CH 15 Summary. Equilibrium is a balance between products and reactants

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1 CH 15 Summary Equilibrium is a balance between products and reactants Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios. Capital K is used to represent the equilibrium constant Products over reactants raised to their stoichiometric coefficients Calculated from balanced equation, subscript designates units used, K c, K p No units used in final written K Equlibrium Calculations and Reaction Quotients ICE tables used to manipulate initial and equilibrium concentrations. Factors influencing K Concentration of chemicals and temperature affect all equilibria T, P, V changes affect K p in gases Catalysis Effect of catalysts and inhibitors, reasons they are used. 1

2 2 Polyatomic Ions Memorize These! Ammonium NH + 4 Nitrate NO - 3 Hydronium H 3 O + Nitrite NO - 2 Acetate CH 3 COO - Phosphate PO 3-4 Carbonate CO 2-3 Cyanide CN - Permanganate MnO - 4 PerchlorateClO - 4 Hydroxide OH - Sulfate SO 2-4

3 3 Chapter 4: Solution Chemistry Molarity Conversions between moles and liters Dilutions Calculate concentration of diluted solution from a stock solution M 1 V 1 =M 2 V 2 Calculate mass of solid needed to make a solution Acid Base Titrations Identifying acids and bases Know properties of acids and bases Know list of specific acids both names and formulas Determine concentration of unknown solutions using titration

4 4 Common Acids to Memorize Strong Hydrochloric Acid: HCl Sulfuric Acid: H 2 SO 4 Nitric Acid: HNO 3 Perchloric Acid: HClO 4 Hydrobromic Acid: Hydroiodic Acid HBr HI Weak Carbonic Acid: H 2 CO 3 Phosphoric Acid: H 3 PO 4 Acetic Acid: CH 3 COOH Hydrofluoric Acid: HF

5 Chapter 16 Acids and Bases 5

6 Bronsted Acids and Bases Acid Compound that loses H + to a base Base Compound that gains H + from an acid SA WB CA CB WB WA CA CB Conjugate Acid-Base Pair Acid-Base pair exchanging the H+ Conjugate acid: Product with the extra H + Conjugate base: Product with 1 less H + ion than reactant 6

7 Carboxylic Acids: -COOH Weak organic acids: COOH group on molecule is acidic Creates resonance structure Stabilizes anion Never fully dissociate in water Will always be an equilibrium reaction 7

8 Reactivity of Weak Acid and Bases Strong Acids and Bases: Full dissociation Conjugate bases and acids form spectator ions Can use Chem 101 stoichiometry in calculations No original reactant or product left in solution Weak Acids and Bases: Partial dissociation Forms an equilibrium: K a or K b Acid/Base strength in aqueous solutions H 3 O + is the strongest acid OH - the strongest base Water acts as weak acid or base in the reaction 8

9 Acid-Base Properties of Water: Autoionization Water is slightly conductive due to the following reaction: 2H 2 O(l) H 3 O + (aq) + OH (aq) Process is called Autoionization Acid-Base Reaction between identical molecules 1 molecule acts as an acid, the other a base 9

10 Calculation of [H 3 O + ] and [OH ] in Water Treat as an equilibrium reaction at 25 C H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) WB WA SA SB K w = [H 3 O + ][OH ] = Ion-Product constant K w is very small: Favors weaker acid-base pair (H 2 O) Make ICE table Products [H 3 O + ] [OH ] Initial 0 0 Change +x +x Equilibrium x x Solve for [H 3 O + ] and [OH - ] 1.0 x = [H 3 O + ][OH ] = [x][x] = x 2 x = [H 3 O + ] = [OH ] = M ph = 7 of pure water 10

11 ph Method of measuring acidity (p)ower of the (H)ydrogen ion Calculations ph = -log[h 3 O + ] [H 3 O + ] = 10 (-ph) poh = -log[oh-] [OH-] = 10 (-poh) K w = [H 3 O + ][OH - ] = 1x10-14 M pk w =ph + poh = 14 Effects in 1M Strong Acid (ph = 0) [H 3 O + ]= 1M then [OH - ]= 1x10-14 M Effects in 1M Strong Base (ph= 14) [OH - ]= 1M then [H 3 O + ]= 1x10-14 M 11

