Chapter 16 Aqueous Ionic Equilibrium
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1 Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall
2 The Danger of Antifreeze each year, thousands of pets and wildlife die ethylene glycol from consuming antifreeze (aka 1,2 ethandiol) most brands of antifreeze contain ethylene glycol sweet taste initial effect drunkenness metabolized in the liver to glycolic acid HOCH 2 COOH if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO 3, causing the blood ph to drop when the blood ph is low, it ability to carry O 2 is compromised acidosis the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol 2
3 Buffers buffers are solutions that resist changes in ph when an acid or base is added they act by neutralizing the added acid or base but just like everything else, there is a limit to what they can do, eventually the ph changes many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion Tro, Chemistry: A Molecular Approach 3
4 Making an Acid Buffer Tro, Chemistry: A Molecular Approach 4
5 How Acid Buffers Work HA (aq) + H 2 O (l) A (aq) + H 3 O+ (aq) buffers work by applying Le Châtelier s Principle to weak acid equilibrium buffer solutions contain significant amounts of the weak acid molecules, HA these molecules react with added base to neutralize it you can also think of the H 3 O + combining with the OH to make H 2 O; the H 3 O + is then replaced by the shifting equilibrium the buffer solutions also contain significant amounts of the conjugate base anion, A - these ions combine with added acid to make more HA and keep the H 3 O + constant Tro, Chemistry: A Molecular Approach 5
6 How Buffers Work new HA H 2 O HA + H 3 O+ HA A Added H 3 O + Tro, Chemistry: A Molecular Approach 6
7 How Buffers Work new A H 2 O HA A HA A + H 3 O+ Added HO Tro, Chemistry: A Molecular Approach 7
8 Common Ion Effect HA (aq) + H 2 O (l) A (aq) + H 3 O+ (aq) adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left this causes the ph to be higher than the ph of the acid solution lowering the H 3 O + ion concentration Tro, Chemistry: A Molecular Approach 8
9 Common Ion Effect Tro, Chemistry: A Molecular Approach 9
10 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0 HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + initial change equilibrium [HA] [A - ] [H 3 O + ] 0 Tro, Chemistry: A Molecular Approach 10
11 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression K a HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + initial change equilibrium [C 2 H [HA] [A - ] [H 3 O + ] Tro, Chemistry: A Molecular Approach 11 3 O x [ HC H O ] 2-2 +x 0 +x x x x ][H O + ] ( x)( x) ( x)
12 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? determine the value of K a since K a is very small, approximate the [HA] eq [HA] init and [A ] eq [A ] init solve for x K a [C 2 H 3 O initial change equilibrium - 2 ][H 3 O [ HC H O ] + ] ( x)( x) ( x) ( 0.100)( ) ( 0.100) Tro, Chemistry: A Molecular Approach H 3 O 2 ][H3O ] x [C K a [ HC H O ] K a for HC 2 H 3 O x x [HA] x x [A - ] x x [H 3 O + ] 0 +x x
13 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init initial change 1 K a for HC 2 H 3 O x 10-5 equilibrium x 1.8 x % 0.018% 5% Tro, Chemistry: A Molecular Approach 13 [HA] x the approximation is valid [A - ] x < [H 3 O + ] 0 +x x
14 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? substitute x into the equilibrium concentration definitions and solve x 1.8 x [ HC H O ] x ( ) M [H 3 initial change equilibrium O + ] x [HA] x Tro, Chemistry: A Molecular Approach 14 -x x 1.8E-5 x M [A - ] x [H 3 O + ] 0 +x ( ) M [C2 H3O2 ] x
15 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? substitute [H 3 O + ] into the formula for ph and solve initial change [HA] x [A - ] x [H 3 O + ] 0 +x equilibrium E-5 ph log 1.8 ( + ) 3O ( 5) log H Tro, Chemistry: A Molecular Approach 15
16 Ex What is the ph of a buffer that is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match initial change K a for HC 2 H 3 O x 10-5 equilibrium K a [HA] [A - ] Tro, Chemistry: A Molecular Approach 16 [C 2 H -x 3 O +x [ HC H O ] ( )( ) 2-2 ( 0.100) ][H O + ] [H 3 O + ] 0 +x 1.8E
17 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? Tro, Chemistry: A Molecular Approach 17
18 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0 initial change equilibrium HF + H 2 O F + H 3 O + [HA] [A - ] [H 3 O + ] Tro, Chemistry: A Molecular Approach 18
19 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression initial change equilibrium K a HF + H 2 O F + H 3 O + [ HF] [HA] 0.14 [A - ] [H 3 O + ] Tro, Chemistry: A Molecular Approach 19 - [F ][H3O x +x 0 +x 0.14 x x x + ] ( x)( x) ( 0.14 x)
20 determine the value of K a since K a is very small, approximate the [HA] eq [HA] init and [A ] eq [A ] init solve for x K K Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? a a 10 pk a initial change equilibrium 3.15 K a for HF 7.0 x 10-4 [ HF] ( x)( x) ( 0.14 x) ( 0.071)( ) ( 0.14) Tro, Chemistry: A Molecular Approach 20 K a [HA] x x - [F ][H3O x ] [A - ] +x K a [H 3 O + ] x 0 +x x 4 x
