# Chapter 17. Acid-Base Equlibria and Solubility Equilibria

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1 Chapter 17 Acid-Base Equlibria and Solubility Equilibria 1

2 Buffer Solutions 2

3 Buffer Solutions Solutions that resist ph change upon addition of small amounts of acid or base Need both acid & conjugate base The acid reacts with added base HA+ OH-H 2 O + A - The base reacts with added acid A - + H 3 O + H 2 O + HA Weak acids & bases only Equilibrium can be established No [H 3 O + ] or [OH - ] added directly Only changes to equilibrium HA+ H 2 OH 3 O + + A - ph controlled by K a 3

4 Buffer Capacity The amount of acid or base that can be neutralized by the buffer components The more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. X must always be negligible in an ICE table Most effective if [acid] = [conjugate base] Can neutralize equally in either direction 4

5 Criteria for Making a Buffer Choose weak acid and conjugate base Must have same anion! HNO 2 & NO 2-, HF & F - Choose acid based on general ph range pka<7 buffer is acidic, pka>7 buffer is basic Can usually adjust within +/1 ph units Buffer salts must be soluble: full dissociation Sodium, potassium salts most common conjugate base NH 4 salts are acidic due to NH 4+ dissociating Concentrations > 0.01M for [HA] & [A-] Must be able to absorb excess acid or base Can always approximate in ICE table!!!!!!! 5

6 A 1.00 L buffer solution is prepared that is M nitrous acid and M sodium nitrite. What is ph? NaNO 2 (aq) NO -( 2 aq) HNO 2 (aq) + H 2 O(l) NO 2- (aq)+ H 3 O + (aq) Initial Change x - + x + x Equilibrium x x x Solve for x using K a from table 7.2x10 [ H3O ][ NO2 ] [ x][0.200] 4 [ HNO ] [0.150] 2 Calculate ph ph = log(5.4 x 10-4 ) = 3.27 x =5.4 x

7 What is the ph after adding 1.00 g HBr? (g HBr [HBr]) 1.00gHBr x mol/80.8g = 0.012mol/1.0L= 0.012M Acid/Base Rxn HBr (aq)+ NO 2- (aq) HNO 2 (aq) + Br - (aq) Strong Acid: [HBr] = -[NO 2- ] = [HNO 2 ] = [Br - ] 0.012M = M = M = 0.012M Old Equilibrium HNO 2 (aq) + H 2 O(l) H 3 O + (aq) +NO 2- (aq) Initial 0.150M M New Equilibrium HNO 2 (aq) + H 2 O(l) H 3 O + (aq) + NO 2- (aq) Initial

8 What is the ph after adding 1.0 g HBr? K a equation HNO 2 (aq) + H 2 O(l) H 3 O + (aq) +NO 2- (aq) Initial Change x - + x + x Equilibrium x - x x Solve for x 7.2x10 2 ] x 4 [ H3O ][ NO [ HNO ] 2 [ ][0.188] [0.162] x =6.2 x 10-4 Calculate ph ph = log(6.2 x 10-4 ) =

9 What is the ph after adding 1.00 g KOH? (g KOH [OH _ ]) 1.00gKOH x mol/46.1g = 0.018mol/1.0L= 0.018M OH - Acid/Base Rxn OH - (aq) + HNO 2 (aq) NO 2- (aq) + H 2 O(l) Strong Base: [OH-] = -[HNO 2 ] = +[NO 2- ] 0.018M = M = M Old Equilibrium HNO 2 (aq) + H 2 O(l) H 3 O + (aq) +NO 2- (aq) Initial 0.150M M New Equilibrium HNO 2 (aq) + H 2 O(l) H 3 O + (aq) + NO 2- (aq) Initial

10 What is the ph after adding 1.00 g KOH? K a equation HNO 2 (aq) + H 2 O(l) H 3 O + (aq) +NO 2- (aq) Initial Change x - + x + x Equilibrium x - x x Solve for x 7.2x10 Calculate ph 2 ] x 4 [ H3O ][ NO [ HNO ] 2 [ ][0.218] [0.132] ph = log( ) = 3.36 x =

11 Without Buffer Buffer Effects With Buffer Water ph = 7.00 Strong Acid (HBr) ph = 1.92 Strong Base (NaOH) ph = Buffer ph = 3.26 Strong Acid (HBr) ph =3.21 Strong Base (NaOH) ph =