12 ph and poh Calculations Strong Acids and Bases 12

13 ph and poh Calculations poh: (p)ower of the (OH - ) ion = pH 1. Find ph and poh of an M HBr at 25 C. HBr (aq) + H 2 O (l) H 3 O + (aq) + Br (aq) [H 3 O + ]= M ph = log[h 3 O + ] = log( ) = 2.30 poh = = Find ph and poh at 25 C of M NaOH NaOH(aq) Na + (aq) + OH (aq) [OH ] = M poh = -log[oh-]= 4.43 ph = poh = = 9.57 Significant Figures Sigfigs in concentration = sigfigs after decimal point in ph [H 3 O + ] = M ph =

14 Acidity Calculations: Strong Acids & Bases Calculate the H 3 O + and OH concentrations at 25 C of an aqueous 0.010M solution of nitric acid? Write reactions: HNO 3 (aq) + H 2 O(l) H 3 O + (aq) + NO 3 (aq) H 3 O + = M 2H 2 O(l) H 3 O + (aq) + OH (aq) H 3 O + = M [H 3 O + ] contributed by water is negligible [H 3 O + ] = ( ) = M [OH - ] =K w [H 3 O + ] = ( ) (0.010) = M Strong acid increases [H 3 O + ] and suppresses [OH - ] ph = -log[h 3 O + ] = -log[0.010m] =

15 Acidity Calculations for Dilute Solutions Calculate the H 3 O + (aq) + OH (aq) of a M solution of NaOH at 25 C? Write reactions: NaOH(aq) Na + (aq) + OH (aq) [OH ] = M 2H 2 O(l) H 3 O + (aq) + OH (aq) [OH ] = M [OH ] < M because of Le Chatelier s Principle Total concentrations: [OH ] = M (10%) NOT less than not negligible Must use an ICE table if solution less than 10-6 M 15

16 Strength of Acids and Bases 16

17 Strength of Acids and Bases Strong acids and bases are strong electrolytes Completely ionized in water, no original compound left Good conductors of electricity. Directional arrow ( ) indicates dissociation is complete Weak acids and bases are weak electrolytes Partial ionization in water, original compound remaining Poor conductors of electricity Double arrow ( ) indicates dissociation is incomplete Governed by an equilibrium constant, K a or K b 17

18 Strong vs. Weak Acids 18

19 Weak Acids and Bases If paired with group 1 cation, will be strong base Stronger acids will dominate over weaker acids HNO 2 (aq)+ CN - (aq) HCN (aq)+ NO 2- (aq) K>1 19

20 Weak Acids and K a 20

21 Weak Acids, K a and pk a ALWAYS write a reaction of weak acid HA in water: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) The equilibrium constant for this reaction is: H O A K = [ HA] + [ 3 ][ ] c = K a K a is the acid ionization constant. Quantitative measure of acid strength Large K a :Stronger acid pk a = -log K a 21

22 K a Values of Common Weak Acids 22

23 K a Calculations mole of HF is dissolved in 1.00 L of water at 25 C. The ph at equilibrium was found to be Calculate K a. HF(aq)+ H 2 O(l) H 3 O + (aq) + F (aq) Make table [HF] [H 3 O + ] [F - ] Initial Change -x +x +x Equilibrium x +x +x [ H3 + O ][ F [ HF] Find [H 3 O + ], [F - ] and [HF] [H 3 O + ] = 10 (-ph) = = 8.3 x 10-3 M = [F - ]=x [HF] = x = M=0.0917M=.092M Calculate K a K a = ] K a [ H O ][ F ] (8.3x10 ) 4 3 = = = 7.5x10 [ HF] [0.092] 23

24 Percent Ionization and K a Measure ph of weak acid of known initial concentration ph gives [H 3 O + ] at equilibrium. Get equilibrium concentrations from table Percent ionization (α) of a weak acid [ + H O ] [ A ] α = 3 x100% = 100% [ HA] [ HA] x Use stoichiometry of chemical equation to calculate equilibrium concentrations in K a 24

