Lecture 8 Professor icks Inorganic Chemistry (CE152) 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 O(l) 4 (aq) 2 3 O + (aq) + SO 2 4 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 1
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 strong bases ionic compounds with O ion KO, NaO, LiO. calculate po from molarity [O ] calculate p p = 14 po Example: A 0.0010 M solution of NaO has a [O ] = 0.0010 M po = log(0.0010) = 3.0 p = 14 3 = 11 2
weak bases react with water to produce O Base + 2 O Base + + O hydrolysis reaction N O(l) + 3 (aq) + 2 N 4 (aq) + O (aq) do not react completely equilibrium constant K b 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) 0.10 0 0 x +x +x 0.10 x x x K b = [O ][CO 3 ] [CO 2 3 ] look up K b for CO 3 2 x 2 1.8 x 10 4 = 0.10x x 2 1.8 x 10 4 = 0.10 x = sqrt{ 0.10* 1.8 x 10 4 } = 0.00424 M % dissociated = 0.00424 0.10 po = log[o ]=log(0.00424)=2.37 p =14pO =142.37=11.62 x100% = 4.2 % less than 5%! approximation is OK dissociation of weak bases Calculate p at equilibrium for a 0.010 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) 0.010 0 0 x +x +x 0.010 x x x K b = [O ][CO 3 ] x 2 1.8 x 10 4 = [CO 2 3 ] 0.010 look up K x = sqrt{ 0.010* 1.8 x 10 4 b for CO 2 3 } = 0.00134 M x 2 1.8 x 10 4 = % dissociated = 0.00134 x100% =13.4% 0.010x 0.010 % dissociated increases as the acid/base is more dilute more than 5%! approximation not good enough! Need to use the quadratic equatiion 3
Le Chateliers Principle breathing controls body p CO 2 dissolved in water forms bicarbonate ion CO 3 2 that is part of the body s defense against sudden changes in p disturbance = decrease CO 2 by exhaling response = increase CO 2 reaction goes towards right shift to right inc [O ] raising the p CO 3 (aq) + 2 O (l) CO 2 (g) + O (aq) overall effect removal of CO 2 by breathing raises body p This counteracts the lowering of p that would occur with the production of acidic intermediates in the metabolism of fats and sugars 4
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 3 2 + SO 2 4 hydrochloric acid nitric acid sulfuric acid Weak acids C 2 3 O 2 F + C 2 3 O 2 + 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 5
conjugate bases of weak acids most conjugate acid/base pairs are both weak exception: conjugates of the strong acids/bases 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 100 0 % dissociated 100 (so weak they do not affect p) alcohols C 2 5 O weaker weak acids Weak acids stronger weak acids 2 O CN ClO 2 Strong Acids Cl, Br, I, NO 3, ClO 4, 2 SO 4 Increasing Acid Strength Strong bases Conjugate bases of weak acids stronger weak bases weaker weak bases C O 2 5 O CN ClO 2 Conjugate bases of Strong Acids (so weak they do not affect p) Cl, Br, I, NO 3, ClO 4, SO 4 Increasing Base Strength weak base N weak bases and the salts of their conjugate acids conjugate acids of bases exist as ionic compounds aka salts compound with lone pairs ammonia often a N containing compound 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 ) 6
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 =? 10 14 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!!!!!!!!! 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 N4 + 4 N 3 a weak base K N 4+ is a weak acid b = 1.76 x 10 5 K a = 5.68 x 10 10 Gibbs Free Energy not on table K a s use K a = 1014 K b N 3 0 %ionized 100 0 %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 2 0.24 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 7
ICE tables for salts of weak acids What is the p of a 0.66 M solution of sodium acetate? initial change equil C 2 3 O 2 (aq) + 2 O C 2 3 O 2 (aq) + O (aq) 0.66 0 0 x +x +x 0.66x +x +x by the usual approximation x = square root (0.66*5.68 x 10 10 ) = 1.936 x 10 5 [O ] = 1.936 x 10 5 po = 4.71 p = 14 4.71 = 9.29 acetate ion K b = 10 14 K a (acetic acid) a basic solution b/c we added the conjugate base of acetic acid 10 = 14 1.76 x10 5 = 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 trends in strengths of acids 2 factors bond strength A weaker bonds break more easily stronger acid electronegativity of conjugate base δ δ+ A higher electronegativity of A makes closer to +1 stronger acid bond more polarized Strengths of Binary Acids 1) the more δ+ X δ 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 8
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 complex ions metal ion + base new complex structure of base stays intact bases are called ligands when they bind to metals as bases 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) 3+ 6 3+ ( 2 O) 5 Al O + 2 O K =14x10 5 a 1.4 10 Al( 2 O) 5 (O) 2+ + 3 O + 9
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 10
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 11
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 10 10 2.86 x 10 11 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 hl charged metal ion weak acid Br is the conjugate base of a strong acid, p neutral solution will be acidic 12
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 10 10 ) (2.8 x 10 11 ) 13
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 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) 0 0.10 0.10 change (M) x x x equilibrium (M) x 0.10 x 0.10 x 1.76 x 105 = (0.10 x)(0.10 x) x = (0.10 ) 2 x 1.76 x 10 5 1.76 x 10 5 = (0.10 x) 2 x = 568!!!!!!!!! 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% 0 100% xlarge xsmall Estimate the p of a 0.10 M N 4 O solution N 3 (aq) + 2 O (l) N 4+ (aq) + O (aq) K b = 1.76 x 10 5 initial (M) 0.10 0 0 reactants and products can be change (M) x +x +x imagined to react into any set of concentrations and then allowed equil (M) 0.10 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 = 0.10 x x = 0.001326 % dissociated = (0.001326/0.010) x100% = 1.3% 14