 + K 1 H 2 – K 1 (F – K 2 )H – 2 K 1 K 2 F = 0 H 3 + K 1 H 2 – K 1 (F – K 2 )H – 2 K 1 K 2 F = 0 Only H remains as an unknown! Only H remains as an unknown!

Solve 1 Cubic Equation Use the dominant solution as first guess in an iteration to the non-dominant one. Use the dominant solution as first guess in an iteration to the non-dominant one. A 3+  B 2+ + H + K 1 = B H / A A 3+  B 2+ + H + K 1 = B H / A K 1 = x 2 / (F – x) K 1 = x 2 / (F – x) x 2 + K 1 x – F K 1 = 0 x 2 + K 1 x – F K 1 = 0 x = ( – K 1 + [K F K 1 ] ½ ) / 2 x = ( – K 1 + [K F K 1 ] ½ ) / 2 x = 5.3  10 –3  H x = 5.3  10 –3  H F = 0.01, K 1 = 6  10 –3, and K 2 = 3  10 – 4 F = 0.01, K 1 = 6  10 –3, and K 2 = 3  10 – 4

H  10 –3 H 2 – 5.8  10 –3 H – 3.6  10 –8 = 0 H out = (3.6  10 –  10 –3 H in – 6  10 –3 H in 2 ) 1/3 H in (the guess) H out (the refinement) 5.3  10 –3 5.6  10 –  10 –3  [H + ] = 5.58  10 – 3 M and by B = H 2 / (H + 2K 2 ), [B 2+ ] = 5.04  10 – 3 M and by C = K 2 B / H, [B 2+ ] = 5.04  10 – 3 M and by C = K 2 B / H, [C + ] = 2.7  10 – 3 M which makes [A 3+ ] = 4.7  10 – 3 M [C + ] = 2.7  10 – 3 M which makes [A 3+ ] = 4.7  10 – 3 M

More Conjugate Calculations HCO 3 –  H + + CO 3 2– pK a2 = HCO 3 –  H + + CO 3 2– pK a2 = pH of 0.10 M Na 2 CO 3 ? (NaOH strong; ignore) pH of 0.10 M Na 2 CO 3 ? (NaOH strong; ignore) CO 3 2– + H 2 O  HCO 3 – + OH – K = K w /K a2 CO 3 2– + H 2 O  HCO 3 – + OH – K = K w /K a2 K = [HCO 3 – ] [OH – ] / [CO 3 2– ] = x 2 / (0.10-x) K = [HCO 3 – ] [OH – ] / [CO 3 2– ] = x 2 / (0.10-x) x  [0.10  1.8  10 –4 ] ½ = 4.2  10 –3 (5% OK) x  [0.10  1.8  10 –4 ] ½ = 4.2  10 –3 (5% OK) pOH = 2.37  pH = – 2.37 = pOH = 2.37  pH = – 2.37 = Sodium carbonate is in laundry soap. Sodium carbonate is in laundry soap.

Acid Structure and Strength Higher polarity liberates protons. Higher polarity liberates protons. While all hydrogen halides are strong acids (save HF), acidity lessens with halide size. While all hydrogen halides are strong acids (save HF), acidity lessens with halide size. Even non-adjacent electronegativity drains e – density, liberating protons. Even non-adjacent electronegativity drains e – density, liberating protons. HClO is weak but HClO 4 is strong. HClO is weak but HClO 4 is strong. (the H is attached to only 1 of the oxygens) (the H is attached to only 1 of the oxygens) Cl 3 CCO 2 H is much stronger than CH 3 CO 2 H Cl 3 CCO 2 H is much stronger than CH 3 CO 2 H

OxoacidsOxoacids German: O 2 = sauerstoff or the stuff that makes sour oxoacids, –X–O–H German: O 2 = sauerstoff or the stuff that makes sour oxoacids, –X–O–H True when X is a fairly electronegative non-metal (halogen, N, P, S) or a metal in a high oxidation state (e.g., H 2 CrO 4 ) True when X is a fairly electronegative non-metal (halogen, N, P, S) or a metal in a high oxidation state (e.g., H 2 CrO 4 ) True as nonadjacent electronegativity aids polarity (e.g., CH 3 CO 2 H) True as nonadjacent electronegativity aids polarity (e.g., CH 3 CO 2 H)

Oxygen-containing Bases M – O – H M – O – H Metals in lower oxidation states tend to give up the OH – and not the H + Metals in lower oxidation states tend to give up the OH – and not the H + Sr(OH) 2  Sr OH – Sr(OH) 2  Sr OH – Al(OH) 3 is basic; Al 3+ is acidic. Al(OH) 3 is basic; Al 3+ is acidic. Water is amphoteric too, acting as a base by becoming H 3 O +, for example. Water is amphoteric too, acting as a base by becoming H 3 O +, for example. –Brønsted definition: water accepts the proton.

Lewis Acids and Bases Water is both acid & base by Lewis too. Water is both acid & base by Lewis too. Base via H 2 O : + H +  H 3 O + Base via H 2 O : + H +  H 3 O + donates lone pair e – to proton donates lone pair e – to proton Acid via H 2 O + S 2–  OH – + HS – Acid via H 2 O + S 2–  OH – + HS – accepts the one of sulfide’s electrons accepts the one of sulfide’s electrons Lewis’ is the most general definition Lewis’ is the most general definition E.g., Cu : NH 3  Cu(NH 3 ) 6 2+ E.g., Cu : NH 3  Cu(NH 3 ) 6 2+ Cu’s the acid and : NH 3 ’s the base. Cu’s the acid and : NH 3 ’s the base.

Acid-Base Algorithm Know Know the significant species for the conditions. Ignore Ignore the rest. For For quantitative reactions, keep only the products. Find Find acids & bases. Solve controlling K Write K expression. Set initial conditions for smallest change. Define all species in terms of x (change). Find x solving K exp. Check any approx. Find pH & [species]