The Chemistry of Acids and Bases Separately Chapter 14 Part I

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1 Page III-14a-1 / Chapter Fourteen Part I Lecture Notes The Chemistry of Acids and Bases Separately Chapter 14 Part I Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK categories. STRONG ACID: HNO 3 (aq) + H 2 O(liq) ---> H 3 O + (aq) + NO 3 -(aq) HNO 3 is about 100% dissociated in water. Acids create hydronium when they react with water. Chemistry 223 Professor Michael Russell Strong and Weak Acids/Bases HNO 3, HCl, HBr, HI and HClO 4 are among the few known strong monoprotic acids. H Cl Strong and Weak Acids/Bases Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H = HOAc H 2 O H 3 O + Cl - HOAc(aq) + H 2 O(liq) qe OAc - (aq) + H 3 O + (aq) OAc - = CH 3 CO 2 - = acetate ion Memorize these five strong acids! Strong and Weak Acids/Bases Strong Base: 100% dissociated in water. NaOH(aq) ---> Na + (aq) + OH - (aq) Other strong monobasic bases: KOH, LiOH Ca(OH) 2 is a strong dibasic system: CaO (lime) + H 2 O --> Strong and Weak Acids/Bases Weak base: less than 100% ionized in water One of the best known weak bases is ammonia, NH3 NH 3 (aq) + H 2 O(liq) qe NH 4 +(aq) + OH - (aq) Ca(OH) 2 (slaked lime) CaO Memorize the three strong monobasic bases! Page III-14a-1 / Chapter Fourteen Part I Lecture Notes

2 Page III-14a-2 / Chapter Fourteen Part I Lecture Notes 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 The Brønsted definition means NH 3 is a BASE in water - and water is itself an ACID NH 3! + H 2 O Base Acid NH OH - Acid Base See Brønsted Acids and Bases Handout ACID-BASE THEORIES NH 3 is a BASE in water - and water is itself an ACID Conjugate Pairs 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. MORE ABOUT WATER H 2 O can function as both an ACID and a BASE. MORE ABOUT WATER Autoionization OH - H 3 O + K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 Pure water undergoes AUTOIONIZATION and so [H 3 O + ] = [OH - ] = 1.00 x 10-7 M Memorize KW! Page III-14a-2 / Chapter Fourteen Part I Lecture Notes

3 Page III-14a-3 / Chapter Fourteen Part I Lecture Notes Calculating [H3O+] & [OH-] You add mol of NaOH to 1.0 L of pure water. Calculate [H3O+] and [OH-]. Solution 2 H2O(liq) qe H3O+(aq) + OH-(aq) Le Chatelier predicts equilibrium shifts to the. [H3O+] < 10-7 at equilibrium. Set up an ICE concentration table. Calculating [H3O+] & [OH-] You add mol of NaOH to 1.0 L of pure water. Calculate [H3O+] and [OH-]. Solution 2 H2O(liq) qe initial change equilib H3O+(aq) + OH-(aq) x +x x x Kw = (x)( x) Because x << M, assume [OH-] = M Kw = (x)( x) (x)(0.0010) = [H3O+](0.0010) Calculating [H3O+] & [OH-] You add mol of NaOH to 1.0 L of pure water. Calculate [H3O+] and [OH-]. Solution 2 H2O(liq) qe H3O+(aq) + OH-(aq) [H3O+] = Kw / = 1.0 x M This solution is because [H3O+] = Kw / = 1.0 x M [H3O+], [OH-] and ph A common way to express acidity and basicity is with ph ph = - log [H3O+] In a neutral solution, [H3O+] = [OH-] = 1.00 x 10-7 at 25 oc ph = -log (1.00 x 10-7) = - (-7) = 7 [H3O+] < [OH-] ph = 7 for neutral solutions! Søren Sørensen, creator of the ph scale [H3O+], [OH-] and ph [H3O+], [OH-] and ph What is the ph of the M NaOH solution? [H3O+] = 1.0 x M If the ph of Diet Coke is 3.12, it is. ph = - log (1.0 x 10-11) = Because ph = - log [H3O+] then log [H3O+] = - ph General conclusion - Basic solution ph > 7 Neutral ph = 7 Acidic solution ph < 7 Public Enemy are not scientists! Take antilog and get [H3O+] = 10-pH [H3O+] = = 7.6 x 10-4 M Page III-14a-3 / Chapter Fourteen Part I Lecture Notes

