Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College

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1 Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College

2 The Common Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 (aq) If acetate ion is added to the solution, Le Châtelier s Principle says the equilibrium will shift to the left. Acetate ion can be added in the form of a strong electrolyte such as sodium acetate, NaC 2 H 3 O 2 NaC 2 H 3 O 2 (aq) Na + (aq) + C 2 H 3 O 2 (aq) Sodium acetate is 100% ionic in solution

3 The Common Ion Effect The extent of ionization of a weak electrolyte can be decreased by adding a strong electrolyte to the solution that has an ion in common with the weak electrolyte.

4 The Common Ion Effect Example: Calculate the fluoride ion concentration and ph of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is Solution: HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) The equilibrium constant expression is [H 3 O + ] [F ] K a = = [HF] 4

5 The Common Ion Effect Because HCl, a strong acid, is also present, HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) the initial [H 3 O + ] is not 0, but rather M. Initial concentration [HF], M [H 3 O + ], M [F ], M (from the HCl) Change x +x +x Equilibrium 0.20 x x concentration Remember, x is small and can be ignored 0 x

6 The Common Ion Effect = (0.10) (x) (0.20) Solve for x. Rearrange to: (0.20) ( ) (0.10) = x M = x

7 The Common Ion Effect Therefore, [F ] = x = M If we add x to 0.10 M [H 3 O + ] = x = ( ) = 0.10 M and ph = log (0.10) ph = 1.00

8 Buffer Solutions A buffer is an application i of the common ion effect Because the common ion shifts the equilibrium in one direction, a buffer is resistant to ph changes, even when small amounts of a strong acid or base is added. d

9 Buffer Solutions If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH to make F and water.

10 Buffer Solutions If acid is added, the F reacts to form HF and water.

11 Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 O K a = a H + 3 O + A [H 3 O + ] [A ] [HA] Rearrange this to separate out the [H 3 O + ] K a = [H 3 O + ] [A ] [HA]

12 Buffer Calculations K + a = [H 3 O ] [A ] [HA] Take the negative log of both sides pk a [A ] [HA] log K = ( log( [H + 3 O ]) + ( log ) log K a base ph acid and [base] pkk a = ph log [acid]

13 Buffer Calculations Rearrange the equation to solve for ph ph = pk + log [base] a [acid] The equation, in this form is known as the Henderson Hasselbalch l equation.

14 Henderson Hasselbalch Hasselbalch Equation Example: What is the ph of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a for lactic acid is Solution: HC 3 H 5 O 3 + H 2 O C 3 H 5 O 3 + H 3 O + Conc M 0.10 M x (What happened to x for the HC 3 H 5 O 3 concentration? )

15 Henderson Hasselbalch Hasselbalch Equation Substitute the concentrations into the Henderson-Hasselbalach equation ph = pk a + log [base] [acid] = log ( )+ log = ( 0.08) ph = 3.77 (0.10) (0.12)

16 Buffer Capacity The buffer capacity is the amount of acid or base that the buffer can neutralize before the ph changes to an appreciable degree. The ph range is the range of ph values over which a buffer system works effectively. It is best to choose an acid with a pk a close to the desired dph.

17 When Strong Acids or Bases Are Added d to a Buffer Assume that all of the strong acid or base is consumed in the reaction.

18 Addition of Strong Acid or Base to a Buffer 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2. Use the Henderson Hasselbalch equation to determine the new ph of the solution.

19 Calculating ph changes when adding strong Sample problem: acid or base to a Buffer A buffer is made by adding mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of thebuffer is4.74. Calculate the ph of this solution after mol of NaOH is added. K 18 5 a HC 2 H 3 O 2 = 1.8 x 10 5

20 Calculating ph changes when adding strong acid or base to a Buffer Initial data, before the reaction: mol HC 2H 3O 2 = mol C 2H 3O 2 = 0.30 M From the acetic acid from the sodium acetate and ph = pk a + log [base] [acid] Since [base] = [acid] ph = pk a = log ( ) =

21 Calculating ph changes when adding strong acid or base to a Buffer The mol NaOH will react with ih0.020 mol of the acetic acid: HC H O + OH C H O 2 3 2( aq) (aq) (aq) + H 2 O (l) This reduces the amount of acetic acid by mole and increases the acetate ion concentration by 0.020mole HC 2 H 3 O 2 C 2 H 3 O 2 OH Before reaction mol mol mol After reaction mol mol mol