25 K a Calculations A M solution of HNO 2 is 19% ionized at equilibrium. Find K a. HNO 2 (aq)+ H 2 O(l) H 3 O + (aq) + NO 2 (aq) Make table [HNO 2 ] [H 3 O + ] [NO 2- ] Initial Change -x +x +x Equilibrium x +x +x Find [H 3 O + ], [NO 2- ] & [HNO 2 ] α = [ H 3O + ] x100% [ HA] 19% [ = [0.0100] X ] 3 X = 1.9x10 M Calculate K a K a [ H3O ][ NO2 ] (1.9x10 ) 4 = = [ HNO2 ] ( x10 ) = x 25

26 Determination of Relative Acidity from K a Which is the stronger acid, HF or HNO 2? HF pk a = log K a = log( ) = 3.12 HNO 2pKa = log K a = log( ) = 3.34 Stronger Acid Higher K a Smaller pk a HF stronger acid 26

27 Find the ph of a M Solution of Hypochlorous Acid [HClO] at 25 o C. ph range from strong acid to pure water Strong acid, ph = -log [0.010] = 2.0 Pure water = ph = -log [1.0 x 10-7 ] = 7.0 K a of hypochlorous acid = Look up in table, not a calculated number Not a strong acid: Incomplete dissociation must use equilibrium calculations Calculate H 3 O + from K a 27

28 Making Approximations in Equilibrium Calculations 28

29 Find the ph of a M Solution of Hypochlorous Acid [HClO] at 25 o C. HClO(aq) + H 2 O(l) H 3 O + (aq) + ClO (aq) Initial Change x - +x +x Equilibrium x - x x Calculate H 3 O + from K a K a = 2.9x10 [ H3 8 + O ][ ClO [ HClO] X = [H 3 O + ] = [HClO] = ~ M ] [ x][ x] = [0.010 x] Calculate ph from [H 3 O + ] ph = -log [H 3 O + ] = -log ( ) = 4.77 Could have approximated! [HClO] eq = initial [HClO] 29

30 Estimating Equilibrium Concentrations Purpose: Simplify equilibrium concentration calculations Avoid the quadratic equation Procedure: Draw up your ICE table Ignore x in the initial acid concentration Solve for x with the ICE table Divide x by the initial acid concentration. If [x]/[ha] x 100< 5%, you can say that [HA] +x = [HA] If not, you cannot make this assumption The error in the number will be over 5% For previous problem estimate [HClO] at equilibrium [x]/[ha] = 1.7x10-5 /0.050 = 0.034% so approximate 30

31 Find the ph of a M Solution of Formic Acid Initial Data [HCHO 2 ] = given in problem [CHO 2- ] and [H 3 O + ]= 0 since you are adding nothing at start Initial HCHO 2 (aq) + H 2 O(l) CHO 2- (aq) + H 3 O + (aq) Change x - + x + x Equilibrium x - x x Look up K a in table and solve for x x =3.0x [ H3O ][ CHO2 ] Ka = 1.8x10 = = [ HCHO ] Calculate ph 2 [ x][ ] [0.050] 4 x x = [H 3 O + ] = M ph =

32 Calculate the ph of a M solution of acetic acid at 25 o C? pk a = 4.77 HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) Initial Change x - +x +x Equilibrium x - x x Convert pk a to K a pk a = 4.77 K a = = 1.7x10-5 With approximation K a + [ H3O ][C2H3O = [ HC H O ] X= 2.54 x ph = ] Solve for [x] with quadratic 5 [ x][ x] K a = 1.7x10 = 5 [3.8x10 x = [H 3 O + ]= Solve for ph ph = log[h 3 O + ] ph= log( ) = 4.74 x] 32

