Acid-Base Chemistry. Brønsted-Lowry Acids & Bases

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1 Acid-Base Chemistry ν There are a couple of ways to define acids and bases ν Brønsted-Lowry acids and bases ν Acid: H + ion donor ν Base: H + ion acceptor ν Lewis acids and bases ν Acid: electron pair acceptor ν Base: electron pair donor Brønsted-Lowry Acids & Bases 1

2 Brønsted-Lowry Acids & Bases Brønsted-Lowry Acids & Bases ν In most acid-base systems, water may play a role as either an acid (H + donor) or a base (H + acceptor) Water as an acid H 2 O(l) + B(aq) OH - (aq) + HB + (aq) Water as a base H 2 O(l) + HA(aq) H 3 O + (aq) + A - (aq) 2

3 Conjugate Acids & Bases ν Acids react with bases and vice versa ν 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 2 O(l) H 3 O + (aq) + Cl - (aq) acid base conjugate conjugate acid base Conjugate Acids & Bases Examples NaOH(aq) + H 2 O(l) OH - (aq) + H 2 O(l) + Na + (aq) base acid conjugate conjugate base acid CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) acid base conjugate conjugate base acid 3

4 Strengths of Acids and Bases ν Strong acids donate H + ions more easily ν The stronger the acid, the weaker the conjugate base associated with that acid ν Strong bases accept H + ions more easily ν The stronger the base, the weaker the conjugate acid associated with that base 4

5 Strengths of Acids and Bases ν Stronger acids will always react to form weaker conjugate bases ν Stronger bases will always react to form weaker conjugate acids Example H 2 SO 4 (aq) + NaOH(aq) + H 2 O(l)??? sulfuric acid can react with either OH - or H 2 O which would it prefer? H 2 SO 4 (aq) + OH - (aq) HSO 4- (aq) + H 2 O(l) H 2 SO 4 (aq) + H 2 O(l) HSO 4- (aq) + H 3 O + (aq) preferred reaction Autoionization of Water ν Water always undergoes some degree of dissociation to form H 3 O + ions and OH - ions 2 H 2 O(l) H 3 O + (aq) + OH - (aq) ν The equilibrium constant for this process at 25 o C is: K w = [H 3 O + ][OH - ] = 1.0 x ν In pure water [H 3 O + ] = [OH - ] = 1.0 x 10-7 M 5

6 Autoionization of Water ν 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 2: Use K w to calculate [H 3 O + ] K w = [H 3 O + ][OH - ] = 1.0 x [H 3 O + ] = K w [OH - ] = 1.0 x = 1.9 x M 6

7 The ph Scale ν ph is a measure of the hydronium ion content of a solution ν 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) ν ph of pure water ([H 3 O + ] = 1.0 x 10-7 M): ph = -log(1.0x10-7 ) = 7.0 ν ph of last example ([H 3 O + ] = 1.9 x M): ph = -log(1.9x10-13 ) = 12.7 The ph Scale ν Neutral is defined as the ph of pure water: ph = 7 ν Acidic solutions have ph lower than 7: ph < 7 acidic ν Basic solutions have ph larger than 7: ph > 7 basic 7

8 Figure 16-2 The ph Scale ν We can also use poh to describe a solution ν poh is defined as: poh = -log[oh - ] ν The sum of ph and poh must equal 14 ph + poh = 14 assuming room temperature (25 o C) 8

9 The ph Scale 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 = 14 poh = = 4.63 Step 2: Determine [H 3 O + ] [H 3 O + ] = 10 -ph = = 2.34 x 10-5 M The ph Scale Example (con t.) Find [H 3 O + ] of a solution that has poh = 9.37 Method 2: Calculate [OH - ], then [H 3 O + ] using K w Step 1: Determine [OH - ] [OH - ] = 10 -poh = = 4.27 x M Step 2: Determine [H 3 O + ] using K w [H 3 O + ] = K w /[OH - ] = (1.0x10-14 )/(4.27x10-10 ) = 2.34 x 10-5 M 9