21 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init initial change 3 1 K a for HF 7.0 x % [HA] 0.14 equilibrium x 1.4 x % the approximation is valid 5% Tro, Chemistry: A Molecular Approach 21 -x < [A 2- ] x [H 3 O + ] 0 +x x
22 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? substitute x into the equilibrium concentration definitions and solve x 1.4 x [ HF] 0.14 x 0.14 ( ) 0.14 M [H 3 initial change equilibrium O + ] x [HA] x Tro, Chemistry: A Molecular Approach 22 -x 3 M [A 2- ] x [H 3 O + ] 0 +x x 1.4E-3 x ( ) M [C2 H3O2 ] x
23 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? substitute [H 3 O + ] into the formula for ph and solve initial change [HA] x [A - ] x [H 3 O + ] 0 +x equilibrium E-3 ph log 1.4 ( + ) 3O ( 3) log H Tro, Chemistry: A Molecular Approach 23
24 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values are close enough initial change equilibrium K a K a for HF 7.0 x 10-4 [HA] [A - ] Tro, Chemistry: A Molecular Approach 24 [F - ][H -x [ HF] +x ( )( ) 3 O ( 0.14) + ] [H 3 O + ] 0 +x 1.4E
25 Henderson-Hasselbalch Equation calculating the ph of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson- Hasselbalch Equation the equation calculates the ph of a buffer from the K a and initial concentrations of the weak acid and salt of the conjugate base as long as the x is small approximation is valid ph pk + a [conjugate base anion] log [weak acid] initial initial Tro, Chemistry: A Molecular Approach 25
26 Deriving the Henderson-Hasselbalch Equation K [H 3 a O + [A ] - ][H 3 [ HA] K a O + ] [HA] - [A ] - + [HA] a log[h 3O ] [HA] log K a [A ] log ph - log[h O ] - [A [HA] ph pk + [A ] ph pk a + log log a - [HA] [A ] ] 3 pk a -[HA] log K a [A ] log log [A ] [HA] Tro, Chemistry: A Molecular Approach 26
27 Ex What is the ph of a buffer that is M HC 7 H 5 O 2 and M NaC 7 H 5 O 2? Assume the [HA] and [A - ] equilibrium concentrations are the same as the initial Substitute into the Henderson-Hasselbalch Equation Check the x is small approximation HC 7 H 5 O 2 + H 2 O C 7 H 5 O 2 + H 3 O + K a for HC 7 H 5 O x 10-5 ( ) pk a log K a log ( ) ( ) - [A0.150 ] ph pk a + + log [HA] O 5 10 ] + 3O ] 2.2 [H [H ph Tro, Chemistry: A Molecular Approach 27 -ph % % < 5%
28 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? Tro, Chemistry: A Molecular Approach 28
29 Practice - What is the ph of a buffer that is 0.14 M HF (pk a 3.15) and M KF? find the pk a from the given K a HF + H 2 O F + H 3 O + Assume the [HA] and [A - ] equilibrium concentrations are the same as the initial Substitute into the Henderson-Hasselbalch Equation Check the x is small approximation - [A ] ph pk a + log [HA] ( 0.071) ph log [H Tro, Chemistry: A Molecular Approach 29 3 O + [H O ] 10 ] 10 -ph % 1% < 5% 3
30 Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? the Henderson-Hasselbalch equation is generally good enough when the x is small approximation is applicable generally, the x is small approximation will work when both of the following are true: a) the initial concentrations of acid and salt are not very dilute b) the K a is fairly small for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of K a Tro, Chemistry: A Molecular Approach 30
31 How Much Does the ph of a Buffer Change When an Acid or Base Is Added? though buffers do resist change in ph when acid or base are added to them, their ph does change calculating the new ph after adding acid or base requires breaking the problem into 2 parts 1. a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A to make more HA added base reacts with the HA to make more A 2. an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A ] Tro, Chemistry: A Molecular Approach 31
32 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? If the added chemical is a base, write a reaction for OH with HA. If the added chemical is an acid, write a reaction for it with A. Construct a stoichiometry table for the reaction HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O mols Before HA A mols added - - mols After OH Tro, Chemistry: A Molecular Approach 32
33 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? Fill in the table tracking the changes in the number of moles for each component HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O mols Before A mols added - - mols After HA OH Tro, Chemistry: A Molecular Approach 33
34 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? If the added chemical is a base, write a reaction for OH with HA. If the added chemical is an acid, write a reaction for it with A. Construct a stoichiometry table for the reaction Enter the initial number of moles for each HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O mols Before mols change mols End HA A - OH Tro, Chemistry: A Molecular Approach 34
35 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O mols Before mols change mols End new Molarity HA A - OH mol [A HA] ] M 1.00 L Tro, Chemistry: A Molecular Approach 35 0
