Acid-Base Chemistry. Brønsted-Lowry Acids & Bases. Conjugate Acids & Bases. Conjugate Acids & Bases 7/6/12

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1 AcidBase Chemistry BrønstedLowry Acids & Bases n There are a couple of ways to define acids and bases n BrønstedLowry acids and bases n Acid: H + ion donor n Base: H + ion acceptor n Lewis acids and bases n Acid: electron pair acceptor n Base: electron pair donor BrønstedLowry Acids & Bases BrønstedLowry Acids & Bases n In most acidbase systems, water may play a role as either an acid (H + donor) or a base (H + acceptor) Water as an acid H O(l) + B(aq) OH (aq) + HB + (aq) Water as a base H O(l) + HA(aq) H 3 O + (aq) + A (aq) Conjugate Acids & Bases n Acids react with bases and vice versa n All acids and bases come with a conjugate pair a base or acid, respectively, that is formed in conjunction with the original species Examples HCl(aq) + H O(l) H 3 O + (aq) + Cl (aq) acid base conjugate conjugate acid base Conjugate Acids & Bases Examples NaOH(aq) + H O(l) OH (aq) + H O(l) + Na + (aq) base acid conjugate conjugate base acid CH 3 COOH(aq) + H O(l) CH 3 COO (aq) + H 3 O + (aq) acid base conjugate conjugate base acid 1

2 Strengths of Acids and Bases n Strong acids donate H + ions more easily n The stronger the acid, the weaker the conjugate base associated with that acid n Strong bases accept H + ions more easily n The stronger the base, the weaker the conjugate acid associated with that base Strengths of Acids and Bases n Stronger acids will always react to form weaker conjugate bases n Stronger bases will always react to form weaker conjugate acids Example H SO 4 (aq) + NaOH(aq) + H O(l)??? sulfuric acid can react with either OH or H O which would it prefer? H SO 4 (aq) + OH (aq) HSO 4 (aq) + H O(l) H SO 4 (aq) + H O(l) HSO 4 (aq) + H 3 O + (aq) preferred rxn because H O is a weaker conjugate acid than H 3 O +. Autoionization of Water n Water always undergoes some degree of dissociation to form H 3 O + ions and OH ions H O(l) H 3 O + (aq) + OH (aq) n The equilibrium constant for this process at 5 o C is: K w [H 3 O + ][OH ] 1.0 x n In pure water [H 3 O + ] [OH ] 1.0 x 10 7 M Autoionization of Water n K w is temperature dependent it increases with increasing temperature Autoionization of Water Example Determine [H 3 O + ] in a M NaOH solution Step 1: since NaOH is a strong base, dissociation is complete [OH ] M Step : Use K w to calculate [H 3 O + ] + 14 Kw [H3O ][OH ] 1.0 x K 1.0 x [H O ] w x 10 M [OH ] 0.053

3 n ph is a measure of the hydronium ion content of a solution n ph is defined as: ph log[h 3 O + ] log is log base 10, not ln (natural log) [H 3 O + ] is given in molar units (M) n ph of pure water ([H 3 O + ] 1.0 x 10 7 M): ph log(1.0x10 7 ) 7.00 n ph of last example ([H 3 O + ] 1.9 x M): ph log(1.9x10 13 ) 1.7 n Neutral is defined as the ph of pure water: ph 7 n Acidic solutions have ph lower than 7: ph < 7 acidic n Basic solutions have ph larger than 7: ph > 7 basic n We can also use poh to describe a solution n poh is defined as: poh log[oh ] n The sum of ph and poh must equal ph + poh assuming room temperature (5 o C) Figure 16 Example Find [H 3 O + ] of a solution that has poh 9.37 Method 1: Calculate ph, then [H 3 O + ] Step 1: Determine ph ph poh Step : Determine [H 3 O + ] [H 3 O + ] 10 ph x 10 5 M Example (con t.) Find [H 3 O + ] of a solution that has poh 9.37 Method : Calculate [OH ], then [H 3 O + ] using K w Step 1: Determine [OH ] [OH ] 10 poh x M Step : Determine [H 3 O + ] using K w [H 3 O + ] K w /[OH ] (1.0x10 14 )/(4.3x10 10 ).3 x 10 5 M 3