4 Page III-14a-4 / Chapter Fourteen Part I Lecture Notes logarithms and sig figs Logarithms are exponents with special sig figs rules. General rule: log of experimentally measured number with N sig figs give numbers with N decimal places after the decimal (digit before decimal only indicates magnitude) Examples: log 3.07*10-3 = (3 sigs, 3 places after decimal) - log 1.1*10-8 = 7.96 (2 sigs, 2 places after decimal) = 7.6 * 10-4 (2 places after decimal, 2 sigs) ph of Common Substances Other px Scales In general px = -log X and: p = - log and so poh = - log [OH - ] K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C Take the log of both sides -log (10-14 ) = - log [H 3 O + ] + (-log [OH - ]) 14 = ph + poh also: 14 = p + pk b ph Scales acidic solutions: ph, [H 3 O + ], [OH - ], poh basic solutions: ph, [H 3 O + ], [OH - ], poh Equilibria Considerations Involving Weak Acids and Bases Acid Conjugate Base acetic, CH 3 CO 2 H CH 3 CO 2 -, acetate ammonium, NH 4 + bicarbonate, HCO 3 - NH 3, ammonia CO 3 2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%). Equilibria Involving Weak Acids and Bases Consider acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O qe H 3 O + + OAc - Acid Conj. base = [H 3 O+ ][OAc - ] = 1.8 x 10-5 [HOAc] (K is designated for ACID) Because [H 3 O + ] and [OAc - ] are SMALL, << 1. Page III-14a-4 / Chapter Fourteen Part I Lecture Notes

5 Page III-14a-5 / Chapter Fourteen Part I Lecture Notes Equilibrium Constants for Weak Acids ( ) Equilibrium Constants for Weak Bases (K b ) Weak acids have < 1 Leads to small [H 3 O + ] and a ph of 2-7 Weak bases have K b < 1 Leads to small [OH - ] and a ph of 12-7 Equilibrium Constants for Weak Acids and Bases HA + H 2 O qe H 3 O + + A - A - + H 2 O qe HA + OH K b Acids Increase strength Ionization Constants for Acids/Bases Conjugate Bases 2 H 2 O qe H 3 O + + OH - *K b = K w Important relations: *K b = K w and p + pk b = 14 Increase strength nd Acid-Base Reactions ACIDS CONJUGATE BASES STRONG weak weak STRONG Reactions always go from the stronger A-B pair (larger K) to the weaker A-B pair (smaller K). nd Acid-Base Reactions A strong acid is 100% dissociated. Therefore, a STRONG ACID - a good H + donormust have a WEAK CONJUGATE BASE - a poor H + acceptor. HNO 3 (aq) + H 2 O(liq) qe 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. Page III-14a-5 / Chapter Fourteen Part I Lecture Notes

6 Page III-14a-6 / Chapter Fourteen Part I Lecture Notes nd Acid-Base Reactions HNO 3! + H 2 O STRONG ACID 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. BASE ACID K for this reaction is large H 3 O + + NO 3 - WEAK BASE nd Acid-Base Reactions Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a WEAK ACID HOAc + H 2 O qe 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 Types of Acid/Base Reactions Strong acid + Strong base reactions: HCl(aq) + NaOH(aq) qe H 2 O(l) + NaCl(aq) H + + Cl - + Na + + OH - qe H 2 O + Na + + Cl - Net ionic equation: H + (aq) + OH - (aq) qe H 2 O(liq) K = 1/K w = 1 x very product favored K! Mixing equal molar quantities of a strong acid and strong base produces a neutral solution. Weak acid + Strong base reactions: CH 3 CO 2 H + OH - qe 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 = 1.8 x 10 9 very product favored K! 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 Strong acid + Weak base reactions: Types of Acid/Base Reactions H 3 O + + NH 3 qe 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/ = 1.8 x 10 9 very product favored K! Mixing equal molar quantities of a strong acid and weak base produces the base's conjugate acid. The solution is acidic. Page III-14a-6 / Chapter Fourteen Part I Lecture Notes Weak acid + Weak base 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. Do not worry about WA + WB reactions in CH 223!