12 Preparing a buffer with a specific ph The Henderson-Hasselbalch Equation ph = pk a + log([conjugate base]/[acid]) A buffer solution is prepared that is M nitrous acid (HNO 2 ) and M sodium nitrite (NaNO 2 ). What is the ph of the solution? K a = 7.2 x 10-4 ph = - log(7.2 x 10-4 ) + log([0.200]/[0.150]) ph = log(1.33) ph = = 3.27 (same as ICE table) 12

13 The Henderson-Hasselbalch Equation A buffer solution of M nitrous acid (HNO 2 ) and M sodium nitrite (NaNO 2 ). K a = 7.2 x 10-4 What is the ph after the addition of 1.00g HBr? ph = log([0.188]/[0.162]) ph = = 3.21 What is the ph after the addition of 1.00g KOH? ph = log([0.218]/[0.132]) ph = =

14 Titration 14

15 Chem 101: Strong Acid-Strong Base Titrations Reaction goes to completion: no equilibrium NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O(l) OH - (aq) + H + (aq) H 2 O(l) 15

16 Solving More Complex Titration Problems 1. Read question carefully to answer what s asked ph at a particular point moles or molarity of original solution ph or volume at equivalence point 2. What chemicals are reacting? Identify all products. Balanced equations: equilibrium or completion Subtract moles to find predominant products 3. Is solution acidic or basic? Is it an equilibrium? You only have 1 equilibrium at most Strong: Direct ph calculations from molarity Weak: equilibrium calculations 4. Volume increases, so watch for dilution factors 16

17 Strong Acid-Strong Base Titrations Reaction goes to completion: no equilibrium NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O(l) OH - (aq) + H + (aq) H 2 O(l) 17

18 Strong Acid-Strong Base Calculations 25.0 ml of 0.100M HCl titrated with 0.100M NaOH 1. At start: 25 ml strong acid only ph log[ H 3 O 2. Add 10 ml NaOH moles O 0.100mol 25.0mLx 1000mL ] log[0.100] x10 3 H 3 mol 0.100mol 3 moles OH 10.0mLx 1.00x10 mol 1000mL 3 3 (2.50x10 mol 1.00x10 mol ) H3O OH molesh O M 3 (0.025L 0.010L ) H 3 O OH ph log[ ]

19 Strong Acid-Strong Base Calculations 25.0 ml of 0.100M HCl titrated with 0.100M NaOH 3. At Equivalence Point: moles acid= moles Base ph=7.0 for SA-SB combination Acid cancels all base 4. Add 35 ml NaOH 0.100mol 3 moles H O 25.0mLx 2.50x10 mol mL 0.100mol 3 moles OH 35.0mLx 3.50x10 mol 1000mL 3 3 (3.50x10 mol 2.50x10 mol ) OH H3O moles M OH (0.035L 0.025L ) OH H O 3 OH poh log[ ] 1.78 ph 14 poh

20 Weak Acid-Strong Base Titrations If [HA]>[OH-], solution is an equilibrium CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) 20

21 Weak Acid-Strong Base Calculations 25.0 ml 0.100M CH 3 COOH titrated with 0.100M NaOH 1. At start: 25 ml acid only: Ka K a 1.8x10 x [ H O 3 ph ] [ H (1.8x10 O ][ CH 3COO [ CH COOH ] ] x [0.100] x x Add 10 ml base: HH calculation moles moles ph CH COOH 3 OH pk a 0.100mol 25.0mLx 1000mL moles CH COO x mol 10.0mLx 1000mL mol 1.00x10 3 mol 3 1.0x10 log[ CH COO ] /[ CH 3COOH ] 4.74 log 3 1.5x

22 Weak Acid-Strong Base Calculations 25.0 ml of 0.100M CH 3 COOH titrated with 0.100M NaOH 3. At Equivalence Point: Moles base= moles acid=0 Use K b for remaining salt K b 4. Add 35 ml base: strong base moles 5.6x10 x [ OH poh CH COOH ] 10 [ OH ][ CH 3COOH ] [ CH COO ] (5.6x10 ph mol 25.0mLx 1000mL 2.50x mol 3.50x10 mol 2.50x10 mol 1.50x10 mol H 3O moles moles 35.0mLx M OH CH 3COO 1000mL 0.035L 0.025L 0.060L poh log[ ] x x x [0.050] 6 3 mol ph 14 OH poh OH 22

23 Strong Acid-Weak Base Titrations If [WB]>[SA], solution is an equilibrium HCl(aq) + NH 3 (aq) NH 4+ (aq) + Cl - (l) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) 23

