Lecture 7. Acids. non-metals form anions. metals form cations H+ - Professor Hicks Inorganic Chemistry (CHE152) + anion. molecular compounds

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1 Lecture 7 Professor icks Inorganic Chemistry (CE152) Acids + + anion + - anion substances that release + ions when dissolved Strong acids Cl NO 3 2 SO 4 + Cl - + NO SO 2-4 hydrochloric acid nitric acid sulfuric acid Weak acids C 2 3 O 2 F + C 2 3 O F - acetic acid hydrofluoric acid non-metals form anions metals form cations molecular compounds ionic compounds all other molecular compounds F ~ 95% (molecules) Acids are molecular compounds dissolve strong acid dissolve acids Cl + + Cl - ~ 100% dissolve + + F - ~5% separated ions (separated ions) Dissolved molecules cations (+ ions) anions (- ions)

2 + Acids - anion Acids are molecular compound because they can dissolve without dissociating into ions Ionic compounds must separate into ions to dissolve Weak acids have a small percentage of molecules separated into + and an anion, the rest stay together as one particle F ~ 95% + and F - 5% Strong acids separate 100% into + and anion in water Cl ~ 0 % + and Cl - ~ 100% has no electrons! ydrogen ion + to bond other atom must provide both electrons (lone pair) + curved arrows used to show electron pair movement N N + 1) start at lone pair 2) end at electron acceptor N 3 (aq) + + N 4+ (aq) carboxylic acids weak acids most common acids in nature R = benzene benzoic acid food preservative O C=O C 3 R = C 3 acetic acid 5% solution = vinegar O C=O R = C, R O - C=O C curved arrows 1) start at lone pair 2) end at electron acceptor

3 hydronium ion 3 O + 3 O + = 2 O form + takes in water reactions of acids in water can be written with + or 3 O + acid ionization constant K a equilibrium constant for acid to hydrolyze acid + 2 O 3 O + + anion larger K a = stronger acid Q = [C 6 5 CO 2 - ][ 3 O + ] [C 6 5 CO 2 ] Benzoic Acid = C 6 5 CO 2 (weak acid) C 6 5 CO 2 (aq) + 2 O (l) C 6 5 CO 2- (aq) + 3 O + (aq) K a = 6.5 x 10-5 reactants products reactants products weak acids much less 100% strong acids almost 100% dissociated dissociated ClO 4 K a = 10 10! Q = K a Gibbs Free Energy Gibbs Free Energy 0 Q 2 x Q Bases Bronsted-Lowry base = proton acceptor proton acceptors must have lone pair no lone pairs not a base Examples of bases hydroxide ion - ionic compounds with O- ion any hydroxide cation ion 3 lone pairs O- C methane + cation O - both proton acceptors = both bases - molecular compounds with lone pairs lone pair N ammonia

4 conjugate acid/base pairs C 2 3 O 2 (aq) + NaO (aq) NaC 2 3 O 2 (aq) + 2 O (l) acid base conjugate base conjugate acid Whenever an acid-base reaction occurs: 1) the product that is acid minus + is called the conjugate base of the acid 2) The product that is base plus + is called the conjugate acid of the base strength conjugate acids/bases the weaker an acid the stronger its conjugate base conjugate acid base Q closer to pure acid Gibbs Free Energy Benzoic acid K a = 6.5 x 10-5 the stronger a base the weaker its conjugate acid Gibbs Free Energy acid conjugate base Q closer to pure conjugate base perchloric acid K a = 10 10! 0 Q 2 x Q 10 10

5 Water autoionizes 2 O (l) + (aq) + O - (aq) Q = [ + ][O - ] or written with hydronium ion 2 2 O (l) 3 O + (aq) + O - (aq) Q = [ 3 O + ][O - ] K eq for this reaction called K w K w = at room temperature Neutral solution has [ + ] =10-7 and [O - ] =10-7