36 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations assuming the [H 3 O + ] from water is 0, and using the new molarities of the [HA] and [A ] HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + [HA] [A - ] [H 3 O + ] initial change equilibrium Tro, Chemistry: A Molecular Approach 36
37 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + initial change equilibrium K a [C [HA] [A - ] [H 3 O + ] Tro, Chemistry: A Molecular Approach 37 2 H 3 O x +x 0 +x x x x ][H O [ HC H O ] ] ( x)( x) ( x)
38 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? K a for HC 2 H 3 O x 10-5 determine the value of K a since K a is very small, approximate the [HA] eq [HA] init and [A ] eq [A ] init solve for x K a [C - + H3O2][H3O ] ( x)( x) [ HC H O ] ( x) - + 2H3O2][H 5 3O( ] ()( x) )( x) [ HC H O ] ( 0.090( ) 0.090) 2 initial change equilibrium [C K a Tro, Chemistry: A Molecular Approach x x [HA] x [A - ] x x [H 3 O + ] 0 +x x
39 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? K a for HC 2 H 3 O x 10-5 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init initial [HA] [A - ] change -x +x equilibrium x 1.47 x 10-5 [H 3 O + ] 0 +x x % 0.016% < 5% the approximation is valid Tro, Chemistry: A Molecular Approach 39
40 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? substitute x into the equilibrium concentration definitions and solve x 1.47 x 10-5 [ ] ( 5 HC H O x ) M [H 3 O initial change equilibrium + ] x [HA] x Tro, Chemistry: A Molecular Approach 40 -x x 1.5E-5 x M [A - ] x [H 3 O + ] 0 +x ( ) M [C2 H3O2 ] x
41 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? substitute [H 3 O + ] into the formula for ph and solve ph initial change equilibrium log 1.47 ( ) H O + 3 ( ) log [HA] x [A - ] x [H 3 O + ] 0 +x 1.5E-5 Tro, Chemistry: A Molecular Approach 41
42 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? K a for HC 2 H 3 O x 10-5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match initial change equilibrium K a [HA] [A - ] Tro, Chemistry: A Molecular Approach 42 [C 2 H 3 -x O [ HC H O ] ( )( ) 2-2 ( 0.090) ][H O +x + ] [H 3 O + ] 0 +x 1.5E-5 5
43 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? find the pk a from the given K a HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + Assume the [HA] and [A - ] equilibrium concentrations are the same as the initial initial change equilibrium [HA] x [A - ] x K a for HC 2 H 3 O x 10-5 [H 3 O + ] 0 +x pk a log K a log ( 5 10 ) Tro, Chemistry: A Molecular Approach 43 x
44 Ex What is the ph of a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L that has mol NaOH added to it? Substitute into the Henderson-Hasselbalch Equation Check the x is small approximation [H [H 3 3 O O + + ] 10 ] 10 -ph HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + 5 pk a for HC 2 H 3 O [A ] ph pk a log [HA] ( 0.110) ph log % 0.016% < 5% Tro, Chemistry: A Molecular Approach 44
45 Ex 16.3 Compare the effect on ph of adding mol NaOH to a buffer that has mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 in 1.00 L to adding mol NaOH to 1.00 L of pure water? HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 + H 3 O + pk a for HC 2 H 3 O ph ph 4.83 p K a [A ] + log [HA] log ( 0.110) mol [ OH ] 1.00 L poh log[oh M ( 10 2 ) log 1.0 ph + poh ph Tro, Chemistry: A Molecular Approach 45 ] poh
46 Basic Buffers B: (aq) + H 2 O (l) H:B + (aq) + OH (aq) buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl H 2 O (l) + NH 3(aq) NH 4 + (aq) + OH (aq) Tro, Chemistry: A Molecular Approach 46
47 Ex What is the ph of a buffer that is 0.50 M NH 3 (pk b 4.75) and 0.20 M NH 4 Cl? find the pk a of the conjugate acid (NH 4+ ) from the given K b Assume the [B] and [HB + ] equilibrium concentrations are the same as the initial Substitute into the Henderson-Hasselbalch Equation Check the x is small approximation NH 3 + H 2 O NH 4+ + OH pk + pk 14 pka 14 - pkb [B] ph pka + log [HB + ] ( 0.50) ph log 9.65 ( 0.20) [H Tro, Chemistry: A Molecular Approach 47 3 O + a ] 10 -ph [H3O ] % <<< 5% 0.20 b 10
48 Ex What is the ph of a buffer that is 0.50 M NH 3 (pk b 4.75) and 0.20 M NH 4 Cl? find the pk b if given K b Assume the [B] and [HB + ] equilibrium concentrations are the same as the initial Substitute into the Henderson- Hasselbalch Equation Base Form, find poh Check the x is small approximation Calculate ph from poh NH 3 + H 2 O NH 4+ + OH poh + [HB + log [B] Tro, Chemistry: A Molecular Approach 48 pk b ( 0.20) ( 0.50) poh log [OH ] poh % ph + poh ph ] % 9.65 < 5 5%
49 Buffering Effectiveness a good buffer should be able to neutralize moderate amounts of added acid or base however, there is a limit to how much can be added before the ph changes significantly the buffering capacity is the amount of acid or base a buffer can neutralize the buffering range is the ph range the buffer can be effective the effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base Tro, Chemistry: A Molecular Approach 49
50 Effect of Relative Amounts of Acid and Buffer mol HA & mol A - Initial ph 5.00 Conjugate Base a buffer is most effective with equal concentrations of acid and base ph pk a (HA) 5.00 p K a - [A ] + log [HA] Buffer mol HA & mol A - Initial ph 4.05 HA + OH A + H 2 O % Change % Change after adding mol NaOH after adding mol NaOH ( ph HA ) 5.09 A - OH ( ph 4.25 HA A ) - OH 100% 100% mols Before mols Before ( 0.030) ( 0.110) mols ph added 1.8% log mols ph added % log mols After mols After