22 Calculating ph changes when adding strong acid or base to a Buffer Use the Henderson Hasselbalch H l equation to calculate the new ph: ph = log (0.320) (0. 280) ph = ph = 4.80

23 Calculating ph changes when adding strong acid or base to a Buffer Sample Problem: What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (before HCl, ph = 7.00) b)1.00 L of buffer that has [HOAc] = M and [OAc ] = M. The ph = 4.68

24 Calculating ph changes when adding strong acid or base to a Buffer Solution to Part (a) HCl is 100% ionic. Calc. [HCl] after adding 100mL 1.00 of 100M 1.00 HCl to 1.00 L of water M 1 V 1 = M 2 V M 1.00 ml = M ml M 100x = M = [H 3 O ] ph = log (1.00 x 10 3 ) ph =

25 Calculating ph changes when adding strong acid or base to a Buffer Solution to Part (b) Step 1 The H 3 O + (from HCl) reacts with OAc (from buffer) to yield HOAc The reaction occurs completely because K is very large. [H 3 O + ] + [OAc ] [HOAc] Before rxn M M M Change M M M After rxn M M

26 Calculating ph changes when adding strong acid or base to a Buffer Solution to Part (b): Step 2 Set up a concentration table (similar to equilibrium typecalculationor or ICE) usingcalculated concentrations [HOAc] and [OAc ] Solve for [H 3 O + ] HOAc + H + 2 O H 3 O + OAc [HOAc] [H 3 O + ] [OAc ] I: Before rxn M M Change x +x +x E: After rxn x M x x M M M

27 Calculating ph changes when adding strong acid or base to a Buffer Solution to Part (b): Step 2 Solve for ph using the Henderson Hasselbalch Equation ph = pk a + log [base] [acid] ph = log (1.8 x 10 5 ) + log [0.599] [0.701] ph = ( 4.74)( 474)+ ( ) 0685) ph = 4.67 The original ii ph was , so the ph is essentially unchanged.

28 Preparing a Buffer How would you prepare a buffer solution with a ph of ? Solution: First, calculate the [H + 3 O ] ph = log [H 3 O + ] = log [H 3 O + ] 4.30 = log [H 3 O + ] [H O + ] = [H 3 O + ] = 5.0 x 10 5 M Choose an acid such that [H 3 O + ] K a

29 Preparing a Buffer Next, consult a list of weak acids: Weak kaid Acid Formula K a Formic acid HCO 2 H 1.8 x 10 4 Benzoic acid HC 6 H 5 COO 6.3 x Acetic acid HC 2 H 3 O x 10 5 Propanoic acid HC 2 H 5 O x 10 5 Since [H 3 O + ] = 5.0 x 10 5 M Ch id h h [H O + ] K Choose an acid such that [H 3 O + ] K a For this problem, I am selecting acetic acid

30 Solution (continued): Preparing a Buffer Use the Henderson HasselbalchHasselbalch equation: ph = pk a + log [base] [acid] Solve for the [base]/[acid] ratio 4.30 = log (1.8 x 10 5 ) + log [base] [acid] 4.30 = log [base] [acid] [base] 0.44 = log [base] [acid]

31 Solution (continued): Preparing a Buffer 3.63 = [base] [acid] A typical laboratory solution of acetic acid is 0.10 M substitute 0.10 M for the [acid] and solve for [base] [base] = M You would need mole of acetate ion added to 1.0 L of 0.10 M acetic acid If you used sodium acetate (MW = 82 g/mol): mol x 82 g/mol = 29.8 g sodium acetate/l

32 Preparing a Buffer In the preparation of a buffer, the important thing to remember is the RATIO OF THE CONJUGATE BASE TO THE ACID determines the ph of the buffer CONCENTRATION of the acid and conjugate base are not important. Diluting a buffer solution does not change its ph

33 Commercial Buffers The solid acid and a salt of the acid (containing the conjugate base) are in a premeasured packet. The solid acid and conjugate base in the packet are mixed with a specified amount of water to give a buffer of a specified ph. Althoughthe the quantityof water does not affect the ph of the buffer, only the recommended volume of buffer should be prepared.

