The Chemistry of Acids and Bases

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The Chemistry of 1 Acids and Bases

2 Acid and Bases

3 Acid and Bases

4 Acid and Bases

5 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O(liq) f H 3 O + (aq) + NO 3- (aq) HNO 3 is about 100% dissociated in water. PLAY MOVIE

Strong and Weak Acids/Bases 6 HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids.

Strong and Weak Acids/Bases Weak acids are much less than 100% ionized in water. 7 One of the best known is acetic acid = CH 3 CO 2 H

Strong and Weak Acids/Bases 8 Strong Base: 100% dissociated in water. NaOH(aq) f Na + (aq) + OH - (aq) Other common strong bases include KOH and Ca(OH) 2. (limited solub.) CaO (lime) + H 2 O f CaO Ca(OH) 2 (slaked lime)

9 Strong and Weak Acids/Bases Weak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O(liq) e NH 4+ (aq) + OH - (aq) PLAY MOVIE

10 ACID-BASE THEORIES The most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theory ACIDS DONATE H + IONS BASES ACCEPT H + IONS

ACID-BASE THEORIES 11 The Brønsted definition means NH 3 is a BASE in water and water is itself an ACID PLAY MOVIE

12 ACID-BASE THEORIES NH 3 is a BASE in water and water is itself an ACID NH 3 / NH 4 + is a conjugate pair related by the gain or loss of H + Every acid has a conjugate base - and vice-versa.

13 Conjugate Pairs

More About Water H 2 O can function as both an ACID and a BASE. 14 In pure water there can be AUTOIONIZATION PLAY MOVIE Equilibrium constant for autoion = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10-14 at 25 o C

More About Water 15 K w = [H 3 O + ] [OH - ] = 1.00 x 10-14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x 10-7 M

Calculating [H 3 O + ] & [OH - ] 16 You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ]. Solution 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) Le Chatelier predicts equilibrium shifts to the. [H 3 O + ] < 10-7 at equilibrium. Set up a ICE table.

Calculating [H 3 O + ] & [OH - ] 17 You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ]. Solution 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) initial 0 0.0010 equilib x 0.0010 + x K w = (x) (0.0010 + x) Because x << 0.0010 M, assume [OH - ] = 0.0010 M [H 3 O + ] = x = K w / 0.0010 = 1.0 x 10-11 M

Calculating [H 3 O + ] & [OH - ] 18 You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ]. Solution 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) [H 3 O + ] = K w / 0.0010 = 1.0 x 10-11 M This solution is because [H 3 O + ] < [OH - ]

[H 3 O + ], [OH - ] and ph 19 A common way to express acidity and basicity is with ph ph = log (1/ [H 3 O + ]) = - log [H 3 O + ] In a neutral solution, [H 3 O + ] = [OH - ] = 25 o C ph = -log (1.00 x 10-7 ) 1.00 x 10-7 at

[H 3 O + ], [OH - ] and ph 20 What is the ph of the 0.0010 M NaOH solution? [H 3 O + ] = 1.0 x 10-11 M ph = - log (1.0 x 10-11 ) = 11.00 General conclusion Basic solution ph > 7 Neutral ph = 7 Acidic solution ph < 7

21 The ph Scale See Active Figure 17.1

[H 3 O + ], [OH - ] and ph 22 If the ph of Coke is 3.12, it is. Because ph = - log [H 3 O + ] then log [H 3 O + ] = - ph Take antilog and get [H 3 O + ] = 10 -ph [H 3 O + ] = 10-3.12 = M 7.6 x 10-4

23 ph of Common Substances

Other px Scales In general px = -log X and so poh = - log [OH - ] K w = [H 3 O + ] [OH - ] = 1.00 x 10-14 at 25 o C 24 Take the log of both sides -log (10-14 ) = - log [H 3 O + ] + (-log [OH - ]) pk w = 14 = ph + poh

25 Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, K a = 3.2 x 10-4

Equilibria Involving 26 Weak Acids and Bases Acid acetic, CH 3 CO 2 H ammonium, NH + 4 bicarbonate, HCO - 3 Conjugate Base CH 3 CO 2-, acetate NH 3, ammonia CO 2-3, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