33 Polyprotic acids 33

34 Polyprotic acids Acids with more than one ionizable H atom 1. Treat first dissociation as monoprotic acids (1H + ) 2. Use successive acid ionization constants 1 for each ionized H atom Number sequentially K a1, K a2, K a3, Each equilibrium constant is less likely to affect ph than the one before it You may not need all of them to solve problems 34

35 35

36 Calculate the ph of a M sulfuric acid. Sulfuric acid is a strong acid: Full dissociation of first H + H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) M H 2 SO M H 3 O M HSO 4 Full dissociation of first H + only! Bisulfate ion, HSO 4- : Weak acid equilibrium with 2 nd H + HSO 4- (aq) + H 2 O(l) H 3 O + (aq) + SO 4 2 (aq) K a = 1.1 x 10-2 for HSO 4- (aq) [HSO 4 ]= [ H 3 O + ]= M Estimate of ph Both H ionize ph = log( ) = 2.00 Ionization1 H ph = log(0.0050) =

37 Calculate the ph of a M sulfuric acid? HSO 4 (aq) + H 2 O(l) H 3 O + (aq) + SO 2 4 (aq) Initial Change x -x + x + x Equilibrium x x x Solve for [H 3 O + ] using quadratic equation K a + 2 [ H 3O ][ SO = 1x10 = [ HSO ] [H 3 O + ] = x = = M Solve for ph ph = log[h 3 O + ] = log(0.0079) = ] [ x][ x] = [ x] x = M 37

38 What is the ph of 0.037M phosphoric acid? H 3 PO 4 has 3 ionizable hydrogen atoms H 3 PO 4 + H 2 O 3H 3 O + + 1PO 4 3- H 3 PO 4 : weak acid each ionization reaction is a separate problem use the results of previous steps Possible ph values All three H + donated, = log( ) = H ionized = log(0.037) = 1.43 H 2 PO 4- and HPO 4 2- have very small K a values ph primarily due to first ionization Lowest ph: log(0.037)= 1.43 not

39 Ionizable Hydrogens and K a: Set up ICE table for each step H 3 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 3 PO 4 =0.037M-0.013M=0.024M X=H 3 O + = H 2 PO 4 = 0.013M Need quadratic! H 2 PO 4- (aq) + H 2 O(l) H 3 O + (aq) + H 3 O + = H 2 PO 4 = 0.013M X= HPO 4 2- = 6.3x10-8 M Used approximation HPO 4 2- (aq)+ H 2 O(l) H 3 O + (aq) + H 3 O + = 0.013M HPO 4 2- = 6.3x10-8 M PO 4 3 = 1.8x10-18 M K a K a K a = H 2 PO 4 (aq) [ H3 4 ] 3 = O ][H2PO [H PO ] 3 HPO 4 2- (aq) [ H3O ][HPO ] 4 8 = = [H PO ] PO 4 3 (aq) [ H3O ][PO ] 4 13 = = [HPO ] 39

40 H 3 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 2 PO 4 (aq) Initial Change x - + x + x Equilibrium x - + x x Approximate H 3 PO 4? [x]/[ha] = 0.016/.037 so no approximation Plug equilibrium concentrations into K a equation K a What is the ph of a M solution of phosphoric acid at 25 C? = [ H = 3 + O ][H2PO [H PO ] 3 40 x = Plug x back into table to get equilibrium concentrations [H 3 PO 4 ] = x = = 0.024M [H 3 O + ] = x = 0.013M use for second ionization [H 2 PO 4 ] = x = 0.013M use for second ionization 4 4 ] [ x][ x] = [0.037 x]

41 What is the ph of a M solution of phosphoric acid at 25 C? H 2 PO 4- (aq) + H 2 O(l) H 3 O + (aq) + HPO 2-4 (aq) Initial Change x - + x +x Equilibrium x x x Plug equilibrium concentrations into K a equation (approximate) Solve for X with quadratic + 2 [ H3O ][HPO ] 8 K a = = = - [H PO ] 2 [0.013][ [0.013] Plug x back into table to get equilibrium concentrations 4 4 x 41 x = 6.3x10-8 [H 2 PO 4 ] = x10-8 = 0.013M [H 3 O + ] = x10-8 = 0.013M No change in ph!!! [HPO 2 4 ] = x = 6.3x10-8 use for next ionization ]