34 Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base). The progress of the reaction is observed using a ph meter and/or an indicator

35 Titration A ph meter or indicators are used to determine when the solution has reached theequivalence equivalence point, at which the stoichiometric amount of acid equals that of base. If an indicator is used, it is chosen so that its color change occurs close to the ph of the equivalence point of the titration.

36 Indicators for Acid Base Titrations

37 Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the ph goes up slowly.

38 Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the ph increases rapidly.

39 Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains ti only water and the salt fromthe cation of the base and the anion of the acid.

40 Titration of a Strong Acid with a Strong Base As more base is added, the increase in ph again levels off.

41 Strong acid titrated with a strong base

42 Titration of a Weak Acid with a Unlike in the previous case, the conjugate base of theacid affects the ph when it is formed. The ph at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations. Strong Base

43 ph Titrant volume, ml

44 How phenolphthalein works

45 Titration of a Weak Acid with a Strong Base In the titration of a weak acid with a strong base, at each point below the equivalence point, the ph of the solution during titration is dt determined d from the amounts of the acid and its conjugate base present at that particular point of the titration. To calculate the ph, treat the solution as a buffer

46 Titration of a Weak Acid with a Strong Base Withweaker acids (lower K a ), the initial ph is higher ph changes near the equivalence point are smaller

47 Titration of a Weak Base with a Strong Aid Acid The ph at the equivalence point in these titrations is < 7. Methyl hlred is a suitable indicator for this titration.

48 Titration of a Weak Base with a Strong Acid Weak base (NH 3 ) titrated with a strong acid (HCl)

49 Titration of Polyprotic Acids Weak diprotic acid (H 2 C 2 O 4 ) titrated with a strong base (NaOH)

50 Titration of Polyprotic Acids For a triprotic acid, H 3 PO 3, phosphorous acid, there is an equivalence point tfor each dissociation

51 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH. K 63x 5 a HBz = a) What is the ph of the solution when the benzoic acid is 50% neutralized? b) What is the ph of the solution at the equivalence point? HBz + NaOH Na + + Bz + H 2 O C 6 H 5 CO 2 H = HBz Benzoate ion = Bz -

52 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH. a) What is the ph of the solution when the benzoic acid is 50% neutralized? b) What is the ph of the solution at the equivalence point? ph at half- way point? ph at equivalence point? Benzoic acid + NaOH ph of initial solution of benzoic acid, a weak acid

53 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NOH NaOH. K 63 5 a HBz = 6.3 x 10 5 a) What is the ph of the solution when the benzoic acid is 50% neutralized? Solution: Amount of benzoic acid reacted = 50.0 ml of a M solution Calculate Volume of NaOH used: M HBz x V HBz = M NaOH NOH x V NaOH NOH M x 50.0 ml = M x V NaOH V NaOH = 12.5 ml Calculate new volume of solution: 100. ml HBz sol n ml NaOH = ml

54 Calculate Molarity of remaining benzoic acid M 1 HBz x V 1 HBz = M 2 HBz x V 2 HBz M x 50.0 ml = M 2 HBz x ml = M Calculate concentration of Bz - HBz + NOH NaOH Na + + Bz + H 2 O Conc M M Use the Henderson-Hasselbalch Hasselbalch equation ph = pk a + log [base] [acid] Since [base] = [acid], log 1 = 0 and ph = pk a ph = -log g( (6.3 x 10-5 ) ph = 4.2

55 Solution: b) What is the ph of the solution at the equivalence point? HBz + NaOH Na + + Bz - + H 2 O At the equivalence point, all the benzoic acid has reacted and is in the form of sodium benzoate. There is no excess of acid or base. (25.0 ml NaOH was used) Bz + H 2 O HBz + OH The benzoate concentrations is: M 1 Bz x V 1 Bz = M 2 Bz x V 2 Bz M x 100 ml = M 2 HBz x 125 ml M 2 Bz Bz = [Bz - ] = M