4 n The extent of dissociation of an acid or base in H O can be quantified using its ionization constant n K a is a specific equilibrium constant n Acids: HA(aq) + H O(l) H 3 O + (aq) + A (aq) acid base c. acid c. base + + [H O ][A ] [H O ][conjugate base] K 3 3 a [HA] [acid] [HA] undissociated acid in solution Example: Acetic acid has a K a 1.8 x 10 5 Determine ph of a 0.0 M acetic acid solution CH 3 COOH(aq) + H O(l) CH 3 COO (aq) + H 3 O + (aq) We can approach this like any equilibrium problem from Chap. 13 CH 3 COOH CH 3 COO H 3 O + initial Δ x x x equil.0 x x x Example (con t.): K a 1.8 x x [CH3COO ][H3O ] [CH COOH] x.0 x 3 + x M ph log[h 3 O + ] log(.0019).7 n K b is a specific equilibrium constant for bases n Bases: B(aq) + H O(l) HB + (aq) + OH (aq) base acid c. acid c. base + [HB ][OH ] [OH ][conjugate acid] Kb [B] [base] [B] unreacted base in solution Example: Determine [B] in a 1.8 x 10 3 M solution of NH 3 NH 3 (aq) + H O(l) NH 4+ (aq) + OH (aq) NH 3 NH 4 + OH initial 1.8x Δ x x x equil 1.8x10 3 x x x 4

5 Example: NH 3 (aq) + H O(l) NH 4+ (aq) + OH (aq) + 5 [NH ][OH ] x K 1.8x10 4 b [NH ] x10 x x 1.7 x 10 4 M [NH 4+ ] [OH ] [NH 3 ] 1.8 x 10 3 M 1.7 x 10 4 M 1.65 x 10 3 M Polyprotic Acids n Some acids contain more than one hydrogen atom that may be donated to form H + ion n These are called polyprotic acids n Examples include: H SO 4 sulfuric acid ( H + ions) H 3 PO 4 phosphoric acid (3 H + ions) H CO 3 carbonic acid ( H + ions) Polyprotic Acids n Each H atom has a unique K a associated with its release to form H + ion n Consider phosphoric acid: H 3 PO 4 (aq) + H O(l) H PO 4 (aq) + H 3 O + (aq) 1 st K a1 7.5 x 10 3 H PO 4 (aq) + H O(l) HPO 4 (aq) + H 3 O + (aq) nd K a 6. x 10 8 HPO 4 (aq) + H O(l) PO 3 4 (aq) + H 3 O + (aq) 3 rd K a3 3.6 x n The first H atom is easiest to pull off, so it has the higher K a value Strengths of Acids n Acid strength is determined by a combination of factors: n Bond polarity the HA bond must be polar in order for the H atom to be transferred to water as H + The H atom in CH 4 is nonacidic because the CH bond is not polar The HCl bond in HCl is polar, and HCl is a strong acid Strengths of Acids n Acid strength is determined by a combination of factors: n Bond strength the stronger the bond, the weaker the acid it is harder to pull away the H atom to form H + Acid Bond Energy K a HF 617 kj/mol 7. x 10 4 HCl 47 kj/mol ~10 6 HBr 36 kj/mol ~10 8 HI 95 kj/mol ~10 9 Strengths of Acids n Oxoacids are those with a HOZ linkage n Electronegativity of Z the higher the electronegativity of Z, the stronger the acid Acid χ Z K a HOCl x 10 8 HOBr.8.5 x 10 9 HOI.5.3 x