24 Strong Acid-Weak Base Titration Calculations 25.0 ml of 0.100M NH 3 titrated with 0.100M HCl 1. At start: 25 ml base only: Use K b calculations K b poh 1.8x10 x [ OH ] [ OH ][ NH [ NH ] (1.8x10 x x Add 10 ml acid: HH equation 5 ph 3 4 ] 2 x [0.100] mol 3 moles NH 25.0mLx 2.50x10 mol mL moles ph H 0.100mol 3 moles 10.0mLx 1.00x10 mol NH4 1000mL 3 1.5x10 pk a log[ NH ]/[ NH 4 ] 9.26 log 3 1.0x

25 Strong Acid-Weak Base Titration Calculations 25.0 ml of 0.100M NH 3 titrated with 0.100M HCl 3. At Equivalence Point: Moles base= moles acid=0 Use K a for remaining weak acid 4. Add 35 ml acid: Strong acid moles K a 0.100mol 3 moles NH 25.0mLx 2.50x10 mol mL H O 3 5.6x10 x [ H ph 3 O 5.28 ] 10 [ H (5.6x mol 35.0mLx 1000mL 3 O ][ NH [ NH ] ] x x x10 mol 0.035L 2 x [0.050] x L 3 3 mol 1.00x10 mol 0.060L H O M H O 3 ph log[ ]

26 Acid-Base Indicators Chemical added during a titration to cause a color change at a particular ph and allow the user to detect the endpoint Equilibrium reaction: HIn (aq) H + (aq) + In - (aq) Red yellow 26

27 Acid-Base Indicators Choose indicator based on range of ph change 27

28 Solubility Equilibria 28

29 Aqueous Salt Solutions & Solubility Chem 101 Ionic compounds used were water soluble Fully dissociated in water: Reaction goes to completion Directly used stoichiometry Solubility defined as: Grams solute/1l saturated solution Chem 112 Most ionic compounds sparingly soluble Exist as an equilibrium reaction: K sp Molar solubility used: Moles solute/1l saturated solution 29

30 Solubility and Molar Solubility Calculations Solubility to K sp g solid /L mol solid /L mol ions /L K sp K sp to Solubility K sp mol ions /L mol solid /L g solid/l 30

31 Solubility Rules (Chapter 4) Anions Solubility Nitrate (NO 3 ) All soluble Perchlorate (ClO 4 ) All soluble Acetate (C 2 H 3 O 2 ) All soluble Halide (Cl, Br, I & F ) Soluble except: Pb 2+, Ag +, Hg 2+ 2, Hg 2+, Cu + Sulfate (SO 2 4 ) Soluble except:ba 2+, Ca 2+,Sr 2+, Pb 2+ 2+, Hg 2 Sulfide (S 2 ) Insoluble except: Group 1 & NH + 4 cations Hydroxide (OH ) Insoluble except: Group 1 & NH + 4 cations Carbonate (CO 2 3 ) Insoluble except: Group 1 & NH + 4 cations Phosphate (PO 3 4 ) Insoluble except: Group 1 & NH + 4 cations 31

32 The Solubility Product Constant, K sp Solubilization of sparingly soluble salt, M a X b : CaCl 2 (s) Ca 2+ (aq) + 2Cl (aq) The equilibrium constant is: K [ Ca ][ Cl ] c [ Ca ][ Cl ] K sp [ CaCl2( s)] K sp is the Solubility Product constant Approximation only: Used to estimate solubility Will never have a denominator Tabulated in Table of Solubility Product Constants 32

33 Qualitative Prediction of Precipitation Precipitation depends upon ion concentration Use the ion reaction quotient, Q c, to decide Calculate Q c with given conditions. Compare Q c to the K sp of the salt. Summary: Remember comparing Q c to K sp Q c > K sp Products too high to stay in solution Precipitation Q c < K sp Not enough ions to precipitate No Precipitation Q c = K sp Reaction at equilibrium: saturated No Precipitation yet 33

34 The Solubility Product Constant, K sp 34

35 Molar Solubility Making 1M BaSO 4 233g solid BaSO 4 (1 mole) 233g BaSO 4 added to 1L volumetric flask Dilute to the 1L mark Mathematically: 1 mole/l = 1M Not everything dissolves!!!!!! Molar solubility is a measure of what does dissolve 35