6 Water is amphoteric (an acid and a base) water acting as a base 2 O (l) + Cl (aq) 3 O + (aq) + Cl - (aq) Conjugate acid of water 3 O + = hydronium ion water acting as an acid 2 O (l) + N 3 (aq) O - (aq) + N 4 + (aq) Conjugate base of water O - = hydroxide ion Le Chateliers principle and K w what if neutral water is perturbed by adding an acid or base? initial M 2 2 O (l) 3 O + (aq) + O - (aq) equilibrium M 10-7 M M disturbance = add NO 3 increase 3 O + to 0.10 M?? Acidic solution has [ 3 O + ] > 10-7 Basic solution has [O - ] > M systems response = to decrease 3 O + is system is at equilibrium? yes b/c Q = K a [ 3 O + ][O - ] = K w = Q = 10-7 x 10-7 = system finds a new equilibrium [ 3 O + ][O - ] = K w 0.10 * [O - ]= [O - ] = / 0.10 =10-13 M Why is [ 3 O + ] ~ 0.10 M? (about) If rxn used no O - [ 3 O + ] = = If rxn used all O- [ 3 O + ] = =0.10 Approximately = 0.10 p scale of [ + ] concentration more convenient than scientific notation p = -log [ 3 O + ] still not sure? take the log of 10 it should be 1 [ 3 O + ]

7 po same idea as p scale of O- concentration po = -log [O - ] [ 3 O + ] p, po, and pk w p + po = 14 Why? [ 3 O + ][O - ] = ) take log both sides 2) log(a*b) = log(a) + log(b) p + po = -log(10-14 ) = 14 p = 14 - po p is what most people think in terms of some problems we get a result [O-] or po use this equation to express it as a p p fun facts! more bacteria that are not harmful grow in acidic conditions (acidophilus strains) more bacteria that are harmful grow in basic conditions blood p about 7.4 stomach p 1.5! + and O - are catalysts for the reactions that hold together, fats, carbohydrates, and proteins!!!! control of p important for life

8 The Strong Acids Cl, Br, I, NO 3, ClO 4, 2 SO 4 molarity of a monoprotic strong acid = molarity of [ 3 O + ] b/c strong acids completely, 100% dissociate For example a 1.0 M solution of Cl has a [ 3 O + ] = 1.0 M

9 Dissociation of weak acids write hydrolysis reaction Acid + 2 O 3 O + + conj base initial (M) change (M) equilibrium (M) K a = Calculate [ 3 O + ] at equilibrium for a 0.55 M solution of F in water. [ 3 O+ ][F-] [F] look up K a for F x x 10-5 = x =??? 0.55-x F (aq) + 2 O (l) 3 O + (aq) + F - (aq) x - +x +x 0.55-x - +x +x to solve for x you could use the quadratic equation a quicker alternative is to look for an approximation if x is very small compared to x 10-5 = x x = sqrt{ 0.55* 3.5 x 10-5 } = M x = change in [F] to reach equilibrium, and final equilibrium molarities [ 3 O + ], [F - ] Percent Dissociated % dissociated = x initial molarity x 100% initial molarity F before dissociation dissolve initial molarity -x F (aq) + (aq) F- (aq) x at equilibrium

10 initial (M) change (M) equilibrium (M) K a = Dissociation of weak acids F (aq) + 2 O (l) 3 O + (aq) + F - (aq) [ 3 O+ ][F-] [F] x +x +x 0.55-x +x +x to solve for x you could use the quadratic equation a quicker alternative is to look for an approximation if x is very small compared to 0.55 x x 10-5 = x =??? 3.5 x 10-5 = 0.55-x x % dissociated x = sqrt{ 0.55* 3.5 x 10-5 } = M is is small compared to 0.55 M? 5% error is widely accepted in science x 100% = 0.79% approximation is valid b/c % dissociated is less than 5% so error could not be larger than 5% The method of successive approximations x 2 x x 10-4 = x x = sqrt{ 0.01* 1.8 x 10-4 } = M 1) substitute this approximation of x back into equation 2) calculate new approximate value of x x = sqrt { 1.8x10-4 ( } = repeat steps 1, 2 x = sqrt { 1.8x10-4 ( } = repeat steps 1, 2 again x = sqrt { 1.8x10-4 ( } = CO 2-3 (aq) + 2 O (l) CO 3- (aq) + O - (aq) initial (M) change (M) -x +x +x equilibrium (M) x +x +x % dissociated = x100% = 13.4% more than 5%! approximation not good! use the method of successive approximations if % dissociation is more than 5% an improved approximation of x a more improved approximation of x the approximations have stopped changing in the 5th decimal place so the approximation good to this precision