6 Strengths of Acids n Oxoacids are those with a HOZ linkage n The more O atoms attached to Z, the stronger the acid Acid O atoms K a HOCl x 10 8 HOClO 1.1 x 10 HOClO 3 ~10 3 HOClO 3 4 ~10 8 Relationship Between K a and K b n The K a of an acid and the K b of its conjugate base are related: HA(aq) + H O(l) A (aq) + H 3 O + (aq) acid c. base + [A ][H O ] K 3 a [HA] A (aq) + H O(l) HA(aq) + OH (aq) base c. acid [HA][OH ] Kb [A ] Relationship Between K a and K b n The K a of an acid and the K b of its conjugate base are related: n The product K a x K b K w + [A ][H O ] [HA][OH ] K K 3 a b [HA] [A ] + [H 3O ][OH ] Kw This is true of any conjugate pair of acid and base n When an acid and a base undergo an exchange reaction, the result is a salt and water: HX(aq) + MOH(aq) MX(aq) + H O acid base salt n If a strong base is neutralized with a strong acid, the resulting solution contains only the salt HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) Determine the ph of 0.84 M solution of sodium acetate, NaCH 3 COO NaCH 3 COO(aq) + H O Na + (aq) + CH 3 COOH(aq) + OH (aq) Complete ionic equation: Na + (aq) + CH 3 COO (aq) + H O Na + (aq) + CH 3 COOH(aq) + OH (aq) [CH3COOH][OH ] Kw 10 K b 5.56 x 10 [CH K 3COO ] a Determine the ph of 0.84 M solution of sodium acetate, NaCH 3 COO [CH 3 COO ] [CH 3 COOH] [OH ] initial Δ x x x equil 0.84 x x x 10 x 5.56 x x assume x is negligible 10 5 x 1.58 x 10 x 1.6 x 10 M [OH ] poh log(1.6x10 5 ) ph

7 Weak acid and salt of a strong base: n If stoichiometric amounts are combined, the solution will be slightly basic Determine ph of a solution with 0.00 M NaOH and 0.00 M acetic acid CH 3 COOH(aq) + NaOH(aq) NaCH 3 COO(aq) + H O Complete ionic equation: CH 3 COOH(aq) + Na + (aq) + OH (aq) Na + (aq) + CH 3 COO (aq) + H O Sodium is a spectator ion and does not actively participate in the reaction CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H O The equilibrium constant, K eq, for this reaction is given by: + [CH3COO ] [CH3COO ][H3O ] Ka Keq + [CH3COOH][OH ] [CH3COOH][OH ][H3O ] K w + [CH3COO ][H3O ] + Ka and K w [H3O ][OH ] [CH3COOH] x 10 9 Keq 1.80 x x 10 CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H O CH 3 COOH OH CH 3 COO initial Δ x x x equil x x 0.00 x CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H O [OH ] [CH 3 COOH] 3.3 x 10 6 M poh log(3.3 x 10 6 ) 5.48 ph [CH3COO ] 0.00 x K 1.8 x 10 [CH3COOH][OH ] x x 1.11x 10 x 3.33 x x 10 assume x is negligible Salts of weak bases: A principle ingredient of chlorine bleach is sodium hypochlorite, NaOCl Determine the ph of a solution containing 1.00% by weight NaOCl MW g/mol K b (ClO ).9 x 10 7 (1.00 g NaOCl)/(74.44 g/mol).0134 mol NaOCl (99.00 g H O)/(.9971 g/ml) 99.9 ml H O [ClO ] (.0134 mol)/(.0999 L).135 M Determine the ph of a solution containing 1.00% by weight NaOCl ClO (aq) + H O HOCl(aq) + OH (aq) [HOCl][OH ] 7 K b.9 x 10 [ClO ] [ClO ] [HOCl] [OH ] initial Δ x x x equil.135 x x x 7

8 Determine the ph of a solution containing 1.00% by weight NaOCl ClO (aq) + H O HOCl(aq) + OH (aq) 7 x 4.9 x 10 x 1.98 x x [OH ] 1.98 x 10 4 M poh log(1.98 x 10 4 ) ph assume x is negligible Lewis Acids and Bases n Remember: n Lewis Acid electron pair acceptor n Lewis Base electron pair donor n A Lewis acid reacts with a Lewis base to form a coordinate covalent bond.. N H H H H + + H + N H H H coordinate covalent bond NH 3 + H + NH 4 + Lewis Acids and Bases n Metal cations are potential Lewis acids n Any empty valence orbitals can accept an electron pair from a Lewis base to form a coordinate covalent compound n If the coordinate covalent compound is an ion, it is called a complex ion Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) Lewis Acids and Bases n One electron pair on the oxygen atom in water can act a Lewis base to hydrated metal ions Fe H O Fe(H O) 6 3+ (aq) n OH is a good Lewis base Al(OH) 3 (s) + OH (aq) Al(OH) 4 (aq) electron pair from OH inserts into the empty p orbital on aluminum to form covalent bond 8