56 This is treated as a hydrolysis reaction K - w h [HBz][OH ] - a K = K [Bz ] 1 x h [x][x] -5 K = 6.3 x 10 [0.020] Solve for ph 2 x = x= poh = log (1.8 x 10 6 ) poh = 5.7 ph = 14 poh = ph = 8.26

57 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH. K a HBz = 6.3 x 10-5 a)what is the ph of the solution when the benzoic acid is 50% neutralized? ph at half-way point = 4.2

58 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH. K a HBz = 6.3 x 10-5 b) What is the ph of the solution at the equivalence point? ph at half-way point = 4.2 Equivalence point ph = 8.25

59 Acetic acid titrated with NaOH ph of base solution, NaOH

60 Solubility Products When an insoluble compound, such as BaSO 4, is added to water, a very small amount of the solid dissolves. The solution is saturated with the BaSO 4 The solubility of the BaSO 4 is treated as an equilibrium i : BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) 4 (s) (aq) 4 (aq) Note that this is a heterogeneous equilibrium

61 Solubility Products The equilibriumconstant expressionfor this equilibrium system is K = [Ba 2+ [SO 2 sp ] 4 ] where the equilibriumconstant is given a special symbol, K sp K sp is called the solubility product.

62 Solubility Products K sp is not the same as solubility. Solubility is most commonly expressed as the mass of solute dissolved in 100 ml (g/100 ml) of solution. The K sp is the product of the concentration of the ions in a saturated solution of an insoluble substance.

63 Sample problem: Solubility Products Cl Calculate l the solubility of BaSO 4. K sp BaSO4 = 1.1 x Solution: Concentration BaSO 4 (s) K sp = [Ba 2+ ] [SO 4 2 ] 1.1 x = [x] [x] = x 2 Ba 2+ (aq) + SO 4 2 (aq) x x 1.05 x 10 5 M = x = [Ba 2+ ] = [SO 4 2 ]

64 Sample problem: Solubility Products Cl Calculate l the solubility of Ca(OH) C(O 2 K sp Ca(OH)2 = 5.5 x 10 5 Solution: Concentration Ca(OH) 2 (s) K sp = [Ca 2+ ] [OH ] x 10 5 = [x] [2x] 2 = 4x x 10 5 M = x x 10 2 M = x = [Ca 2+ ] Ca 2+ (aq) + 2 OH (aq) x 2x

65 Factors Affecting Solubility The Common Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. BaSO (s) Ba (aq) + SO 4 (aq)

66 Factors Affecting Solubility Sample problem: Cl Calculate l the solubility of BaSO 4 in a M solution of H 2 SO 4 K sp BaSO4 = 1.1 x Solution: BaSO 4(s) Ba 2+ (aq) + SO 4 2 (aq) Concentration x 0.050M K = [Ba 2+ [SO 2 sp ] 4 ] 1.1 x = [x] [0.050] 2.2 x 10 9 M = x = [Ba 2+ ]

67 Factors Affecting Solubility ph If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. K sp Mg(OH)2 = 5.6 x 10-12

68 Factors Affecting Solubility Complex Ions Mtli t L i id d f Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

69 Factors Affecting Solubility Complex Ions The formation of these complex ions increases the solubility of these salts. K sp AgCl = 1.8 x soluble

70 Factors Affecting Solubility Amphoterism Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Examples of such cations are Al 3+, Zn 2+, and Sn 2+. Addition of base

71 Will a Precipitate Form? Using the reaction quotient, Q, in a solution, If Q = K sp, the system is at equilibrium and the solution is saturated. (both solid and ions are present) If Q < K sp, more solid will dissolve until Q = K sp. (only ions are present) If Q > K sp, the salt will precipitate until Q = K sp. (there are excess ions present in solution, so a precipitate will form)

72 Selective Precipitation of Ions In the process of qualitative analysis, we use the differences in solubilities of salts to separate ions in a mixture.