27 Equilibria Involving Weak Acids and Bases Consider acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O e H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) Because [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

Equilibrium Constants for Weak Acids 28 Weak acid has K a < 1 Leads to small [H 3 O + ] and a ph of 2-7

Equilibrium Constants for Weak Bases 29 Weak base has K b < 1 Leads to small [OH - ] and a ph of 12-7

Ionization Constants for Acids/Bases 30 Acids Increase strength Conjugate Bases Increase strength

31 Relation of K a, K b, [H 3 O + ]

K and Acid-Base Reactions 32 A strong acid is 100% dissociated. Therefore, a STRONG ACID a good H + donor must have a WEAK CONJUGATE BASE a poor H + acceptor. HNO 3 (aq) + H 2 O(liq) e H 3 O + (aq) + NO 3- (aq) STRONG A base acid weak B Every A-B reaction has two acids and two bases. Equilibrium always lies toward the weaker pair. Here K is very large.

33 K and Acid-Base Reactions We know from experiment that HNO 3 is a strong acid. 1. It is a stronger acid than H 3 O + 2. H 2 O is a stronger base than NO 3-3. K for this reaction is large

34 K and Acid-Base Reactions Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a WEAK ACID HOAc + H 2 O e H 3 O + + OAc - WEAK A base acid STRONG B Because [H 3 O + ] is small, this must mean 1. H 3 O + is a stronger acid than HOAc 2. OAc - is a stronger base than H 2 O 3. K for this reaction is small

Types of Acid/Base Reactions 35 Strong acid (HCl) + Strong base (NaOH) H + + Cl - + Na + + OH - e H 2 O + Na + + Cl - Net ionic equation H + (aq) + OH - (aq) e H 2 O(liq) K = 1/K w = 1 x 10 14 Mixing equal molar quantities of a strong acid and strong base produces a neutral solution.

Types of Acid/Base Reactions 36 Weak acid (acetic ac.) + Strong base (NaOH) CH 3 CO 2 H + OH - e H 2 O + CH 3 CO 2 - This is the reverse of the reaction of CH 3 CO 2 - (conjugate base) with H 2 O. OH - stronger base than CH 3 CO 2 - K = 1/K b = 1/(5.6 x 10-10 ) = 1.8 x 10 9 Mixing equal molar quantities of a weak acid and strong base produces the acid s conjugate base. The solution is basic.

Types of Acid/Base Reactions 37 Strong acid (HCl) + Weak base (NH 3 ) H 3 O + + NH 3 e H 2 O + NH 4 + This is the reverse of the reaction of NH 4 + (conjugate acid of NH 3 ) with H 2 O. H 3 O + stronger acid than NH 4 + K = 1/K a = 1.8 x 10 9 Mixing equal molar quantities of a strong acid and weak base produces the bases s conjugate acid. The solution is acid.

Types of Acid/Base Reactions 38 Weak acid + Weak base PLAY MOVIE Product cation = conjugate acid of weak base. Product anion = conjugate base of weak acid. ph of solution depends on relative strengths of cation and anion.

Types of Acid/Base Reactions: Summary 39

a ph meter 40 0.0001 M 0.003 M 0.06 M Calculations with Equilibrium Constants ph of an acetic acid solution. What are your observations? 2.0 M

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the ph. Step 1. Define equilibrium concs. in ICE table. [HOAc] [H 3 O + ] [OAc - ] 41 initial change equilib

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the ph. 42 Step 1. Define equilibrium concs. in ICE table. [HOAc] [H 3 O + ] [OAc - ] initial 1.00 0 0 change -x +x +x equilib 1.00-x x x Note that we neglect [H 3 O + ] from H 2 O.

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the ph. 43 Step 2. Write K a expression This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A).

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the ph. 44 Step 3. Solve K a expression First assume x is very small because K a is so small. Therefore, Now we can more easily solve this approximate expression.

Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the ph. 45 Step 3. Solve K a approximate expression x = [H 3 O + ] = [OAc - ] = [K a 1.00] 1/2 x = [H 3 O + ] = [OAc - ] = 4.2 x 10-3 M ph = - log [H 3 O + ] = -log (4.2 x 10-3 ) = 2.37

Equilibria Involving A Weak Acid Consider the approximate expression 46 For many weak acids [H 3 O + ] = [conj. base] = [K a C o ] 1/2 where C 0 = initial conc. of acid Useful Rule of Thumb: If 100 K a < C o, then [H 3 O + ] = [K a C o ] 1/2

Equilibria Involving A Weak Acid Calculate the ph of a 0.0010 M solution of formic acid, HCO 2 H. 47 HCO 2 H + H 2 O e HCO - 2 + H 3 O + K a = 1.8 x 10-4 Approximate solution [H 3 O + ] = [K a C o ] 1/2 = 4.2 x 10-4 M, ph = 3.37 Exact Solution [H 3 O + ] = [HCO 2- ] = 3.4 x 10-4 M [HCO 2 H] = 0.0010-3.4 x 10-4 = 0.0007 M

48 Weak Bases

Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the ph. 49 NH 3 + H 2 O e NH + 4 + OH - K b = 1.8 x 10-5 Step 1. Define equilibrium concs. in ICE table [NH 3 ] [NH 4+ ] [OH - ] initial change equilib

Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the ph. 50 NH 3 + H 2 O e NH + 4 + OH - K b = 1.8 x 10-5 Step 1. Define equilibrium concs. in ICE table [NH 3 ] [NH 4+ ] [OH - ] initial 0.010 0 0 change -x +x +x equilib 0.010 - x x x

Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the ph. 51 NH 3 + H 2 O e NH 4 + + OH - K b = 1.8 x 10-5 Step 2. Solve the equilibrium expression Assume x is small (100 K b < C o ), so x = [OH - ] = [NH 4+ ] = 4.2 x 10-4 M and [NH 3 ] = 0.010-4.2 x 10-4 0.010 M The approximation is valid!

Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the ph. 52 NH 3 + H 2 O e NH + 4 + OH - K b = 1.8 x 10-5 Step 3. Calculate ph [OH - ] = 4.2 x 10-4 M so poh = - log [OH - ] = 3.37 Because ph + poh = 14, ph = 10.63

53 Acid-Base Properties of Salts MX + H 2 O f acidic or basic solution? Consider NH 4 Cl NH 4 Cl(aq) f NH 4+ (aq) + Cl - (aq) (a) Cl - ion is a VERY weak base because its conjugate acid is strong. Therefore, Cl - f neutral solution Cl - does not hydrolyze, i.e. does not react with water appreciably So, conjugates of weak acids and bases determine ph of salt solutions because

54 NH 4 Cl(aq) f NH 4+ (aq) + Cl - (aq) (b) Acid-Base Properties of Salts Reaction of NH 4 + with H 2 O NH 4 + + H 2 O f NH 3 + H 3 O + acid base base acid NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH 4 + f acidic solution See TABLE 17.4 for a summary of acidbase properties of ions.

55 Acid-Base Properties of Salts

Acid-Base Properties of Salts 56 Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O f neutral CO 3 2- + H 2 O e HCO 3 - + OH - base acid acid base K b = 2.1 x 10-4 Step 1. Set up ICE table [CO 2-3 ] [HCO 3- ] [OH - ] initial change

Acid-Base Properties of Salts 57 Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O f neutral CO 3 2- + H 2 O e HCO 3 - + OH - base acid acid base K b = 2.1 x 10-4 Step 1. Set up ICE table [CO 3 2- ] [HCO 3- ] [OH - ] initial 0.10 0 0 change -x +x +x equilib 0.10 - x x x

Acid-Base Properties of Salts 58 Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O f neutral CO 2-3 + H 2 O e HCO - 3 + OH - base acid acid base K b = 2.1 x 10-4 Step 2. Solve the equilibrium expression Assume 0.10 - x 0.10, because 100 K b < C o x = [HCO 3- ] = [OH - ] = 0.0046 M

Acid-Base Properties of Salts 59 Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O f neutral CO 2-3 + H 2 O e HCO - 3 + OH - base acid acid base Step 3. Calculate the ph [OH - ] = 0.0046 M poh = - log [OH - ] = 2.34 ph + poh = 14, so ph = 11.66, and the solution is.