51 Effect of Absolute Concentrations of Buffer mol HA & 0.50 mol A - Initial ph 5.00 Acid and Conjugate Base a buffer is most effective when the concentrations of acid and base are largest ph pk a (HA) 5.00 p K a - [A ] + log [HA] Buffer mol HA & mol A - Initial ph 5.00 HA + OH A + H 2 O % Change % Change after adding HA mol A NaOH - OH after adding ( 5.18 HA A mol ) - NaOH ( OH 100% ph ) % ph 5.18 mols Before mols Before ( 0.060) ( 0.51) mols ph 0.4% added log molsph added % + log mols After mols After
52 Effectiveness of Buffers a buffer will be most effective when the [base]:[acid] 1 equal concentrations of acid and base effective when 0.1 < [base]:[acid] < 10 a buffer will be most effective when the [acid] and the [base] are large Tro, Chemistry: A Molecular Approach 52
53 Buffering Range we have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum ph at which the buffer will be effective - [A ] ph pk a + log [HA] ph ph Lowest ph pk pk a a + log 1 ( 0.10) ph ph Highest ph pk pk a a ( ) log 10 therefore, the effective ph range of a buffer is pk a ±1 when choosing an acid to make a buffer, choose one whose is pk a is closest to the ph of the buffer 53
54 Ex. 16.5a Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with ph 4.25? Chlorous Acid, HClO 2 pk a 1.95 Nitrous Acid, HNO 2 pk a 3.34 Formic Acid, HCHO 2 pk a 3.74 Hypochlorous Acid, HClO pk a 7.54 Tro, Chemistry: A Molecular Approach 54
55 Ex. 16.5a Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with ph 4.25? Chlorous Acid, HClO 2 pk a 1.95 Nitrous Acid, HNO 2 pk a 3.34 Formic Acid, HCHO 2 pk a 3.74 Hypochlorous Acid, HClO pk a 7.54 The pk a of HCHO 2 is closest to the desired ph of the buffer, so it would give the most effective buffering range. Tro, Chemistry: A Molecular Approach 55
56 Ex. 16.5b What ratio of NaCHO 2 : HCHO 2 would be required to make a buffer with ph 4.25? Formic Acid, HCHO 2, pk a 3.74 [CHO - log 2 [A ] [HCHO ] ph pk a + log [HA] [CHO [CHO ] 2 ] log [HCHO2] [HCHO2] [CHO ] 0.51 log 2 [HCHO2] Tro, Chemistry: A Molecular Approach 56 ] 0.51 to make the buffer with ph 4.25, you would use 3.24 times as much NaCHO 2 as HCHO 2
57 Buffering Capacity buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness the buffering capacity increases with increasing absolute concentration of the buffer components as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves buffers that need to work mainly with added acid generally have [base] > [acid] buffers that need to work mainly with added base generally have [acid] > [base] Tro, Chemistry: A Molecular Approach 57
58 Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer Tro, Chemistry: A Molecular Approach 58
59 Titration in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete when the reaction is complete we have reached the endpoint of the titration an indicator may be added to determine the endpoint an indicator is a chemical that changes color when the ph changes when the moles of H 3 O + moles of OH, the titration has reached its equivalence point Tro, Chemistry: A Molecular Approach 59
60 Titration Tro, Chemistry: A Molecular Approach 60
61 Titration Curve a plot of ph vs. amount of added titrant the inflection point of the curve is the equivalence point of the titration prior to the equivalence point, the known solution in the flask is in excess, so the ph is closest to its ph the ph of the equivalence point depends on the ph of the salt solution equivalence point of neutral salt, ph 7 equivalence point of acidic salt, ph < 7 equivalence point of basic salt, ph > 7 beyond the equivalence point, the unknown solution in the burette is in excess, so the ph approaches its ph Tro, Chemistry: A Molecular Approach 61
62 Titration Curve: Unknown Strong Base Added to Strong Acid Tro, Chemistry: A Molecular Approach 62
63 Tro, Chemistry: A Molecular Approach 63
64 Titration of 25 ml of M HCl with M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (aq) initial ph -log(0.100) 1.00 initial mol of HCl L x mol/l 2.50 x 10-3 before equivalence point added 5.0 ml NaOH mol HCl excess0.100 mol NaOH NaOH L ofmolenaoh added 1molHCl L x NaOH NaOH 1mol HCl mol x NaOH 10-3 mol mol HCl HCl L HCl + L 1molNaOH 1 L L 1mol 1 L 4 HCl L NaOH moles moleshclused added NaOH M [H O ] mol HCl NaOH used M HCl [H O ] initial mol HCl - mol HCl used mol HCl mol HCl used + mol phhcl -log[h excess -3 3O ] pH mol HCl -log excess ( ) Tro, Chemistry: A Molecular Approach 64
65 Titration of 25 ml of M HCl with M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (aq) at equivalence, 0.00 mol HCl and 0.00 mol NaOH ph at equivalence 7.00 after equivalence point added 30.0 ml NaOH 5.0 x 10-4 mol NaOH xs mol NaOH mol NaOH L of added NaOH L NaOH mol NaOH added - initial mol + K-4 1 L mol3.00 NaOH w mol NaOH excess 1 LHCl + [H3O ] mol mol 10 NaOH mol NaOH moles [ H [OH ] NaOH used 3NaOH Oadded ][OH excess NaOH ] K L HCl + L NaOH L HCl L NaOH mol NaOH excess M NaOH [OH ] M NaOH [OH ] w ph - log[h3o + ] ph - log 11. ( ) Tro, Chemistry: A Molecular Approach 65