10 Ionization Constants ν The extent of dissociation of an acid or base in H 2 O can be quantified using its ionization constant ν Acids: HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid base c. acid c. base K a = [H 3O + ][A - ] [HA] = [H 3O + ][conjugate base] [acid] [HA] = undissociated acid in solution Ionization Constants ν K a is a specific equilibrium constant ν Acids: HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid base c. acid c. base K a = [H 3O + ][A - ] [HA] = [H 3O + ][conjugate base] [acid] [HA] = undissociated acid in solution 10

11 Ionization Constants Example: Acetic acid has a K a = 1.8 x 10-5 Determine the ph of a 0.2 M acetic acid solution CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) We can approach this like any equilibrium problem from Chap. 15 CH 3 COOH CH 3 COO - H 3 O + initial x x x equil.2 x x x Ionization Constants Example (con t.): K a = 1.8 x 10-5 = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] 1.8 x 10-5 = x 2.2 x x = M ph = -log[h 3 O + ] = -log(.0019) =

12 Ionization Constants ν K b is a specific equilibrium constant for bases ν Bases: B(aq) + H 2 O(l) HB + (aq) + OH - (aq) base acid c. acid c. base K b = [HB+ ][OH - ] [B] = [OH- ][conjugate acid] [base] [B] = undissociated base in solution Ionization Constants Example: Determine [B] in a 1.82 x 10-3 M solution of NH 3 NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH - (aq) NH 3 NH 4 + OH - initial 1.82x x x x equil 1.82x10-3 x x x 12

13 Ionization Constants Example: NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH - (aq) K b = 1.8x10-5 = [NH + 4][OH - ] [NH 3 ] = x x x x = 1.72 x 10-4 M = [NH 4+ ] = [OH - ] [NH 3 ] = 1.82 x 10-3 M 1.72 x 10-4 M = 1.65 x 10-3 M Polyprotic Acids ν Some acids contain more than one hydrogen atom that may be donated to form H + ion ν These are called polyprotic acids ν Examples include: H 2 SO 4 sulfuric acid (2 H + ions) H 3 PO 4 phosphoric acid (3 H + ions) H 2 CO 3 carbonic acid (2 H + ions) 13

14 Polyprotic Acids ν Each H atom has a unique K a associated with its release to form H + ion ν Consider phosphoric acid: H 3 PO 4 (aq) + H 2 O(l) H 2 PO 4- (aq) + H 3 O + (aq) 1 st K a1 = 7.5 x 10-3 H 2 PO 4- (aq) + H 2 O(l) HPO 4 2- (aq) + H 3 O + (aq) 2 nd K a2 = 6.2 x 10-8 HPO 4 2- (aq) + H 2 O(l) PO 4 3- (aq) + H 3 O + (aq)3 rd K a3 = 3.6 x ν The first H atom is easiest to pull off, so it has the higher K a value Strengths of Acids ν Acid strength is determined by a combination of factors: ν Bond polarity the H-A bond must be polar in order for the H atom to be transferred to water as H + The H atom in CH 4 is non-acidic because the C-H bond is not polar The H-Cl bond in HCl is polar, and HCl is a strong acid 14

15 Strengths of Acids ν Acid strength is determined by a combination of factors: ν 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 7.2 x 10-4 HCl 427 kj/mol ~10 6 HBr 362 kj/mol ~10 8 HI 295 kj/mol ~10 9 Strengths of Acids ν Oxoacids are those with a H-O-Z linkage ν Electronegativity of Z the higher the electronegativity of Z, the stronger the acid Acid χ Z K a HOCl x 10-8 HOBr x 10-9 HOI x

16 Strengths of Acids ν Oxoacids are those with a H-O-Z linkage ν The more O atoms attached to Z, the stronger the acid Acid O atoms K a HOCl x 10-8 HOClO x 10-2 HOClO 2 3 ~10 3 HOClO 3 4 ~10 8 Relationship Between K a and K b ν The K a of an acid and the K b of its conjugate base are related: HA(aq) + H 2 O(l) A - (aq) + H 3 O + (aq) acid c. base K a = [A- ][H 3 O + ] [HA] A - (aq) + H 2 O(l) HA(aq) + OH - (aq) base c. acid K b = [HA][OH ] [A - ] 16