36 Molar solubility of BaSO 4 1. Solubilization: BaSO 4 Ba 2+ (aq) + SO 4 2- (aq) 2. Find K sp from table: K sp = [Ba 2+ ][SO 4 2- ]= Make Table BaSO 4 (s) Ba 2+ SO 2-4 Initial: 1 Mole/L water 0 0 Change: -x +x +x Equilibrium: 1-x(but it s solid!) x x 4. Plug in values: = [Ba + ][SO 4 2- ]= [x][x] = x 2 x = Determine molar solubility (1 mole salt/1l soln) BaSO 4 dissociated = x = mole/1l 36

37 Solubility and Molar Solubility of Barium Sulfate BaSO 4 Ba 2+ (aq) + SO 4 2- (aq) Molar Solubility = 1 x10-5 M = 1 x10-5 mol/l Solubility = 1 x10-5 mol/l x 233g/mol= 2.33x10-3 g/l Ba 2+ = 1 x10-5 M = 1 x10-5 moles SO 4 2- = 1 x10-5 M = 1 x10-5 moles BaSO 4 = 1mole - 1 x10-5 moles BaSO 4 still approximately 1mole Mass = 233g g= 233g 37

38 Molar Solubility of Mg 3 (AsO 4 ) 2 1. Solubilization: Mg 3 (AsO 4 ) 2 3Mg 2+ (aq) + 2AsO 3-4 (aq) 2. Find K sp K sp = [Mg 2+ ] 3 [AsO 3-4 ] 2 = Table Mg 3 (AsO 4 ) 2 (s) 3Mg 2+ 2AsO 4 3- Initial: Change: - +3x +2x Equilibrium: - 3x 2x 4. Fill in values = [Mg 2+ ] 3 [AsO 4 3- ] 2 = [3x] 3 [2x] 2 = 108x 5 x = M 5. Determine Molar solubility MgAsO 4 dissociated = M 38

39 Comparison of Molar Solubility vs. K sp Compound K sp Molar Solubility BaSO M Mg 3 (AsO 4 ) M Solubility Mg 3 (AsO 4 ) 2 solubility is 5X greater than BaSO 4 K sp BaSO 4 K sp is ten billion times greater than Mg 3 (AsO 4 ) 2 Can t always use K sp directly to compare solubilities! Molar solubility is a function of the # of ions present Can only be used if # ions produced is identical 39

40 Reaction Quotients & Solubility Q vs. K sp 40

41 Quantitative Prediction of Precipitation Precipitation depends upon ion concentration Use the ion reaction quotient, Q c, to decide Calculate Q c with given conditions. Compare Q c to the K sp of the salt. Summary: Remember comparing Q c to K sp Q c > K sp Ions too high to stay in solution Precipitation Q c < K sp Q c = K sp Not enough ions to precipitate No Precipitation at equilibrium: saturated No Precipitation yet 41

42 50.0 ml of a M solution Ca(NO 3 ) 2 of is mixed with 25.0 ml of a M solution of K 2 SO 4. Will precipitation occur? Ca(NO 3 ) 2(aq) + K 2 SO 4(aq)? CaSO 4 (s) +2K + (aq) +2NO 3(aq) CaSO 4 (s) Ca 2+ (aq) + SO 4 2 (aq) K sp = [Ca 2+ ] = M x 0.050L = 6.7 x 10-4 M [SO 4 2 ] = 8.3 x 10-4 M L Q c = [Ca 2+ ] [SO 4 2 ] = [ ][ ] = < so Q c < K sp No precipitation, too dilute 42

43 Add solid K 2 SO 4 to 50.0 ml of M Ca(NO 3 ) 2 solution. How many grams of K 2 SO 4 are needed to start precipitation? Ca(NO 3 ) 2 (aq) Ca 2+ (aq) + 2NO 3 (aq) CaSO 4 (s) Ca 2+ (aq) + SO 4 2 (aq) CaSO 4 (s) Ca 2+ SO 4 2 Initial: Change: - +x +x Equilibrium: x Solve for x K sp = [Ca 2+ ][SO 4 2 ] = = [0.0010][x] x = M Calculate g K 2 SO x10 mol L 50.0mL 174.3g gk2so4 x x x g L 1000mL 1 mol 43

44 How many g of K 2 SO 4 are needed to precipitate 99% of the calcium in the previous solution? CaSO 4 (s) Ca 2+ SO 2 4 Initial: Change: - - ( x.99) +x Equilibrium: M x Solve for [X] K sp = [Ca 2+ ][SO 2 4 ] = = [ ][x] x = 0.91M = [SO 2 4 ] Calculate g K 2 SO mol L 50.0mL 174.3g gk2so4 x x x 7. 9g L 1000mL 1 mol 44