66 Titration of 25 ml of M HCl with M NaOH HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (aq) at equivalence, 0.00 mol HCl and 0.00 mol NaOH ph at equivalence 7.00 after equivalence point mol NaOH mol NaOH L ofmol added NaOH NaOH excess L NaOH mol NaOH added - initial mol 1 LHCl -4 1 L L HCl + L mol3.00 NaOH mol mol 10 NaOH mol NaOH moles NaOH added excess NaOH NaOH used M NaOH [OH ] L HCl L NaOH mol NaOH excess M NaOH [OH ] poh -log[oh ] ph poh added 30.0 ml NaOH 5.0 x 10-4 mol NaOH xs ( 3 10 ) poh -log ph Tro, Chemistry: A Molecular Approach 66
67 Adding NaOH to HCl 25.0 added ml ml M NaOH HCl equivalence mol point NaOH HCl ph added ml NaOH mol HCl NaOH ph added ml NaOH mol HCl NaOH ph Tro, Chemistry: A Molecular Approach 67
68 ph Titration Curve of Strong Acid with NaOH Volume NaOH Added, ml ph Derivative Titration of 25.0 ml of M HCl with M NaOH The 1st derivative of the curve is maximum at the equivalence point Since the solutions are equal concentration, the equivalence point is at equal volumes Tro, Chemistry: A Molecular Approach 68
69 Species Distribution in a Strong Acid vs ph Fraction of Species in Mixture HA A ph After about ph 3, there is practically no HCl left, it has all been reacted and become NaCl + H 2 O Tro, Chemistry: A Molecular Approach 69
70 Titration of 25 ml of M HCHO 2 with M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) Initial ph: [HCHO 2 ] initial change -x equilibrium x ph - log[h O [CHO 2- ] [H 3 O + ] x +x x x ( ) log ] x [H K a 1.8 x 10-4 ( x)( x) ( x) Tro, Chemistry: A Molecular Approach 70 K a 3 4 O + ] [CHO2 ][H [HCHO O ] 3 + ] 3 x M 100% 4.2% < 5% 2
71 Titration of 25 ml of M HCHO 2 with M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) initial mol of HCHO L x mol/l 2.50 x 10-3 before equivalence mols Before mols added mols After HA 2.50E E-3 pk a - logka -log 1.8 A E-4 ( ) OH 0 5.0E-4 0 added 5.0 ml NaOH mol NaOH L NaOH 2 ph pk a + log 1 L 4 mol HCHO mol NaOH 3.14 Tro, Chemistry: A Molecular Approach 71 ph ph log mol CHO
72 Titration of 25 ml of M HCHO 2 with M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) initial mol of HCHO L x mol/l 2.50 x 10-3 at equivalence initial mols Before change mols added equilibrium mols After K b, CHO [HCHO 2 ] 4 HA 2.50E-3 0 +x x K K w a - 0 [CHO 2- ] x 5.00E-2-x CHO 2 (aq) + H 2 O added (l) 25.0 HCHO ml 2(aq) NaOH + OH (aq) A[OH - ] OH K b 5.6 x mol CHO mol NaOH L NaOH 0 0 [HCHO [OH 0 2][OH ] K b L HCHO - ] 1.7 x L NaOH -6 M 1 L [H [CHO + 2 ] K w 3M 10 O CHO mol NaOH - +x 2.50E-3 ] 2 [OH-3 ] ( 14 x10 )( x) mol CHO2 x 2.50E-3 x ( L HCHO x ) M CHO 62 x [OHpH] log[h 10 + OM] 11 ( ) log L NaOH 72
73 Titration of 25 ml of M HCl with M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) after equivalence point mol NaOH L of added NaOH 1 L mol moles NaOH NaOH added added NaOH excess - initial mol HCHO2 mol LNaOH HCl + + excess L NaOH [ H O ][OH ] K w M NaOH [OH ] ph - log[h3o + ] mol NaOH L x NaOH 10-4 mol NaOH xs 1 L mol -4 10K w [H NaOH mol NaOH 3O ] -3 mol NaOHL HCl [OH ] mol NaOH L NaOH used 14 mol NaOH M 1 10excess NaOH [OH ] Tro, Chemistry: A Molecular Approach added 30.0 ml NaOH ph - log 11. ( )
74 Titration of 25 ml of M HCl with M NaOH HCHO 2(aq) + NaOH (aq) NaCHO 2(aq) + H 2 O (aq) after equivalence point mol NaOH L of added NaOH 1 L mol moles NaOH NaOH added added NaOH excess - initial mol HCHO2 mol LNaOH HCl + excess L NaOH M NaOH [OH ] poh -log[oh ] ph poh mol NaOH mol mol NaOH NaOH used mol NaOHLexcess HCl L NaOH M poh -log( NaOH ) [OH ] 2.04 ph Tro, Chemistry: A Molecular Approach added 30.0 ml NaOH mol NaOH L x NaOH 10-4 mol NaOH xs 1 L mol NaOH -4-3
75 Adding NaOH to HCHO 2 initial added HCHO ml 2 solution NaOH equivalence mol point NaOH HCHO 2 xs ph mol CHO 2 added [CHO added ] 40.0 init ml NaOHM ml NaOH [OH ] eq mol 1.7 mol HCHO x 10 NaOH -6 2 xs ph ph pk a half-neutralization added ml NaOH mol HCHO NaOH 2 xs ph added 20.0 ml NaOH mol HCHO 2 ph 4.34 Tro, Chemistry: A Molecular Approach 75
76 14 Titration Curve of Weak Acid with NaOH ph Volume NaOH Added, ml Tro, Chemistry: A Molecular Approach 76
77 ph Titration Curve of Weak Acid with NaOH Volume NaOH Added, ml ph Derivative ph at equivalence 8.23 Titration of 25.0 ml of M HCHO 2 with M NaOH The 1st derivative of the curve is maximum at the equivalence point Since the solutions are equal concentration, the equivalence point is at equal volumes Tro, Chemistry: A Molecular Approach 77
78 Species Distribution in a Weak Acid vs ph Fraction of Species in Mixture HA A ph at ph 3.74, the [HCHO 2 ] [CHO 2 ]; the acid is half neutralized half-neutralization occurs when ph pk a Tro, Chemistry: A Molecular Approach 78
79 Titrating Weak Acid with a Strong Base the initial ph is that of the weak acid solution calculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 before the equivalence point, the solution becomes a buffer calculate mol HA init and mol A init using reaction stoichiometry calculate ph with Henderson-Hasselbalch using mol HA init and mol A init half-neutralization ph pk a Tro, Chemistry: A Molecular Approach 79