11 1) Break into groups of each group will be assigned an acid 2) Determine the p and % dissociated. acid K a M assigned 1) ClO x ) CO x E-02 3) C 7 6 O x E-02 4) C 2 3 O x E-02 5) ClO 2.90 x E-05 6) CN 4.90 x E-07 7) C 6 5 O 1.30 x E-08 8) F 3.50 x E-01 9) NO x E-01 Determine the p and % dissociated for a 1.5 x 10-4 M solution of acetic acid using the method of successive approximations Le Chateliers Principle and % dissociated F + 2 O F + 3 O + K a = 3.5 x 10-4 ~100% 90% 10% 40% 50% 60% 70% 80% 1% 30% at infinite dilution percent dissociated molarity declines with dilution disturbance = increase 2 O (dilute acid) response = decrease [ 2 O] 1.0 M 0.10 M M M etc

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13 Polyprotic Acids often acid molecules have more than one ionizable these are called polyprotic acids 1 = monoprotic, 2 = diprotic, 3 = triprotic Cl = monoprotic, 2 SO 4 = diprotic, 3 PO 4 = triprotic ionizable s have different K a s polyprotic acids ionize in steps each ionizable removed sequentially removing of the first + makes removal of the second + harder 2 SO 4 is a stronger acid than SO 4 3 PO 4 is a stronger acid than 2 PO 4-2 PO 4- is a stronger acid than PO 4 2- Sulfuric acid is a diprotic acid 2 SO 4 (aq) 2 O (l) 3 O + (aq) + SO 4 - strong acid ~ 100% SO 4- (aq) 2 O (l) 3 O + (aq) + SO 4 2- weak acid much less 100% a 1.0 M solution of 2 SO 4 has [ 3 O + ] = 1.0 M (first step) + a little more 3 O + from second dissociation Estimate the concentration of sulfate ion in a 5.0 M solution of sulfuric acid. ints: In the first hydrolyses of sulfuric acid it acts as a strong acid For the second hydrolyses K a = 1.0 x10-2

14 Strong bases ionic compounds with O - ion KO, NaO, LiO. calculate po from molarity [O - ] calculate p p = 14 - po Example: A M solution of NaO has a [O - ] = M po = -log(0.0010) = 3.0 p = 14-3 = 11 Weak bases react with water to produce O - Base + 2 O Base- + + O - hydrolysis reaction N 3 (aq) + 2 O (l) N 4+ (aq) + O - (aq) do not react completely equilibrium constant K b

15 Dissociation of weak bases Calculate p at equilibrium for a 0.10 M solution of Na 2 CO 3 in water. write hydrolysis reaction base + 2 O conj acid + O - initial (M) change (M) equilibrium (M) CO 2-3 (aq) + 2 O (l) CO 3- (aq) + O - (aq) x +x +x x x x K b = [O- ][CO 3- ] [CO 3 2- ] look up K b for CO 3 2- x x 10-4 = 0.10-x x x 10-4 = 0.10 x = sqrt{ 0.10* 1.8 x 10-4 } = M % dissociated = po = -log[o - ]=-log( )=2.37 p =14-pO = =11.62 x100% = 4.2 % less than 5%! approximation is OK dissociation of weak bases Calculate p at equilibrium for a M solution of Na 2 CO 3 in water. write hydrolysis reaction base + 2 O conj acid + O - initial (M) change (M) equilibrium (M) CO 2-3 (aq) + 2 O (l) CO 3- (aq) + O - (aq) x +x +x x x x K b = [O- ][CO 3- ] [CO 3 2- ] look up K b for CO 3 2- x x 10-4 = x % dissociated increases as the acid/base is more dilute x x 10-4 = x = sqrt{ 0.010* 1.8 x 10-4 } = M % dissociated = x100% =13.4% more than 5%! approximation not good enough! Need to use the quadratic equation or successive approximations method

16 acids + + anion + - anion if an acid is uncharged its conjugate base is negatively charged conjugate bases of acids exist as ionic compounds aka salts often from group I since all group I salts are soluble salts of acids replace + any cation + anion Na + - anion K + - anion Strong acids Cl NO 3 2 SO 4 + Cl - + NO SO 2-4 hydrochloric acid nitric acid sulfuric acid Weak acids C 2 3 O 2 F + C 2 3 O F - acetic acid hydrofluoric acid Salts of Strong acids LiCl NaNO 3 K 2 SO 4 Li + Cl - Na + NO - 3 2K + SO 2-4 lithium chloride sodium nitrate potassium sulfate Salts of Weak acids Mg(C 2 3 O 2 ) 2 CsF Mg 2+ 2C 2 3 O - 2 Cs + F - magnesium acetate cesium fluoride conjugate bases of weak acids most conjugate acid/base pairs are both weak exception: conjugates of the strong acids and strong bases (O - ) are weak bases/acids acetic acid Gibbs Free Energy a weak acid acetate ion acetate ion Gibbs Free Energy conjugate base acetate ion is a weak base acetic acid 0 % dissociated % dissociated 100