42 HPO 4 2- (aq) + H 2 O(l) H 3 O + (aq) + 42 PO 4 3 (aq) Initial 6.3x Change x - + x + x Equilibrium 6.3x10-8 x x x [H 3 O + ] 3rd ionizationk a3 = < K a2 = 6.3x10-8 If second ionization didn t change ph, neither will 3 rd Use [H 3 O + ] from last ionization to calculate ph Calculate ph What is the ph of a M solution of phosphoric acid at 25 C? ph due only to first ionization ph = log( ) = 1.89 Original estimate was 1.43

43 Weak Bases and K b 43

44 Weak Bases Brønsted Lowry reaction for a weak base: B(aq) + H 2 O(l) HB + (aq) + OH (aq) + Equilibrium constant : [ HB ][OH ] K = = [B] K b is the base ionization constant. K b defines amount of dissociation On a log scale: pk b = log K b Strong bases: High[OH - ], low [H 3 O + ] Large K b & small pk b pk a + pk b = 14 3 Categories c K b Metal hydroxides, Most anions, & Organic amines 44

45 Metal Hydroxides Strong bases; Group 1A ex: NaOH, KOH Weak bases: Anything else withoh - ex: Ca(OH) 2 Ca 2+ Fe(OH) 3 Fe 3+ Dissociation affected by solubility Sparingly soluble salts Will act as weak bases (K b ) Use equilibrium chemistry (K sp ) Solubility Product Constant (next chapter.) 45

46 Anions Act as H + acceptors in water The negative charge attracts H + CO 3 2-, CH 3 COO -, etc. Exceptions: Anions of strong acids: 100% dissociation, no equilibrium established Cl, Br, I, ClO 4, NO 3, HSO 4 Anions with ionizable hydrogen ion Amphiprotic H 2 PO 4, etc. 46

47 Organic Amines General structure of R 3 N: R contains H,C,O, (main group atoms) Lone pair on nitrogen accepts H + R 3 N:(aq) + H 2 O(l) R 3 N H + (aq) + OH (aq) Conjugate acid: Ammonium ion, NH + 4 Common amines: CH 3 H 3 C N H 2 N CH 3 trimethylamine aniline 47

48 Common Weak Bases 48

49 Calculations Using K b 49

50 Relationship between K a & K b Base & water NH 3 + H 2 O NH 4+ + OH K b = 1.8X10-5 pk b = 4.74 Conjugate acid & water: NH 4+ + H 2 O H 3 O + + NH 3 K a = 5.6X10-10 pk a = 9.26 Add two reactions together: multiply K values Multiple equilibria (Ch. 15) 2H 2 O H 3 O + + OH K w =K a x K b = 1.00x10-14 pk a + pk b = (at 25 C) so pk w = = Only true for conjugate acid/base pairs in water! 50

51 Find the ph of a 0.10 M solution of ammonia at 25 o C. Reaction: NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH (aq) Initial Change x - + x + x Equilibrium 0.10 x - x x Fill in K b expression K b = 1.8x10 5 = [ NH + 4 [ NH ][OH 3 ] ] = [ x][ x] [0.10] x = [OH - ] = 13. x10 3 M Approximate [NH 3 ] Solve for ph poh = log[ OH ] = log[1.3x10 ph = 14 poh = = ] =

52 Find the percent ionization of a 0.10 M solution of ammonia at 25 o C. Calculate % Ionization [OH - ] = 1.3 x 10-3 from previous problem % Ionization = [OH [ NH 3 ] ] x100% = 1.3x x100% = 1.3% 52

53 Molecular Structure and Strength of Acids 53

54 Measuring Acid Strength Strongest Acid: Weakest bond to acidic H 3 Criteria in order of importance Charge: High negative charge on anion: weaker acid Leaves stronger bond to any remaining hydrogens Structure: Larger anions will create longer, weaker H-X bonds Presence of oxoatoms increase electronegativity Electronegativity: High electronegativity of anion: stronger acid Withdraws electrons from acidic H stronger acid 54