80 Titrating Weak Acid with a Strong Base at the equivalence point, the mole HA mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established mol A original mole HA calculate the volume of added base like Ex 4.8 [A ] init mol A /total liters calculate like a weak base equilibrium problem e.g., beyond equivalence point, the OH is in excess [OH ] mol MOH xs/total liters [H 3 O + ][OH ]1 x Tro, Chemistry: A Molecular Approach 80
81 Ex 16.7a A 40.0 ml sample of M HNO 2 is titrated with M KOH. Calculate the volume of KOH at the equivalence point Write an equation for the reaction for B with HA. Use Stoichiometry to determine the volume of added B HNO 2 + KOH NO 2 + H 2 O L 40.0 ml L 1mL L NO L KOH mol NO 1L NO 2 2 1mol KOH 1mol NO 1mL L L Tro, Chemistry: A Molecular Approach L KOH mol KOH 20.0 ml
82 Ex 16.7b A 40.0 ml sample of M HNO 2 is titrated with M KOH. Calculate the ph after adding 5.00 ml KOH Write an equation for the reaction for B with HA. Determine the moles of HA before & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A made HNO 2 + KOH NO 2 + H 2 O L mol HNO ml mol HNO 2 1mL 1L L mol KOH 5.00 ml 1mL 1L mols Before mols added mols After HNO NO mol KOH OH Tro, Chemistry: A Molecular Approach 82
83 Ex 16.7b A 40.0 ml sample of M HNO 2 is titrated with M KOH. Calculate the ph after adding 5.00 ml KOH. Write an equation for the reaction of HA with H 2 O Determine K a and pk a for HA Use the Henderson- Hasselbalch Equation to determine the ph HNO 2 + H 2 O NO 2 + H 3 O + Table 15.5 K a 4.6 x 10-4 pk a log K a ph HNO NO ph log mols Before mols added - - mols After log ( ) OH Tro, Chemistry: A Molecular Approach 83 pk a NO + log HNO 2 2 0
84 Ex 16.7b A 40.0 ml sample of M HNO 2 is titrated with M KOH. Calculate the ph at the half-equivalence point Write an equation for the reaction for B with HA. Determine the moles of HA before & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A made HNO 2 + KOH NO 2 + H 2 O L mol HNO ml mol HNO 2 1mL 1L at half-equivalence, moles KOH ½ mole HNO 2 mols Before mols added mols After HNO NO OH Tro, Chemistry: A Molecular Approach 84
85 Ex 16.7b A 40.0 ml sample of M HNO 2 is titrated with M KOH. Calculate the ph at the half-equivalence point. Write an equation for the reaction of HA with H 2 O Determine K a and pk a for HA Use the Henderson- Hasselbalch Equation to determine the ph HNO 2 + H 2 O NO 2 + H 3 O + Table 15.5 K a 4.6 x 10-4 pk a log K a ph HNO NO ph log mols Before mols added - - mols After log ( ) OH Tro, Chemistry: A Molecular Approach 85 pk a NO + log HNO 2 2 0
86 Titration Curve of a Weak Base with a Strong Acid Tro, Chemistry: A Molecular Approach 86
87 Titration of a Polyprotic Acid if K a1 >> K a2, there will be two equivalence points in the titration the closer the K a s are to each other, the less distinguishable the equivalence points are titration of 25.0 ml of M H 2 SO 3 with M NaOH Tro, Chemistry: A Molecular Approach 87
88 Monitoring ph During a Titration the general method for monitoring the ph during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ] using a probe that specifically measures just H 3 O + the endpoint of the titration is reached at the equivalence point in the titration at the inflection point of the titration curve if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator Tro, Chemistry: A Molecular Approach 88
89 Monitoring ph During a Titration Tro, Chemistry: A Molecular Approach 89
90 Indicators many dyes change color depending on the ph of the solution these dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l) Ind (aq) + H 3 O + (aq) the color of the solution depends on the relative concentrations of Ind :HInd when Ind :HInd 1, the color will be mix of the colors of Ind and HInd when Ind :HInd > 10, the color will be mix of the colors of Ind when Ind :HInd < 0.1, the color will be mix of the colors of HInd Tro, Chemistry: A Molecular Approach 90
91 Phenolphthalein 91
92 Methyl Red H C H C H C H C C C C CH C H C H (CH 3 ) 2 N N N N H H 3 O+ OH- NaOOC C C H H C H C H C H C (CH 3 ) 2 N C C N N N C CH C H C H NaOOC C C H Tro, Chemistry: A Molecular Approach 92
93 Monitoring a Titration with an Indicator for most titrations, the titration curve shows a very large change in ph for very small additions of base near the equivalence point an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in ph pk a of HInd ph at equivalence point Tro, Chemistry: A Molecular Approach 93
94 Acid-Base Indicators 94
95 Solubility Equilibria all ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water Tro, Chemistry: A Molecular Approach 95
96 Solubility Product the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp for an ionic solid M n X m, the dissociation reaction is: M n X m (s) nm m+ (aq) + mx n (aq) the solubility product would be K sp [M m+ ] n [X n ] m for example, the dissociation reaction for PbCl 2 is PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) and its equilibrium constant is K sp [Pb 2+ ][Cl ] 2 Tro, Chemistry: A Molecular Approach 96