7 Page III-14a-7 / Chapter Fourteen Part I Lecture Notes Types of Acid/Base Reactions: Summary M M 0.06 M Equilibria Involving A Weak Acid Determining the ph of an acetic acid solution 2.0 M a ph meter 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 ICE equilibrium concs. [HOAc] [H 3 O + ] [OAc - ] initial 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. Step 2. Write expression =1.8 x 10-5 = [H 3O + ][OAc - ] [HOAc] This is a quadratic. Solve using quadratic formula.... but there's a better way, sometimes! = x x 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 3. Solve expression =1.8 x 10-5 = [H 3O + ][OAc - ] [HOAc] First assume x is very small because is so small. And so x = [H 3 O + ] = [OAc - ] = [ 1.00] 1/2 = =1.8 x 10-5 = x x x 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 3. Solve approximate expression =1.8 x 10-5 = x x = [H 3 O + ] = [OAc - ] = [ 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 Page III-14a-7 / Chapter Fourteen Part I Lecture Notes

8 Page III-14a-8 / Chapter Fourteen Part I Lecture Notes Equilibria Involving A Weak Acid Consider the approximate expression =1.8 x 10-5 = x For many weak acids [H 3 O + ] = [conj. base] = [ C a ] 1/2 where C a = initial conc. of acid Useful Rule of Thumb: If 100 < C a, then [H 3 O + ] = [ C a ] 1/2 or ph = - log [ C a ] 1/2 x = [H 3 O + ] = [ 1.00] 1/2 Equilibria Involving A Weak Acid Calculate the ph of a M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O qe HCO H 3 O + = 1.8 x 10-4 Approximate solution [H 3 O + ] = [ C a ] 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] = x 10-4 = M ph = * is not less than C a! Weak Bases Equilibria Involving A Weak Base You have M NH 3. Calc. the ph. NH 3 + H 2 O qe NH OH - K b = 1.8 x 10-5 Step 1. Define equilibrium concs. [NH 3 ] [NH 4 +] [OH - ] initial change -x +x +x equilib x x x Equilibria Involving A Weak Base You have M NH 3. Calc. the ph. Equilibria Involving A Weak Base You have M NH 3. Calc. the ph. NH 3 + H 2 O qe NH OH - K b = 1.8 x 10-5 Step 2. Solve the equilibrium expression K b = 1.8 x 10-5 = [NH + 4 ][OH - ] x 2 = [NH 3 ] x Assume x is small (100 K b < C b ), so x = [OH - ] = [NH 4 +] = 4.2 x 10-4 M and [NH 3 ] = x M The approximation is valid! NH 3 + H 2 O qe NH 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 = or ph = 14 + log [K b C b ] 1/2 = Page III-14a-8 / Chapter Fourteen Part I Lecture Notes

9 Page III-14a-9 / Chapter Fourteen Part I Lecture Notes Overview: Calculating ph of Acids & Bases Strong acid: ph = - log C a = - log [H 3 O + ] Strong base: ph = 14 + log C b = 14 + log [OH -1 ] Weak acid: ph = - log [ C a ] 1/2 (100 * Ka < Ca) Weak base: ph = 14 + log [K b C b ] 1/2 (100 * Kb < Cb) Memorize! Acid-Base Properties of Salts MX + H 2 O ----> acidic or basic solution? Consider NH 4 Cl NH 4 Cl(aq) ----> NH 4 +(aq) + Cl - (aq) (a) Reaction of Cl - with H 2 O Cl - + H 2 O ----> HCl + OH - base acid acid base Cl - ion is a VERY weak base because its conjugate acid is strong. Therefore, Cl > neutral solution Acid-Base Properties of Salts Acid-Base Properties of Salts NH 4 Cl(aq) ----> NH 4 +(aq) + Cl - (aq) (b) Reaction of NH 4 + with H 2 O NH H 2 O ----> NH 3 + H 3 O + acid base base acid NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH > acidic solution Acid-Base Properties of Salts Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral Acid-Base Properties of Salts Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO H 2 O qe HCO OH - base acid acid base K b = 2.1 x 10-4 Step 1. Set up ICE concentration table [CO 3 2-] [HCO 3 -] [OH - ] initial change -x +x +x equilib x x x CO H 2 O qe HCO OH - base acid acid base K b = 2.1 x 10-4 Step 2. Solve the equilibrium expression K b = 2.1 x 10-4 = [HCO - 3 ][OH - ] = [CO 2 3 ] x x Assume x 0.10, because 100 K b < C b x = [HCO 3 -] = [OH - ] = M Page III-14a-9 / Chapter Fourteen Part I Lecture Notes