55 Effect of Charge H 3 AsO 4 H 2 AsO 4 HAsO 4 2 Charge: HAsO 4 2 has highest negative charge Higher charge: weaker acid Harder to pull H away, stronger bond Acidity based on charge difference H 3 AsO 4 > H 2 AsO 4 > HAsO 4 2 Structure: No differences Electronegativity: No differences 55

56 Effect of Structure: Bond Length vs. Electronegativity for Binary acids Direct bond between H and electronegative atom Two competing forces Large Atomic radius Long bonds are weaker High Electronegativity High electronegativity pulls electrons to more electronegative atom; weakens bond Experimentally measured acid strength HI>HBr>HCl>>HF Bond length is dominating force 56

57 Oxoacids More double bonded oxygens: Stronger acid H 2 SO 4 H 2 SO 3 Two terminal oxo atoms More resonance stabilizes ion More e- withdrawal from O-H H is more positive: more acidic One terminal oxo atom Fewer resonance structures Less e- withdrawal from O-H H is less positive: less acidic 57

58 Charge: No difference Both neutral Effect of Electronegativity Structure: No difference in terminal oxo atoms 1 oxo atom each Electronegativity: S more electronegative than P H 2 SO 3 more acidic than H 3 PO 4 S more electronegative than Se H 2 SO 4 more acidic than H 2 SeO 4 58

59 Acid Strength Bond length vs Electronegativity Charge Difference Structure: Terminal O Electronegativity 59

60 Acid-Base Properties of Salts Hydrolysis 60

61 Hydrolysis of Ionic Salts Ionic salts dissolved in water affect ph NaNO 3 (s) + H 2 O(aq) Na + (aq) + NO 3- (aq) Undergo Hydrolysis Hydrolysis: Reaction of an ionic salt with water May change the ph of the solution Both cations and anions may undergo hydrolysis Not all ions hydrolyze Examine both ions to determine acid/base character 3 step process to predict acidity of a salt solution 1. Write the reaction that dissociates salt into its ions 2. Check the cation for acid hydrolysis: produce H + 3. Check the anion for base hydrolysis: produce OH - 61

62 Is NaCl (aq) Acidic, Basic or Neutral 1. Dissociate the salt into ions: NaCl(aq) Na + (aq) + Cl (aq) 2. Check the cation for hydrolysis: Na + (aq) + H 2 O(l) NaOH(aq) + H + (aq) No Hydrolysis: NaOH strong base, not acidic 3. Check the anion for hydrolysis: Cl (aq) + H 2 O(l) HCl (aq) + OH -(aq) No Hydrolysis: HCl is a strong acid, not basic If neither acidic or basic, solution is neutral 62

63 Is Na 2 CO 3 Acidic, Basic or Neutral 1. Dissociate the salt into ions: Na 2 CO 3 (aq) 2Na + (aq) + CO 3 2- (aq) 2. Check the cation for hydrolysis: Na + (aq) + H 2 O(l) NaOH(aq) + H + (aq) No Hydrolysis: NaOH would be a strong base No H + generated: Solution is not acidic 3. Check the anion for hydrolysis: CO 2-3 (aq) + H 2 O(l) HCO 3- (aq) + OH - (aq) Anion hydrolyzes: HCO 3- (aq) weak acid Some OH - is generated: Solution may be basic If not acidic but possibly basic, solution is basic 63

64 Is Fe(NO 3 ) 2 Acidic, Basic or Neutral? 1. Dissociate the salt into ions: Fe(NO 3 ) 2 (aq) Fe 2+ (aq) + 2NO 3 (aq) 2. Check the cation for hydrolysis: Fe 2+ (aq) + H 2 O(l) FeOH + (aq) + H + (aq) Reaction occurs: FeOH + is a weak base H + generated: Acidic solution 3. Check the anion for hydrolysis: NO 3 (aq) + H 2 O(l) HNO 3 (aq) + OH - (aq) No OH- generated: Nitric is a strong acid, not basic Overall: the solution is acidic 64