97 Tro, Chemistry: A Molecular Approach 97
98 Molar Solubility solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature the molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction M n X m (s) nm m+ (aq) + mx n (aq) molar solubility ( n+ m) n n K sp m m Tro, Chemistry: A Molecular Approach 98
99 Ex 16.8 Calculate the molar solubility of PbCl 2 in pure water at 25 C Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K sp [Pb 2+ ][Cl ] 2 [Pb 2+ ] [Cl ] Initial 0 0 Change +S +2S Equilibrium S 2S Tro, Chemistry: A Molecular Approach 99
100 Ex 16.8 Calculate the molar solubility of PbCl 2 in pure water at 25 C Substitute into the K sp expression K sp [Pb 2+ ][Cl ] 2 K sp (S)(2S) 2 Find the value of K sp from Table 16.2, plug into the equation and solve for S Initial 3 Change K sp 4 S Equilibrium K [Pb 2+ ] sp 0 4S S +S S 2 10 M [Cl ] S 2S Tro, Chemistry: A Molecular Approach 100
101 Practice Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10-2 M Tro, Chemistry: A Molecular Approach 101
102 Practice Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10-2 M Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) K sp [Pb 2+ ][Br ] 2 [Pb 2+ ] [Br ] Initial 0 0 Change +(1.05 x 10-2 ) +2(1.05 x 10-2 ) Equilibrium (1.05 x 10-2 ) (2.10 x 10-2 ) Tro, Chemistry: A Molecular Approach 102
103 Practice Determine the K sp of PbBr 2 if its molar solubility in water at 25 C is 1.05 x 10-2 M Substitute into the K sp expression K sp [Pb 2+ ][Br ] 2 K sp (1.05 x 10-2 )(2.10 x 10-2 ) 2 plug into the equation and solve K Initial K sp Change sp [Pb 2+ ] 0 +(1.05 x 10-2 ) [Br ] ( 2 )( ) (1.05 x 10-2 ) Equilibrium (1.05 x 10-2 ) (2.10 x 10-2 ) 0 2 Tro, Chemistry: A Molecular Approach 103
104 K sp and Relative Solubility molar solubility is related to K sp but you cannot always compare solubilities of compounds by comparing their K sp s in order to compare K sp s, the compounds must have the same dissociation stoichiometry Tro, Chemistry: A Molecular Approach 104
105 The Effect of Common Ion on Solubility addition of a soluble salt that contains one of the ions of the insoluble salt, decreases the solubility of the insoluble salt for example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) addition of Cl shifts the equilibrium to the left Tro, Chemistry: A Molecular Approach 105
106 Ex Calculate the molar solubility of CaF 2 in M NaF at 25 C Write the dissociation reaction and K sp expression Create an ICE table defining the change in terms of the solubility of the solid CaF 2 (s) Ca 2+ (aq) + 2 F (aq) K sp [Ca 2+ ][F ] 2 [Ca 2+ ] [F ] Initial Change +S +2S Equilibrium S S Tro, Chemistry: A Molecular Approach 106
107 Ex Calculate the molar solubility of CaF 2 in M NaF at 25 C Substitute into the K sp expression assume S is small Find the value of K sp from Table 16.2, plug into the equation and solve for S Initial Change K sp [Ca 2+ ][F ] 2 K sp (S)( S) 2 K sp (S)(0.100) 2 K sp Equilibrium S S S [Ca 2+ ] ( 0.100) S ( 0.100) S M [F ] S S Tro, Chemistry: A Molecular Approach 107
108 The Effect of ph on Solubility for insoluble ionic hydroxides, the higher the ph, the lower the solubility of the ionic hydroxide and the lower the ph, the higher the solubility higher ph increased [OH ] M(OH) n (s) M n+ (aq) + noh (aq) for insoluble ionic compounds that contain anions of weak acids, the lower the ph, the higher the solubility M 2 (CO 3 ) n (s) 2 M n+ (aq) + nco 2 3 (aq) H 3 O + (aq) + CO 2 3 (aq) HCO 3 (aq) + H 2 O(l) Tro, Chemistry: A Molecular Approach 108
109 Precipitation precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound if we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur Q K sp, the solution is saturated, no precipitation Q < K sp, the solution is unsaturated, no precipitation Q > K sp, the solution would be above saturation, the salt above saturation will precipitate some solutions with Q > K sp will not precipitate unless disturbed these are called supersaturated solutions Tro, Chemistry: A Molecular Approach 109
110 precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added Tro, Chemistry: A Molecular Approach 110
111 Selective Precipitation a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others a successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different Tro, Chemistry: A Molecular Approach 111
112 Ex What is the minimum [OH ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? precipitating may just occur when Q K sp [ [OH Q [Mg Q 2+ ][OH K ( 0.059) ][OH ] ( ) ] sp 2 ( 0.059) ] Tro, Chemistry: A Molecular Approach 112
113 Ex What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH ] 1.9 x 10-6 M [ [OH Q [Ca Q 2+ ][OH K ( )][OH ] ( ) ] sp 2 ( 0.011) ] Tro, Chemistry: A Molecular Approach 113
114 Ex What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH ] 1.9 x 10-6 M precipitating Ca 2+ begins when [OH ] 2.06 x 10-2 M [Mg [Mg ] Q ][ [Mg when 2+ Q K sp ( 2 ) ] ( ) ( ) 2 ][OH ] Tro, Chemistry: A Molecular Approach 114 M when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from M to 4.8 x M