45 Common Ion Effect 45

46 The Common Ion Effect in Solubility (Buffers in Acid-Base Chemistry) PbCl 2 (s)pb 2+ (aq) + 2Cl (aq) Add NaCl: NaCl(s) Na + (aq) + Cl (aq) Added Cl pushes reaction towards reactants Create more PbCl 2 (s) Removing some Pb 2+ from solution. Common ion (Cl-) reduces the solubility Effect of LeChatelier's Principle System shifts to remove added stress to 1 side Used to remove ions from solution. 46

47 Molar Solubility of PbCl 2 1. Solubilization: 2Cl - (aq) PbCl 2 Pb 2+ (aq) + 2. Find K sp in table K sp = [Pb 2+ ] [Cl - ] 2 = 3. Make Table PbCl 2 (s) Pb 2+ Cl x 2x Initial: 1 Mole/L water 0 Change: -x +x Equilibrium: 1-x x 4. Plug in values = [Pb 2+ ][Cl - ] 2 = [x][2x] 2 = 4x 3 47

48 How much is the solubility of lead(ii) chloride changed in the presence of 0.85 M NaCl? PbCl 2 has a molar solubility of M in pure water NaCl(s) Na + (aq) + Cl (aq) PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) PbCl 2 Pb 2+ 2Cl - Initial: Change: - x +2x Equilibrium: - x x Solve for [x] Approximation: 0.85 > , so ( x) ~ 0.85 K sp = [Pb 2+ ][Cl ] 2 = = [x][ x)] 2 = [x][0.85] 2 x = = [Pb 2+ ] = M The concentration is reduced about 1000x. 48

49 Complex Ion Formation 49

50 Complex-Ion Formation Polyatomic cation or anion consisting of a central metal ion that has other groups called ligands bonded to it. NH 3 2+ Cu 2+ (aq) + 4NH 3(aq) NH 3 Cu NH 3 NH 3 (aq) The metal center is a Lewis acid that accepts electron pairs The ligands are Lewis bases that donate electron pairs. The formation reaction of a complex ion is reversible The equilibrium is described by a formation constant, K f. 50

51 Complex Ion Formation Equilibrium Cu 2+ (aq) + 4NH 3(aq) [Cu(NH 3 ) 4 ] 2+ K f (aq) [ Cu( NH3) 4] 2 [ Cu ][ NH ] Find K f in Table of Complex Ion Formation Constants gives the stoichiometry of the complex ion Common Lewis bases that give complex ions NH 3, CN, OH, Cl, Br, and I 51

52 K f Values for Complex Ions 52

53 Calculate the molar solubility of zinc carbonate in pure water at 25 C. ZnCO 3 (s) Zn 2+ CO 3 2 Initial: Change: - +x +x Equilibrium: - x x Solve for [X] K sp = [Zn 2+ ][CO 3 2 ] = = [x][x] = x 2 x = ZnCO 3 (s)zn 2+ (aq) + CO 3 2 (aq) The molar solubility of ZnCO 3 in water is M 53

54 Calculate the molar solubility of zinc carbonate in 1.0M NH 3 at 25 C. Dissociation (K sp ): ZnCO 3 (s)zn 2+ (aq) + CO 2 3 (aq) Dissolution ( K f ): Zn 2+ (aq) + 4NH 3 (aq) Zn(NH 3 ) 2+ 4 (aq) Together (K c )= K sp xk ZnCO f 3(s) + 4NH 3(aq) Zn(NH 3 ) 2+ 4 (aq) +CO 2 3 (aq) ZnCO 3 (s) 4NH 3 (aq) Zn(NH 3 ) 4 2+ CO 3 2 Initial: Change: - -4x +x +x Equilibrium: x x x K c = K f K sp = ( ) ( ) = Solve for [X] 5.7x10 3 K c 2 [ Zn( NH3) 4 ][ CO [ NH ( aq)] ] [ x][ x] [1.0 4x] 4 x = M = molar solubility : by a factor of

55 Qualitative Analysis 55

56 Qualitative Analysis of Metal ions Determination of ions present in solution Present or not present only Selective precipitation 1 compound has low K sp, the others very high or none Complex formation Only 1 complex forms May dissolve a precipitate Flame Tests Flame color used as ID 56

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