10 Page III-14a-10 / Chapter Fourteen Part I Lecture Notes Acid-Base Properties of Salts Calculate the ph of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO H 2 O qe HCO OH - base acid acid base K b = 2.1 x 10-4 Step 3. Calculate the ph [OH - ] = M poh = - log [OH - ] = 2.34 ph + poh = 14, so ph = 11.66, and the solution is. or ph = 14 + log [K b C b ] 1/2 = Lewis acid = electron pair acceptor (BF 3 ) Lewis base = electron pair donor (NH 3 ) New bond formed using electron pair from the Lewis base. Coordinate covalent bond Notice geometry change on reaction. The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH > COMPLEX IONS All metal ions form complex ions with water and are of the type [M(H 2 O) x ] n+ where x = 4 and 6. [Cu(NH 3 ) 4 ] 2+ Page III-14a-10 / Chapter Fourteen Part I Lecture Notes

11 Page III-14a-11 / Chapter Fourteen Part I Lecture Notes Add NH3 to light blue [Cu(H2O)4] > The Fe2+ in heme can interact with O2 or CO in a Lewis acid-base reaction. light blue Cu(OH)2 and then deep blue [Cu(NH3)4]2+ Lewis acid & base theory explains Many complex ions containing water undergo HYDROLYSIS to give acidic solutions. AMPHOTERIC nature of some metal hydroxides. [Cu(H2O)4]2+ + H2O ---> [Cu(H2O)3(OH)]+ + H3O+ Al(OH)3(s) + 3 H+ --> Al H2O Here Al(OH)3 is a Brønsted base. Al(OH)3(s) + OH- --> Al(OH)4Here Al(OH)3 is a Lewis acid. This explains why water solutions of transition metals are acidic. Many complex ions are very stable. Cu NH3 qe [Cu(NH3)4]2+ K for the reaction is called O H Al3+ Formation of complex ions explains why you can dissolve a precipitate by forming a complex ion. AgCl(s) + 2 NH3 qe Kformation Ag(NH3)2+ + Cl- or a "formation constant" Here Kf = 6.8 x Reaction is strongly product-favored. Kf = [Cu(NH 3 )4 ] AgCl(s) 2+ 4!"Cu 2+ #$!"NH #$ 3 Page III-14a-11 / Chapter Fourteen Part I Lecture Notes

12 Page III-14a-12 / Chapter Fourteen Part I Lecture Notes Formation of complex ions explains why you can dissolve a precipitate by forming a complex ion. AgCl(s) qe Ag + + Cl - K sp = 1.8 x Ag NH 3 qe Ag(NH 3 ) 2 + K form = 1.6 x AgCl(s) + 2 NH 3 qe Ag(NH 3 ) Cl - K net = K sp K form = 2.9 x 10-3 Hints for This Chapter ph (strong acid) = - log C a ph (strong base) = 14 + log C b ph (weak acid) = - log [ *C a ] 1/2 ph (weak base) = 14 + log [K b *C b ] 1/2 14 = ph + poh = p + pk b K w = 1.00 * = [H 3 O + ][OH - ] = *K b (25 C) Know equivalence point ph values for different titrations Know how to use formation constants Understand Lewis acid/base theory End of Chapter 14 Part I See: Chapter Fourteen Part I Study Guide Chapter Fourteen Part I Concept Guide Types of Equilibrium Constants Page III-14a-12 / Chapter Fourteen Part I Lecture Notes