17 Relationship Between K a and K b ν The K a of an acid and the K b of its conjugate base are related: ν The product K a x K b = K w K a K b = [A- ][H 3 O + ] [HA] [HA][OH ] [A ] + - 3O ][OH ] Kw = [H = This is true of any conjugate pair of acid and base Salts of Acids and Bases ν When an acid and a base undergo an exchange reaction, the result is a salt and water: HX(aq) + MOH(aq) MX(aq) + H 2 O acid base salt ν If a strong base is neutralized with a strong acid, the resulting solution contains only the salt HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 17

18 Salts of Acids and Bases Determine the ph of M solution of sodium acetate, NaCH 3 COO NaCH 3 COO(aq) + H 2 O Na + (aq) + CH 3 COOH(aq) + OH - (aq) Complete ionic equation: Na + (aq) + CH 3 COO - (aq) + H 2 O Na + (aq) + CH 3 COOH(aq) + OH - (aq) K = [CH 3 COOH][OH- ] [CH 3 COO - ] = K w K a = 5.56 x Salts of Acids and Bases Determine the ph of M solution of sodium acetate, NaCH 3 COO [CH 3 COO - ] [CH 3 COOH] [OH - ] initial x x x equil x x x 5.56 x = x x assume x is negligible x 2 = 1.58 x x = 1.26 x 10-5 M = [OH - ] poh = -log(1.26x10-5 ) = 4.90 ph =

19 Salts of Acids and Bases Weak acid and salt of a strong base: ν If stoichiometric amounts are combined, the solution will be slightly basic Determine ph of a solution with M NaOH and M acetic acid CH 3 COOH(aq) + NaOH(aq) NaCH 3 COO(aq) + H 2 O Complete ionic equation: CH 3 COOH(aq) + Na + (aq) + OH - (aq) Na + (aq) + CH 3 COO - (aq) + H 2 O Sodium is a spectator ion and does not actively participate in the reaction Salts of Acids and Bases CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O CH 3 COOH OH - CH 3 COO - initial x x -x equil x x x K = 1.8 x 10 9 = [CH 3 COO - ] [CH 3 COOH][OH - ] = x x 2 assume x is negligible x 2 = = 1.11 x x 10 x = 3.33 x

20 Salts of Acids and Bases CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O [OH - ] = [CH 3 COOH] = 3.3 x 10-6 M poh = -log(3.3 x 10-6 ) = 5.5 ph = = 8.5 Salts of Acids and Bases 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 - ) = 2.9 x 10-7 (1.00 g NaOCl)/(74.44 g/mol) =.0134 mol NaOCl (99.00 g H 2 O)/(.9971 g/ml) = ml H 2 O [ClO - ] = (.0134 mol)/( L) =.135 M 20

21 Salts of Acids and Bases Determine the ph of a solution containing 1.00% by weight NaOCl ClO - (aq) + H 2 O HOCl(aq) + OH - (aq) K b = [HOCl][OH- ] [ClO - ] = 2.9 x 10-7 [ClO - ] [HOCl] [OH - ] initial x x x equil.135 x x x Salts of Acids and Bases Determine the ph of a solution containing 1.00% by weight NaOCl ClO - (aq) + H 2 O HOCl(aq) + OH - (aq) 2.9 x 10-7 = x x x = 1.98 x 10-4 assume x is negligible [OH - ] = 1.98 x 10-4 M poh = -log(1.98 x 10-4 ) = 3.70 ph =

22 Lewis Acids and Bases ν Remember: ν Lewis Acid = electron pair acceptor ν Lewis Base = electron pair donor ν A Lewis acid reacts with a Lewis base to form a coordinate covalent bond.. N H H H H + + H + N H H H NH 3 + H + NH 4 + coordinate covalent bond Lewis Acids and Bases ν Metal cations are potential Lewis acids ν Any empty valence orbitals can accept an electron pair from a Lewis base to form a coordinate covalent compound ν If the coordinate covalent compound is an ion, it is called a complex ion Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2+ (aq) 22

23 Lewis Acids and Bases ν One electron pair on the oxygen atom in water can act a Lewis base to hydrated metal ions Fe H 2 O Fe(H 2 O) 6 3+ (aq) ν OH - is a good Lewis base Al(OH) 3 (s) + OH - (aq) Al(OH) 4- (aq) electron pair from OH - inserts into the empty 2p orbital on aluminum to form covalent bond 23

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