60 Polyprotic acids Polyprotic acids have more than one ionizable proton. The protons are removed in steps, not all at once. It is always easier to remove the first proton in a polyprotic acid than the second. Therefore, K a1 > K a2 > K a3 etc.

Lewis Acids & Bases 61 Lewis acid a substance that accepts an electron pair PLAY MOVIE Lewis base a substance that donates an electron pair PLAY MOVIE

62 Reaction of a Lewis Acid and Lewis Base PLAY MOVIE New bond formed using electron pair from the Lewis base. Coordinate covalent bond Notice geometry change on reaction.

Lewis Acids & Bases 63 Formation of hydronium ion is also an excellent example. Electron pair of the new O-H bond originates on the Lewis base.

64 Lewis Acid/Base Reaction

Lewis Acids & Bases 65 Other good examples involve metal ions.

Lewis Acids & Bases 66 The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 leads to COMPLEX IONS

67 Lewis Acids & Bases [Ni(H 2 O) 6 ] 2+ + 6 NH 3 f [Ni(NH 3 ) 6 ] 2+ + DMG DMG = dimethylglyoxime, a standard reagent to detect nickel(ii)

Lewis Acid-Base Interactions in Biology 68 PLAY MOVIE The heme group in hemoglobin can interact with O 2 and CO. The Fe ion in hemoglobin is a Lewis acid O 2 and CO can act as Lewis bases Heme group

69 Lewis Acids & Bases Many complex ions containing water undergo HYDROLYSIS to give acidic solutions. [Cu(H 2 O) 4 ] 2+ + H 2 O f [Cu(H 2 O) 3 (OH)] + + H 3 O +

70 Lewis Acids & Bases This explains why water solutions of Fe 3+, Al 3+, Cu 2+, Pb 2+, etc. are acidic. This interaction weakens this bond Another H 2 O pulls this H away as H +

Amphoterism of Al(OH) 3 71

Lewis Acids & Bases 72 This explains AMPHOTERIC nature of some metal hydroxides. Al(OH) 3 (s) + 3 H 3 O + f Al 3+ + 6 H 2 O Here Al(OH) 3 is a Brønsted base. Al(OH) 3 (s) + OH - f Al(OH) - 4 Here Al(OH) 3 is a Lewis acid.

Lewis Acids & Bases 73 Formation of complex ions explains why you can dissolve a ppt. by forming a complex ion. AgCl(s) e Ag + + Cl - K sp = 1.8 x 10-10 Ag + + 2 NH 3 f Ag(NH 3 ) 2 + K form = 1.6 x 10 7 ------------------------------------- AgCl(s) + 2 NH 3 e Ag(NH 3 ) 2 + + Cl - K net =

74 Why? Why are some compounds acids? Why are some compounds bases? Why do acids and bases vary in strength? Can we predict variations in acidity or basicity?

75 Why is CH 3 CO 2 H an Acid? +0.12 +0.24 See Figure 17.12 0.32 1. The electronegativity of the O atoms causes the H attached to O to be highly positive. 2. The O H bond is highly polar. 3. The H atom of O H is readily attracted to polar H 2 O.

76 Acetic acid Trichloroacetic acid K a = 1.8 x 10-5 K a = 0.3 Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl. Cl withdraws electrons from the rest of the molecule. This makes the O H bond highly polar. The H of O H is very positive.

77 Hydrohalic Acids Acid strength of HX increases as you go down the group; HI>HBr>HCl>>HF This is generally attributable to decreasing HX bond strength as you go down a group Going across a period, increasing acidity is generally attributable to increasing electronegativity. HF>H 2 O>NH 3 >CH 4

78 Oxoacids Generally when comparing homologous acids, the greater the number of oxygen atoms, the greater the acidity So nitric acid is stronger than nitrous acid, perchloric is stronger than chloric, etc. Due to more oxygens withdrawing electron density from the O-H bond.

Basicity of Oxoanions 79 NO 3 - CO 3 2- PO 4 3- These ions are BASES. They become more and more basic as the negative charge increases. As the charge goes up, they interact more strongly with polar water molecules.