17 (so weak they do not affect p) alcohols C 2 5 O 2 O weaker weak acids CN Weak acids stronger weak acids ClO 2 Strong Acids Cl, Br, I, NO 3, ClO 4, 2 SO 4 Increasing Acid Strength Strong bases C 2 5 O - O - Conjugate bases of weak acids stronger weak bases CN - weaker weak bases ClO 2 - Conjugate bases of Strong Acids (so weak they do not affect p) Cl -, Br -, I -, NO 3-, ClO 4-, Increasing Base Strength weak base conjugate acids of bases exist as ionic compounds aka salts compound with lone pairs often a N containing compound N weak bases and the salts of their conjugate acids ammonia N R if a base is uncharged its conjugate acid is positively charged amines N salt of its conjugate acid + - anion and N R + - anion when they act as bases gaining + they become positively charged examples N 4 Cl C 3 N 3 (ClO 4 ) K a K b = K w for acetic acid the hydrolysis reaction is C 2 3 O 2 (aq) + 2 O(l) C 2 3 O 2- (aq) + 3 O + (aq) K a =1.76 x10-5 for acetate ion the hydrolysis reaction is C 2 3 O 2- (aq) + 2 O C 2 3 O 2 (aq) + O - (aq) K b =5.68 x10-10 notice if you add them the conjugate acid and base cancel overall reaction becomes 2 2 O(l) 3 O + (aq) + O - (aq) K w =? when reactions are added the overall K eq is the product of the K eq s K a K b = 1.76 x10-5 x 5.68 x10-10 = 10-14!!!!!!!!!

18 conjugate acids of weak bases most weak bases conjugate acids are weak the conjugate acid of the strong base O - = 2 O is a weak acid Gibbs Free Energy N N + 3 N N 3 N 3 a weak base K N 4+ is a weak acid b = 1.76 x 10-5 K a = 5.68 x Gibbs Free Energy not on table K a s use K a = K b 0 %ionized %ionized 100 same free energy bowl looked at from other side salts of weak acids/bases if soluble fully dissociate into ions initial molarity calculated from chemical formula Example 0.33 M NaC 2 3 O 2 a solution 0.33 M in C 2 3 O M Ca(C 2 3 O 2 ) 2 a solution 0.48 M in C 2 3 O 2-2 C 2 3 O 2- per 1 Ca(C 2 3 O 2 ) 2 ICE tables for salts of weak acids What is the p of a 0.66 M solution of sodium acetate? C 2 3 O 2- (aq) + 2 O C 2 3 O 2 (aq) + O - (aq) initial change -x +x +x equil 0.66-x +x +x by the usual approximation x = square root (0.66*5.68 x ) = x 10-5 [O - ] = x 10-5 acetate ion K b = = -14 K a (acetic acid) 1.76 x10-5 po = 4.71 a basic solution b/c we added p = = 9.29 the conjugate base of acetic acid = 5.68 x10-10 this problem is setup like other weak base problems but you will not find acetate in table of bases you must recognize it as the conjugate base of a weak acid and calculate its K b

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20 trends in strengths of acids 2 factors bond strength A weaker bonds break more easily stronger acid electronegativity of conjugate base - + A higher electronegativity of A makes closer to +1 stronger acid bond more polarized Strengths of Binary Acids 1) the more + -X - polarized the bond, the more acidic the bond 2) the weaker the -X bond, the stronger the acid binary acid strength increases to the right across a period example -C < -N < -O < -F binary acid strength increases down the column example -F < -Cl < -Br < -I Strengths of Oxoacids more oxygen atoms stronger acid helps polarize the -O bond more oxygen atoms in the chemical formula, like adding a single atom of greater electronegativity used to compare similar acids Example: 2 SO 4 is a strong acid but 2 SO 3 is a weak acid