73 Analysis of the Silver Group The silver group consists of three ions: Ag +, Hg 2+ 2 and Pb 2+ All three ions form precipitates p with Cl Ag + (aq) + Cl (aq) AgCl (s) Hg 2+ 2 (aq K sp = 1.8 x aq) + 2 Cl (aq) Hg 14 2 Cl 2 (s) K sp = 1.4 x Pb 2+ (aq aq) + 2 Cl (aq aq) PbCl 2 (s) () K sp sp = 1.7 x 10 5

74 Analysis of the Silver Group 6 M HCl is added dropwise to a solution containing M concentrations of Ag +, Hg 2+ and Pb 2+ 2 a) What is the concentration of Cl when the first ion precipitates? b) When the second ion precipitates, what is the concentration of the ion that precipitated first? c) What is the concentration of Cl needed to precipitate i 99% of the Pb 2+? Ag + (aq) + Cl (aq) AgCl (s) Hg 2+ 2 (aq 2+ aq) + 2 Cl Cl (aq Pb 2+ (aq) + 2 Cl (aq) PbCl K sp = 1.8 x aq) Hg 2 Cl 2 (s) K sp = 1.4 x PbCl 2 (s) K sp sp = 1.7 x 10 5

75 Analysis of the Silver Group 6 M HCl is added dropwise to a solution containing M concentrations of Ag +, Hg 2+ 2 and Pb 2+ a) What is the concentration of Cl when the first ion precipitates? Solution: Generally, the compound with the smallest K sp precipitates first, so calculate concentration of Cl necessary to precipitate the Hg 2+ 2 Hg 2 Cl 2 (s) Hg Hg 2+ 2 (aq aq) + 2 Cl Concentrations M 2x K sp Hg2Cl2 Hg2Cl2 = 1.4 x = [Hg 2+ 2 ] [Cl ] x = [0.025] [2x] 2 4x 2 = 5.6 x x 2 = 1.4 x x = x M [Cl ] = 2x = 7.4 x 10 9 M Cl (aq)

76 Analysis of the Silver Group 6 M HCl is added dropwise to a solution containing M concentrations of Ag +, Hg 2+ 2 and Pb 2+ a) Whatisthe the concentration of Cl when the first ion precipitates? Solution Part a) (continued): Next, calculate the Cl concentration needed to precipitate Ag + (Although the K sp for AgCl is larger than the K sp for Hg 2 Cl 2, the difference in reaction stoichiometry does have an effect.) AgCl (s) Ag + (aq) + Cl (aq) K sp = 1.8 x K sp AgCl AgCl = 1.8 x = [Ag + ] [Cl ] 1.8 x = [0.025] ][] [x] x = 7.2 x 10 9 M The concentration of Cl needed to precipitate the Ag + is less than that for Hg 2+ 2, so, Ag + precipitates first.

77 Analysis of the Silver Group 6 M HCl is added dropwise to a solution containing M concentrations of Ag +, Hg 2+ 2 and Pb 2+ b) When the second ionprecipitates precipitates, whatisthe the concentration of the ion that precipitated first? Solution, Part b): Hg 2+ secondwhen 74x 2 precipitates [Cl ] = M Calculate the concentration of Ag + K sp AgCl AgCl = 1.8 x = [Ag + ] [Cl ] 1.8 x = [Ag + ] [7.4 x 10 9 ] [Ag + ] = 2.4 x 10 2 M

78 Analysis of the Silver Group 6 M HCl is added dropwise to a solution containing M concentrations of Ag +, Hg 2+ 2 and Pb 2+ c) Whatisthe the concentration of Cl needed to precipitate 99% of the Pb 2+? Solution Part c): If 1% of the Pb 2+ will remain in solution, calculate that amount (Note 1% = 0.01) 01) 0.01 x M = 2.5 x 10 4 M Calculate the concentration of Cl K sp PbCl2 PbCl2 = 1.7 x 10 5 = [Pb 2+ ] [Cl ] x 10 5 = [2.5 x 10 4 ] [ [Cl ] 2 = 6.8 x 10 2 [Cl ] = 0.26 M ] [Cl ] 2

79 PbCl 2 Pb(CO 3 ) 2 PbI 2 PbCrO 4 Lead Chemistry Illustrates the transformation of one insoluble compoundinto an even less soluble compound. K sp PbCl2 = 1.7 x 10 5 K sp PbI2 = 9.8 x 10 9 K 74x 14 sp Pb(CO3)2 = K sp PbCrO4 = 2.8 x K sp PbCrO4 Pb(OH)2 = 1.8 x 10 32

80 Separations by Difference in K sp

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