65 Is ZnF 2 Acidic, Basic or Neutral 1. Dissociate the salt into ions: ZnF 2 (aq) Zn 2+ (aq) + 2F (aq) 2. Check the cation for hydrolysis: Zn 2+ (aq) + H 2 O(l) ZnOH + (aq) + H + (aq) Reaction Occurs: ZnOH + (aq) is a weak base H 3 O + generated: Acidic Solution 3. Check the anion for hydrolysis: F (aq) + H 2 O(l) HF (aq) + OH - (aq) Reaction Occurs: HF is a weak acid OH - is generated: Basic Solution Can t tell acidity : Both cation and anion hydrolyze Both H 3 O + (aq) and OH - (aq) possible 65

66 Is ZnF 2 Acidic, Basic or Neutral 1. Dissociate the salt into ions: ZnF 2 (aq) Zn 2+ (aq) + 2F (aq) 2. Cation Hydrolysis: K a based Zn 2+ (aq) + H 2 O(l) ZnOH + (aq) + H + (aq) 3. Anion Hydrolysis: K b based F (aq) + H 2 O(l) HF (aq) + OH - (aq) Can t tell acidity : Both cation and anion hydrolyze Both H + (aq) and OH - (aq) possible Higher K value determines acidity 66

67 Determine Acidity Look up K a s for both acids K a (Zn 2+ ) = for acid reaction K a (HF) = for base reaction Calculate K b for the anion reaction K b (F ) = K w /K a = / = Larger equilibrium constant wins K a (Zn 2+ )> K b (F ) > H + created > OH - created Cation hydrolysis dominates : Solution is acidic. 67

68 Find the ph of a solution with M CHOONa. First determine if acidic, basic or neutral Anion hydrolysis: CHOOH is a weak acid, so basic Need to use K b HCOO - (aq) + H 2 O(l) HCOOH(aq) + OH - (aq) Initial Change x - + x + x Equilibrium x - x x Initial Data [HCOO - ] = from problem [HCOOH] and [OH - ]= 0 Look up K a in table and calculate K b K b [ K = [ K w a ] ] 1.0x10 = 1.8x = 5.6x

69 Find the ph of a solution with M CHOONa. HCOO - (aq) + H 2 O(l) HCOOH(aq) + OH - (aq) Initial Change x - + x + x Equilibrium x - x x Solve for x [ OH ][ HCOOH ] [ x][ x] x 10 = = [ HCOO ] [0.05] x =[OH-]= 1.7x10-6 Calculate ph poh = -log[1.7x10-6 ] = 5.77 ph = 14-pOH =

70 Hydrolysis of Anions from Polyprotic Acids NaHCO 3, NaH 2 PO 4, NaHSO 4 Both acid and base hydrolysis occur Cation Hydrolysis: K a = 4.8x10-11 HCO 3- (aq) + 2H 2 O(l) CO 3 2- (aq) + H 3 O + (aq) Anion Hydrolysis: K b = 2.4x10-8 HCO 3- (aq) + H 2 O(l) H 2 CO 3 (aq) + OH (aq) K b = Kw/Ka = 1x10-14 /4.2x10-7 = 2.4x10-8 Compare equilibrium constants Larger K determines equilibrium reaction solution is basic 70

71 Hydrolysis of Complexed Metal Ions Metal Ions form complexes with water (hydrated) Pull electrons towards metal ion Polarize O-H bond in attached waters H+ ions dissociate from oxygen Al Cation hydrolysis only Al(H 2 O) 6 3+ (aq) + H 2 O(l) Al(OH)(H 2 O) 5 2+ (aq) + H 3 O + (aq) K a = 1.3x10-5 O H H 71

72 Lewis Acids and Bases BF 3 (g)+ NH 3 (g) F 3 B-NH 3 (g) Lewis Acid: Accepts a pair of electrons BF 3 No ionizable H Not bronsted acid Lewis Base: Donates a pair of electrons :NH 3 No H to accept Not bronsted base F F B F H N H H 72

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