21 Lewis Acids Any substance that can accept an electron pair to form a new bond is called a Lewis acid + qualifies as a Lewis acid because it has no electrons it must accept electron pairs to form bonds Many metal ions accept electron pairs to form Coordinate Covalent Bonds Electron deficient species in general - such as boron compounds that violate the octet rule are Lewis acids N B N B complex ions metal ion + base new complex structure of base stays intact metal ion is acting as a Lewis Acid bases are called ligands when they bind to metals -O- Al 3+ repeat 6x Al( 2 O) 6 3+ ligand metal ion complex ion complex ions metal ion + base new complex water becomes more acidic when bound to metal Al( 2 O) ( 2 O) 5 Al O + 2 O K a = 1.4 x 10-5 Al( 2 O) 5 (O) O +

22 acidity of complex ions increases as metal ion becomes smaller and/or more highly charged not acidic metal ion + water complex ion charge increases both increase acidity size decreases Classifying Salt Solutions as Acidic, Basic, or Neutral cations of group 1 (Li +, Na +, K +, etc) will not change the p anions that are conjugate bases of strong acids are such weak bases they will not change the p NaCl LiNO 3 KBr neutral solutions Cl NO 3 Br group 1 ions conjugate bases of strong acids

23 Classifying Salt Solutions as Acidic, Basic, or Neutral if the anion is the conjugate base of a weak acid, it will form a basic solution NaF KNO 2 solution will be basic Na + K + group I ions neutral F NO 2 - conjugate bases of weak acids basic Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution acidic N 4 Cl weak acid conjugate base of a strong acid neutral Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is a small / highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution acidic Al(NO 3 ) 3 weak acid conjugate base of a strong acid neutral

24 Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the p of the solution depends on the relative strengths of the acid and base solution will be acidic N 4 F K is larger than K b of the F a of N + 4 ; 5.68 x x Example: Determine whether a solution of the following salts is acidic, basic, or neutral a) SrCl 2 Sr 2+ slightly acidic Cl is the conjugate base of a strong acid p neutral solution will be slightly acidic Example: Determine whether a solution of the following salts is acidic, basic, or neutral b) AlBr 3 Al 3+ is a small, highly charged metal ion weak acid Br is the conjugate base of a strong acid, p neutral solution will be acidic

25 Example: Determine whether a solution of the following salts is acidic, basic, or neutral c) C 3 N 3 NO 3 C 3 N 3+ conjugate acid of a weak base acidic NO 3 is the conjugate base of a strong acid, p neutral solution will be acidic Example: Determine whether a solution of the following salts is acidic, basic, or neutral d) NaCO 2 Na + is in group I, neutral CO 2 base of a weak acid basic solution will be basic Example: Determine whether a solution of the following salts is acidic, basic, or neutral e) N 4 (CO 2 ) N 4+ conjugate acid of a weak base acidic CO 2 conjugate base of a weak acid basic K a (N 4+ ) > K b (F ); solution will be acidic (5.68 x ) (2.8 x )

26 Classifying Salt Solutions as Acidic, Basic, or Neutral N 4 O weak acid Basic strong base Estimate the p of a 0.10 M N 4 O solution (when bad approximations go good) Forms N 4 + and O - conjugate acid of N 3 base will the solution be basic or acidic? Write down an equilibrium reaction that includes O - and N + 4 N 3 (aq) + 2 O (l) N 4+ (aq) + O - (aq) K b = 1.76 x 10-5 initial (M) change (M) x -x -x equilibrium (M) x x x 1.76 x 10-5 = (0.10 x)(0.10 x) x = (0.10 ) 2 x 1.76 x x 10-5 = (0.10 x) 2 x = 568!!!!!!!!! Very bad approximation!!!!!!!! If approximation is very bad then x is large Gibbs Free Energy Gibbs Free Energy placing the ball in a different initial position does not change where it will end up at equilibrium 0 100% x large Estimate the p of a 0.10 M N 4 O solution 0 100% x small N 3 (aq) + 2 O (l) N 4+ (aq) + O - (aq) K b = 1.76 x 10-5 initial (M) reactants and products can be change (M) -x +x +x imagined to react into any set of concentrations and then allowed equil (M) x x x to move to equilibrium like the ball in the bowl x 2 x = sqrt {0.10 x 1.76 x10-5 } 1.76 x 10-5 = x x = % dissociated = ( /0.010) x100% = 1.3%

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