115 Qualitative Analysis an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry a sample containing several ions is subjected to the addition of several precipitating agents addition of each reagent causes one of the ions present to precipitate out Tro, Chemistry: A Molecular Approach 115
116 Qualitative Analysis Tro, Chemistry: A Molecular Approach 116
117 117
118 Group 1 group one cations are Ag +, Pb 2+, and Hg 2 2+ all these cations form compounds with Cl that are insoluble in water as long as the concentration is large enough PbCl 2 may be borderline molar solubility of PbCl x 10-2 M precipitated by the addition of HCl Tro, Chemistry: A Molecular Approach 118
119 Group 2 group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+ all these cations form compounds with HS and S 2 that are insoluble in water at low ph precipitated by the addition of H 2 S in HCl 119
120 Group 3 group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides all these cations form compounds with S 2 that are insoluble in water at high ph precipitated by the addition of H 2 S in NaOH 120
121 Group 4 group four cations are Mg 2+, Ca 2+, Ba 2+ all these cations form compounds with PO 4 3 that are insoluble in water at high ph precipitated by the addition of (NH 4 ) 2 HPO 4 Tro, Chemistry: A Molecular Approach 121
122 Group 5 group five cations are Na +, K +, NH 4 + all these cations form compounds that are soluble in water they do not precipitate identified by the color of their flame 122
123 Complex Ion Formation transition metals tend to be good Lewis acids they often bond to one or more H 2 O molecules to form a hydrated ion H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l) Ag(H 2 O) 2+ (aq) ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H 2 O) 2 + the attached ions or molecules are called ligands e.g., H 2 O Tro, Chemistry: A Molecular Approach 123
124 Complex Ion Equilibria if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) + 2 (aq) + 2 NH 3(aq) Ag(NH 3 ) + 2 (aq) + 2 H 2 O (l) generally H 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) + 2 (aq) Tro, Chemistry: A Molecular Approach 124
125 Formation Constant the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) + 2 (aq) the equilibrium constant for the formation reaction is called the formation constant, K f + [Ag(NH3) 2 ] K f + 2 [Ag ][NH ] 3 Tro, Chemistry: A Molecular Approach 125
126 Formation Constants Tro, Chemistry: A Molecular Approach 126
127 Ex ml of 1.5 x 10-3 M Cu(NO 3 ) 2 is mixed with ml of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Write the formation reaction and K f expression. Look up K f value Determine the concentration of ions in the diluted solutions Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq) [Cu [NH K f 2+ 4 ] 4 3] [Cu(NH3) [Cu ][NH ] L 1L mol ( L L) L ] L Tro, Chemistry: A Molecular Approach mol ( L L) M M
128 Ex ml of 1.5 x 10-3 M Cu(NO 3 ) 2 is mixed with ml of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Create an ICE table. Since K f is large, assume all the Cu 2+ is converted into complex ion, then the system returns to equilibrium Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq) K f [Cu 2+ [Cu(NH ][NH 3 ) 4 3] 2+ 4 ] [Cu 2+ ] [NH 3 ] [Cu(NH 3 ) 2+ 2 ] Initial 6.7E Change - 6.7E-4-4(6.7E-4) + 6.7E-4 Equilibrium x E
129 Ex ml of 1.5 x 10-3 M Cu(NO 3 ) 2 is mixed with ml of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Substitute in and solve for x confirm the x is small approximation Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 2 2+ (aq) Initial x K f 2+ 4 ] 4 3] [Cu(NH3) [Cu ][NH [Cu 2+ ] ( 6.7E ) ( ) [NH 3 ] ( x)( 0.11) 0.11 Change ( - 6.7E (6.7E-4) + 6.7E )( 0.11) Equilibrium x E-4 since 2.7 x << 6.7 x 10-4, the approximation is valid [Cu(NH 3 ) 2 2+ ] 13 0 Tro, Chemistry: A Molecular Approach 129
130 The Effect of Complex Ion Formation on Solubility the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl (s) Ag + (aq) + Cl (aq) K sp 1.77 x Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K f 1.7 x 10 7 adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag + Tro, Chemistry: A Molecular Approach 130
131 131
132 Solubility of Amphoteric Metal Hydroxides many metal hydroxides are insoluble all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH some metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion substances that behave as both an acid and base are said to be amphoteric some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+ Tro, Chemistry: A Molecular Approach 132
133 Al 3+ Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O+ (aq) addition of OH drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH (aq) Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH (aq) Al(H 2 O) 3 (OH) 3(s) + H 2 O (l) Tro, Chemistry: A Molecular Approach 133
134 Tro, Chemistry: A Molecular Approach 134
Lecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc.
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